Sep
08

# ROBERT BROWN, “BACKRADIATION” AND MATHEMATICS

Written by Alberto Miatello

A few months ago Professor Robert G. Brown of Duke University wrote at WUWT and elsewhere that members of Principia Scientific International (PSI) don’t know Mathematics (sic). Brown then defied PSI to prove that the inner core of a hollow metal sphere is not heated by “backradiation” ( a junk science concept alien to thermodynamics experts).

Then, as if to “prove” his claim that “backradiation” is “real” Brown  wrote that if we wrapped a hot light bulb up with an aluminum foil, then in a very short time the bulb would be  overheated. To Brown the ensuing overheating of the light bulb was by “backradiation” generated from the aluminum foil (applied scientists may be heard sniggering at the very idea!).

However, Brown was unable to show any actual relevant calculations (only a long and boring array of meaningless algebraic gymnastics) to support what he said. Nonetheless, I took very seriously his challenge and I tried to calculate, in the most precise way, what really happens whenever you wrap a light bulb up with an aluminum foil.

What I present herein are typical, sound and established equations that any serious applied physicists, engineers, technicians, etc., will often use in the course of their work in what is euphemistically known as the “real world”. This is perhaps why academics like Brown, residing in their (“unreal world”) ivory towers, so often fail? For instance, it is a routine task for applied scientists and engineers to, say, calculate how much an electric wire may heat surrounding plastic insulation cable, or define the temperature reached by the casing of car engines in close contact with pistons, etc.

Clearly, it comes as a surprise for Mr. Brown (but not for PSI!) that in the fields of applied science it is impossible to find ANY manuals, technical textbooks, etc., using climate science's mysterious “backradiation” to carry out  such practical (“real world”) calculations.

“Backradiation” is regarded as a sort of “Arabian Phoenix” in the scientific and technical community. Of course, the academics and global warming believers are saying it exists, but nobody has ever actually found it! Maybe Professor Brown believes otherwise, but I have never met an engineer devising an engine, an industrial plant, or an electric device using “backradiation” to calculate how much heat is passing through the materials!

But let us go back to our calculations.

We can take a small (traditional) light bulb (not new halogen type) with incandescent tungsten filament, 2 centimeters (cm) radius, reaching 100° C Temperature (T) on the glass surface, and very thin aluminum foil (0.01 cm thickness) applied all around the glass surface of the bulb.

Problem: what will be the new temperature reached by the glass surface of the bulb beneath the aluminum foil?

Now, to describe and formalize the above problem in the best mathematical way we can write down a formula like this:

Now, to describe and formalize in the best mathematical way the above problem, we can write down a formula like this:

1)

100 2.01

dT = - Q dx/x

T 2.00

No need to be afraid!

This equation simply means, in rigorous mathematical symbolism, that the outflowing (-Q) thermal energy from an unit (1 cm.) zone of a spherical surface 2π (implicit: cm²), of a light bulb whose radius is 2.00 cm., and whose surface temperature is 100° C, is passing through an aluminum foil, whose thickness is 0.01 cm. (2.01 – 2.00 = 0.01).

Integral sign  from “T” bottom to 100 up, simply means: “let us calculate how much the surface temperature of bulb, in close contact with aluminum foil, changes (dT) from an unknown value T (to be discovered) of surface (= 2.00 cm. radius), up to 100° C, corresponding to 2.01 cm. radius.

Not so difficult, is it?

Now, aluminum conductivity on a metric scale is 240 W/m.k, so if we reduce it to a cm. scale, we have to divide per 100, and so we have

K = 2.4 W/cm.k

It is very important to see that the equation above is totally in accordance with 2nd Law of Thermodynamics, and we have – Q = k*… (or Q = -k*…) meaning that heat is flowing from hotter to colder bodies, and NOT the contrary!

As you can see, in the equation above there’s no symbol of “backradiation” at all, only heat flowing from inside the light bulb, to outside.

Now, since – Q = k * A (cm²) *dT/dx (= 0.01cm), we have the following Q rate:

-Q = 2.4/0.01 = 240 joule/sec.°C

Thus, we have:

2)

100 2.01

6.28 dT = 240 dx/x

T 2.00

Namely:

6.28 * (100 - T) = 240 * (ln 2.01 – ln 2.00) =

628 = 6.28T + (240 * ln (2.01/2.00)) =

628 = 6.28T + (240 * 0.005) =

628 = 6.28T + 1.2 =

T = 626.8/6.28 and finally

T = 99.8° C

So we have that the glass surface of the light bulb decreases its surface temperature very quickly, at a rate of 0.2°C/sec. (100° C – 99.8° C) = 12° C/min., while the aluminum foil surface, in close contact with, it increases it at the same rate, of 0.2° C/sec. But, as the hot electric filament and home electric plant is constantly releasing new electric energy to the glass surface of the bulb, of course there will be no decrease of temperature of the glass surface (that will remain constant at 100° C), but just a rapid increase of temperature of the aluminum foil.

This means that we will record an average increase of surface temperature of aluminum foil around 60° C in just 5 minutes!

But, as in high conductive metals (such as iron, copper, tungsten, silver) the increase of temperature is very rapid, especially when the thickness is very small, then it is experimentally proved that the increase of temperature in aluminum foil will follow a sigmoid function, as the error Gaussian function (erf), namely:

3)

x

2/√π e ^-t² (dt)

0

Therefore, it will take much less than 5 minutes for thin aluminum foil to increase the temperature for 50°/60° C.

And then, the aluminum will start to heat the surface of the glass, but not from “backradiation”, but simply by conduction, i.e. the hotter material (aluminum foil) will transmit heat to the boundary of the “colder” material (glass), according to 2nd Law of Thermodynamics.

Now, the question could be: “what would change the surface temperature of the glass light bulb, in close contact with an insulator like polyurethane (and not anymore with a highly conductive metal like aluminum)?”

The answer is not so difficult, as we have learnt from formulas above how to handle this kind of problems.

It is sufficient to replace – in the equations 1) and 2) – the k of aluminum (240), with the k of polyurethane (0.032)

4)

100 2.01

6.28 dT = 0.032 dx/x

T 2.00

Thus, we have:

6.28 * (100 - T) = 0.032 * (ln (2.01/ 2.00))

628 = 6.28T +( 0.032 * 0.005)

628 = 6.28T + 0.00016

T = (628 – 0.00016)/6.28

T = 99.999974° C

This means that the surface of the glass light bulb, in close contact with a thin polyurethane foil, shall “yield” to the polyurethane foil just a very modest amount of energy, corresponding to just 0.000026° C/sec. (100° C – 99.999974° C), namely 0.001° C/min.

But this is not – as anyone can clearly see – the outcome of any fanciful “backradiation”.

The increase in temperature from the aluminum foil surrounding the glass of the light bulb is simply a result of the very high thermal conductivity of aluminum (2.4 W/cm.K). It is the same way as the metal of your car (also with high thermal conductivity) – it too will soon become overheated if you leave your car outside in the hot sun of a summer's day.

But energy ( –Q) – as you can see - is NOT “backradiated” at all because thermal energy is only flowing from hotter to colder bodies (from inside the light bulb to outside). Then the aluminum will start to heat the boundary surface of the glass in close contact with it, but by conductivity, NOT by “backradiation” which doesn’t exist at all!

Conductivity, and NOT “backradiation” is the key-concept to understand regarding the heating mechanism of that aluminum foil.

And again PSI (and applied science) comes to prove this with rigorous, sound, and mathematically indisputable equations. Sadly, for the uninformed academic, Robert Brown, it seems the requisite Math took a little vacation from his statements.

Alberto Miatello

#1 Joel Shore 2013-09-08 18:04
"Clearly, it comes as a surprise for Mr. Brown (but not for PSI!) that in the fields of applied science it is impossible to find ANY manuals, technical textbooks, etc., using climate science's mysterious “backradiation” to carry out such practical (“real world”) calculations."

Really, so you are saying that applied scientists never use the Stefan-Boltzmann Equation for the net rate of radiative transfer between an object and its surroundings: P = sigma*epsilon*A*(T^4 - T_0^4) where sigma is the Stefan-Boltzmann constant, epsilon is the emissivity, A is the surface area of the object, T is the temperature of the object, and T_0 is the temperature of the surroundings?

What this equation tells you is that the rate at which heat is transferred by radiation between the object and its surroundings depends not only on the temperature of the object but also the temperature of the surroundings. In particular, if you apply it to a case where the object is generating thermal energy at a certain rate P, then in the steady-state, the object will have to be radiating at that same rate P. It then follows that if the temperature of the surroundings is higher then the temperature of the object will have to be higher in order for this radiative emission to occur. (And, this is true even though the net radiative heat flow is from the hotter object to the cooler surroundings, as the 2nd Law requires it must be.)

The term involving T_0^4 is universely interpreted as being radiation emitted by the cooler surroundings that is absorbed by the warmer object, i.e., your verboten back-radiation. Even if you don't interpret it this way, it doesn't change the mathematics that I have described above.

"As you can see, in the equation above there’s no symbol for “backradiation” at all - only heat flowing from inside the light bulb to the outside."

You aren't even considering radiative transfer. You are just considering heat transfer via conduction in the foil.

Yes, if you do a calculation that is completely irrelevant to the radiative transfer then there is no back-radiation because there is no radiation, period.
#2 Greg House 2013-09-08 18:38
Quoting Joel Shore:
...Equation for the net rate of radiative transfer between an object and its surroundings: P = sigma*epsilon*A*(T^4 - T_0^4) where ... T_0 is the temperature of the surroundings... The term involving T_0^4 is universely interpreted as being radiation emitted by the cooler surroundings that is absorbed by the warmer object, i.e., your verboten back-radiation. Even if you don't interpret it this way, it doesn't change the mathematics that I have described above.

Such an interpretation is completely false. The notion "temperature of the surroundings" implies only that the body cools to that temperature. No "radiation from colder surroundings to the warmer body" is involved there.

It is funny how you have been repeatedly trying to promote an absurd non-existing process, which "warming by back radiation" ("greenhouse effect") is, by taking a formula out of the context and misinterpreting the terms.
#3 Joel Shore 2013-09-08 20:08
As I said, even if you don't interpret it this way (and I have yet to be shown one physics textbook that doesn't), the mathematics is the same and the mathematics says exactly what I described: that the net rate at which a body radiates to its surroundings depends not only on the temperature of the body but also the temperature of the surroundings.

Ergo, if the body is generating thermal energy at a constant rate and you increase the temperature of the surroundings, the temperature of the body must increase. This really isn't that complicated. The fact that it is at all controversial is testimony to how strongly people will deny science that disagrees with their strongly-held ideological preferences.
#4 Greg House 2013-09-08 20:38
Quoting Joel Shore:
the net rate at which a body radiates to its surroundings depends not only on the temperature of the body but also the temperature of the surroundings. Ergo, if the body is generating thermal energy at a constant rate and you increase the temperature of the surroundings, the temperature of the body must increase.

The heat transfer from a body depends on the initial temperature of the body and, of course, on it's temperature when the body stops cooling. This second temperature, of course, coincides with the so called "temperature of surroundings", at which the body stops cooling.

If the temperature of the surroundings increases, the body would have stop cooling at this higher temperature, logically. The body's temperature would not increase, the body would only in the process of it's cooling transfer less energy to the surroundings, because the cooling would stop at a relative higher temperature.

There is nothing more in that formula, no "back radiation", no "transfer from cold to hot", nothing like that.

You have simply taken the formula out of the context and misinterpreted the terms.
#5 ewiljan 2013-09-08 21:14
Quoting Joel Shore:
"Clearly, it comes as a surprise for Mr. Brown (but not for PSI!) that in the fields of applied science it is impossible to find ANY manuals, technical textbooks, etc., using climate science's mysterious “backradiation” to carry out such practical (“real world”) calculations."

That is correct not even one demonstration of "back radiation" ever!!!

Really, so you are saying that applied scientists never use the Stefan-Boltzmann Equation for the net rate of radiative transfer between an object and its surroundings: P = sigma*epsilon*A*(T^4 - T_0^4) where sigma is the Stefan-Boltzmann constant, epsilon is the emissivity, A is the surface area of the object, T is the temperature of the object, and T_0 is the temperature of the surroundings?

Real scientists often use the non-Climate Clown Stefan-Boltzmann Equation for the maximum rate of radiative transfer between two flat infinate parallel emissive surfaces. That is all it was ever formulated to do.

P/m^2 = sigma*epsilon*(Ta^4 - Tb^4) where:
Ta is the absolute temperature of the higher temperature surface.
Tb is the absolute temperature of the lower temperature surface.
sigma is Stefan's constant, Which inclused the solid angle for Lam,bertian surfaces
epsilon is the product of the effective emissivity of each surface.

The radiative heat flux is from surface Ta to surface Tb. The terms in the parens (), must be evaluated first, and since T^4 is a radiative potential nether can be zero as the radiative potential for zero Kelvin is (not a number). The P/m^2 is a maximum, that may be further deminished by any loss due to transmissivity of the intervening medium.
The S-B equation has been demonstrated to be correct for radiative heat transfer, independent of any other heat transfer between the two surfaces.
Epsilon has also been demonstrated as correctly determined by this equation, by compatison to detailed emissivity measurement of each surface, paying attention to all of the details of Kerchhoff's law of radiation.

There is a Stefan's constant.
There is a Boltzmann's constant.
There is no Stefan-Boltzmann constant.
All of what you claimed is but Climate Clown nonsense.
#6 Joel Shore 2013-09-08 21:48
In your discussion, you are applying the equation to a body that has no source of thermal energy. Of course, in such a case, the temperature of the body would decrease...It would not increase. Nobody claims that it would!

However, now consider the case where the body is producing or receiving thermal energy (e.g., by conversion from electrical energy in the case of a light bulb, or by receiving it from the sun in the case of the Earth). Now, it will no longer cool down to the temperature of its surroundings.

So, the question becomes: What temperature does it reach in steady-state (where it is radiating away as much thermal energy as it is receiving or generating) AND is this temperature higher if the surroundings are warmer than if the surroundings are cooler? The Stefan-Boltzmann Equation gives a very clear and unequivocal answer to this question!
#7 Joel Shore 2013-09-08 22:00
"Real scientists often use the non-Climate Clown Stefan-Boltzmann Equation for the maximum rate of radiative transfer between two flat infinate parallel emissive surfaces. That is all it was ever formulated to do."

From hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html :

Stefan-Boltzmann Law

...

If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form

P = e*sigma*A*(T^4 - Tc^4).

[Of course, this is also basically the same thing that every physics textbook I have ever seen says.]

"There is a Stefan's constant.
There is a Boltzmann's constant.
There is no Stefan-Boltzmann constant."

From en.wikipedia.org/wiki/Stefan_boltzmann_constant :

The Stefan–Boltzmann constant (also Stefan's constant), a physical constant denoted by the Greek letter σ, is the constant of proportionality in the Stefan–Boltzmann law
#8 Joel Shore 2013-09-08 22:06
By the way, just to clear away a confusion you seem to have: The S-B equation gives the RATE of cooling, as any physics textbook that discusses it would tell you. It does not give the total amount of energy transferred in the process of cooling from one temperature to another; it does not even have the right units to do that: It has units of power, not units of energy.
#9 Greg House 2013-09-08 22:21
Quoting Joel Shore:
However, now consider the case where the body is producing or receiving thermal energy (e.g., by conversion from electrical energy in the case of a light bulb, or by receiving it from the sun in the case of the Earth). Now, it will no longer cool down to the temperature of its surroundings. So, the question becomes...

I am not inclined to change the subject to save you from further embarrassment.

You derived a non-existing process from a formula by misinterpreting the terms, let's stick to it.
#10 ewiljan 2013-09-08 22:25
Quoting Joel Shore:
As I said, even if you don't interpret it this way (and I have yet to be shown one physics textbook that doesn't), the mathematics is the same and the mathematics says exactly what I described: that the net rate at which a body radiates to its surroundings depends not only on the temperature of the body but also the temperature of the surroundings.

This just means that the texts you reviewed
were written and published since 1972, when the kinetic theory of heat, and the theory of statistical mechanics, became popular. All theory, no demonstration at all. Therefore such college texts cannot be claimed to be correct or truthful.

Ergo, if the body is generating thermal energy at a constant rate and you increase the temperature of the surroundings, the temperature of the body must increase.

Yes indeed known since the late 1600's, and need not have anything to do with thermal radiation at all, which thermal radiation does obey all the "old" laws of thermodynamics. This is classical thermodynamics, and has nothing to do with the silly Climate Clown claims of GHE, back radiation, increasing atmospheric CO2 doing anything, sea levels rising, anthropoids, or any other catastrophic event.

This really isn't that complicated. The fact that it is at all controversial is testimony to how strongly people will deny science that disagrees with their strongly-held ideological preferences.

It is not complicated at all, it is silly!
It is a vain attempt by Climate Clowns to distance themselves from GHE, and the required "back radiation", which only worked until, informed folk figured out the immense Fraud and Scam that you Climate Clowns are still trying to pull on the people and governments of the world. Joel, I can understand your panic, say or do anything that may keep you (SNIPPED) ( Was that acceptable Tommy? ), he will not say.
#11 ewiljan 2013-09-08 23:05
Mr. Miatello,
Your article is correct for most household incandecent lamps that maintain a reasonable
filament temperature by thermal conduction through the supports and the argon backfill.
As the luminous efficacy of such lamps is very low most of the input electrical power is dissipated via thermal conductivity from the outer glass envelope to the surrounding air.
Such a lamp thermally isolated in a vacuum would destroy itself from the overheating and evaporation of the filiment, within minutes.

We still have the problem, of the nonsense
in the most confusing way.
IMHV the only way to defeat the mnistaken claim of Professor Robert G. Brown of Duke University is to demonstrate that there is no power for such back radiation, and also that there is absolutly no physical reason for such radiation to be generated by thermal means.
#12 ewiljan 2013-09-08 23:37
Quoting Joel Shore:
"Real scientists often use the non-Climate Clown Stefan-Boltzmann Equation for the maximum rate of radiative transfer between two flat infinate parallel emissive surfaces. That is all it was ever formulated to do."

That is correct, Please discuss any errors!
Climate Clowns do nothing but claim everything.

From hyperphysics.phy-astr.gsu.edu/hbasethermo/stefan.html :
Stefan-Boltzmann Law
If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form.
P = e*sigma*A*(T^4 - Tc^4).

Please show where Mr. Stefan, or Mr. Boltzmann,
ever claimed such a rediculous thing.
Nothing of objects only emissive surfaces. 2ea.
I reject all of your Nuevo Science Climate Clown nonsense. Please show any reference pre 1970 that even claims a Stefan-Boltzmann Law.
There has been a Stefan-Boltzmann equation for a very long time. It always works when used properly.

There is a Stefan's constant.
There is a Boltzmann's constant.
There is no Stefan-Boltzmann constant.
No matter what wikipedia say!!!
#13 ewiljan 2013-09-09 00:16
Quoting Joel Shore:
In your discussion, you are applying the equation to a body that has no source of thermal energy. Of course, in such a case, the temperature of the body would decrease...It would not increase. Nobody claims that it would!

That is quite an admision from such as Joel Shore. . How does that square with your Climate Clown Second law of thermodynamics v.s.
Clausius's real 2LTD?

However, now consider the case where the body is producing or receiving thermal energy (e.g., by conversion from electrical energy in the case of a light bulb, or by receiving it from the sun in the case of the Earth). Now, it will no longer cool down to the temperature of its surroundings.

Why such a question is this Thermodynamics 101?

So, the question becomes: What temperature does it reach in steady-state (where it is radiating away as much thermal energy as it is receiving or generating) AND is this temperature higher if the surroundings are warmer than if the surroundings are cooler?

Ah! Switching now to only radiative heat transfer with a fixed power source, What other tricks do you have to confuse others?

The Stefan-Boltzmann Equation gives a very clear and unequivocal answer to this question!

The S-B equation is not even applicable as it only relates to two isothermal emissive infinite parallel surfaces and the power per unit area transfered between the surfaces. With
the infinite plates the power supplied by the hot surface, and dissipated by the cold surface, to maintain both isotherms is also infinite. So much for your misuse of the S-B equation.
#14 ewiljan 2013-09-09 02:14
(SNIPPED) = from jail Why are you not talking Tommy? See John, He may have some advise!!!

You are banned for 2 days.
#15 Guest 2013-09-09 09:36
This comment has been deleted by Administrator
#16 John Marshall 2013-09-09 10:02
I have had a long email exchange with Dr Brown and as I insisted the 2nd law cannot be broken he maintained that this law applied to a system not to parts of that system. Complete crap. I gave up.

A cooling object will lose heat only if its surroundings are cooler than it is. Otherwise the object will warm up to the temperature of the surroundings. Not back radiation just normal heat flow.

For those above who still insist on this mythical process please explain why liquids placed into a vacuum flask still loose heat, albeit slowly because one method of heat transfer has been removed, ie. conduction through the mirrored walls. If back radiation worked the temperature of the liquid would increase.
#17 Lonny Eachus 2013-09-09 11:44
Even if you were correct, Joel, it would have no bearing on the "Greenhouse Effect", because at best you are showing objects reaching equilibrium. Your own examples do not show any radiative transfer from cold to warm, which the greenhouse models DO rely on.

Now take the actual case of the hypothesized atmospheric "back radiation", in which the surroundings are KNOWN TO BE, from simple observation, cooler than the supposed initial source. (The ground being the "initial source" in the models, in an attempt to isolate variables.)

Certainly the S-B equation gives different answers to the equation as the initial temperature of the surroundings varies. This is not in dispute. But where does that imply "back radiation"? THAT is the question here, not equilibrium temperature, which makes your argument a straw man.

What you are missing is that even if the "warmists" arrived at the right equilibrium temperature, if they did so using incorrect methods their models are still wrong.
#18 Robert Brown 2013-09-09 11:50
Very amusing. We have John Marshall stating a truism for a passively cooling unheated sphere as if it somehow applies to a heated sphere. At least get the problem right, John. The same thing is true, after all, for the example worked out in such detail above -- the bulb and foil don't warm up to the temperature of their surroundings, they get warmer still because they have to reject net heat to their surroundings at the exact rate the bulb is heated.

The really fun bit is that I have not real quarrel with the algebra above, except for the fact that it is incomplete. In physics class they teach -- in fact, I teach -- that heat can be transferred by conduction, convection, and radiation. Last time I looked, aluminum foil was highly reflective. It is also a simple fact that energy loss from light bulbs is not dominated by conduction, or conduction+convection, but by radiation:

www.posterus.sk/?p=9400

Ballpark of 83% leaves via one form or another of radiation. Therefore the biggest problem with the discussion above is that it focused on the least important mechanism of heat transfer, one that is less than 1/5 of the whole, while not addressing the primary transfer mechanism at all. Aluminum is, in fact, a pretty good reflector of light visible and invisible. It's sort of pointless to get all mathematical and everything, don't you think, and leave out the most important factor in a cooling process?

No?

Well then, back to rationalizing your absurd beliefs that increasing the effective impedance of energy flow in ANY cooling channel between two reservoirs will have no effect on the system, even if one of the two reservoirs isn't held at constant temperature, it is held at constant power. Even the example above demonstrates that THAT is not true. Why do you think it would be true for conduction but not equally true for radiation?

rgb
#19 Lonny Eachus 2013-09-09 11:59
More accurately, I should have written "Where does that imply further heating of warmer objects via back radiation?" The question is not its existence, but its effects.
#20 Robert Brown 2013-09-09 13:23
Wow, Joel, I gotta say (after reading some of the replies on this thread) that this really is pointless. These folks have no conception of the FIRST law of thermodynamics, let alone the second. The argument for warming doesn't even require mentioning the SBE, it only requires the first law, the second law, and a monotonic relation between temperature difference in ANY channel and the rate of energy transfer in that channel, subject to very broad constraints.

But seriously, just a waste of time. When people just make stuff up and reject the contents of ELEMENTARY textbooks on the subject because they just don't like the conclusion those contents lead to, how can you argue with them? If somebody tries to solve the light bulb problem while pretending that it doesn't primarily cool via radiation and completely ignoring radiation, what can you do?

Get them to say "oops"?

Never happen. It's a religious issue, not a scientific one.

rgb
#21 Robert Brown 2013-09-09 13:44
So you have empirical evidence to back that up? All of the studies of light bulb operation and efficiency I've ever read (and I've actually read a few, where you are just making this stuff up as you go to sound wise) suggest that around 83% of the heat loss from an incandescent bulb is via radiation, and only 17% from conduction/convection from a bulb's surface. 17% is not a trivial fraction, but it is not the dominant mode of energy loss, and neglecting it in the analysis is such a serious error that it makes all of the conclusions basically ignorable -- time to do the problem over, this time actually taking into account that the aluminum foil will reflect a large fraction of the radiant energy back into the bulb.

Evidence for that is, of course, the fact that the room gets dark when the bulb is surrounded by the foil, so gee, some energy isn't making it directly out into the room. Nor is it being absorbed by the foil to any significant extent, certainly not in all or most wavelengths.

Since aluminum is a good conductor of heat (better than the air it replaces), a thin layer of aluminum foil would actually behave very slightly like a HEAT SINK for the light bulb if it were true that it only cooled by conduction. Compare the conductivity of aluminum to that of air, and account for its larger surface in contact with the air. But in any event its purely conductive effect is almost completely negligible if it smoothly clings to the bulb and has almost exactly the same area. Try using a thin piece of aluminum foil to take a pot roast out of the oven if you don't believe me -- you might as well be grabbing the pot with your fingers.

But I have to say I was most amused by your assertion that the only way to defeat me is to prove that "there is no power for such back radiation". You are absolutely correct. I wish you well with that enterprise.

I do have to wonder, though, how you are going to prove it given that all of the devices you might use to measure it rely on, errr, there being power in back radiation? Is the power going to be magic power, that appears in the radiant energy thermometers but that doesn't carry energy to anything else?

I look forward to seeing your microscopically justified physical theory proving that atoms that emit radiation have knowledge of the temperature of medium in which the atoms that eventually absorb that radiation reside, and/or the equally interesting demonstration that those absorbing atoms have some knowledge of the temperature of the source of the photons they are absorbing. I imagine that you'll need to modify quantum field theory, in particular quantum statistical electrodynamics, more than a bit to accomplish that.

Good luck with that! It's an ambitious undertaking. Personally, I'm content to trust the existing theory of Q.E.D. as it is amazingly successful at describing everything from classical electrodynamic phenomena through hard quantum phenomenal like photon absorption and emission, consistently, and in agreement with many experiments. But no doubt you have better insight.
#22 Joel Shore 2013-09-09 16:42
***I'm*** changing the subject?

No...The subject has always been a case where the sphere has a source of thermal energy, either because it is receiving it from the sun in the case of the Earth, or because it is generating thermal energy from some other form of energy like electrical energy in the case of the light bulb.

It is Slayers who constantly go back to a case where the sphere has no source of thermal energy but is simply cooling, hoping that their audience won't notice the obvious misdirection. Then, they can have great fun by claiming that those who they argue against are claiming that heat flows from cold to hot and other such nonsense.

In reality, of course, heat (which the macroscopic net energy flow) is always from hot to cold but in the case where the atmosphere has greenhouse gases, it requires a higher temperature at the Earth's surface to sustain the rate of energy flow necessary for the Earth system to be emitting energy at the same rate as it is absorbing it from the sun.
#23 Greg House 2013-09-09 16:45
Quoting Robert Brown:
I look forward to seeing your microscopically justified physical theory proving that atoms that emit radiation have knowledge ... that those absorbing atoms have some knowledge of ...

This is the most ridiculous sort of argumentation warmists ever used. In short, it means that ANY non-existing process exists because the physical objects involved have no knowledge of the process being impossible.

Warming the source of radiation by back radiation ("greenhouse effect") is impossible for the simple reason that the assumption of it being possible leads inevitably to production of energy out of nothing. Even without numbers it must be clear that in certain cases a body warmed by back radiation would produce even more radiation and thus even more back radiation and get warmer and warmer and so on, and more energy would come out of the system than has entered it, which is absurd.
#24 Joel Shore 2013-09-09 16:46
"Ah! Switching now to only radiative heat transfer with a fixed power source, What other tricks do you have to confuse others?"

As I noted to Greg, it is only the sophistry of the Slayers that constantly misdirects us to the case where there is not a fixed power source, hence talking about what the Earth would do if there was no sun, rather than what happens in the presence of the sun.

"The S-B equation is not even applicable as it only relates to two isothermal emissive infinite parallel surfaces and the power per unit area transfered between the surfaces."

...Except in every physics textbook, which all talk about it in the context of the radiative transfer between an object and its surroundings.
#25 Joel Shore 2013-09-09 16:56
"Certainly the S-B equation gives different answers to the equation as the initial temperature of the surroundings varies. This is not in dispute."

Great...That means the greenhouse effect is not in dispute and there is nothing to argue about. The greenhouse effect follows from the mathematics of the S-B Equation. Whether you call the 2nd term in that equation "back radiation" or "magical mystery radiation from the planet Zyrcon" makes absolutely no difference in the results you get from the mathematics.

The reason that the Slayers are so focused on "back radiation" is because they realize that this is where their sophistry is most effective: Many people do not understand the distinction between heat (net macroscopic energy flow) that must go from hot to cold and the combination of the radiative energy flows in both directions that contribute to this macroscopic quantity. In fact, most people don't even know that thermodynamics is a description of the microscopic world that follows from a more fundamental description at the microscopic level based on statistical mechanics.

The net effect of "back radiation" is to reduce the net amount of heat flowing from a hot object to its cooler surroundings in the case when the cooler object is no longer as cool. But, if you don't want to believe (despite all the evidence) that there can be radiation in both directions, you don't have to. You just have to apply the S-B Equation and you will get the result that the Earth's surface is warmer when surrounded by a radiatively-active (absorbing and emitting atmosphere) than when it is surrounded by an atmosphere transparent to radiation (and hence, what might as well just be space).
#26 Joel Shore 2013-09-09 17:00
So, now you are putting yourself into the physics-denier camp. You are denying an entire huge branch of physics (statistical mechanics) that has countless successes in describing our world because you don't like that physics contradicts your ideological views.

And, the irony of it is that even denying statistical mechanics isn't enough as long as you accept that the S-B Equation describes an object and its surroundings, as every textbook on physics I can find, says it does.

This is what denial looks like.
#27 Joel Shore 2013-09-09 17:02
Robert,

I am afraid you are right. It is sad to see people so wedded to complete and utter nonsense just because they can't accept science that goes against their world view. But, I guess it is the same story that has happened throughout history.
#28 Joel Shore 2013-09-09 17:07
To correct a one-letter typo in my post that makes a big difference, the last sentence in the 2nd-to-last paragraph should read:

"In fact, most people don't even know that thermodynamics is a description of the MACROSCOPIC world that follows from a more fundamental description at the microscopic level based on statistical mechanics."
#29 Joel Shore 2013-09-09 17:10
John,

Are you aware that the Earth is heated by the sun? Just wondering.

What is your liquid placed in the vacuum flask being heated by?
#30 Joel Shore 2013-09-09 17:33
The point, by the way, that Robert was trying to make to you is that HEAT is a macroscopic quantity. In fact, thermodynamics is a description of the universe at a MACROSCOPIC level. It was never formulated to be a description of what happens microsopically because those who came up with it did not even know the most basic facts about the microscopic world...Thermodynamics predates the discovery of the electron by many decades.

The modern understanding of thermodynamics...and the 2nd Law in particular...is that the irreversibility implied by the 2nd Law (the future is different from the past because, for example, heat flows only from hot to cold) arises from completely reversible behavior at the microscopic level. That in fact is what makes physics interesting and deep, rather than just a collection of silly arbitrary and capricious rules, as you want to reduce it to with your incorrect ideas that photons can only go from a hotter object to a colder object.
#31 Greg House 2013-09-09 17:46
Quoting Joel Shore:
The modern understanding of thermodynamics...and the 2nd Law in particular...

It can be easily demonstrated without any reference to the 2nd Law that no back radiation warming is possible. I have already said it in my previous comment, but can repeat it for you.

Even without numbers it must be clear that in certain cases a body warmed by back radiation would produce even more radiation and thus even more back radiation and get warmer and warmer and so on, and more energy would come out of the system than has entered it, which is absurd.

As I said before, the IPCC knew it apparently, so in their second report they cooked the calculations and let "greenhouse gases" radiate in only one direction (to the surface, of course) to avoid the absurdity of their concept (energy production out of nothing) to become obvious.
#32 Greg House 2013-09-09 17:51
The "greenhouse effect" you and some others have been promoting is not science. It is an absurd concept, such a process does not exist.
#33 ewiljan 2013-09-09 19:38
Quoting ewiljan:
(SNIPPED) = from jail Why are you not talking Tommy? See John, He may have some advise!!!

You are banned for 2 days.

OK, I really need this Blog. Have you checked with John? Have you checked your own feedback Tommy?
#34 ewiljan 2013-09-09 19:48
Quoting Joel Shore:
So, now you are putting yourself into the physics-denier camp. You are denying an entire huge branch of physics (statistical mechanics) that has countless successes in describing our world because you don't like that physics contradicts your ideological views.

And, the irony of it is that even denying statistical mechanics isn't enough as long as you accept that the S-B Equation describes an object and its surroundings, as every textbook on physics I can find, says it does.

This is what denial looks like.

Try some real science. Please demonstrate any of what you claim of global warming!
#35 ewiljan 2013-09-09 19:56
Quoting Robert Brown:
Wow, Joel, I gotta say (after reading some of the replies on this thread) that this really is pointless. These folks have no conception of the FIRST law of thermodynamics, let alone the second. The argument for warming doesn't even require mentioning the SBE, it only requires the first law, the second law, and a monotonic relation between temperature difference in ANY channel and the rate of energy transfer in that channel, subject to very broad constraints.

But seriously, just a waste of time. When people just make stuff up and reject the contents of ELEMENTARY textbooks on the subject because they just don't like the conclusion those contents lead to, how can you argue with them? If somebody tries to solve the light bulb problem while pretending that it doesn't primarily cool via radiation and completely ignoring radiation, what can you do?

Get them to say "oops"?

Never happen. It's a religious issue, not a scientific one. rgb

Gee we got the heavy hitters to help Joel.

OK Bobby, Demonstrate "any" of Your Climate Clown nonsense! No Climate Clown textbooks
Just physical proof. Non-physical effects are denied in science!
#36 Joel Shore 2013-09-09 20:28
All that you have demonstrated is that Dunning and Kruger ( en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect ) are worthy of a Nobel Prize for their insights into human psychology.
#37 Tim Folkerts 2013-09-09 20:31
"it must be clear ..."

No, it is not clear that anything would "get warmer and warmer". In fact the opposite is quite clear -- back-radiation (in conjunction with an active heating source like the sun -- will lead to definite but finite warming.

If you think it is quite clear, then give us some equations and numbers for a 'simple' case where this would happen. Perhaps the old "heated sphere surrounded by a thin shell'. You know, the one where the sphere will be 2^(1/4) times warmer with the shell than without the shell. 2^(1/4) times warmer in agreement with every physics text, not infinitely warmer in agreement with Greg's physics.
#38 Greg House 2013-09-09 20:48
Quoting Tim Folkerts:
In fact the opposite is quite clear -- back-radiation (in conjunction with an active heating source like the sun -- will lead to definite but finite warming.

No way.

"Finite warming" would mean that at certain temperature the "warming by back radiation" would not work any longer. The logical consequence would be that for ANY temperature B temperature A can be found so that the back radiation warming started at A and ended at B, which means there would be no warming by back radiation at B. In other words, the assumption of "FINITE warming by back radiation" is equivalent to "no warming by back radiation is possible at ANY temperature".
#39 Tim Folkerts 2013-09-09 21:35
Wow. What convoluted logic. Lets start at the beginning.

"The logical consequence would be that for ANY temperature B temperature A can be found so that the back radiation warming started at A and ended at B..."

Yes. For the case I suggested, the temperature will start at A = B*2^(-1/4) = 0.84 B and raise to B. So if it ended at 300 K< it started at 252 K (for those who prefer concrete numbers rather than equations.)

"... which means there would be no warming by back radiation at B"
No. Which means there would be no FURTHER warming once the temperature B was reached. The surface would have warmed as much as possible based on the converging infinite series that brought the temperature from A to B.

"In other words, the assumption of "FINITE warming by back radiation" is equivalent to "no warming by back radiation is possible at ANY temperature"."
Let me correct you one more time. The assumption of FINITE warming is equivalent to ... ummm .. .finite warming. That no matter what temperature you start with, the back radiation can only be so effective, so there is some upper limit to the warming.
#40 ewiljan 2013-09-09 22:35
Quoting Robert Brown:
So you have empirical evidence to back that up?

I do indeed.

suggest that around 83% of the heat loss from an incandescent bulb is via radiation, and only 17% from conduction/convection from a bulb's surface.

Typical Climate Clown getting the data reversed. The manufacturers, both GE and Osram say that 16% of the power of an 60 Watt lamp is radiated from the filament through then glass envelope. The rest is conducted to the base and envelope, to be dissipated via convection to the surrounding air.

But I have to say I was most amused by your assertion that the only way to defeat me is to prove that "there is no power for such back radiation". You are absolutely correct. I wish you well with that enterprise.

The no power is easy, Your Lukewarm friend,
Dr. Roy Spencer, demonstrates just that with his steel shell nonsense. All radiative power from the inner sphere to the steel shell every pico-watt, "must be", simultaneously dissipated to a colder sink. If this where
not the case, The temperature of the shell "must change" to support such a condition.
This involves "all" of the known laws of physical science. This demonstrates that absolutely "no" energy/power is available for some sort of non-physical "back radiation".

I also say, that there is absolutely no physical reason for such radiation to be generated by thermal means.

This one is harder, Such a proof demands a absolute definition of both "temperature" and "heat" in physical, not conceptual, terms!
No such definition is available, from Climate Clowns, sceptics, deniers, or anyone else.

I look forward to seeing .. blah blah blah.
Oh again the Climate Clown argument that measurable physical observations, "must"
have some controlling intelligence.

Good luck with that! It's an ambitious undertaking. Personally, I'm content to trust the existing theory of Q.E.D. as it is amazingly successful at describing everything. from classical electrodynamic phenomena through hard quantum phenomenal

I agree, if you would ever try to understand QED!
To me, QED is a earthling painful attempt to demonstrate, that all physical observations have the same result, but from an entirely different point of view.
#41 Greg House 2013-09-09 22:43
Quoting Tim Folkerts:
"... which means there would be no warming by back radiation at B"
No. Which means there would be no FURTHER warming once the temperature B was reached.

Come on, your "no FURTHER warming at B" is the same as "no warming at B". In your process the body at temperature B will not get warmed by back radiation, exactly as you wish. The rest is simply a matter of logic. You concept is contradictory, a sheer nonsense. Warming by back radiation can not exist.
#42 ewiljan 2013-09-09 22:48
#43 Tim Folkerts 2013-09-09 22:55
"the body at temperature B will not get warmed by back radiation... "

It already HAS been warmed! It used to be at 0.84B and now it is at B. You are saying "the object that will get warmed will not get warmed"!

If I give you \$50, but then give you "no FURTHER money", is that the same as giving you "no money at all"???
#44 ewiljan 2013-09-09 23:05
Quoting Joel Shore:
The point, by the way, that Robert was trying to make to you is that HEAT is a macroscopic quantity.

Robert refuses to say what heat is,at any micro or macro observation. What is heat?

In fact, thermodynamics is a description of the universe at a MACROSCOPIC level. It was never formulated to be a description of what happens microsopically because those who came up with it did not even know the most basic facts about the microscopic

So what?

The modern understanding of thermodynamics...and the 2nd Law in particular...is that the irreversibility implied by the 2nd Law (the future is different from the past because, for example, heat flows only from hot to cold) arises from completely reversible behaviour at the microscopic level. That in fact is what makes physics interesting and deep, rather than just a collection of silly arbitrary and capricious rules, as you want to reduce it to with your incorrect ideas that photons can only go from a hotter object to a colder object.

What complete Climate Clown nonsense.
Show me even one of your non-physical photons.
#45 Greg House 2013-09-09 23:14
Quoting Tim Folkerts:
"the body at temperature B will not get warmed by back radiation... "

Right, in your story that is under investigation here the body indeed a)has been warmed but b)at the temperature B no warming by back radiation occurred.

I took the "b" part and logically demonstrated that it is equivalent to "no warming by back radiation occurs at ANY temperature.

I fully understand that you do not like it, but it is simply your ridiculous contradictory concept, sorry. It is just because you are trying to make an infinite (fictional, absurd) process finite to save it from the apparent absurdity. This won't work, as you can see.

Your "method" is of the same quality as the one of the IPCC, who let the "greenhouse gases" radiate in only one direction to make the apparent production of energy out of nothing as a result of the "greenhouse effect" invisible.
#46 Tim Folkerts 2013-09-09 23:22

The manufacturers, both GE and Osram say that 16% of the power of an 60 Watt lamp is radiated from the filament AS VISIBLE LIGHT.

The sun emits ~ 50% of its energy as IR and only 40% as visible. For a filament that is half as hot, the visible would drop as a % of the total, and the IR would increase, to something like 10% visible (or maybe 16% depending on the actual filament temperature) and 90% IR (of which ~ 60-70% would be 0.7- 2 um for which glass it transparent). So well over half of the power into the bulb is radiated right out of the bulb as IR.
#47 Robert Brown 2013-09-09 23:46
Ah, yes, but you have to actually know how to read graphs and stuff like that to figure all of that out.

I also say, that there is absolutely no physical reason for such radiation to be generated by thermal means.

This one is harder, Such a proof demands a absolute definition of both "temperature" and "heat" in physical, not conceptual, terms!
No such definition is available, from Climate Clowns, sceptics, deniers, or anyone else.

Gosh, really? So the zeroth law of thermodynamics is lost on you, eh? Equipartition theorem, statistical mechanics, all just whoosh, right over your head?

Well, that's certainly one way to argue. I have to wonder -- do they make everybody who posts on this list take special (SNIPPED OUT INSULTS AGAINST PSI)

(Administrator: I am one of the members of PSI who don't take drugs at all and think your statement is way out of line)

In the meantime, it is good to know that you say that there is absolutely no physical reason for radiation to be generated by thermal means. Really, that says it all. More than all.

Sheesh, I gotta say, it's worse than I thought. Y'all are tempting me to become a warmist just because you can't fix stupid.

rgb
#48 ewiljan 2013-09-09 23:48
Quoting Tim Folkerts:

The manufacturers, both GE and Osram say that 16% of the power of an 60 Watt lamp is radiated from the filament AS VISIBLE LIGHT.

The amount "radiated" at long distances can be measured by wideband bolometers. From the 60 Watt lamp Less than 10 watts is measured as radiated. the other 50+ Watts is dissipated by other means.

The sun emits ~ 50% of its energy as IR and only 40% as visible. For a filament that is half as hot, the visible would drop as a % of the total, and the IR would increase, to something like 10% visible (or maybe 16% depending on the actual filament temperature) and 90% IR (of which ~ 60-70% would be 0.7- 2 um for which glass it transparent). So well over half of the power into the bulb is radiated right out of the bulb as IR.

So you claim, where are your measurements?
#49 Robert Brown 2013-09-10 00:00
Even without numbers it must be clear that in certain cases a body warmed by back radiation would produce even more radiation and thus even more back radiation and get warmer and warmer and so on, and more energy would come out of the system than has entered it, which is absurd.

It is indeed absurd, so why do you assert this? There are absolutely zero cases where this happens, and if you write down the actual differential equations, they are quite solvable with clearly and trivially derivable steady state solutions. By that I mean that if you start the central sphere/bulb, heated with a fixed constant power, too warm in temperature (which is an initial condition independent of the differential equations per se), it will cool to the dynamic equilibrium temperature determined by the "back radiation" of its environment, and if you start it at an initial temperature that is too cold, it will warm to exactly the same temperature.

This happens for the plain old heat equation OR for the Stefan-Boltzmann equation. In both cases heat is transferred from warmer to colder. In the case of heat conduction, the rate of transfer is proportional to the difference in temperature. In the SBE, it is proportional to the difference in the fourth power of the temperature.

You, and Joe Postma, continually WANT to assert that if there is such a thing as backradiation, the process of radiative heat transfer is dynamically unstable, but that is simply nonsense. But I'm sure you know this -- on other threads I've actually written out all of the differential equations and directly proven that a steady state solution exists, is stable, involves net heat flowing only from hot to cold, and is in detailed balance as far as energy flow is concerned.

Also, the IPCC doesn't have the "greenhouse gases" radiate in only one direction -- that too is an absurd allegation. Don't get me wrong -- I don't think they get the greenhouse computation right, but their physics can be mistaken in some of its assumptions without being stupid. In no case is backradiation creating energy out of nothing.

rgb
#50 Robert Brown 2013-09-10 00:12
The modern understanding of thermodynamics...and the 2nd Law in particular...is that the irreversibility implied by the 2nd Law (the future is different from the past because, for example, heat flows only from hot to cold) arises from completely reversible behaviour at the microscopic level. That in fact is what makes physics interesting and deep, rather than just a collection of silly arbitrary and capricious rules, as you want to reduce it to with your incorrect ideas that photons can only go from a hotter object to a colder object.

Excuse me? Nobody I have heard asserted any such thing, although you came close up above. The Stefan-Boltzmann equation describes the NET radiative power transferred between a body and its surroundings when the body and surroundings are in thermal equilibrium and are at least approximately "black bodies" (usually with an emissivity assumed to be near unity). It contains an outgoing term for the object that arises from integrating over the Planck distribution, and an incoming term (which is equally outgoing from the surroundings). Photons go both ways, and are simply more likely to be emitted and in general are emitted with more energy from warmer black bodies than from cooler ones, resulting in a biased transfer of radiative power from warmer to cooler. This does, incidentally, satisfy the second law.

So I'm not sure what your point is. Most of the nuts on this list spend all of their time ranting about how colder objects cannot transfer radiation at all to warmer ones, or asserting that the radiation thus transferred somehow doesn't "count" in the energy budget on the receiving end. If your point is that the correct description of the system is statistical mechanics (of open systems, no less) rather than thermodynamics, well sure. OTOH, for simple problems like the one posed, it is pretty obvious that one doesn't NEED real stat mech to very precisely explain and compute what is going on, any more than I need to use quantum theory to describe the motion of a baseball.

rgb
#51 Robert Brown 2013-09-10 00:20
OK Bobby, Demonstrate "any" of Your Climate Clown nonsense! No Climate Clown textbooks
Just physical proof. Non-physical effects are denied in science!

I have no idea what this even means.

Truly, you have a dizzying intellect. I will wait until you get going... then I'm certain I'll be even dizzier.

rgb
#52 Greg House 2013-09-10 00:46
Quoting Robert Brown:
Even without numbers it must be clear that in certain cases a body warmed by back radiation would produce even more radiation and thus even more back radiation and get warmer and warmer and so on, and more energy would come out of the system than has entered it, which is absurd.

It is indeed absurd, so why do you assert this? There are absolutely zero cases where this happens, and if you write down the actual differential equations, they are quite solvable with clearly and trivially derivable steady state solutions. By that I mean that if you start the central sphere/bulb, heated with a fixed constant power, too warm in temperature (which is an initial condition independent of the differential equations per se), it will cool to the dynamic equilibrium temperature determined by the "back radiation" of its environment,...

What you have written is both unrelated to what you have quoted stuff and false, because why would anything "heated with a fixed constant power" be "too warm" and cool to the "temperature of environment". Complete nonsense.

I guess you just write pseudo-scientific stuff to impress the readers who do not see through it, so primitive.

Again, "warming by back radiation" must be a cyclical process, and in case where the body is at a stable temperature initially "warming by back radiation" would run infinitely and very soon more energy would leave the system that has entered it. A clearly absurd concept.
#53 Greg House 2013-09-10 00:51
Quoting Robert Brown:
Also, the IPCC doesn't have the "greenhouse gases" radiate in only one direction -- that too is an absurd allegation.

This is exactly what they did in their 2nd assessment report.

Again, (www.ipcc.ch/publications_and_data/publications_and_data_reports.shtml#1) the IPCC second assessment report, 1995, "Working Group I: The Science of Climate Change" page 58, (docs.google.com/open?id=0B1gFp6Ioo3aka3NsaFQ3YlE3XzA). Or imgur.com/gDRQL15.

In the picture "greenhouse gases" back-radiate 324 W/m² only in one direction to the surface. If the same 324 W/m² radiation from "greenhouse gases" to space were not "missing", the system would radiate away more energy than enters the system, which would be equivalent to creating energy out of nothing and clearly absurd. The absurd "greenhouse effect" can "work" only on paper if cooked properly.
#54 ewiljan 2013-09-10 01:05
[quote name="Robert Brown"]Ah, yes, but you have to actually know how to read graphs and stuff like that to figure all of that out./quote]

(SNIPPED)

[b[I also say, that there is absolutely no physical reason for such radiation to be generated by thermal means.
Yes indeed here I was writing of your claims of non-physical "back radiation", To repeat
There is absolutely no physical reason for such non-physical radiation from cold to hot
be generated by thermal means.

This one is harder, Such a proof demands a absolute definition of both "temperature" and "heat" in physical, not conceptual, terms!
No such definition is available, from Climate Clowns, sceptics, deniers, or anyone else.
From the above, my statement

Gosh, really? So the zeroth law of thermodynamics is lost on you, eh?
The zeroth law of thermodynamics "only"
says that this value of temperature is "what the thermometer says it is.; All of the laws of thermodynamics and all the laws
of physical observation demand that all constants and variables be defined. Temperature, and heat have never been defined.
All is only "ya know what we mean"

All Climate Clowns take advantage, intentionally of such lack of physical, definition.

Well, that's certainly one way to argue. I have to wonder -- do they make everybody who posts on this list take special stupid medications, or does PSI membership come with a lifetime supply of serious narcotics?

(Administrator:Since you didn't show it as a quote you have added plagiarizing to your growing list of problems?)

Tommy where are you?

(Administrator: Watching YOU closely since you are the most troublesome person we have)

In the meantime, it is good to know that you say that there is absolutely no physical reason for radiation to be generated by thermal means. Really, that says it all. As you must agree, I stated no such thing, ALgore.
#55 ewiljan 2013-09-10 01:22
Quoting Robert Brown:
OK Bobby, Demonstrate "any" of Your Climate Clown nonsense! No Climate Clown textbooks
Just physical proof. Non-physical effects are denied in science!

I have no idea what this even means.

Truly, you have a dizzying intellect. I will wait until you get going... then I'm certain I'll be even dizzier.
You betcha!!!! idiot!

rgb

Cool! Do you have any evidence if what you claim? Where is it?

dO YOU HAVE ANY EVIDENSE OF WHAT YOUN CLAIM?l
#56 Robert Brown 2013-09-10 01:40
So you claim, where are your measurements?

Gosh, did you miss the link I included above?

Here's another. And yes, actual measurements:

archive.iypt.org/iypt.../2011_13_Light_bulb_Iran_RMN_AT_v1.pdf‎

Note well the measured radiant efficiency curve, which saturates at around 90%. That is, this MEASUREMENT shows that in fact, incandescent light bulbs run anywhere near their design voltage (or higher) lose some 90% of their energy in the form of radiation, and only around 10% to conduction, even if the bulb is suspended in a water bath where water is a far better conductor/convector of heat than air.

So the interesting thing is that you were perfectly willing to pull numbers right out of your nether regions without any experimental evidence whatsoever to back them up. I've now presented you with two papers (and I'm sure there are plenty more) that work through the theory, one of which compares theory to experiment quite successfully.

It is, as I noted, a simple point of fact that the entire article up above is pure nonsense because it assumes without proof and in fact correctly that the light bulb loses the input power predominantly through conduction and convection, and error that you blindly and foolishly supported. In fact, over 80% -- as much as 90% -- of the power input to an incandescent light bulb as electrical Joule heating emerges as radiant energy, although as you do correctly note only a fraction of that -- perhaps a sixth, is visible light (which is why incandescent light bulbs are lousy ways to make light).

One has to ask -- where is the author of the top article? Does he have any comment to make on this egregious error? Apparently not. Where is John O' Sullivan, who once again posted a triumphant crow to his list over how stupid I am and how this article proves it, in spite of the fact that it cannot even get the basic physics of the LIGHT BULB right. Then there is the eternal problem that this problem is not the same as the heated sphere in a vacuum precisely BECAUSE even 10% conduction is an annoying, confounding effect.

What about it, John? Prepared to go "oops" once again? Oh, sorry, these are MAGIC light bulbs, not real light bulbs, the ones that do not cool via radiation at all.

Never happen, of course. It's all about the headlines and the warm fuzzy feeling of having a dizzying intellect and a medium that permits you to advance a political agenda and the hell with reason, math, or science.

Including, my friend ewiljan, measured science, since you seem to want to evade any sort of algebra like the plague, consider it OK to just make stuff up like "a bulb placed in a vacuum will overheat in minutes" when you have no idea whether or not this is true EITHER on sound theoretical grounds OR on the "experiment" or "measurement" grounds you want to hold everybody else to when you don't want to think about theories you don't understand.

rgb
#57 ewiljan 2013-09-10 03:04
[N
ote well the measured radiant efficiency curve, which saturates at around 90%. That is, this MEASUREMENT shows that in fact, incandescent light bulbs run anywhere near their design voltage (or higher) lose some 90% of their energy in the form of radiation, and only around 10% to conduction, even if the bulb is suspended in a water bath where water is a far better conductor/convector of heat than air.
Show your claimed numbers, All yack and no data!
Ib] Interesting Climate Clown Bullshit. I must admit that you Bobby are an expert on misdirection.

b]Including, my friend ewiljan, measured science, since you seem to want to evade any sort of algebra like the plague, consider it OK to just make stuff up like "a bulb placed in a vacuum will overheat in minutes" when you have no idea whether or not this is true. You Bobby Lie I indeed have done that (1971) and recorded the result. EITHER on sound theoretical grounds OR on the "experiment" or "measurement" grounds you want to hold everybody else to when you don't want to think about theories you don't understand.

I wish never to consider rabid warmest theory.
I have associates that do, They call me "No fun at all"

Have you ever done that one (SNIPPED), I thought not. Please show any of your measurements,and al other measurementsl that you have verified. You can only post the results of others, those likely to be more stupid than you, and cannot write legibly.
#58 John Marshall 2013-09-10 08:00
My liquid in my vacuum flask is hot on entry, stopper inserted and left alone. NO extra heating because that was not the conditions that normally exist, unless your flask has an internal heating coil.

The sun is the only source of heat for the earth. Geothermal heat does not have the flux to have much, if any input. But what has this to do with what I wrote above

Also Joel please explain why deserts are hotter than rainforest at the same latitude. IF the GHE existed the reverse would apply given the high volume of GHG's in the rainforest atmosphere, water vapour. So called GHG's COOL the atmosphere by expediting heat loss.
#59 John Marshall 2013-09-10 08:03
TOTAL rubbish. If the 2nd law failed at the microscopic level it would fail at all levels. You cannot bend the laws to fit your ill informed knowledge.

I suggest you look up the real meaning of entropy.
#60 John Marshall 2013-09-10 08:11
NO Dr brown the idea of back radiation is to continue the idea that CO2 drives climate by increasing the energy that comes from the sun. The model used by, and published in AR4 by the IPCC is of a flat earth with 24/7 sunlight whereas reality is a spherical planet revolving on its axis once every 24 hours. reality needs no GHE to work. Look at the K&T figures, TOA insolation spread over the whole planet which does not happen because we, in reality, have a night time.
#61 John Marshall 2013-09-10 08:18
No Dr Brown my sphere is heated by the sun I have never mentioned an internally heated sphere. by an internally heated sphere would radiate heat to the level of the applied power and no further.
#62 Tim Folkerts 2013-09-10 08:29
"In the picture "greenhouse gases" back-radiate 324 W/m² only in one direction ..."

No, quite clearly the greenhouse gases radiate up and down -- the diagram shows 199 W/m^2 of thermal IR going up from the atmosphere.

For an extended 'object' like the atmosphere, there can be different temperatures at different locations, and hence emit different amounts of IR in different directions.

PS. The radiation -- whether going up or down -- is from both the GHGs AND the clouds.
#63 Greg House 2013-09-10 09:08
Quoting Tim Folkerts:
"quite clearly the greenhouse gases radiate up and down -- the diagram shows 199 W/m^2 of thermal IR going up from the atmosphere.
For an extended 'object' like the atmosphere, there can be different temperatures at different locations, and hence emit different amounts of IR in different directions.

No, again, in the IPCC picture the "greenhouse gases" radiate only in one direction to the surface. Regardless of what in the same picture they meant by "emitted by atmosphere" (also in one direction only!), it is much less than what the "greenhouse gases" radiate to the surface (165 is less than 324), so even if they meant just "greenhouse gases" when they wrote "atmosphere" in the same picture (absurd, but let's assume it), the correction would still lead to energy production (324-165=159) out of nothing. More energy comes out of the system than enters the system. Absurd.

As for your "emit different amounts of IR in different directions", according to the IPCC the "greenhouse gases" radiate in all directions, hence differences in thickness and temperature in the atmosphere can not play any part and the overall radiation out of the atmosphere must be the same in both directions. Both the thicker and the thinner, warmer and colder parts, every single part radiate in both directions.
#64 Joel Shore 2013-09-10 10:35
Ah...Someone who has uncritically bought into the nonsense that Postma writes.\

Here are the facts:

(1) The climate models used don't assume a flat Earth. The picture that you refer to is a just a graphic showing the energy balanced averaged over the entire Earth. It represents about the simplest model you can write down, not the more complex models actually used to do serious quantitative studies.

(2) To understand the global energy balance, you don't have to understand the details of how the sunlight is distributed. You just have to know the TOTAL power from the sun that is absorbed by the Earth system (including atmosphere). The total amount of power coming into the Earth is what places a bound on what the average surface temperature can be under the assumption of a radiatively-inactive atmosphere (i.e., an atmosphere that allows all the radiation emitted by the surface to escape to space). If you want more details like the actual temperature distribution then, yes, obviously you need a model that includes details like the diurnal and seasonal distributions of sunlight and the effective heat capacities of various elements of the system...And, as I noted above, those are the sort of details included in more complex models. However, they are not necessary to understand how warm the Earth's surface could be on average given the amount of sunlight it receives (and assuming the surface emits all this energy back into space without the atmosphere absorbing any of it).
#65 Joel Shore 2013-09-10 10:46
There is no intrinsic difference between a body receiving a fixed amount of power from another (hotter) source like the sun vs one where it is being produced internally.

And, Dr Brown's point was that your example of liquids in a vacuum flask is fundamentally different because there is no such source. So, no, nothing regarding back-radiation would predict that the liquid would warm instead of cool. The back-radiation from the cooler surroundings would, by the 2nd Law, always be less than the radiation the object is emitting and it would cool.

In the case of the Earth or an internally-heated object, is not the back-radiation that is warming the object; it is that source of energy that is. However, the back-radiation is preventing the object at a given temperature from cooling as rapidly as it would at that same temperature if it were instead surrounded by space at 2.7 K. And, since it must be in net emitting heat at the same rate as thermal energy is being produced (or received from the sun) in order to be in a steady-state, the steady-state temperature necessary for this to occur will be higher.

It is really not rocket science. It is the same thing that would happen if your drafty house with the furnace running full-steam was only able to keep your house at 15 C when it was -20 C outside. If you then added insulation to the attic and walls, better windows, etc., the furnace would now be able to keep your house at a toastier temperature.
#66 Tim Folkerts 2013-09-10 10:49
"No, again, in the IPCC picture ..."

Seriously? You are basing your understanding on a SINGLE picture? And from that you are imagining what the paper actually says? And from that you are extrapolating to the entirety of science?

"hence differences in thickness and temperature in the atmosphere can not play any part and the overall radiation ... "

Seriously? Difference in temperature cannot play a roll in the amount of thermal IR emitted?
#67 Joel Shore 2013-09-10 10:55
There is no intrinsic difference between whether you have an internal heat source or have a hotter body, like the sun, supplying the heat. The point is that in our examples, this is always the case and in your example it is not the case.

That is a fundamental difference and is why nobody would argue that the liquid would warm instead of cool. You are just making up an argument that we, nor anybody else who understands, physics would ever give.

As for deserts vs rainforest, the real difference between them is in the diurnal temperature range. You would have to demonstrate that the deserts were really hotter on average. There are also other important differences involving cloud albedo. (Clouds have a cooling effect due to albedo and a warming effect due to the greenhouse effect. However, their albedo effect is larger ON AVERAGE than their greenhouse effect. ("On average" because the relative magnitudes of the effects vary with cloud type...and also with latitude due to variation insolation...and so forth.) There are other issues too: There are huge transports of energy around the surface of the Earth. This doesn't matter for the global energy balance, but does matter when you are trying to do energy balance on a local scale without accounting for them.
#68 David Cosserat 2013-09-10 14:19
I do wish you guys would stop quarrelling.:-)

You are never going to convince one another using angry words. Instead I propose that you all carefully study the following experiment, which is designed to provide EMPIRICAL PROOF one way or the other.

1. A solid cylindrical core hangs inside a vacuum chamber, suspended by extremely thin copper wires which are connected through the top of the vacuum chamber wall via a vacuum-sealed entry point.

2. The outer surface of the core and the inner surface of the vacuum chamber are painted matt black with a measured emissivity of ~1.

3. Embedded inside the core is a heating coil, connected to an external power supply via the copper suspension wires.

4. A miniature electronic temperature sensor is mounted on the cylinder's surface, connected to a temperature reading device via the copper suspension wires.

5. Another electronic temperature sensor is attached to the vacuum chamber wall.

6. The chamber is pumped to hard vacuum (
#69 David Cosserat 2013-09-10 14:20
6. The chamber is pumped to hard vacuum (
#70 David Cosserat 2013-09-10 14:21
I do wish you guys would stop quarrelling.:-)

You are never going to convince one another using angry words. Instead I propose that you all carefully study the following experiment, which is designed to provide EMPIRICAL PROOF one way or the other.

1. A solid cylindrical core hangs inside a vacuum chamber, suspended by extremely thin copper wires which are connected through the top of the vacuum chamber wall via a vacuum-sealed entry point.

2. The outer surface of the core and the inner surface of the vacuum chamber are painted matt black with a measured emissivity of ~1.

3. Embedded inside the core is a heating coil, connected to an external power supply via the copper suspension wires.

4. A miniature electronic temperature sensor is mounted on the cylinder's surface, connected to a temperature reading device via the copper suspension wires.

5. Another electronic temperature sensor is attached to the vacuum chamber wall.

6. The chamber is pumped to hard vacuum (less than 1mbar).

7. The power supply is switched on and delivers exactly 2 watts of power.

8. The surface temperature of the core rises slowly until it eventually reaches a steady value Tc1.

9. At this point the vacuum chamber's wall temperature is measured as Tv1.

The whole experiment is now repeated but this time the vacuum chamber is first placed inside a freezer. When it reaches steady state temperatures, the vacuum chamber's wall temperature is measured as Tv2 and the core's surface temperature is measured as TC2.

RESULT: Tv2 is ~30degC below its previous value, Tv1.

QUESTION: Will Tc2 be significantly different from Tc1?

A. If I am correct in my understanding of the Slayer position they say that the core temperature will not have changed significantly.

B. If I am correct in my understanding of the Warmist position it is that the core temperature will have dropped by ~30degC.

So, no squabbling and no colluding, what do all you gentlemen predict?
#71 ewiljan 2013-09-10 16:47
Quoting Tim Folkerts:
"it must be clear ..."

No, it is not clear that anything would "get warmer and warmer". In fact the opposite is quite clear -- back-radiation (in conjunction with an active heating source like the sun -- will lead to definite but finite warming.

If you think it is quite clear, then give us some equations and numbers for a 'simple' case where this would happen. Perhaps the old "heated sphere surrounded by a thin shell'. You know, the one where the sphere will be 2^(1/4) times warmer with the shell than without the shell. 2^(1/4) times warmer in agreement with every physics text, not infinitely warmer in agreement with Greg's physics.

The sphere increased in temperature spontaneously so it would be able to radiate its fixed power to a warmer surface, the shell,
The shell simultaneously is radiating exactly that same power outward to a colder environment. There is no power to be "back radiated to the sphere. The sphere radiares
the inner surface of the shell absorbs all of that radiation but radiates nothing at all,
It is the outside surface of the shell that radiates all of the power absorbed by the inner surface. This proves that radiation does not emit from the inner surface of the shell as its temperature is lower than that of the sphere Thermal radiation is not a process if the fake statistical mechanics.
#72 ewiljan 2013-09-10 18:10
Quoting Robert Brown:
Even without numbers it must be clear that in certain cases a body warmed by back radiation would produce even more radiation and thus even more back radiation and get warmer and warmer and so on, and more energy would come out of the system than has entered it, which is absurd.

It is indeed absurd, so why do you assert this? There are absolutely zero cases where this happens,
Thank you! you have just asserted "no back radiation".

if you write down the actual differential equations, they are quite solvable with clearly and trivially derivable steady state solutions.
What differential equations would you be using to solve a steady state thermodynamic equilibrium?

By that I mean that if you start the central sphere/bulb, heated with a fixed constant power, too warm in temperature (which is an initial condition independent of the differential equations per se), it will cool to the dynamic equilibrium temperature
True but not by the "back radiation" of its environment,. You can not demonstrate any
"back radiation", such is a Climate Clown fallacy. Please explain how your central sphere/bulb got to a temperature above the dynamic equilibrium temperature?

and if you start it at an initial temperature that is too cold, it will warm to exactly the same temperature.
Indeed, again with no such "back radiation"

This happens for the plain old heat equation OR for the Stefan-Boltzmann equation. In both cases heat is transferred from warmer to colder. In the case of heat conduction, the rate of transfer is proportional to the difference in temperature. In the SBE, it is proportional to the difference in the fourth power of the temperature.
Thank you again for pointing out that it is the "difference" in thermal potential that determines both the value and direction.
The Ta^4 is indeed the radiative potential from any emissive surface. Again you demonstrate the complete lack of "back anything" in thermodynamics.

You, and Joe Postma, continually WANT to assert that if there is such a thing as backradiation, the process of radiative heat transfer is dynamically unstable,
Yes and both are insisting that the concept of"back radiation' is un-physical and Climate Clown nonsense

Thermal radiation only emits as determined by Kirchhoff's law of radiation, Likely different at each wavelength and in each direction. Your statistical mechanics Bull Shit when applied to thermal radiation is just that Bull Shit. !
#73 ewiljan 2013-09-10 18:30
Quoting Robert Brown:
The modern understanding of thermodynamics...and the 2nd Law in particular...is that the irreversibility implied by the 2nd Law (the future is different from the past because, for example, heat flows only from hot to cold) arises from completely reversible behaviour at the microscopic level. That in fact is what makes physics interesting and deep, rather than just a collection of silly arbitrary and capricious rules, as you want to reduce it to with your incorrect ideas that photons can only go from a hotter object to a colder object.

Excuse me? Nobody I have heard asserted any such thing, although you came close up above. The Stefan-Boltzmann equation describes the NET radiative power transferred between a body and its surroundings when the body and surroundings are in thermal equilibrium and are at least approximately "black bodies" (usually with an emissivity assumed to be near unity). It contains an outgoing term for the object that arises from integrating over the Planck distribution, and an incoming term (which is equally outgoing from the surroundings). Photons go both ways, and are simply more likely to be emitted and in general are emitted with more energy from warmer black bodies than from cooler ones, resulting in a biased transfer of radiative power from warmer to cooler. This does, incidentally, satisfy the second law.

So I'm not sure what your point is. Most of the nuts on this list spend all of their time ranting about how colder objects cannot transfer radiation at all to warmer ones, or asserting that the radiation thus transferred somehow doesn't "count" in the energy budget on the receiving end. If your point is that the correct description of the system is statistical mechanics (of open systems, no less) rather than thermodynamics, well sure. OTOH, for simple problems like the one posed, it is pretty obvious that one doesn't NEED real stat mech to very precisely explain and compute what is going on, any more than I need to use quantum theory to describe the motion of a baseball.

rgb

From your buddy and fellow Climate Clown. In this thread Joel Shore 2013-09-09 16:33
"The modern understanding of thermodynamics...and the 2nd Law in particular...is that the irreversibility implied by the 2nd Law (the future is different from the past because, for example, heat flows only from hot to cold) arises from completely reversible behavior at the microscopic level. That in fact is what makes physics interesting and deep, rather than just a collection of silly arbitrary and capricious rules, as you want to reduce it to with your incorrect ideas that photons can only go from a hotter object to a colder object."

#74 ewiljan 2013-09-10 18:51
So I'm not sure what your point is. Most of the nuts on this list spend all of their time ranting about how colder objects cannot transfer radiation at all to warmer ones, or asserting that the radiation thus transferred somehow doesn't "count" in the energy budget on the receiving end. If your point is that the correct description of the system is statistical mechanics (of open systems, no less) rather than thermodynamics, well sure. OTOH, for simple problems like the one posed, it is pretty obvious that one doesn't NEED real stat mech to very precisely explain and compute what is going on, any more than I need to use quantum theory to describe the motion of a baseball.

statistical mechanics has anything to do With orthogonal time varying electromagnetic fields? Please show how statistical mechanics
can show how several such EM fields from several locations may or may not influence the magnitude and direction of such electromagnetic radiation? Tell me about using the wrong version of science to deliberately confuse others.
#75 ewiljan 2013-09-10 19:32
Your reference ignores the tremendous convective
heat transfer from the filament to the glass envelope (via argon). and that from the envelope to the 20 C air mass. This can easily be 100x the thermal conductivity of the gasses,
This is the balance to limit both the filament temperature and the envelope temperature. You offer no reference from the only ons that know.
These are the folk that produce the things and do know "why" they are built that way!
The aluminium foil does little or nothing to affect this thermal convection. Again your bookish analysis is nonsense, trust the measurements. Alberto's analysis remains correct. Thermal radiation is not involved.
#76 ewiljan 2013-09-10 19:54
This one is harder, Such a proof demands a absolute definition of both "temperature" and "heat" in physical, not conceptual, terms!
No such definition is available, from Climate Clowns, sceptics, deniers, or anyone else.

Gosh, really? So the zeroth law of thermodynamics is lost on you, eh? Equipartition theorem, statistical mechanics, all just whoosh, right over your head?

Are you claiming that the zero'th law of thermodynamics either defines in physical terms, or even assists in understanding what temperature "is" or what heat "is" , forget your statistical mechanics Bull Shit. If in isotherm has temperature "T" That "T' has nothing to do with your fake non-Newtonian "kinetic energy". It is heat, a form of energy, different from Newtonian kinetic energy, different from electrical energy, but never defined in physical terms.
Temperature may best be defined as a measure of "noise" which has energy but much different than other forms of energy that have "direction"!
#77 Joel Shore 2013-09-10 20:55
It amazing how people who understand so little physics can have such strong opinions. The Second Law is something that emerges from microscopic reversibility plus the statistics of large numbers of particles.

The idea that different physics emerging at different levels is fundamental to our understanding of everything from heat flow, phase transitions, the fractional quantum Hall Effect, and all sorts of fascinating phenomena.

It is excusable to be ignorant. It is inexcusable to let your ideology oppose any attempt to dispel your ignorance and to combine your ignorance with an arrogance that you know better than people who know orders of magnitude more than you do.
#78 Tim Folkerts 2013-09-10 21:18
B) The core would warm up if the wall warms.

NOTE1: The warming of the core would always be less than the warming of the wall. The exact amount depends on the specific numbers chosen for the initial temperature of the walls and the power of hte heater.

NOTE2: Paint outgasses badly, so no one would want to paint their vacuum chamber.
#79 ewiljan 2013-09-10 22:18
Quoting Joel Shore:
It amazing how people who understand so little physics can have such strong opinions. The Second Law is something that emerges from microscopic reversibility plus the statistics of large numbers of particles.

The second law of thermodynamics was established way before your fake theory of statistical mechanics. Why do you insist of corrupting the laws of thermodynamics?

Did you even read what you wrote?

It is excusable to be ignorant. It is inexcusable to let your ideology oppose any attempt to dispel your ignorance and to combine your ignorance with an arrogance that you know better than people who know orders of magnitude more than you do.

If it is inexcusable to let your ideology oppose any attempt to dispel your ignorance,
Why do you do that?

Then to combine your ignorance with an arrogance that you know better than people who know orders of magnitude more than you do, which is almost anyone, is even more inexcusable.
#80 ewiljan 2013-09-10 22:59
And, Dr Brown's point was that your example of liquids in a vacuum flask is fundamentally different because there is no such source. So, no, nothing regarding back-radiation would predict that the liquid would warm instead of cool. The back-radiation from the cooler surroundings would, by the 2nd Law, always be less than the radiation the object is emitting and it would cool.

The Second Law of thermodynamics by Rudolf Julius Emmanuel Clausius (1867). Specifically prohibits your fake "back radiation". That law has never been falsified. You Climate Clowns can provide no demonstration of this so called radiation.

All of what you claim in this reply is easily done with no "back radiation" and no violation of 2LTD. The proper use if the S-B equation between the temperature of the surface and the temperature of the atmosphere will eliminate any claim of "back radiation". The earth is partially radiating only to the intermediate temperature atmosphere, not to cold space, and certainly not to zero Kelvin, which is the Climate Clown ridiculous claim and the source of the need for your ridiculous claim of "back radiation".
#81 ewiljan 2013-09-10 23:43
B. almost but only "initially" Tc2 will never drop by Tv2-Tv1. Eventually both temperatures
reach "T3" with Tv2 < Tv3 < Tv1, and Tc2 < Tc3 < Tc1 All of the 2 watts is absorbed by the walls of the freezer. There is never any of the fake "back radiation". The actual equilibrium temperatures depend on the convective heat transfer in the surrounding environment of the vacuum chamber. Why increase the complexity of the original,and secondary situations?
#82 Tim Folkerts 2013-09-11 00:12
"You Climate Clowns can provide no demonstration of ___________"

You are not in a discussion with "Climate Clowns" -- you are in a discussion with people who earned PhDs in physics LONG before we worried about climate change. We represent 'main-stream physics' and our views mirror those of our physics colleagues in colleges and universities around the world. We represent the "textbook answers'.

Here is your challenge (same goes for Greg or Joe Postma). Go into (or send an email to) ANY Physics Department in ANY university and find the professor who teaches thermodynamics (or ask for the oldest prof who got his PhD long before "climate change' might have 'corrupted' his thinking). Ask for an opinion on this thread. I guarantee that prof will side with Dr Brown & Dr Shore and Dr Folkerts.

Give us the name of ANY physics prof in ANY university who will side with you and your interpretation of thermodynamics. Currently the list is 3-0 against you.
#83 Greg House 2013-09-11 00:22
Quoting Joel Shore:
In the case of the Earth or an internally-heated object, is not the back-radiation that is warming the object; it is that source of energy that is. However, the back-radiation is preventing the object at a given temperature from cooling as rapidly as it would at that same temperature if it were instead surrounded by space at 2.7 K.

Back radiation can warm neither a cooling object nor a warming object. The assumption that back radiation can supply the source with energy is false, because such a process would lead to the body getting warmer and warmer infinitely and to energy production out of nothing, which becomes obvious if we consider the body initially at a stable temperature. I have already said that on this thread. The "greenhouse effect" is an absolutely absurd concept.
#84 Greg House 2013-09-11 00:26
Quoting Tim Folkerts:
Seriously? Difference in temperature cannot play a roll in the amount of thermal IR emitted?

I did not say that. Please, do not distort my point.

I said "overall radiation out of the atmosphere must be the same in both directions", despite internal differences in thickness and temperature.
#85 Greg House 2013-09-11 00:35
Quoting Tim Folkerts:

Seriously? You are basing your understanding on a SINGLE picture? And from that you are imagining what the paper actually says? And from that you are extrapolating to the entirety of science?

"Entirety of science"? Come on, I am talking about the "greenhouse effect" based on the absolutely absurd notion of "warming the source by back radiation", and the IPCC who presented this nonsense as science in their 2nd assessment report on the page 58.

Their presentation is apparently cooked, it becomes obvious if one takes a closer look at their "energy balance" they presented in that picture.
#86 Greg House 2013-09-11 00:51
Quoting Tim Folkerts:
You are not in a discussion with "Climate Clowns" -- you are in a discussion with people who earned PhDs in physics

I do not see a contradiction here, although I do not like the term, because it is insulting to real clowns who are highly skilled professionals.

People who earned PhDs in physics and promote the absurd so called "greenhouse effect" can only be either stupid (rare but possible) or liars. OK, crazy would be another option, but that is all.
#87 ewiljan 2013-09-11 01:30
Quoting Tim Folkerts:
"You Climate Clowns can provide no demonstration of ___________"

You are not in a discussion with "Climate Clowns" -- you are in a discussion with people who earned PhDs in physics LONG before we worried about climate change. We represent 'main-stream physics' and our views mirror those of our physics colleagues in colleges and universities around the world. We represent the "textbook answers'.

Here is your challenge (same goes for Greg or Joe Postma). Go into (or send an email to) ANY Physics Department in ANY university and find the professor who teaches thermodynamics (or ask for the oldest prof who got his PhD long before "climate change' might have 'corrupted' his thinking). Ask for an opinion on this thread. I guarantee that prof will side with Dr Brown & Dr Shore and Dr Folkerts.

Give us the name of ANY physics prof in ANY university who will side with you and your interpretation of thermodynamics. Currently the list is 3-0 against you.

No mater if Dr Brown & Dr Shore and Dr Folkerts.are Climate Clowns or not any of you can provide even one demonstration of this so called "back radiation".
Not any of you can provide even one demonstration that objects with a temperature must radiate in all directions. Case in point the earth radiated "no" energy to the Sun.
Not any of you can provide the reason for thermal electromagnetic radiation at all!
So much for your collective PHD's,
#88 Greg House 2013-09-11 01:45
Quoting ewiljan:
Not any of you can provide the reason for thermal electromagnetic radiation at all!

That's why I think we should not try going very deep into the matter. For practical reasons it is sufficient to demonstrate that if there is back radiation it can not warm the source, because if there is no back radiation it can not warm anything anyway. Since there no third option, the "greenhouse effect" would be fully explored this way. There is no need for deep theoretical discussions.
#89 GHEbreaker 2013-09-11 03:12
@Joel Shore + Robert Brown etc.

I repeat, no serious engineer, or physicist, or technician is using the S-B ALONG WITH the Fourier equation of conduction, to calculate the temperatures of two closely connected bodies (aluminum foil + glass of light bulb), because otherwise you would get TOTALLY WRONG RESULTS, because you would "create" new further Thermal energy out of nothing.

The point is that both Robert Brown and Joel Shore are just saying words, words, words...a lot of annoying words, but they are unable to make precise calculations that engineers and technicians are making every days, getting correct results.

For instance, last year I showed with precise calculations that in Wood's experiment of 1909, with greenhouse boxes, that Pratt tried to repeat in 2009, the difference of 15°-20° C between the 2 boxes that Pratt found after 1 hour of solar heating, and that NEITHER Robert Wood nor Nasif Nahle (in 2011) evidenced in their experiments, was NOT caused by the fancy "backradiation", but simply by a small difference of humidity between the 2 boxes, that changed the air CONDUCTIVITY.
So, sorry for you, but if you want to to persuade the persons that - for instance - in the Kiehl and Trenberth diagram it could be possibile to invent afresh a "backradiation" from sky, coming from nowhere for 390 W/m^2, forgetting that air emissivity is not 1, but it is 0.2, and therefore 390 W/m^2 would be like 1950 W/m^2 black body, corresponding to 430K namely 157° C of air temperature(!) you are perfectly free to do that.

But please, don't do that with me. I'm working since 1990 with technicians and engineers making TRUE calculations, projecting industrial plants and machinery, and NONE of us ever used your fancy "backradiation" or similar, to calculate heat transmission of plants and surfaces.
And sorry for you, our calculations have ALWAYS been confirmed by ACTUAL RESULTS, because we did not create energy out of nothing as you.

You believe in "backradiation", and probably in flames going downward, or rivers flowing from outlet to the source.
I don't! I see that flames are going upwards, because heat is flowing from more dense to less dense regions of air, and according to a negative lapse-rate, and rivers are following Pascal-Bernoulli Law, going from source to outlet.
Sorry, once you'll be able to show me flames and rivers following a reverse path, maybe I can start to believe in your magic "backradiation"!
#90 Greg House 2013-09-11 03:25
Quoting GHEbreaker:
You believe in "backradiation", and probably in flames going downward, or rivers flowing from outlet to the source. I don't!

Let me guess how they could explain the rivers: the water flows in both directions, but the NET flow is from the source to the outlet.
#91 GHEbreaker 2013-09-11 03:35
@Joel Shore, Robert Brown, etc.

Ah, I forgot...

You are totally confusing the actual process of heating inside a light bulb, and misusing the equations.

First of all, the tungsten filament is reaching a very high temperature, nearly 3000K, and, since it is transmitting energy to the inside of glass surface, in a vacuum chamber, it is correct to use the SB to calculate the temperature of the INTERNAL SURFACE OF GLASS.

But then, heat transmission from glass internal surface to the aluminum foil in CLOSE CONTACT with it, is following JUST the Fourier equation of CONDUCTION, WITH HEAT (Thermal energy released by the filament in its turn receiving electric energy from electric household plant) FLOWING JUST FROM INSIDE TO OUTSIDE, FROM GLASS TO ALUMINUM FOIL.
AND THEN, WHEN ALUMINUM FOIL GET HEATED MORE AND MORE QUICKLY THAN GLASS SURFACE, FOR HIGHER CONDUCTIVITY OF ALUMINUM, IT STARTS TO HEAT THE GLASS SURFACE OF BULB, BUT FOR CONDUCTION, NOT FOR "BACKRADIATION". 2ND LAW OF THERMODYNAMICS AND CONDUCTION, NOT BACKRADIATION.
#92 ewiljan 2013-09-11 03:47
Quoting Greg House:
Quoting ewiljan:
Not any of you can provide the reason for thermal electromagnetic radiation at all!

That's why I think we should not try going very deep into the matter. For practical reasons it is sufficient to demonstrate that if there is back radiation it can not warm the source, because if there is no back radiation it can not warm anything anyway. Since there no third option, the "greenhouse effect" would be fully explored this way. There is no need for deep theoretical discussions.

While "back radiation" lives, so will Greenhouse effect. There is certainly a need for deep theoretical discussions. There thyree are the folk that teach fake science to unsuspecting high school graduates.Theswen are the folk that make the scam of AGW possible!

8
#93 ewiljan 2013-09-11 04:05
Boars go in both directions . Why not rivers?
?Err I guess the river going to the outlet ,allows, the bots to go through the locks while actually gaining energy. See boats "can gain energy and reduce entropy" spontaneously almost.
Is there no Second Law of Tugboating?
#94 David Cosserat 2013-09-11 04:40
Tim,

On your NOTE1: I would actually expect (on my current understanding of Warmist physics, but I'm happy to be corrected) that Tc [measured in K] would change in proportion to the change in Tv [measured in K]. But it would certainly change significantly.

On your NOTE2: This is an interesting point that could indeed confound a practical experiment. Are you suggesting that black paint outgassing would be an initial effect or do you think the paint would go on outgassing until all the black particulate matter had disappeared, thus confounding the assumption that the emissivity of the two surfaces was close to 1?
#95 David Cosserat 2013-09-11 04:42
ewiligan,

Hi, I thought you has resigned from PSI because you felt they were not extreme enough for you.

You say: Eventually both temperatures reach "T3" where T3 lies somewhere between Tc2 and Tc1.

But if that is true (at steady state) in the freezer, it must, by straightforward logic, also be true (at steady state) in the first case, so Tv1=Tc1.

From this it follows logically that it would always be the case (at steady state) that Tv=Tc, whatever the value of Tv.

I will allow that as an option C and see what other people have to say.
#96 Joel Shore 2013-09-11 10:26
"The second law of thermodynamics was established way before your fake theory of statistical mechanics."

That is why it is impressive that statistical mechanics allows us to understand how the laws of thermodynamics arise from behaviors on the microscopic scale.

"Why do you insist of corrupting the laws of thermodynamics?"

I don't. I just correctly understand the concept of the thermodynamic limit and that heat is a macroscopic quantity, as any reasonable presentation of the subject explains, for example here en.wikipedia.org/wiki/Heat : "Transfers of energy as heat are macroscopic processes. Kinetic theory explains them as the microscopic motions and interactions of microscopic constituents such as molecules and photons."

"Then to combine your ignorance with an arrogance that you know better than people who know orders of magnitude more than you do, which is almost anyone, is even more inexcusable."

So, what are you qualifications for claiming that you know orders of magnitude more than Robert Brown, Tim Folkerts, and myself about this subject? What courses have you taken in thermal and statistical physics and at a what level? Where have you taught physics and out of what textbook? For what companies have you worked and done radiative transfer calculations?
#97 Joel Shore 2013-09-11 10:28
A reference for the concept of the thermodynamic limit: en.wikipedia.org/wiki/Thermodynamic_limit
#98 Joel Shore 2013-09-11 10:40
Fluids are complex because the motions of the molecules are highly-correlated, but let's stick to a simple case of a gas: You have two containers of gas separated by an impermeable partition, with one at twice the pressure of the other. (Although not necessary, for simplicity and greater analogy to photons, let's assume that the pressures are low enough that the gases are rarified and collisions between gas molecules are relatively infrequent.)

You then poke a hole in the partition. We both presumably agree that the net flow of molecules will be from the higher pressure to the lower pressure container until the two pressures equalize.

However, do you believe that if you looked at the hole, you would find gas molecules only going through in one direction (from high to low pressure) or do you believe that you would find molecules going through in both directions but with more going from high to low pressure than going from low to high pressure for the simple reason that more molecules will be impinging on the hole from the high pressure side than the low pressure side?

And, if you believe the latter, how do you reconcile this with the 2nd Law of Thermodynamics that clearly predicts (from entropy considerations) flow from higher pressures to lower pressures, just as it predicts heat flow from higher temperatures to lower temperatures?
#99 Tim Folkerts 2013-09-11 11:40
"Let me guess how they could explain the rivers: the water flows in both directions, but the NET flow is from the source to the outlet."

YES! EXACTLY CORRECT!

On a microscopic level, water molecules are moving at a few 100 m/s. At any given instant, nearly half of the water molecules in a river are indeed moving upstream.

The slight imbalance of water molecules moving downstream is the net current of ~ 1 meter per second that we observe macroscopically.

Nearly half of the electrons in a wire are moving from + to -. The slight imbalance is the net movement of electrons we observe macroscopically flowing from - to +.

Nearly half of the photons in the atmosphere are moving from colder to hotter. The slight imbalance is the net heat we observe macroscopically flowing from hot to cold.
#100 Greg House 2013-09-11 12:00
For the 5th time: the assumption that cold supplies warm with energy and has a warming effect on it is equivalent to production of energy out of nothing. This easily becomes obvious if the warm body is at a stable temperature initially. There is no way around it.
#101 Tim Folkerts 2013-09-11 12:17
I just meant the paint would mess with the vacuum in the chamber and with the pumps that are trying to remove all those VOCs.
#102 Robert Brown 2013-09-11 12:49
For the umpty-zillionth time. No, it is nothing of the sort. You continue to pretend that there is no energy source present in the posited problem, so that it does not describe passive cooling of an object to equilibrium with a surrounding colder reservoir.

The present problem is to describe what happens to an actively heated, finite sized reservoir, in thermal contact with some sort of thermal impedance (resistance to the flow of heat) with an effectively infinite sized cold reservoir that can absorb basically any amount of heat without noticeably changing temperature (in the context of the original problem, the surrounding night sky at 3 K, but the surrounding environment of an air conditioned laboratory room at a nearly constant temperature works fine too), will do when the impedence of the thermal connection is increased.

The answer is bone simple thermodynamics. The temperature of the heated reservoir will then increase from the temperature required to keep it in a constant-temperature steady state where the input energy from the heat source precisely balances the rate at which energy flows through the original impedance of the thermal connection to the cold reservoir to what is required to lose exactly the same amount of heat per unit time to the cold reservoir through the increased impedance.

Surely this isn't so very difficult to understand. It is why you put on a coat when it is cold outside, and why you insulate your house. The coat impedes the flow of energy from your heated body to the colder outdoors, or from your heated house to the colder outdoors. After you increase the insulation, the exact same rate of energy release in the house causes the house to get warmer until the increased temperature differential pushes heat through the insulation at a rate that matches the input and the house is again at equilibrium, only hotter.

Did the insulation "heat the house"? Of course not. The heater heated the house. But the dynamic equilibrium temperature of the heated house without question depends on the rate that the house cools through all channels to the surrounding environment.

Note well all channels -- conduction, convection and radiation all cool the house. Altering the rate of energy flow through any of these channels will alter the temperature at which the power delivered to heat the house is in dynamic balance with its rate of loss to the surrounding cold reservoir.

The easiest way to see that this is true for radiative physics is to consider a system in a vacuum, with no conductive or convective heat transfer at all (which conveniently enough, describes the Earth surrounded by 3 K outer space) that cools only via radiation. Surrounding it with a reflective or absorptive layer simply decreases the rate energy can leave via radiation, and consequently causes the system to warm until it re-establishes power balance through the obstacle.

rgb
#103 Tim Folkerts 2013-09-11 12:51
So basically the recent discussion has been ...

GREG: The idea of back radiation equivalent to the idea of water flowing upstream. Both are stupid.

TIM: Water (at least some of the molecules) does flow upstream all the time.

GREG: OK, water molecules do flow upstream. So it is not stupid for some of the water molecules to go the 'wrong way'. But it is still stupid for some of the photons to go the 'wrong way' because ... ummm ... well, because I said so.
#104 Robert Brown 2013-09-11 13:09
Ah, apparently you missed the part where I presented one of many places where the measured rates of energy loss of a light bulb are presented, measurements that indicate that over 80% of the energy loss is in electromagnetic radiation, with as little as 10% lost via conduction and convection combined. Of the loss in radiation, only a quarter or fifth of it is in the visible part of the spectrum. I don't really care much whether the radiation is from the filament, the bulb, the gas in between -- I'm happy enough to draw a little surface just outside of the bulb itself and consider only energy flow through that surface. I'm even happier to consider a bulb in a vacuum, eliminating conduction and convection altogether.

Now, without the foil you can easily measure energy coming out of that surface in the form of visible light. In fact, your eye is a very lovely detector for that energy. When we surround the bulb (in a vacuum to eliminate that pesky conduction even though even in air loss via conduction is smaller than loss via visible light alone) with a thin, smooth, opaque reflector, gosh, the room gets dark! We've interrupted the energy flowing out of the bulb in the visible part of the spectrum, and our eyes can easily experimentally verify that this interruption occurred. The energy that was flowing out is now being reflected by the shiny foil back into the closed volume containing the bulb. This all happens instantly.

Now, it is left as a simple exercise. You are dumping power P into the bulb. All of P was being lost to the surrounding dark room via radiation in one channel or another -- IR, visible, whatever -- simply because we stuck it in a vacuum and it has no choice -- in dynamic equilibrium, power into the bulb equals power out in radiation, period. We surround the bulb with a reflective shell that reflects 100% of the power in at least the visible part of the spectrum back into the closed volume containing the bulb. The vacuum itself, of course, cannot heat. The foil/shell will eventually heat, but it has a trivial heat capacity (it is very thin). All of the reflected power is directly incident on the bulb itself, and as plain old radiation, it interacts with the bulb, which is hardly perfectly reflective or transparent.

What exactly do you think happens to the temperature of the bulb? Note that we've nicely eliminated conduction altogether.

Incidentally, I'd point out that aluminum has a much higher thermal conductivity than glass or air, so that surrounding the bulb with foil in perfect thermal contact with the glass and with a larger external surface area than the glass would ordinarily be strictly expected to cool the glass an bulb by reducing the thermal impedance to the surrounding air rather than increasing it, but I'm guessing that even qualitatively assessing the actual heat equation is probably beyond you, eh? So let's stick to a vacuum.

rgb
#105 Robert Brown 2013-09-11 13:42
I see. So you are then prepared to conclude that the sun transmits its energy to the earth via conduction, rather than cooling via radiation and in the process warming the earth? Because the light bulb in a vacuum can apparently (according to you) no longer cool.

I suppose you didn't think that just maybe a light bulb's glass contains gas at a pressure (very low pressure at room temperature) that -- when heated -- is designed to be at just under 1 atmosphere of pressure, just under so that the bulb will IMPLODE instead of EXPLODE if it is broken. The glass is, of course, as thin as it can be to be stable under this range of variation, with a persistent overpressure on the outside where spheroidal surfaces are considerably more stable under compression than extension, hence the invention of the archectural arch and dome, etc. So when you put a bulb in a vacuum, and turn it on, you -- for the first time every create a pressure close to 1 atm inside the bulb and zero outside the bulb. This is, if you prefer, equivalent to taking the bulb and filling it with gas at 2 atm in the first place. I suspect that a glass bulb would have a hard time supporting a water column ten meters high in air, especially as the glass is in the process of thermally expanding at what is doubtless a differential rate.

But I'm sure that you -- at least -- are certain that your observation of explosion had nothing to do with pressure as opposed to the overheating of the filament. I agree that the filament and bulb would get warmer, BTW -- you've blocked 10% to 20% of the heat flow out of the lamp, of course it will get warmer to increase the rate of loss in the radiative channel. That if anything will further increase the pressure inside a bulb designed only to withstand a negative pressure to a positive overpressure greater than 1 atm.

I'm just curious -- even you agree that the bulb loses at least 16% or so of its energy in visible light, right (in part because there is really no doubting the published numbers on that, that's bone simple to measure). If you surround the bulb with any opaque reflector, all of that power is delivered back onto the bulb. What exactly do you think happens to that power? It isn't going out into the room any more -- the room is now dark. It's hard to read by the light emitted by a foil covered bulb.

The foil if anything decreases the thermal impedance of the bulb by replacing a layer of air with a layer of higher thermal conductivity foil (the argument in the top article gets the sign wrong because it doesn't take into account the fact that the foil DISPLACES existing air with a lower thermal conductivity, and since it completely neglects radiation and since we AGREE that visible light radiation is a nontrivial component of the bulb's energy loss (the reason for having the bulb in the first place, after all) its answer isn't going to be even close to correct.

What do you think happens to the bulb? Obviously it will warm -- or at least I would have said it is obvious -- but you want to claim otherwise.

rgb
#106 Robert Brown 2013-09-11 14:03
Really, Greg?

Stupid? Let's see, I lecture introductory physics cold (no notes) because I don't need notes, I know the subject. I never prepare solutions to the physics problems my students struggle with, because I can solve any introductory physics problem cold and get it right within careless errors essentially 100% of the time, without looking, on the spot. I've written textbooks on both introductory physics (both semesters), advanced graduate level electrodynamics (all online on my website, feel free to check them for errors). I'm a theorist, have completed basically a masters worth of math, have written a book on computer science, am an expert systems administrator and and ubercoder with over a gigabyte of code I've authored in a variety of languages. I've got a patent pending in advanced bayesian predictive modeling in the context of multiple databases with privacy boundaries. I think you'd find it very difficult to prove me "stupid" in any defensible meaning of the term.

So that leaves us with "liars" or "crazy". Well, if somebody way smarter than you and far better educated than you who actually knows the relevant equations, laws, and theorems of physics without having to look them up and who can do algebra, geometry, calculus through functional analysis and differential equations disagrees with your strictly verbal arguments, unsupported by anything like a formal argument based on some sort of mutually agreed on physical principles I suppose you could indulge in ego-protecting cognitive dissonance and call them a liar to make yourself feel better, but is that really productive?

Once again I have to remark that this site is one giant hotbed of narcissistic personality disorder and gradiosity. Ew-jan cannot seem to make a single post without ad-hominem in it, even to people he's never met. Somehow he thinks that calling somebody a clown is something other than a logical fallacy in rational argumentation.

Now, I'm happy to have a rational, ad-hominem free civilized discussion based on the accepted laws of physics, a certain amount of algebra, even some well-done conceptual reasoning, provided that at the end of it both parties agree that solutions that at the end of the day the unique correct solution has to quantitatively incorporate all of the relevant physics. But it never seems to happen, because the only physical principle I ever hear invoked is that "cold can never heat warm" or the like, which is actually not what the second law of thermodynamics says and which is not what happens in the greenhouse effect (a rather complex process) or the much simpler toy problems discussed to help people see that surrounding a radiatively cooled heated object with ANY passive absorber between it and a surrounding cold reservoir will result in its temperature increasing.

In the meantime, I guess calling me crazy or a liar is an easy way to avoid having to actually present an algebraically consistent argument while protecting your ego, so sure, go for it.

rgb
#107 Robert Brown 2013-09-11 14:09
Obviously simply measuring downward directed radiative flux by pointing a detector up doesn't count, I guess. In fact, not even directly measuring the entire downward directed radiative spectrum as presented in e.g. Grant Petty's book doesn't count. This radiation, I suppose, just disappears into thin air if it happens to encounter a warmer object.

First law of thermodynamics? We don't need no stinkin' first law of thermodynamics...

rgb
#108 Robert Brown 2013-09-11 14:26
One day you really might want to try writing that down using algebra, differential equations, and so forth. Because when you do, and actually solve the system of differential equations involved, you know, it turns out that the warming is self-limiting and under no circumstances diverges.

But of course to even attempt this you'd have to understand calculus and you'd have to quantitatively incorporate the underlying equations of physics. You obviously cannot do either one, so you are reduced to making absurd and untrue verbal statements that directly contradict the actual solutions.

Again, at some point you might want to turn on your inner rational self. As I said, you would have a very, very hard time proving objectively that I am stupid. You would find it impossible to prove that I am ignorant of the relevant physics -- quite the contrary -- or incompetent in mathematics or computation or logical or quantitative reasoning. If you are honest with yourself you really need to acknowledge that, while acknowledging that you will never equal me in any one of these abilities in your lifetime.

That leaves us with liar or crazy. Well, I'm a catastrophic anthropogenic global warming skeptic, and write extensively against it. It would seem that I have little motivation to lie about the GHE or the physics in this case. You therefore have to not only think I'm a liar, but I'm some sort of "deep mole" liar, planted in the skeptic community to pretend to be a skeptic so I can shoot down the well-founded arguments of experts in climate physics like yourself by distorting the real equations and laws in ways you are supposed to be too dumb to detect.

Seriously? Does this sound even vaguely plausible to you? You have to construct an entire Universe filled with shady payoffs, personal perversity, and so on to even think of convincing yourself that this is truly so, granted that I am competent in physics and mathematics and hence without question more competent to formulate a consistent physical argument concerning these matters than you are.

This leaves crazy. If I'm well-educated, skilled in the field and related fields, competent, and honest, you are down to crazy. Crazy is a matter of judgment. I don't think I'm crazy. I'm stably married, have never been treated for a psychological problem, never been arrested, have held a job that requires considerable organization and rational thought for 31 years straight, and still learn things even in introductory physics from my students from time to time and praise them when I do so I'm not even particularly an egotist or personality disordered. I'm happy. Not a lot of traction for "crazy".

Think on this. In the end, liar or crazy don't matter. What matters is being able to present a coherent algebraic solution to the problem based on commonly accepted physical principles. I can and have done so, and the solution proves you incorrect. No amount of verbal argument or ad hominem changes that, and you have yet to advance a similarly founded algebraic counterargument.

rgb
I think the non-physicists in this conversation need to better understand that temperature does not directly correspond to energy, or even heat. Temperature is measurement tool we employ to get a better understanding of energy flow/transfer (heat).

Physicists (correctly) see energy as a quantity that cannot be created or destroyed. It can just be moved. And laws of thermodynamics provide a framework of how it (energy) moves.

People that don't have training in physics (and/or anybody that doesn't see energy as a quantity) tend to use terms like temperature, heat, and energy interchangeably. They don't understand that temperature is an indirect measurement of heat, and heat deals with transfer of energy and/but does not necessarily correspond with energy.

I have found that the potential for confusion for terms like temperature, heat, and energy is vast amongst alarmists scientists. sometimes they make statements that would insinuate that they consider temperature a quantitative measurement of heat or energy.

I think alarmists fail in these conversations because they lack the ability and training to compartimentalize diffferent concepts related to thermodynamics. I have found that if we help alarmists better compartimentalize their thinking by imploring them to avoid using ambiguous and vague terminology they eventually realize that they really don't have an argument/point, and they will forever hate you for the favor.
#110 Robert Brown 2013-09-11 14:48
Actually, microscopically temperature is defined primarily through your choice of the equipartition theorem (sufficient for a decently quantitatively correct description of most simple systems at the introductory level) or through the partition function in stat mech. It has meaning only in a well-known coarse grained limit -- atoms don't have a temperature, but a collection of atoms can, just like matter does not have a continuous mass distribution but after you coarse-grain blocks of atoms made of elementary particles, the discrete nature of particles becomes unresolvable noise on a macroscopic scale.

Everybody actually trained in physics understands this, and I actually did research in computational simulations of critical phenomena for a decade or two that involved discrete markov chains and importance sampling Monte Carlo, so I understand it better than most. Which has absolutely nothing to do with whether or not hot objects radiate energy in at least an approximation of a blackbody spectrum, whether or not electromagnetic radiation emitted by an object at some temperature can be absorbed by another object at a different temperature, and so on.

It has even less to do with the argument at hand, which actually can be formulated without specifying the Stefan-Boltzmann equation, although to do the computation of the actual temperature rise one would need a quantitative description of the bidirectional radiation processes as a function of temperature, and the usual assumptions of large numbers of particles in good thermal contact and sufficient time to populate the available energy states according to a thermal distribution (partition function) -- all of which are perfectly reasonable in the context of this discussion, especially since I am restricting my remarks to the description of dynamic equilibrium of objects with good thermal conductivity, finite heat capacity, and only radiative coupling to a physically reasonable surrounding cold reservoir that can easily enough be constructed experimentally or that is a very good approximation of deep space at 3K and shielded from the sun.

As for heat -- surely you know better than to conflate heat with internal energy. The usual definition of heat in the first and second law is marked with the little dash because it is not an extensive quantity. The zero'th law helps define what temperature is from the point of view of permitting us to construct thermometers. Equipartition or stat mech give increasingly precise meaning to the concept in terms of energy distributed in the most probable way among many degrees of freedom AND the probability of that energy transferring between systems with differential distributions of energy in their available degrees of freedom. "Heat" is usually referenced in terms of entropy or energy made unavailable for doing work.

Is there any real need to work through an intro thermodynamics textbook here? None of this matters in the context of the discussion, where the concepts of thermodynamics and reality of radiative energy transfer more than suffice to support the conclusion of warming.
#111 Tim Folkerts 2013-09-11 15:56
Richard Feynmann once said: "Nature uses only the longest threads to weave her patterns, so that each small piece of her fabric reveals the organization of the entire tapestry."
This quote epitomizes how physicists view the world. A few broad principles weave their way thoughout the universe. Everything is connected.

What I see from many people here is desire to do the opposite -- to comparmentalize the universe into little bits. To cut out tiny patches of fabric and willfully ignore the rest of the picture. Certainly, knowing more about lots of individual bits is good and important, but that is just a starting point. You have to learn how your patch interconnects with the rest of the tapestry.

You can't "cut out" one little bit and pretend the rest does not exist, but I see that again and again.

ewiljan praises Clausius but calls statistical mechanics "bullshit". But they are part of the same thread. If one is bullshit, then so is the other!

Greg looks at one diagram (ie Kiehl & Trenberth) and thinks that is sufficient, without even bothering to take even the first short trip along that thread to read the actual paper. The diagram DOES have IR going both up and down, and that happens in a self-consistent way.

GHEbreaker points out that we don't need to worry about radiant heat exchange for closely connected opaque objects. True enough -- but there is much more to it than closely connected opaque objects. The land, the sun, the sky and space are NOT closely connect opaque objects. For his work, I suspect simple heat conduction works well enough, but it is easy to find links to all sorts of "real engineering" that does deal with radiant energy transfer.

It is the long threads that are fascinating.
* Seeing the connections between water molecules flowing "upstream" and photons moving from cool to warm.
* seeing that not every case of thermal transfer is conduction OR convetion OR radiation.
* seeing that the 2nd Law of Thermodynamics is intimately tied to probabilities.
* learning the interconnections between energy flows each way and net energy flows; between "warming" and "heating"; between conservation of energy and the K-T diagram.

Stopping short of that mean never enjoying the beauty that is nature.
#112 Joel Shore 2013-09-11 16:30
Robert: You've summarized what really puzzles me about these Slayer folks! I could sort of understand how they would imagine that I could be lying to them, since I believe that AGW is a problem and hence that action needs to be taken, and are hence part-and-parcel with the people with the Black Helicopters who want a 1-world government, and so on and so forth.

But, here we have folks like you, and Roy Spencer, and Willis Eschenbach desperately trying to convince these folks, with a strong motivation to do so since these folks do not present a good face for the skeptic movement (to put it mildly!)...and yet they still persist in their ignorant and demonstrably wrong arguments.

What this proves is just how willing people are to believe what they want to believe, despite all attempts to try to provide them with some small amount of scientific literacy on this matter.
#113 Tim Folkerts 2013-09-11 17:52
rgb says: "If you are honest with yourself you really need to acknowledge that, while acknowledging that you will never equal me in any one of these abilities in your lifetime."

Unfortunately, there is a well-known cognitive bias whereby people with poor understanding mistakenly rate their own ability as above average.

DUNNING-KRUGER EFFECT
The Dunning–Kruger effect is a cognitive bias in which unskilled individuals suffer from illusory superiority, mistakenly rating their ability much higher than average. This bias is attributed to a metacognitive inability of the unskilled to recognize their mistakes. ...

In a series of studies, they examined the subjects' self-assessment of logical reasoning skills, grammatical skills, and humor. After being shown their test scores, the subjects were again asked to estimate their own rank: the competent group accurately estimated their rank, while the incompetent group still overestimated theirs.

en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
#114 ewiljan 2013-09-11 18:00
Quoting David Cosserat:
ewiligan,
Hi, I thought you has resigned from PSI because you felt they were not extreme enough for you.
.

Naw, They are only not aggressive enough at putting down Academic Arrogance, that we have three (3) fine examples in this thread.
These Professors are teaching their fake science to innocent kids!
I tried but that was rejected. That was all about sunsettommys deleting part of my reply, but refusing to give me reason for such. I could find no pattern to his deletes, even though he has the right to do such, as PSI is not my blog.

You say: Eventually both temperatures reach "T3" where T3 lies somewhere between Tc2 and Tc1. Your "eventually" is my "steady state". So you are simply saying that at steady state, Tv2=Tc2.

Never If Tc has a source of power and Tv con only dissipate that power, Tc > Tv, in every case

But if that is true (at steady state) in the freezer, it must, by straightforward logic, also be true (at steady state) in the first case, so Tv1=Tc1. From this it follows logically that it would always be the case (at steady state) that Tv=Tc, whatever the value of Tv.

Not at all, read again carefully.
B. almost, but only "initially" Tc2 will never drop by Tv2-Tv1. Eventually both temperatures
reach "T3" with:
Tv2 < Tv3 < Tv1,and
Tc2 < Tc3 < Tc1
All of the 2 watts is absorbed by the walls of the freezer. There is never any of the fake "back radiation". The actual equilibrium temperatures depend on the convective heat transfer in the surrounding environment of the vacuum chamber. There is "no' necessary equality, or even proportionality. All temperatures depend on the differing "local" geometry! All temperatures can be calculated, but only to "back of the envelope" accuracy. Your ill defined thought problem has way to many undefined variables. To learn, build such and measure! Thought problems "always" give a false (non-physical) result!
#115 Greg House 2013-09-11 18:22
Quoting Robert Brown:
a system in a vacuum, with no conductive or convective heat transfer at all which conveniently enough, describes the Earth surrounded by 3 K outer space) ... Surrounding it with a reflective or absorptive layer simply decreases the rate energy can leave via radiation, and consequently causes the system to warm until it re-establishes power balance through the obstacle.

This is false.

Firstly, the body would still radiate to the reflector and secondly, what the reflector reflects to the body can not have any effect on the body's temperature. Back radiation can not warm the source, this is physically impossible.

Reflector works only one way. It prevents colder objects behind it from getting warmed by the body on the other side, namely but not letting the radiation pass through. That is all.
#116 Greg House 2013-09-11 18:40
Quoting Robert Brown:
In the meantime, I guess calling me crazy or a liar is an easy way to avoid ...

It is not that I just called you a liar or stupid or crazy, please do not distort my point.

Let me illustrate this idea with an easy example. Calling someone a thief just like that would be, of course, name calling. Calling a thief someone with certain police record or/and convicted for theft is not name calling. I hope you understand the difference.

My point is that anyone including you who promote the non-existing physically absurd "greenhouse effect" can only be either stupid or liar or crazy. Maybe there are more options, I am not a professional psychologist.

This, however, should not affect the discussion on the subject, it is merely a reaction to attempts to argue from authority when presenting pseudo-scientific bupkis.
#117 David Cosserat 2013-09-11 18:42
Robert,

Talk about slaying the Slayers! Trouble is you and Joel Shore and Tim Folkerts are all supreme professionals and excellent communicators, against which are pitted two or three hapless opponents who apparently have no math, no proper understanding of how to conduct a polite scientific debate without making ad-hominem attacks when they feel outflanked, and, in the case of Greg House, always end up calling people liars when he disagrees with them (which is why I never respond to him having received exactly that treatment - and me, an arch skeptic!).

As you wisely suggest, the great irony is that the Slayers are wasting their time trying to fight people like you who are at the forefront of fighting the CAGW crowd - and are doing so rather more effectively than the Slayers themselves are achieving.

Having said all that, this still does NOT nail the issue of whether or not the GHG effect is a reality (however mild an un-alarming the phenomenon may turn out to be if it exists). May I suggest, therefore, that rather than trying to convince Slayers by intellectual argument (which is clearly not working) it would be better to demonstrate whether (or not) they are wrong by experiment.

Up-thread I have suggested a simple empirical test. Depending on its outcome this should at least give either Slayers or Warmists pause for thought. Astonishingly, I have on record previous comments from both sides of this debate to the effect that "experiments are not needed because the physics are certain". With both sides saying exactly the same thing, I now know that I am dead right to insist on an experimental decider.

So I repeat my suggested empirical test here at top nesting level on this blog trail so that it hopefully doesn't get lost in all the sub-discussions...
#118 David Cosserat 2013-09-11 18:43
So here is the experiment...

1. A solid cylindrical core hangs inside a vacuum chamber, suspended by extremely thin copper wires which are connected through the top of the vacuum chamber wall via a vacuum-sealed entry point.

2. The outer surface of the core and the inner surface of the vacuum chamber are painted matt black with a measured emissivity of ~1.

3. Embedded inside the core is a heating coil, connected to an external power supply via the copper suspension wires.

4. A miniature electronic temperature sensor is mounted on the cylinder's surface, connected to a temperature reading device via the copper suspension wires.

5. Another electronic temperature sensor is attached to the vacuum chamber wall.

6. The chamber is pumped to hard vacuum (less than 1mbar).

7. The power supply is switched on and delivers exactly 2 watts of power.

8. The surface temperature of the core rises slowly until it eventually reaches a steady value Tc1.

9. At this point the vacuum chamber's wall temperature is measured as Tv1.

The whole experiment is now repeated but this time the vacuum chamber is first placed inside a freezer. When it reaches steady state temperatures, the vacuum chamber's wall temperature is measured as Tv2 and the core's surface temperature is measured as TC2.

RESULT: Tv2 is, let us say, ~30degC below its previous value, Tv1.

QUESTION: Will Tc2 be significantly different from Tc1?

A. If I am correct in my understanding of the Slayer position they say that the core temperature will not have changed significantly.

B. If I am correct in my understanding of the Warmist position it is that the core temperature will have dropped significantly.

So, no squabbling and no colluding, what do all you gentlemen predict?
#119 David Cosserat 2013-09-11 19:00
ewiljan,

There is no T3 in my proposed experiment - it was you that introduced that unnecessary complication. I have just Tv1/Tc1 in phase 1 and then Tv2/Tc2 in phase 2, both measured after the system has achieved steady state temperatures.

And my problem is not ill defined , it is precise. If you feel there are "too many undefined variables" getting in the way of answering my simple question at the end, please enlighten me on what they are. But don't just bluster and obfuscate.
#120 ewiljan 2013-09-11 19:05
Actually, microscopically temperature is defined primarily through your choice of the equipartition theorem (sufficient for a decently quantitatively correct description of most simple systems at the introductory level) or through the partition function in stat mech. It has meaning only in a well-known coarse grained limit -- atoms don't have a temperature, but a collection of atoms can, just like matter does not have a continuous mass distribution but after you coarse-grain blocks of atoms made of elementary particles, the discrete nature of particles becomes
unresolvable noise on a macroscopic scale.

What total nonsense, do to theoretical statistical mechanics, plus The Kinetic theory of heat, a corruption of the Clausius Kinetic theory of gas.
The discrete nature of particles becomes unresolvable noise on a macroscopic scale.
Again Bull shit! All is unresolvable at any micro-micro-microscopic scale. Show where your molecules, atoms, or particles of atoms can wiggle at a rate, so to produce the time varying electromagnetic field required for such to emit at the frequency of electromagnetic UV radiation (lots at 8000 Kelvin). The obvious conclusion is that you and your two also "Academic Arrogant" buddies Have "not a clue". Thanks for teaching fake science to innocent kids.

None of the earth's temperature regulating
process need any statistical analysis except
Do not change The GHE definition from depending on "back radiation" (false),to that of "only reducing the emissivity of the earth surface"(true). The Earth's atmosphere at its own temperature radiates to cold space at least twice the power that the surface could radiate.
The radiative gasses in the atmosphere, keep this planet at a low liveable temperature.
#121 ewiljan 2013-09-11 20:30
Quoting David Cosserat:
ewiljan,

There is no T3 in my proposed experiment - it was you that introduced that unnecessary complication. I have just Tv1/Tc1 in phase 1 and then Tv2/Tc2 in phase 2, both measured after the system has achieved steady state temperatures.

Yes indeed, the T3 was my addition to indicate that your ill defined demonstration has more than one "steady state" results or measurements. depending on the thermal mass and geometry of your vacuum chamber. Close to your B will occur quickly. T3 conditions may take days or weeks. They certainly do involve
that nasty out-gassing problem with the paint. If you had measured, rather than imagined,the physical situation, you would have now learned at least four "Aw Shits" that are never written down in books.

And my problem is not ill defined , it is precise. Nonsense you claim no thermal conductivity of copper wires. You propose only a thought problem, not a physical problem

If you feel there are "too many undefined variables" getting in the way of answering my simple question at the end, please enlighten me on what they are. But don't just bluster. and obfuscate.

Your difficulty is that you think you can demand, rather than measure, what is!
Just another form of "Academic Arrogance"
#122 ewiljan 2013-09-11 20:48
Quoting Robert Brown:
Obviously simply measuring downward directed radiative flux by pointing a detector up doesn't count, I guess. In fact, not even directly measuring the entire downward directed radiative spectrum as presented in e.g. Grant Petty's book doesn't count. This radiation, I suppose, just disappears into thin air if it happens to encounter a warmer object.

First law of thermodynamics? We don't need no stinkin' first law of thermodynamics...

rgb
So true But the other Mexican Law,of some number, states that all idiots like Dr. R Brown
must be eaten now!~
#123 Joel Shore 2013-09-11 21:20
I predict that one will find the same result as Pictet found when he did an experiment 200 years ago: www2.ups.edu/faculty/jcevans/Pictet%27s%20experiment.pdf , namely that the temperature of a colder body that is in radiative communication (but no significant convective or conductive communication) will affect the steady-state temperature of a warmer body, hence bringing us right up to the start of the science as of 1800! (See p. 741 for Pictet's own description of the expt and pp. 749-750 for a modern understanding of it, along with a description of how to set it up.
#124 Joel Shore 2013-09-11 21:35
"The Earth's atmosphere at its own temperature radiates to cold space at least twice the power that the surface could radiate.
The radiative gasses in the atmosphere, keep this planet at a low liveable temperature."

And, yet, satellite spectra of the Earth from space show the opposite: www.acs.org/content/acs/en/climatescience/atmosphericwarming/_jcr_content/articleContent/columnbootstrap_1/column0/image.img.jpg/1374178157948.jpg and can be fit very well using the very same radiative transfer theory equations and understanding that we accept and you call "Bull shit".
#125 ewiljan 2013-09-11 22:13
Gee Timmy assistant something,
Why is it that that you and your two other cLIMATE cLOWENS, Can provide no demonstration of what you gleefully claim? Demonstrate any "back radiation" ever. Demonstrate any of the AR22 claims of any "warmings" by anthropoids or not.
Where is the imaginary Greenhouse Effect? Where is any "back radiation"? Where is anything that you "pontificate" about?

SunsetTommy, have I missed insulting anyone significant? I am trying to be an equal opportunity asshole. MY assholism is not to reflect upon PSI, but only me,a visitor!
You have seen it all, Please share, I must learn to be better!
#126 ewiljan 2013-09-11 22:44
Quoting Robert Brown:
Obviously simply measuring downward directed radiative flux by pointing a detector up doesn't count, I guess. In fact, not even directly measuring the entire downward directed radiative spectrum as presented in e.g. Grant Petty's book doesn't count. This radiation, I suppose, just disappears into thin air if it happens to encounter a warmer object.

First law of thermodynamics? We don't need no stinkin' first law of thermodynamics...

rgb

Why does up radiate on not? I have often measured the value of thermal radiation. Please describe in physical terms what that measured thermal radiation may be?

Gee we have Robert,Greg and Tim, Who have no clue, still trying to promote the fake and
falsified Greenhouse gas theory! Get a life folk!
Well, you are right that photons do travel both ways. The second law refers to net flow, it doesn't alter reality such that the flow of photons only goes one way. It just says that the preponderance of flow will be from hotter to colder.
#128 David Cosserat 2013-09-12 03:28
Joel,

Thank you for a clear answer.

Unfortunately we really cannot proceed further until Slayers sign up. So far all I have had is incomprehensible and ungrammatical obfuscation from ewiljan. And he is hardly a good advertisement for the Slayer position.

All that having been said, it does NOT prove that Slayers are wrong and that Warmists are right. Only an experiment can do that. However until we can get both sides of this discussion to agree on an experiment that will clearly differentiate between the two positions, we cannot proceed.

I am astonished that more Slayers are not weighing in. Where the heck are they all?
"All that having been said, it does NOT prove that Slayers are wrong and that Warmists are right. Only an experiment can do that. However until we can get both sides of this discussion to agree on an experiment that will clearly differentiate between the two positions, we cannot proceed."

Kudos for attempting to bring the level of the discussion above rhetoric and planting it firmly in the soil of experimentation. Let's say the experiment is run and the outcome clearly indicates the Warmists position, "B". And from this you/we conclude that "back radiation" does exist. IOW, the experiment proves/demonstrates that in the first part of the experiment (before it is put in the freeze) photons (or, some kind of electro-magnetic radiation) is coming off the inner wall of the vacuum chamber and being recieved by Tc1 and instantly converted to kinetic energy movement of the molecules in Tc1 and instantly re-radiated (assuming Tc1 is at some kind of equillibrium or threshhold).
So, from this you can IMO, correctly claim to have demonstrated the existence of "back radiation." (So, IMO, the debate as to whether back radiation exists is over. It exists.)

But here's the big questions, from this can you correctly, accurately, and honestly claim to have defeated the thinking/position of the "dragon slayers?"
I don't think you can. The dragon slayers are *not* saying that the photons don't flow both ways. They are not saying that the "back radiation" that you describe does not exist. They are saying, however, that if you are going to employ math equations you have to be careful to apply them correctly. Along those lines it's important to properly conceptualize what temperature is and what it isn't. You warmists have dropped the ball on this issue. Warmists think of/conceptualize temperature as/to being something that it isn't. If you warmists were to endeavor to gain a better understanding that temperature is a proxy measurement, an indirect method for accessing the realities that are taking place with energy and flow thereof then you would be less inclined to make the conceptual error that you are making here.

Allow me to put it in a nutshell . . .

(Comment too long. Comment continued below.)
(Comment continued from above: Allow me to put it in a nutshell . . .)

. . . exactly what conceptual error you warmists are making. You are counting the energy (and/or flow of energy, heat) of the equation twice. Let me put it in the context of your example, when you are measuring Tv1 temperature the number you recieve is a proxy indicator of energy (and/or flow of energy, heat). That some of the energy is actually reflected or simply radiated back (back radiation) is not something that will be reflected in the temperature measurement. So, to maintain accuracy you would want to subtract the amount of back radiation from the equation, at this juncture. This is the step you warmist are missing. And when that back radiation is again reflected back out again you warmists are adding it to the equation (correctly) and fooling themselves in thinking the process achieves a net gain:

Legend:
T - temperature (not a direct measure of energy or heat, a proxy for energy and/or energy flow)

Warmists:
T(proxy for energy flow) + Brr = Net increase in energy

Dragon Slayers:
T(proxy for energy flow) - Br + Brr = No Net increase/decrease in energy

Note that Br and Brr are the same magnitude but have opposite signs, thus they cancel each other out. Consequently dragon slayers are correct to ignore it and Warmists are incorrect to consider it additive and not both additive and subtractive to overall energy and/or energy flow.

Jim McGinn
Convection has been refuted.
#131 ewiljan 2013-09-12 16:30
Thank you Jim McGinn,
You are correct Thermal Flux can "only" be measured by measuring Flux not temperature.
Such a Fluxmeter may be constructed by mounting back to back thermocouples on the opposite ends of a some thermal resistance/conductance. The thermocouples will now measure delta temperature across the thermal resistance with the sign of the output indicating the "direction" of the flux. Such instruments are made with an accuracy of two milliwatts, over the range of -100 C to 150C
If the vacuum chamber were well insulated and all flux from the chamber forced through the fluxmeter all flux. generated by the internal power (2 Watts) must exit the apparatus through the meter. If this flux is also 2 Watts, All power is accounted for and none available for any "back radiation".
To counter the "warmest" insistence on such "back radiation" a more complex demo involving three or more spherical concentric shells, with measured emissivity, and multiple fluxmeters is required. Such meters will show the total thermal power as it progresses from power at innermost to outermost and some isothermal.
The "warmest" position will then become is that only "net" flux is measured. This is true, but the measured flux is the only physical flux, All others claimed by some T^4, are imaginary and non-physical.
#132 ewiljan 2013-09-12 16:44
Joel, Thank you for a clear answer.

David, Are you sure you wishy to provide "all" geometric details for your demonstration, including measured emissivity? I would suggest a simplified geometry so all can understand the physical mechanism.
The size and shape of both your cylinder and the vacuum chamber will affect the answer. This is what I meant by ill defined.Your insistence on only two outcomes is also ill-conceived..
#133 Kelvin Vaughan 2013-09-12 17:01
How does an electromagnetic wave of lower frequency from a colder object speed up the vibrating particles of a hotter object which are moving at a faster rate? Surely they would slow the vibrations down! i.e. the radiation from a colder object would cool the hotter object.
#134 Joel Shore 2013-09-12 17:39
As I have noted several times before, your argument about whether the term represents radiation going from the colder to the warmer body (which it does) is completely irrelevant: You get the greenhouse effect regardless of whether you interpret the term in this way.

You seem to be avoiding discussing the steady-state temperature of the body, which is what in the end is most relevant.
#135 Joel Shore 2013-09-12 17:50
Your conjecture doesn't even satisfy the 1st Law of Thermodynamics, i.e., it doesn't conserve energy: A photon has energy so when an object absorbs a photon, its energy must increase by an amount equal to the energy of the photon. It cannot decrease!

As for where your naive picture goes astray, for one thing, objects emit at a distribution of frequencies, so while the AVERAGE frequency of a photon from a colder object will be lower than the AVERAGE frequency of a photon from a warmer object, it is only on average. Furthermore, the details of the actual absorption and emission of photons and the relation to the thermal vibrations is actually complex (and at some level, quantum mechanical), so your naive picture is also hampered by that.

In the end, you are simply grasping at straws to desperately try to have nature work according to your ideological preferences rather than how it actually works.
#136 Tim Folkerts 2013-09-12 18:00
Jim, I laud your effort here, but there are two things in particular that I take issue with.

1) Your effort "to properly conceptualize what temperature is" doesn't work well. Temperature is a measure of average kinetic energy of the particles. In many cases temperature can indeed be used as a proxy for other things like internal energy or heat, but it is important to keep the three ideas separate.

So rather than
T(proxy for energy flow) + Brr = Net increase in energy
I would write out the actual energy flows -- which is then a concise statement of conservation of energy
(energy in) - (energy out) = change in energy.
[More later]

2) You suggest that incoming radiation is "instantly converted to kinetic energy movement of the molecules in Tc1 and instantly re-radiated"

That is not conceptually correct. If the energy is "instantly re-radiated" then you are talking about "reflection". If, on the other hand, the energy is indeed converted to KE of the solid, then it becomes an indistinguishable part of the KE of the solid. There is no way for the atoms to decide "I must immediately radiate the KE from one source, but I can hold onto the KE from other sources."

This misconception leads to your conclusion that "Br and Brr are the same magnitude but have opposite signs". Here you are describing perfect reflection. And, yes, perfect reflection will NOT help warm the object.

***********************

(energy in) - (energy out) = net increase in energy.

(energy from heater + Bra)
- (Fre + conduction loss + convection loss)
= Net increase in energy

Where
Bra = "back" radiation that is ABSORBED
Fre = "forward" radiation that is EMITTED

Since we have eliminated conduction & convection in this example, then we can simplify to

(energy from heater) + Bra - Fre = Net increase in energy

Note that Bra & Fre are NOT the same. The temperature of the WALLS in a proxy for Bra. The temperature of the OBJECT is a proxy for Fre.
Quoting Lonny Eachus:
Even if you were correct, Joel, it would have no bearing on the "Greenhouse Effect", because at best you are showing objects reaching equilibrium. Your own examples do not show any radiative transfer from cold to warm, which the greenhouse models DO rely on.

Now take the actual case of the hypothesized atmospheric "back radiation", in which the surroundings are KNOWN TO BE, from simple observation, cooler than the supposed initial source. (The ground being the "initial source" in the models, in an attempt to isolate variables.)

Certainly the S-B equation gives different answers to the equation as the initial temperature of the surroundings varies. This is not in dispute. But where does that imply "back radiation"? THAT is the question here, . . .

Let me take a shot at clarifying the thinking that underlies this notion.

I think photons (electromagnetic energy) do (does) move both from warmer entities to colder and from colder to warmer. Alarmist (warmies) refer to the photons (electromagnetic energy) that is flowing from colder entities to warmer entities as, "back radiation." They seem to insinuate that modern physics (or, at least, the dragon slayers) have ignored the existence of back radiation and go on to suggest all kinds of evil motives therof.

In order to protect the reputation of dragon slayers I would like to recommend a solution. How about if we let them keep their designation of back radiation if they will allow us to keep our designation of Front radiation: the photons (electromagnetic energy) that is flowing from warmer entities to colder entities. And the following formula:

Then we just need to explain to them that methods of modern thermodynamics/physics were not cast based on the assumption that only front radiation is significant. Nor did/do they ignore back radiation. Because here's the thing, the methods of modern thermodynamics/physics were/are cast on the assumption that net radiation gain/loss (Front radiation minus back radiation) is significant. This is standard. This is the consensus. This assumption is so basic it is hardly mentioned by physicists.

IOW, sorry to burst your bubble, warmies. The thing you are not getting is that the methods of modern thermodynamics/physics has never ignored the existence of "back radiation". It's just that it's existence has been subsumed into our understanding (and our calculatory procedures) at a fundamental level.

Quoting Lonny Eachus:

Lonny continued:
. . . not equilibrium temperature, which makes your argument a straw man.

What you are missing is that even if the "warmists" arrived at the right equilibrium temperature, if they did so using incorrect methods their models are still wrong.

Jim McGinn
Find out more about the vortex phase of water.
#138 Lonny Eachus 2013-09-12 18:09
Quote: "Great...That means the greenhouse effect is not in dispute and there is nothing to argue about."

Hardly.

Quote: "Whether you call the 2nd term in that equation "back radiation" or "magical mystery radiation from the planet Zyrcon" makes absolutely no difference in the results you get from the mathematics."

Not at all. You are conflating two very different things.

Quote: "The net effect of "back radiation" is to reduce the net amount of heat flowing from a hot object to its cooler surroundings in the case when the cooler object is no longer as cool."

I understand all that. But in order for that to happen, as the system (more closely) approaches equilibrium, SOMETHING must absorb that heat. The whole problem with your argument here is that the "back radiation" model REQUIRES THE EARTH'S SURFACE to be what is absorbing that heat. Not "less heat loss", but actual warming.

Your scenario would require the atmosphere to warm as the surface cools. Which is not unreasonable as a hypothesis but unfortunately for that hypothesis it is contrary to observation. The warming of the atmosphere is already accounted for in the models. So even if your hypothesis were correct, the models would still be wrong.

The "back radiation" models are attempting to explain the warming of the surface IN THE ABSENCE of the atmospheric warming that would have to take place if it were a system simply moving toward equilibrium in the way you describe.

You can't have it both ways. I believe you are attempting to explain the observed phenomenon by arguing that it works in ways that are contrary to that same observation.
#139 Lonny Eachus 2013-09-12 18:23
You appear to be arguing against yourself.

Why don't you simply conduct the experiment? Stick a thermocouple to the bulb, let it warm to equilibrium "on" temperature, and measure it.

Then cover it all with a thin, smooth layer of aluminum foil, let it warm to equilibrium "on" temperature again, and measure it?

Dr. Latour has spent a large part of his adult life working with systems that control heat transfer. Do you honestly think he is completely ignorant of how it works?

I'm not taking anybody's word, but I do tend to lean toward the person with practical experience in the field. But as I say: I'm willing to suspend judgement until someone actually PERFORMS the experiment, and shows us the results.
#140 Lonny Eachus 2013-09-12 18:30
I say once again: why don't you simply refute the argument, rather than making personal insults? The latter doesn't serve to convince anybody.
#141 Lonny Eachus 2013-09-12 18:45
Tim, take a step back and look at what this discussion is actually about.

THIS is the image published by the IPCC to explain the climate models their reports were based on.

The IPCC intended this to be an accurate, if slightly simplified, diagram representing the actual climate models used as a basis for the reports.

In the diagram, you can clearly see that the "greenhouse gases" are shown to be emitting 169 W/m^2 upward (plus 30 W/m^2 to account for albedo), while at the same time emitting 333 W m^2 downward. Add the 40 W/m^2 representing direct escaping emission from surface and you get a TOTAL outward radiation of 239 W/m^2 outward, while still having (supposedly) 333 W/m^2 "back radiating" inward.

This is the discrepancy Greg is referring to. If you don't see a problem there, you simply aren't looking.
#142 Lonny Eachus 2013-09-12 18:46
#143 David Cosserat 2013-09-12 18:50
Jim,

Thank you for your kind words about my attempt to replace rhetoric with experiment.

I try to avoid the term 'back radiation' because it seems to have a negative impact on some people who adopt the Slayer position. Contrary to your assertion, my experiment is NOT designed to prove or disprove 'back radiation' (doesn't even mention it in fact) if by that term you simply mean a radiative potential exerted by a cooler body in the direction of a warmer body. The real question is whether that radiative potential "does work" - i.e. whether energy flows from cooler to hotter body and then gets thermalised at the receiving higher temperature surface; or alternatively is offset by an equal and opposite radiation potential from the hotter surface towards the cooler one so that no such energy flow occurs. That as I understand it is the crux of the difference between Warmist physics and Slayer physics.

My experiment is exactly designed to test that crucial difference:

(1) In the Slayer case, with no energy flowing back from the colder to the warmer surface, the temperature of the colder surface (and hence its radiative potential) obviously cannot influence the temperature Tc of the warmer surface. So we would expect the outcome to be that the warmer object remains at the same temperature irrespective of the temperature Tv, since Tc is dictated solely by the fixed 2 watt power input it is receiving and the emissivity and area of the emitting surface. Thus, if the Slayers are right, putting the whole experiment in the freezer and repeating the test will not change Tc.

(2) In the Warmist case on the other hand, when the experiment is repeated in the freezer, the energy flowing from cooler to warmer body will be less that it was in the first case and so we would expect Tc to be reduced significantly.

The above is my best attempt at articulating my current understanding of the difference between the Slayer and Warmist positions. Please either confirm that you agree with this or, if you disagree, please let me know exactly where you think I am in error.

+++++++++++++

In your further explanation, you simply compare the Warmist and Slayer rationales for their different views on energy flows and then state your own opinion that the Warmist physics is wrong and the Slayer physics is right. You are entitled to your opinion but it is precisely because such rationales are manifestly failing to persuade people (either way) that I am proposing to carry out an experiment rather than resorting to yet more ineffectual rhetoric!

But thanks once again for your considered response to my experiment. It is the first coherent comment from a Slayer that I have received so far on this thread.
#144 Lonny Eachus 2013-09-12 18:54
Again, you are completely ignoring the actual observed temperature differentials between the surface and the atmosphere.

If the atmosphere actually worked like a blanket "back radiating" and thus warming the earth's surface (remember this is what it's about... the surface), or KEEPING it warmer, as you stated elsewhere, then the atmosphere itself would be much closer to the temperature of the surface. But it isn't. We don't see the warming of the atmosphere that would have to take place if that were true.
#145 Lonny Eachus 2013-09-12 19:01
Quoting Robert Brown:
Most of the nuts on this list spend all of their time ranting about how colder objects cannot transfer radiation at all to warmer ones, or asserting that the radiation thus transferred somehow doesn't "count" in the energy budget on the receiving end.

And who would those be? I don't recall seeing anybody making quite that argument.

True, people have said "cold objects cannot make warmer objects even warmer", but I don't believe I've seen anyone intentionally arguing that a photon cannot travel from a colder body to a warmer one. The whole argument is about net transfer.

But maybe you've been paying more attention than I have.
[quote name="ewiljan"]Thank you Jim McGinn,
You are correct Thermal Flux can "only" be measured by measuring Flux not temperature.
Such a Fluxmeter may be constructed by mounting back to back thermocouples on the opposite ends of a some thermal resistance/conductance. The thermocouples will now measure delta temperature across the thermal resistance with the sign of the output indicating the "direction" of the flux. Such instruments are made with an accuracy of two milliwatts, over the range of -100 C to 150C
If the vacuum chamber were well insulated and all flux from the chamber forced through the fluxmeter all flux. generated by the internal power (2 Watts) must exit the apparatus through the meter. If this flux is also 2 Watts, All power is accounted for and none available for any "back radiation".
To counter the "warmest" insistence on such "back radiation" a more complex demo involving three or more spherical concentric shells, with measured emissivity, and multiple fluxmeters is required. Such meters will show the total thermal power as it progresses from power at innermost to outermost and some isothermal.
The "warmest" position will then become is that only "net" flux is measured. This is true, but the measured flux is the only physical flux,

Well, I'd even go so far as to suggest that NET (emphasis mine) flux is the only flux we can measure anyway. (I mean, it's not like we even have a choice. Simply put, our instruments cannot/do not differentiate between radiation being the result of front radiation and "back radiation." Instruments either can or cannot detect radiation.
#147 Lonny Eachus 2013-09-12 19:13
Quoting Joel Shore:

It is really not rocket science. It is the same thing that would happen if your drafty house with the furnace running full-steam was only able to keep your house at 15 C when it was -20 C outside. If you then added insulation to the attic and walls, better windows, etc., the furnace would now be able to keep your house at a toastier temperature.

This is an egregiously bad -- even false -- example.

Again you are conflating two completely different concepts. Typical house insulation (such as fiberglass) prevents heat loss via CONDUCTION and has little if anything to do with radiative loss. Empty space would do the job just as well, probably much better.

By throwing insulation into the radiation argument you simply confuse the issue.
#148 David Cosserat 2013-09-12 19:17
ewiljan,

You are one of those people I have often had to deal with in my professional life who insist on responding inappropriately to a different question from the one asked.

I am not asking you what the value of Tc will be, just whether or not it will be lower when Tv gets lower. You do not need to know any geometric details or anything else about the physical experiment in order to answer that simple yes/no question.

My description was carefully framed in order to get prior agreement from Slayers that they really do believe that Tc will not change when Tv is lowered. Unless and until that agreement is forthcoming we simply do not have an experiment that is worth doing. Once we do have agreement, and the experiment has been done and the results have been published, all experimental details and procedures will of course be published, as you would expect from a professional engineer who has spent a significant part of his career constructing working electronic and software systems.
#149 Tim Folkerts 2013-09-12 19:21
David,

I'm working on an actual experiment along these line -- hopefully I can tell you the results soon.

I am NOT using vacuums because 1) I want others to be able to easily replicate the experiment and 2) ummm ... well ... I don't have a vacuum chamber handy that I can put into a freezer.

I am limiting conduction & convection other ways, while maximizing the effects of radiation.

PS I am ALSO working on an experiment to confirm or refute the TOP post -- you know, the one that no one is actually discussing!
#150 David Cosserat 2013-09-12 19:36
Tim,

Great, we can collaborate as we work on our separate empirical approaches. Do contact me at .
#151 Lonny Eachus 2013-09-12 19:46
Added later: actually, it DOES add up, if you count the 80 W/m^2 they refer to as "latent heat" absorbed by the atmosphere.

The part that doesn't actually add up, is that their "back radiation" model shows the surface absorbing 494 W/m^2 of radiation, while TOTAL average incoming solar radiation is only 341 W/m^2. Granted... they also show the surface radiating, so the net NUMBERS add up, but the physics doesn't.

This energy budget, by the way, as shown in the diagram can be traced back to Hansen, and has been pushed since by Trenberth et al.

At any given moment, according to the models represented by the diagram, the surface is absorbing 494 W/m^2 while radiating 396 W/m^2, leaving a net difference of 98. So the net shown is fine.

But then, what causes the surface to radiate more than the net incoming radiation? I would like that one explained to me.
#152 Ollie Hughes 2013-09-12 19:49
You will get no agreement, until you propose a simple geometry that measures heat flux, with the resultant temperatures. Measuring temperature alone says nothing of heat transfer.;
Quoting Tim Folkerts:
Jim, I laud your effort here, but there are two things in particular that I take issue with.

1) Your effort "to properly conceptualize what temperature is" doesn't work well.
Temperature is a measure of average kinetic energy of the particles. In many cases temperature can indeed be used as a proxy for other things like internal energy or heat, but it is important to keep the three ideas separate.

Jim McGinn:
Temperature is what we use to sample reality. It is not reality.

So rather than
T(proxy for energy flow) + Brr = Net increase in energy
I would write out the actual energy flows -- which is then a concise statement of conservation of energy
(energy in) - (energy out) = change in energy.
[More later]

2) You suggest that incoming radiation is "instantly converted to kinetic energy movement of the molecules in Tc1 and instantly re-radiated"

That is not conceptually correct.

Jim McGinn:
I concede the point. It's a big oversimplification.

If the energy is "instantly re-radiated" then you are talking about "reflection". If, on the other hand, the energy is indeed converted to KE of the solid, then it becomes an indistinguishable part of the KE of the solid. There is no way for the atoms to decide "I must immediately radiate the KE from one source, but I can hold onto the KE from other sources."

This misconception leads to your conclusion that "Br and Brr are the same magnitude but have opposite signs". Here you are describing perfect reflection.

And, yes, perfect reflection will NOT help warm the object.

Jim McGinn:
True, but don't berate me for giving your argument the benefit of the doubt.

***********************

(energy in) - (energy out) = net increase in energy.

(energy from heater + Bra)
- (Fre + conduction loss + convection loss)
= Net increase in energy

Where
Bra = "back" radiation that is ABSORBED
Fre = "forward" radiation that is EMITTED

Jim McGinn
IMO, they are both emitted and absorbed.

Since we have eliminated conduction & convection in this example, then we can simplify to

(energy from heater) + Bra - Fre = Net increase in energy

Note that Bra & Fre are NOT the same. The temperature of the WALLS in a proxy for Bra. The temperature of the OBJECT is a proxy for Fre.

Jim McGinn:
Interesting, even though I'm not sure to what degree I understand all of what you are saying here.

Jim McGinn
Find out more about the vortex phase of water.
#154 Lonny Eachus 2013-09-12 19:56
Well, if you teach physics then, perhaps you can explain the following to me:

According to the diagram that sparked this whole debate (see the link given up above), the Earth's surface is radiating an average of 494 W/m^2, while total incoming solar radiation is 341 W/m^2.

Now, the NET energy budget shown in that diagram adds up. I'm not disputing that.

But I have never heard of a system in which a body being heated by an outside source actually radiates MORE energy, on average, than the net average incoming radiation.

I understand that the body in question (Earth) is wrapped by an atmosphere. Still, I do not recall encountering any other situations in my physics classes in which a heated body is radiating more energy than it is receiving from the heat source, regardless of whether it is surrounded by some other material.

Can you give me another example where that is true?
#155 Lonny Eachus 2013-09-12 20:09
Very clearly the core temperature would be lower if the outer temperature is lower.

But this is NOT an equivalent problem to the "back radiation" question.

In the "back radiation" models (which make up the vast majority of greenhouse warming models), the system is being heated from the OUTSIDE at a rate of 341 W/m^2. But according to the energy budget diagram used by those models (backed by the data cited in the papers), the Earth is actually radiating outward at 494 W/m^2, with some of that radiation "reflected" (actually re-radiated) back to it and absorbed.

I am wondering how the heated object -- the Earth -- regardless of whether it is wrapped by an atmosphere, can radiate more energy than the radiation incoming to the system. I am unaware of any other systems that behave in that way.

In order to be an experiment that is even halfway representative of the Earth system, your outer shell would have to be semi-transparent, and the radiative heat source must be on the outside.

I would like someone to give me an example of any such system in which the core radiates more strongly than the source.
#156 Lonny Eachus 2013-09-12 20:15
Please show me a system in which the heated object (in this case Earth) radiates more energy from its surface (not net outflow) than there is energy incoming from the initial source (in this case the sun).

The "back radiation" model requires the Earth to be radiating approximately 50% more energy into the atmosphere than it is receiving from total insolation.

I am only asking for an example of any other system that behaves this way: in which the surface of the "core" radiates about 50% more energy, on average, per unit area than the radiation that is incoming.
#157 Squid2112 2013-09-12 20:24
Mr. Shore, as a casual observer to this conversation, observing with great interest, I respectfully request that you stop being an asshole. Mr. Vaughan posed a good question and good point. You, on the other hand, replied like an asshat. I have tolerated a lot of back and forth BS from both sides in this debate, but I find your response to Mr. Vaughan rather offensive and quite demeaning.

Thank you.
#158 Lonny Eachus 2013-09-12 20:27
There is a simple experiment that can prove the greenhouse gas warming hypothesis, as shown in the IPCC diagram (never forget that is what this debate is about):

Create a system with a "core" (analogous to Earth), wrapped in some kind of gas or blanket (atmosphere... I'm not particular what kind of material you choose here, but it must be semi-transparent).

In order to match the "back radiation" greenhouse models, the surface of the "core" in the experiment must radiate outward into its "blanket" approximately 50% more energy, per unit area, than the total radiation being received into the system from the heat source.

(Per the IPCC: Earth's surface radiates at 494 W /m^2 while total insolation is only 341 W/m^2.)

If you can successfully perform this experiment, and show me an example of a "core" that radiates more than the total radiation coming into the system, then I may concede that the "back radiation" model may have some credibility.

Here is a link to the picture that is behind this ENTIRE debate. IPCC intended this to be an accurate representation of the AVERAGE energy budget, and this is the energy budget that the majority of greenhouse models are based on:

3.bp.blogspot.com/_N4XvePkmKjA/TUY7Q_dLomI/AAAAAAAAAUk/aIoVuXHQ_y4/s1600/Atmospheric+Energy+Diagram.gif

Note that the NET numbers do add up. Yet the surface is required to radiate at 494 W/m^2 while insolation is 341 W/m^2.

Some of that radiation, according to the model, is re-radiated back and absorbed. I grant this. Yet outgoing surface radiation is still required to be 494 W/m^2.

Please show me another system in which the "core" is 50% hotter than the outside heat source at equilibrium.
#159 Lonny Eachus 2013-09-12 20:30
Just to make clear what I'm saying: I am aware that the NET energy radiating outward from the Earth system is not in dispute. What I am asking is for an example of a system in which the SURFACE of the "core" of the system (analogous to Earth surface wrapped in atmosphere) is hotter than the incoming radiation.

Others have said here that internal heating of the Earth must be insignificant. So no internal heating allowed.
#160 Squid2112 2013-09-12 20:32
Mr. Eachus,

Even as a relative lay person in this discussion, I believe I can provide for you the short answer. There is no such system as you describe, as even to a 5th grader, such a system is painfully obviously impossible, else-wise, there would be work-free, never ending energy for all.

Nobody will show you this system because it does not and can not exist! If one purports to know of such a system, call a cop, they are a fraudster.
#161 Lonny Eachus 2013-09-12 20:38
Oh, crap. My mistake. I don't know what the hell I was thinking.

INCOMONG is 341. SURFACE radiation is 396. Surface ABSORPTION (again according to the diagram) is 494.

Even so, the surface radiation is still 55 W/m^2 higher than the insolation. I do not know of any other examples of systems that behave this way.
#162 Ollie Hughes 2013-09-12 20:39
Quoting Lonny Eachus:
Quoting Joel Shore:

It is really not rocket science. It is the same thing that would happen if your drafty house with the furnace running full-steam was only able to keep your house at 15 C when it was -20 C outside. If you then added insulation to the attic and walls, better windows, etc., the furnace would now be able to keep your house at a toastier temperature.

This is an egregiously bad -- even false -- example.

Again you are conflating two completely different concepts. Typical house insulation (such as fiberglass) prevents heat loss via CONDUCTION and has little if anything to do with radiative loss. Empty space would do the job just as well, probably much better.

By throwing insulation into the radiation argument you simply confuse the issue.

Thank you, All of the CAGW is based on thermal
radiation alone. None of the proponents of AGw have shown any knowledge of thermodynamics, and the physical difference in conductive and radiative heat transfer.
#163 Lonny Eachus 2013-09-12 20:42
Ignore the above. I must be having a bad day... the numbers incorrect.

According to the diagram, the surface is ABSORBING 494 W/m^2, not radiating that much. The actual difference between surface radiation and total insolation is only 55 W/m^2.

But even so, I still do not know of any systems in which the core object's surface is radiating hotter than the incoming radiation.
#164 Lonny Eachus 2013-09-12 20:43
Correction: radiating outward at 396 W/m^2, not 494.

That's still 55 W/m^2 more than the incoming radiation from the sun.
#165 Lonny Eachus 2013-09-12 20:45
Again, I must correct my original comment. It is not 50% more, it is more like 14%.

Still, 14% is significant. How can the surface be radiating 14% more energy than exists in the incoming radiation?
#166 Ollie Hughes 2013-09-12 20:51
Quoting Joel Shore:
Your conjecture doesn't even satisfy the 1st Law of Thermodynamics, i.e., it doesn't conserve energy: A photon has energy so when an object absorbs a photon, its energy must increase by an amount equal to the energy of the photon. It cannot decrease!

As for where your naive picture goes astray, for one thing, objects emit at a distribution of frequencies, so while the AVERAGE frequency of a photon from a colder object will be lower than the AVERAGE frequency of a photon from a warmer object, it is only on average. Furthermore, the details of the actual absorption and emission of photons and the relation to the thermal vibrations is actually complex (and at some level, quantum mechanical), so your naive picture is also hampered by that.

In the end, you are simply grasping at straws to desperately try to have nature work according to your ideological preferences rather than how it actually works.

All of your Bull Shit is based on false stastical mechanics. Why not take a look around and try to figure what is?
#167 Joel Shore 2013-09-12 21:03
Venus is one example...except the numbers are much more dramatic. It's called "recycling."

Let's give another example. Suppose we lived on a planet where 90% of our new aluminum was made from recycled aluminum. Then, with only 1 million tons of aluminum created fresh from newly mined bauxite per year, we could nonetheless produce 10 million tons of new aluminum products a year.

Another example is the money supply: The federal reserve may only print, say, \$1 million in new \$1 bills each year, but if you monitor the number of times these new bills as passed back and forth between buyers and seller in the economy, you'd find much more than \$1 million in exchanges.

What you are really telling us is that you don't have good intuition about how a conserved quantity behaves, so you have invented some rule that you think must be satisfied ... And, you are simply incorrect.
#168 Sunsettommy 2013-09-12 21:04

I notice that people hardly use the QUOTE button would make you posts much easier to follow.

Quoting Mr. Shore,

Quote:
In the end, you are simply grasping at straws to desperately try to have nature work according to your ideological preferences rather than how it actually works.
It is the eighth box from the left at the top of the comment box you write in.

Please use it when you are quoting,to make it clear it is a quote and not part of your own comment.
#169 Ollie Hughes 2013-09-12 21:06
Quoting ewiljan:
Quoting Tim Folkerts:
"You Climate Clowns can provide no demonstration of ___________"

You are not in a discussion with "Climate Clowns" -- you are in a discussion with people who earned PhDs in physics LONG before we worried about climate change. We represent 'main-stream physics' and our views mirror those of our physics colleagues in colleges and universities around the world. We represent the "textbook answers'.

Here is your challenge (same goes for Greg or Joe Postma). Go into (or send an email to) ANY Physics Department in ANY university and find the professor who teaches thermodynamics (or ask for the oldest prof who got his PhD long before "climate change' might have 'corrupted' his thinking). Ask for an opinion on this thread. I guarantee that prof will side with Dr Brown & Dr Shore and Dr Folkerts.

Give us the name of ANY physics prof in ANY university who will side with you and your interpretation of thermodynamics. Currently the list is 3-0 against you.

No mater if Dr Brown & Dr Shore and Dr Folkerts.are Climate Clowns or not any of you can provide even one demonstration of this so called "back radiation".
Not any of you can provide even one demonstration that objects with a temperature must radiate in all directions. Case in point the earth radiated "no" energy to the Sun.
Not any of you can provide the reason for thermal electromagnetic radiation at all!
So much for your collective PHD's,

Humm! we hear nothing from the Professors
on how to demonstrate their false claims!
They all seem to have only imaginary (non-physical) answers.
#170 Joel Shore 2013-09-12 21:09
If you deny any physics after 1850 that you don't like, there is really nothing to talk about. You are just a denier of modern science that doesn't suit your ideological preferences.

And, by the way, the 1st Law of Thermodynamics predates statistical mechanics.
#171 Joel Shore 2013-09-12 21:13
What is demeaning about pointing out that someone's very simplistic and naive picture of a complicated process conflicts with well-established facts about that process?
#172 Ollie Hughes 2013-09-12 21:24
Quoting Lonny Eachus:
Again, I must correct my original comment. It is not 50% more, it is more like 14%.

Still, 14% is significant. How can the surface be radiating 14% more energy than exists in the incoming radiation?

Lonny,
None of the objects in this universe have ever been in thermodynamic equilibrium. All is trying to catch up! All is cyclic.
This earth has at least 107 cycles, none of them co-measurant, not even day and night with seasons.
#173 Tim Folkerts 2013-09-12 21:35
I would point people to this fascinating video from Richard Feynmann on the nature of 'understanding' in science.

Thermal radiation exists. It is an experimental fact. "Why" is a very different issue.

It can be 'explained' using the same rules (eg Maxwell's equations) that self-consistently explain any number of other fascinating phenomena. I could say "any accelerating electric charge radiates EM energy, and the vibrating atoms in a solid are accelerating" as an explanation. If you know have worked through that topic in E&M, that should suffice. If not, you need to study E&M.

There are no "short-cut answers" to such a question.
#174 Tim Folkerts 2013-09-12 21:43
Joel, your last line about "grasping at straws to desperately try to have nature work according to your ideological preferences" is a bit much. It is quite possible that the question was indeed simply naive, rather than ideological.

This could be a corollary to Hanlon's Razor, like "Never attribute to ideology that which is adequately explained by lack of knowledge."
#175 Joel Shore 2013-09-12 21:50
Squid2112: Perhaps my response was a bit jaded. But, I did not detect the sort of humility in Kelvin Vaughan's question that would be to the effect of "What is wrong with the simple picture I have in my head of this?" Instead, it seemed to be the arrogance of having so much self-assuredness in one's own understanding that one is willing to make broad pronouncements like "the radiation from a colder object would cool the hotter object" because this conclusion suits one's fancy more than the opposite conclusion supported by way more evidence.

There seems to be an incredible amount of over-confidence around here about one's own ability to do science combined with an incredible disrespect for the scientific knowledge of the scientific community. In the face of such arrogance, is it demeaning to point out to people that maybe their extreme confidence in their own abilities relative to others who has displayed much more expertise is perhaps misplaced?
#176 Tim Folkerts 2013-09-12 22:08
Quote:
How can the surface be radiating 14% more energy than exists in the incoming radiation?
But it doesn't. The incoming radiation to the surface consists of ~ 161 W.m^2 of incoming solar radiation and ~ 333 W.m^2 of incoming thermal radiation from the atmosphere & clouds. The atmosphere & clouds have a large store of thermal energy (billions of joules per square meter), and can easily part with 333 Joules/m^2 each second down toward the surface.

You already agree that the net numbers add up (ie energy is conserved), so there is really no conflict here. The net flow is always from hotter to colder. The 1st and 2nd Laws of Thermodynamics are upheld.
#177 Ollie Hughes 2013-09-12 22:12
Quoting Tim Folkerts:
I would point people to this fascinating video from Richard Feynmann on the nature of 'understanding' in science.

Thermal radiation exists. It is an experimental fact. "Why" is a very different issue.

It can be 'explained' using the same rules (eg Maxwell's equations) that self-consistently explain any number of other fascinating phenomena. I could say "any accelerating electric charge radiates EM energy, and the vibrating atoms in a solid are accelerating" as an explanation. If you know have worked through that topic in E&M, that should suffice. If not, you need to study E&M.
There are no "short-cut answers" to such a question.

Indeed, Folk here have studied well, We all undersand Maxwell's equations. We ask from the the three illustrious professors here, to describe in physical terms "how a temperature of 8000 Kelvin can produce that much electromagnetic energy emitted from each square centimeter"?
#178 Tim Folkerts 2013-09-12 23:01
With all due respect, if you "understood Maxwell's equations" (and Planck's work with "cavity radiation"), then you would not need an explanation -- you could work it out yourself. There is a reason that this is "textbook physics" -- generations of scientists have studied thermal radiation. There are both experimental and theoretical reasons to believe
P/A = ε σ [(T_1)^4 - (T_2)^4]
which exactly answers your question "how a temperature of 8000 Kelvin can produce that much electromagnetic energy"

If you don't understand, then a 3 minute tutorial will not do the job.
#179 Ollie Hughes 2013-09-13 00:04
Quoting Tim Folkerts:
With all due respect, if you "understood Maxwell's equations" (and Planck's work with "cavity radiation"), then you would not need an explanation -- you could work it out yourself. There is a reason that this is "textbook physics" -- generations of scientists have studied thermal radiation. There are both experimental and theoretical reasons to believe
P/A = ε σ [(T_1)^4 - (T_2)^4]
which exactly answers your question "how a temperature of 8000 Kelvin can produce that much electromagnetic energy"

If you don't understand, then a 3 minute tutorial will not do the job.

We all know how much energy (as measured)
is produced. Can any professor say in physical terms why "temperature" produces that much electromagnetic energy?b
#180 Ollie Hughes 2013-09-13 00:39
Quoting Sunsettommy:

I notice that people hardly use the QUOTE button would make you posts much easier to follow.

Quoting Mr. Shore,

Quote:
In the end, you are simply grasping at straws to desperately try to have nature work according to your ideological preferences rather than how it actually works.

It is the eighth box from the left at the top of the comment box you write in.

Please use it when you are quoting,to make it clear it is a quote and not part of your own comment.
Hi Tommy,
The now banned ewiljan is my neighbor, I am glad that there is only one of him, I could not tolerate two. In his defense, he belives that anyone that demands he should think in the demanders way need be destroyed immidiately if not sooner. onky after that can polite discourse take place, I agree!
#181 Ollie Hughes 2013-09-13 02:51
Why no comment from the professors?
#182 Ollie Hughes 2013-09-13 02:57
It is only your misuse of the theory of statistical mechanics, that I disagree with!7sm7g
#183 David Cosserat 2013-09-13 04:29
Ollie,

Your comment shows that you have no real understanding of the issue at stake. You are simply trying to change my experiment to another experiment that you, apparently, would prefer. I suggest that is just a bit impertinent.

Or perhaps you are just obfuscating because you actually don't know the answer to my question and don't like to admit it?

Whatever your views on temperature (whether or not it is a good a proxy for energy) is anyway beside the point. It does not prevent you from answering my very simple question:

If Tv is reduced, will Tc stay the same or will it reduce?

But thank you, nevertheless, for at least responding. Slayers in their droves should be coming here to confirm that Tc will not change when Tv is lowered, but they are curiously absent.

This concerns me.
#184 John Marshall 2013-09-13 06:23
So we are still arguing about a foil covered light bulb failing as proof that the 2nd law can be violated. Not so, failure does not prove anything of the sort.
Lamp manufacturers go to great lengths to ensure the brightest light for your buck. their light filaments are working at the very edge of light flux/coil strength envelope. The enclosure is filled with argon to ensure that he excess heat is convected away from the coil for the glass to radiate/convect this away to the surrounding area. This is why the bulb gets hot. Covering it with foil interferes with this process and so the coil gets hotter due to lack of cooling NOT back radiation. Energy input to the coil continues as before and the coil fails due to melting.

So think of it as insulation.
#185 Ollie Hughes 2013-09-13 06:44
Quoting David Cosserat:
Ollie,

Your comment shows that you have no real understanding of the issue at stake. You are simply trying to change my experiment to another experiment that you, apparently, would prefer. I suggest that is just a bit impertinent.

Or perhaps you are just obfuscating because you actually don't know the answer to
my question and don't like to admit it?
If Tv is reduced, will Tc stay the same or will it reduce?

Is that Tc will always go lower, but not as much as Tv is lowered.

Whatever your views on temperature (whether or not it is a good a proxy for energy) is anyway beside the point. It does not prevent you from answering my very simple question:

If Tv is reduced, will Tc stay the same or will it reduce?

But thank you, nevertheless, for at least responding. Slayers in their droves should be coming here to confirm that Tc will not change when Tv is lowered, but they are curiously absent.
This concerns me.
I also!

David I do not want to confuse, I wish to help! Your demo will show the decrease in temperature. It will not show "why" Can we not work together to form a bullet proof demonstration of your wishes and my wishes?
They are damn near the same. I wish to show that there is no power avalible for any "back radiation".
#186 Joel Shore 2013-09-13 08:34
And, your deep understanding of statistical mechanics (such that you are qualified to judge that I am misusing it) comes from where exactly?

How many papers have you published in the area of statistical mechanics and in what journals?
#187 Joel Shore 2013-09-13 08:37
Whether or not there is back-radiation is irrelevant. Slayers are using the argument that there is no back-radiation to argue that there is no greenhouse effect, i.e., that greenhouse gases don't increase the temperature at the surface of the Earth.
#188 Squid2112 2013-09-13 08:41
Mr. Shore,

Again, condescension! Mr. Vaughan posed a very good question. He quite humbly conveyed his thoughts in a succinct manner. You, on the other hand, replied with crassness and authoritarian narcissism. And to make matters worse, you did it twice more. For purporting to be such an intellect and subject matter expert, you sure are stupid. This was a very simple question and paradigm, for which I was eager to read a coherent and thoughtful response. I am disappointed. However, as always, I am learning new things. Today's lesson for me; Pay no attention to Mr. Shore's responses, he is an asshat.

Now, is this really your intended goal? You continue to drop by digits on the credibility scale. If you can not, in a professional and polite manner, answer a simple question, provide useful information in a succinct manner, why the hell do you prance around on these blogs? Needing attention? Not getting enough at home?

I don't believe I will pay much attention to anymore of your comments. You lose credibility with me.

Respectfully,
Squid

Now, perhaps someone else out there can take a "thoughtful" stab at explaining what Mr. Vaughan was alluding to. I would very much like to understand this subject better, and I enjoy learning from those who ACTUALLY project intellect upon the subject matter. I would like to understand how something (matter, radiation, whatever) with a lower frequency can add energy to something (matter, radiation, whatever) possessing a higher frequency. Please forgive me Mr. Vaughan if I am not conveying your thoughts accurately, but I believe this IS the core gist of the question, and this DOES speak directly to the topic and subject matter at hand.

Thank you,
Squid
#189 Joel Shore 2013-09-13 08:42
"This is why the bulb gets hot. Covering it with foil interferes with this process and so the coil gets hotter due to lack of cooling NOT back radiation. Energy input to the coil continues as before and the coil fails due to melting.

So think of it as insulation."

I thought Slayers whole argument is that the getting-hotter due to lack of cooling process can't occur. The back-radiation is not the major issue. I don't personally give a crap if you believe in back-radiation or not. If you understand that the greenhouse gases interfere with the cooling of the Earth and hence result in the Earth's surface being at a higher temperature than it would be in their absence (and that more greenhouse gases lead to a warmer Earth's surface), then you understand the greenhouse effect.
#190 Squid2112 2013-09-13 08:46
P.S. Mr. Shore,

Please quit with your "detecting" bullshit. You are obviously not good at it, nor does it have any usefulness to these discussions. Simply participate in intelligent, polite discussion and quit being an arrogant asshat.

Thank you.
#191 David Cosserat 2013-09-13 09:27
Ollie,

Thank you very much for responding so positively.

By suggesting that Slayer physics also says that Tc will drop when Tv is lowered you are putting yourself outside what I believe to be the standard Slayer position.

However I may well be wrong about this, in which case it will be necessary to establish a different agreed Slayer position around which it may or may not be necessary to design a modified experiment.

If you think you can contribute to resolving this issue, welcome aboard. First, may I suggest that you explain what you think is the proper Slayer position in quantitative terms. There are not many variables involved: Tv, Tc and P (along with the constants: surface areas Ac and Av, emissivity and S-B constant). So I guess my question is: given P and Tv (and constants Ac, Av, etc.) what is your formula for Tc?

If you prefer to communicate privately in the first instance my email address is .

I look forward to working with you!

Rgds
David
#192 John Marshall 2013-09-13 10:16
Is not a lack of cooling a result of insulation?

Greenhouse gasses are supposed to adsorb outgoing LIR and re-radiate it to the surface to increase the surface heat, that is the GHE!!!! Not in any way, shape or form how a greenhouse works. so the label is a stupid one because it means the opposite of the fact of a greenhouse working. The GHE is not to do with interfering with cooling. but heating with re-radiated energy. I am convinced that your understanding of the GHE is wrong.

In fact the so called GHG's cool the surface because of their reaction with SIR thus reducing insolation to the surface. water vapour in fact acts as a heat carrier from the surface to high in the atmosphere due to latent heat transfer.
Well, I think you mean to say they are moving at highter frequency. (Their speed is always the same, they move at the speed of light.) I don't think the frequency (hotness) of an object has any effect on whether or not a subsequent wave (electro magnetic wave) regardless of frequency, will be absorbed or reflected. (As to whether or not a hot object will be [as a result of its hotness] more likely to reflect an incoming wave or more likely to absorb and quickly re-emit a incoming wave is not a distinction that we are capable of making, to the best of my knowledge. But I think all of this is much ado about nothing in that thermodynamics has never been about absolute flow of energy from one object to another (something that is very difficult to isolate and measure). It has always been about NET flow of electro magnetic energy.

So, why the confusion and who is to blame? It seems to me that people on both sides of the issue have failed to understand and/or failed to communicate that thermodynamic principles were cast in the context of NET quantities (ie. front radiation minus back radiation).
Physicists (dragon slayers) know this innately. Of course some dragon slayers may not know this innately and might actually believe that the reason we ignore back radiation is because of some physical principle, which is what Kelvin seems to be expressing here. But I think Kelvin is mistaken on this point. The reason we ignore "back radiation" is because back radiation is already included in our understanding. (And, it's not like our instruments have the ability to distinguish between front radiation and back radiation anyways.)

due to the problems with isolating and entity and, on a smaller scale, Heisenbergian uncertainty)
#194 Joel Shore 2013-09-13 16:11
(1) But...the net effect of the GHE's are to return to the surface only SOME of the energy that is radiated by the surface, thus reducing the cooling of the surface. The net flow of energy, i.e., the heat, is still from the warmer surface to the cooler atmosphere.

(2) The GHG's do not absorb as strongly in the SIR region as in the LIR region, so their net effect is warming of the surface. This is shown in radiative transfer calculations that very successfully reproduce empirical data, such as the emission of radiation from the Earth as seen by satellites in space.

(3) Yes, water vapor has multiple effects, the convective effect acting to reduce the surface temperature and the radiative effect acting to increase it. However, the net effect is a surface temperature larger than it could possibly be in the absence of greenhouse gases.
#195 Joel Shore 2013-09-13 16:17
One simple answer is that if you take a system oscillating at a high frequency and superpose an oscillation at a lower frequency, you don't reduce the energy of the oscillator; you increase it.

Given your posting here, however: principia-scientific.org/latest-news/301-robert-brown-backradiation-and-mathematics.html#comment-1656 I question whether you are really as open-minded as you seem to be portraying yourself to be. I guess we'll see.
#196 Ollie Hughes 2013-09-13 16:45
[quote name="Joel Shore"]And, your deep understanding of statistical mechanics (such that you are qualified to judge that I am misusing it) comes from where exactly?

How many papers have you published in the area of statistical mechanics and in what journals?[/quote
Ewiljan was correct nothing but academic arrogance.
#197 Tim Folkerts 2013-09-13 18:14
Let me take a stab at this ...

Quote:
...from a colder object
This is immaterial. A 10 um photon is a 10 um photon. It behaves the same way whether it came from a block of ice or from the sun.

Quote:
... a hotter object which are moving at a faster rate?
"Moving at a faster rate" is a bit ambiguous. Hotter gas molecules move at a higher speed, but that is not really the issue. Molecules and solids vibrate at various frequencies. For example, CO2 molecules vibrate at a frequency of about
2E13 Hz -- which corresponds to a 15 um IR photon. So CO2 can absorb and emit 15 um photons.

CO2 also vibrates (& rotates) at other specific frequencies as explained by quantum mechanics. Bigger molecules tend to have more possible frequencies. Solids tend to have a lot of possible vibrational frequencies.

Quote:
How does an electromagnetic wave of lower frequency from a colder object speed up the vibrating particles ...
The absorption & emission of photons doesn't speed up the vibrations -- it increases the amplitude. Furthermore, the photon has to be at a frequency that the molecule / solid /liquid can absorb/emit. If it is the wrong frequency, it will not get absorbed.

Much like a child on a swing, a small push at the right time adds energy and increases the amplitude (like absorbing a photon of the right frequency). A small push at the 'wrong time' removes energy and decreases the amplitude (like emitting a photon).

Typically the vibrations can have lots of possible amplitudes, gaining energy occasionally and losing energy occasionally.
#198 Tim Folkerts 2013-09-13 18:22
There is a bit of 'academic arrogance' in Joel, but you are also rather arrogant here. You, on your own authority, proclaim that Joel is misusing statistical mechanics.

Rather than simply calling "bullshit" you should point out a specific error. You say you disagree -- so what is the statistical mechanical error in anything Joel said that you disagree with?
Quoting Joel Shore:
(1)
But...the net effect of the GHE's are to return to the surface only SOME of the energy that is radiated by the surface,

Is GHE even a scientifically concise (measured, tested) notion/concept. We're expected to believe that GHE's have some kind of dramatic thermal property yet there is zero evidence in the literature of these qualities having been demonstrated/tested. I think if they did rigorously test this notion and actually quantify the GHE characteristics of molecules, these result would reveal that there is little difference in the GHE characteristics of GHEs (CO2, methane) and the normal constituents of the atmosphere N2, O2. Note that after all these years global warming believing scientists has bothered to do the work to make GHE a scientifically concise notion. I guess nobody wants to kill the goose that laid the golden egg.

Quoting Joel Shore:
(2) . . . thus reducing the cooling of the surface. The net flow of energy, i.e., the heat, is still from the warmer surface to the cooler atmosphere.

Yes, the net flow.

Quoting Joel Shore:
(3)
(2) The GHG's do not absorb as strongly in the SIR region as in the LIR region, so their net effect is warming of the surface.

Sir, Lir? What in the world are you talking about?

Quoting Joel Shore:
(4)
This is shown in radiative transfer calculations

What, exactly, is "shown?" Please provide a reference to said, "radiative transfer calculations."

Quoting Joel Shore:
(5)
that very successfully reproduce empirical data, such as the emission of radiation from the Earth as seen by satellites in space.

Do you have a reference for this, "empirical data." Or are you saying we should take your word on this. (And, in that case, who are you?)

[quote name="Joel Shore"](6)
(3) Yes, water vapor has multiple effects, the convective effect acting to reduce the surface temperature and the radiative effect acting to increase it. However, the net effect is a surface temperature larger than it could possibly be in the absence of greenhouse gases.

Well, that's great and all. But if that's what you believe then one can only wonder why you don't put forth a detailed and referenced argument to that effect.

If you ever become aware of any laboratory data that indicates GHE qualities ever having been quantified,tested/measured please let me know.
#200 Tim Folkerts 2013-09-13 19:41
Quote:
I think if they did rigorously test this notion and actually quantify the GHE characteristics of molecules, these result would reveal that there is little difference in the GHE characteristics of GHEs (CO2, methane) and the normal constituents of the atmosphere N2, O2.

Simply stating an opinion ("I think") is not very scientific of you! The data is readily available; the quantum mechanical theory is well-founded. This is not something we need to have opinions about.

Try looking at Spectralcalc www.spectralcalc.com/spectral_browser/db_intensity.php
For "Species" select H2O, CO2, O2, & N2.
For "Spectral Range" try somerthing like 1-40 microns.
Under "Options" select 'Scale by Atmospheric Abundance'

Then plot the results. The GHGs are clearly orders of magnitude stronger absorbers. AND they absorb in the important wavelengths (around 10 um)

Quote:
Sir, Lir? What in the world are you talking about?

Joel was simply following the abbreviations that John Marshall had introduced. From the context, it is pretty clear these mean "Solar infrared radiation" (0.7-4 um) and "Longwave infrared radiation" (4-100 um)

Quote:
Specralcalc does some of this. MODTRAN is a common calculator for such things. You can play with a user-friendly interface here:
forecast.uchicago.edu/Projects/modtran.html
Quote:

I think if they did rigorously test this notion and actually quantify the GHE characteristics of molecules, these result would reveal that there is little difference in the GHE characteristics of GHEs (CO2, methane) and the normal constituents of the atmosphere N2, O2.

Simply stating an opinion ("I think") is not very scientific of you!

IMO, it's unscientific to not ask these questions. And it's even more unscientific to not have asked these questions yourselves and not have attempted to answer them and not have tested your answers a long time ago. It's not like the premise of AGW was discovered last week. Thirty years and xxxx billions of dollars and they still have no empirical data underlying their basic premise that GHE's have dramatic thermal properties that the rest of the astmosphere does not possess. I mean, think about it. Why is it that you can't turn to hard data and state something like, CO2 has a greenhouse gas coefficient of xx and O2 only has xx?

The data is readily available; the quantum mechanical theory is well-founded. This is not something we need to have opinions about.

I agree. The data is available. There is no drama in the data.

Try looking at Spectralcalc www.spectralcalc.com/spectral_browser/db_intensity.php
For "Species" select H2O, CO2, O2, & N2.
For "Spectral Range" try somerthing like 1-40 microns.
Under "Options" select 'Scale by Atmospheric Abundance'

Then plot the results. The GHGs are clearly orders of magnitude stronger absorbers. AND they absorb in the important wavelengths (around 10 um)

This is "spectral range." We need something that indicates greenhouse thermal characteristics. Remember this is the greenhouse warming theory, not the spectral range theory.

Quote:

Sir, Lir? What in the world are you talking about?
Joel was simply following the abbreviations that John Marshall had introduced. From the context, it is pretty clear these mean "Solar infrared radiation" (0.7-4 um) and "Longwave infrared radiation" (4-100 um)

We're expected to pretend it means something significant?

Quote:

Specralcalc does some of this. MODTRAN is a common calculator for such things. You can play with a user-friendly interface here:
forecast.uchicago.edu/Projects/modtran.html

You are expecting your audience to decypher your thinking on all of this?
Tim Folkerts:
GHGs are clearly orders of magnitude stronger absorbers. AND they absorb in the important wavelengths (around 10 um)

Jim McGinn:
If it is so clear to you then why do you think you're having such a difficult time making it clear to everybody else?
#203 Lonny Eachus 2013-09-14 01:47
Quoting Tim Folkerts:
But it doesn't. The incoming radiation to the surface consists of ~ 161 W.m^2 of incoming solar radiation and ~ 333 W.m^2 of incoming thermal radiation from the atmosphere & clouds. The atmosphere & clouds have a large store of thermal energy (billions of joules per square meter), and can easily part with 333 Joules/m^2 each second down toward the surface.

You already agree that the net numbers add up (ie energy is conserved), so there is really no conflict here. The net flow is always from hotter to colder. The 1st and 2nd Laws of Thermodynamics are upheld.

You conveniently sidestepped my actual question.

According to the model -- I didn't make this up, I'm looking right at the IPCC's own diagram -- the SURFACE of the earth is radiating outward at 396 W/m^2.

Yes, the diagram shows "back radiation" which makes the NET add up. BUT THAT'S NOT THE POINT. Regardless of the NET budget at the top of the atmosphere, those models still require the SURFACE to be radiating more W/m^2 that total incoming insolation (341 W/m^2).

THE FIGURES ARE AVERAGE, so your argument that it's not at equilibrium is irrelevant. And regardless of "back radiation" making the net add up, the SURFACE is still hotter than incoming insolation. This is not a misunderstanding; it is explicit in their model.

I say again: show me any other system in which the warmed object -- surrounded by a blanket or not -- radiates more energy at its surface ON AVERAGE than the total incoming radiation. REGARDLESS of the temperature at the OUTSIDE of the blanket. As I stated before, in this little gedankeneksperiment I'm not questioning the NET at the top of the atmosphere. Only the surface radiation.

In fact, show me any "semi-closed" system (i.e., communicating with its surroundings only via gravity and radiation) in which a large part of the system is hotter than the outside source of that heat. I am intensely curious as to how that could work. I suppose it might be made possible by some kind of focusing reflector but such is not the case on Earth, and even if it were, it would require the cooling of other parts.

To put it in fewer words: directly on its face, without any need to refer to S-B equations, the model requires the Earth's surface to be "hotter" than the incoming radiation doing the heating.

I don't recall hearing of any other system that worked quite that way.
#204 Tim Folkerts 2013-09-14 02:43
First of all, I will say once again, the incoming radiation is 5780 K, so it is warmer than the surface. Your concern is about the magnitude of the energy, not the temperature of the energy.

Your previous wording was better: "How can the surface be radiating 14% more energy than exists in the incoming radiation?"

Analogies are fraught with challenges, but let me use one anyway. Suppose I receive \$161 every day from some outside source. Can I pay you more than \$161 per day? Sure, I can take \$200 out of my savings account, and give you \$361. You can take that \$361, give \$200 back to me (which I put back into savings) and you can spend \$161 somewhere else. You and I can quite happily do this indefinitely without either of us running out of money. I am "radiating" more money than exists in the income, but no one would complain that I am creating money.

************************

As for examples ... it is tough to come up wit something familiar that would work. Near as I can think right now (and I am a bit sleepy), any example basically has to be a variation of the greenhouse effect.
#205 GHEbreaker 2013-09-14 02:48
@ Robert Brown

AH AH AH, you're fantastic!

Now you changed again your statements!!

Now you're saying that surrounding the glass of the bulb with the aluminum foil is expetected to reduce (correct!) the temperature of the glass surface, as the more conductive aluminum is quickly absorbing a lot of Thermal energy from the glass.

BUT THAT'S WHAT I WROTE IN MY EQUATION AND ARTICLE, SIR!! YOU EVEN DID NOT READ WHAT I WROTE!!

Moreover, as the light bulb tungsten filament is constantly receiving ELECTRIC ENERGY from the household electric plant, actually there will be NO reduction of temperature in the surface of glass of light bulb, but just a rapid INCREASE of temperature in the aluminum foil, that then will start to heat the surface of glass below, THROUGH CONDUCTION, not through "backradiation".

Furthermore, Mr. Brown , wrote that 83-90% of heat transmission in the light bulb + aluminum foil is through radiation, BUT WE ALL AGREE ON THAT!

Of course I wrote above that the tungsten filament in the core of light bulb reaches 2800-3000K, and then it is trasmitting Thermal energy to the glass of bulb in the vacuum THROUGH RADIATION, that's evident!
And that's the same heating mechanism of our Sun toward our Earth, radiation in vacuum.

But then, the heat exchange between the glass surface of bulb and the aluminum foil takes place JUST through CONDUCTION, not through "backradiation"!

Mr. Robert Brown is trying to hide that it was HE, and not I, to use the word "backradiation"!!

Backradiation , according to the AGW/GHE supporters, should mean that the core (tungsten filament) of the light bulb would increase its temperature, because it would receive further Thermal energy, and THAT'S CRAZY!!

IT WOULD BE LIKE SAYING THAT A BURNING WOOD (500°C) IN THE FIREPLACE WOULD INCREASE STILL FURTHER ITS TEMPERATURE, BECAUSE IT WOULD RECEIVE THERMAL ENERGY FROM YOUR BODY (37° C) NEAR THE FIREPLACE.

According to Robert Brown, if backradiation exists, then the tungsten filament should change its temperature from 3000K to 3100K, by receiving Thermal energy from the aluminum foil, and then the 3100K filament should increase still further the temperature of aluminum foil, up - say - 473K, and then the aluminum foil should "backradiate" energy to the filament increasing still further the temperature up to 3300K, and so on...like a perpetuum motion, or like an endless tennis game.

and that's simply CRAZY!!

Like a tennis match in the US Open: body B receives heat from body A, and then increase the temperature of body A with bakradiation, and then body A increases still further the temperature of body B, and again and again, like a tennis ball whose velocity increases after any exchanges!

#206 Tim Folkerts 2013-09-14 02:51
I can think of three possible that explanations for difficulty in making an idea understood.
1) The material is intrinsically difficult
2) Some people have insufficient background knowledge.
3) I am a lousy teacher.
4) Others are lousy students.

I will leave it up to each person to decide which one(s) they think are correct here.
#207 David Cosserat 2013-09-14 03:39
Tim,

I think all 3 of those 4 explanations are correct but the remaining one (whichever it is) is certainly not correct.
#208 Kelvin Vaughan 2013-09-14 05:34
Thanks Squid.
#209 Lonny Eachus 2013-09-14 05:53
Quoting Joel Shore:
Venus is one example...except the numbers are much more dramatic. It's called "recycling."

I would have to see the numbers. I have been looking but haven't found them yet.

According to NASA...

nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html

... Venus has very close to twice the insolation (W/m^2) as Earth. I know the approximate average surface temperature but I do not know the energy radiated from the surface. Without knowing more about other physical parameters it is pointless for me to try to calculate it.

Your examples do not appear to be very analogous to this situation. First off, in order for Venus' temperature to be due to "recycling", you would apparently have to subscribe to the hypothesis that Venus is in fact subject to a "runaway greenhouse effect", but that is a subject that is very much in debate. See for example:

fgservices1947.wordpress.com/2010/02/22/venus-gate-%E2%80%93-or-there-is-no-runaway-greenhouse-effect-on-venus/

That is only an example and not the only source: there are quite a few that say essentially the same thing.

So I cannot accept Venus as a valid example, because that would require assuming the same effect you are arguing for, which means it would be a circular argument. I'm not criticizing, you understand: I'm simply saying that for that reason I cannot accept Venus as a valid example.

Your aluminum example simply isn't analogous. I am not aware of anything that absorbs and re-radiates aluminum. I'm not trying to be a smartass here; I see what you're trying to say but the analogy seems pretty weak.

Quote:

What you are really telling us is that you don't have good intuition about how a conserved quantity behaves, so you have invented some rule that you think must be satisfied ... And, you are simply incorrect.
No. I was simply asking for an example of any other physical system that behaved the same way. Venus cannot be accepted as a valid example, and aluminum recycling and money supply are obviously not physical systems similar to the one under discussion. They might make great analogies, but that's all they are. I didn't ask for an analogy, I asked for an example.

So: if you want to convince me that your point is valid, please point me to a demonstration or example, other than Venus, of a body warmed by an outside source that radiates, on average and at its surface, significantly more energy per unit area than is incoming. Internal heat source not allowed.
#210 Lonny Eachus 2013-09-14 06:01
Quoting Joel Shore:

There seems to be an incredible amount of over-confidence around here about one's own ability to do science combined with an incredible disrespect for the scientific knowledge of the scientific community. In the face of such arrogance, is it demeaning to point out to people that maybe their extreme confidence in their own abilities relative to others who has displayed much more expertise is perhaps misplaced?

Interesting. Who do you feel is being "disrespectful", and who do you consider to be "the scientific community"?

I will not assume what you meant by that, but it could easily be interpreted as an astoundingly arrogant thing to say.
#211 Lonny Eachus 2013-09-14 06:20
Quoting Joel Shore:

I thought Slayers whole argument is that the getting-hotter due to lack of cooling process can't occur. The back-radiation is not the major issue. I don't personally give a crap if you believe in back-radiation or not. If you understand that the greenhouse gases interfere with the cooling of the Earth and hence result in the Earth's surface being at a higher temperature than it would be in their absence (and that more greenhouse gases lead to a warmer Earth's surface), then you understand the greenhouse effect.

No. In fact, if you had actually read LaTour's treatise on the subject, you would know that in fact back-radiation *IS* the essential topic under discussion. His whole position was that back-radiation (as described and used by nearly all AGW climate models) is thermodynamically impossible.

This is NOT the same as back-radiation preventing cooling, you understand. If you look at the energy budget as published by the IPCC (which IS intended to be a fairly accurate average), you see that the models actually call for the back-radiation to be absorbed BY THE SURFACE.

If on the other hand it were a matter back-radiation preventing the surface from cooling as the system moves toward equilibrium, we would see a lower temperature differential between the surface and the lower atmosphere. (Not that it would stay that way in the real world. Convection could carry it away from the surface... but then you would still have to explain why that heat isn't seen higher in the atmosphere either.)

Other means of warming the surface might well be possible, but even the one suggested earlier (back-radiation preventing cooling) would result in the models STILL being wrong.

If the models are wrong, the models are wrong. Even if they manage to arrive at the "right" answer occasionally (though they consistently have not), it would mean that they are based on bad science. Surely you understand this.
#212 Tim Folkerts 2013-09-14 12:41
Actually, I just did this experiment yesterday. I glued a small thermistor to a light bulb. The glass warmed to ~ 55 C when uncovered. With Al foil, covering the bulb, the temperature of the thermistor shot up to over 100 C within a couple minutes.
Quoting Joel Shore:
Squid2112:
There seems to be an incredible amount of over-confidence around here about one's own ability to do science combined with an incredible disrespect for the scientific knowledge of the scientific community. In the face of such arrogance, is it demeaning to point out to people that maybe their extreme confidence in their own abilities relative to others who has displayed much more expertise is perhaps misplaced?

I agree. People that think they got it all figured out really make it difficult for those of us that do.
#214 Tim Folkerts 2013-09-14 15:46
Quote:
2π ∫dT = - Q ∫dx/x
No need to be afraid!
Not afraid – just a bit confused. It turns out this equation has serious errors.

Quote:
– Q = k * A (cm²) *dT/dx
Ah, now we can start making some sense of this. This is just heat flow. Rearranging and integrating gives something like the first integral
∫dT = - Q/k  ∫ (1/A) dx

With a little 'reverse engineering, we can see that the original equation is for a CYLINDER (A = 2πxh), not a sphere (A = 4πx^2). And a factor of “1/k” is missing from the right side. Not an auspicious start for “rigorous mathematical symbolism”.

Quote:
T = 99.8° C
So we have that the glass surface of the light bulb decreases its surface temperature very quickly, at a rate of 0.2°C/sec. (100° C – 99.8° C) = 12° C/min.
No, what we have is that -- once steady-state is achieved -- the outer surface of the aluminum cylinder will be 99.8 C. To know anything about the RATES, you would need to include information about the masses and specific heats of the foil and bulb.
#215 Ollie Hughes 2013-09-14 17:08
Quoting Tim Folkerts:
I can think of three possible that explanations for difficulty in making an idea understood.
1) The material is intrinsically difficult
2) Some people have insufficient background knowledge.
3) I am a lousy teacher.
4) Others are lousy students.

I will leave it up to each person to decide which one(s) they think are correct here.

Quoting Tim Folkerts:
I can think of three possible that explanations for difficulty in making an idea understood.
1) The material is intrinsically difficult
2) Some people have insufficient background knowledge.
3) I am a lousy teacher.
4) Others are lousy students.

I will leave it up to each person to decide which one(s) they think are correct here.

Strange, Thermal radiation from hot to cold
has been known and measured for at least 150 years, such thermal radiation is measured even
when the transmission medium is a vacuum. All three of the professors claim that is how the earth receives the energy from the Sun. None of the professors can come up with one measurement of thermal radiation from cold to hot, Not one can explain why such electromagnetic radiation is created by the "Temperature" and why it must be the same in all directions. Can they at least give som time frame for the "invention" of the non-measurable "back-radiation"? Was the term used before the AGW mess? Why has GHE changed from being dependent on the missing "back radiation" to be just an claimed, but never demonstrated, reduction of the electromagnetic radiation from the warm earth to cold space? Can a time frame for this latest change in GHE be established?
#216 Ollie Hughes 2013-09-14 17:17

Great lets go measure some Djokovic! If it also cannot be measured, can it be physical?
Those with a PHD in physics must know!
#217 Tim Folkerts 2013-09-14 18:20
Not being facetious -- could you tell me what sort of specific measurement you are talking about? How exactly would you measure the magnitude of the thermal radiation from a hot surface to a cold surface?

Only by knowing exactly what sort of measurement would satisfy you can we move ahead.
Quoting Ollie Hughes:
Strange, Thermal radiation from hot to cold
has been known and measured for at least 150 years, such thermal radiation is measured even when the transmission medium is a vacuum. All three of the professors claim that is how the earth receives the energy from the Sun. None of the professors can come up with one measurement of thermal radiation from cold to hot, Not one can explain why such electromagnetic radiation is created by the "Temperature" and why it must be the same in all directions. Can they at least give som time frame for the "invention" of the non-measurable "back-radiation"? Was the term used before the AGW mess? Why has GHE changed from being dependent on the missing "back radiation" to be just an claimed, but never demonstrated, reduction of the electromagnetic radiation from the warm earth to cold space? Can a time frame for this latest change in GHE be established?

Jim McGinn
I don't doubt that this is true: "none of the professors can come up with one measurement of thermal radiation from cold to hot." But isn't it also true that none of the professors can come up with one measurement of thermal radiation from hot to cold either? More specifically, instruments don't have the ability to distinguish between back radiation and front radiation. They just measure radiation/temperature.

Nevertheless warmies assertion that electromagnetic waves do move in both directions between hotter and colder objects is correct! Right?

But here's what they don't get:
Back radiation is already assumed in thermodynamics. Warmies don't get this. I'm not completely sure if dragon slayers are completely clear on this either.
Tim,

One of the most difficult things with respect to conveying a scientific idea is dealing with the fact that your audience brings with it all kinds of baggage. Unless one is open to and skilled at dealing with other people's baggage you really can't make any progress in conveying a scientific idea. IMO.
Quoting Tim Folkerts:
Not being facetious -- could you tell me what sort of specific measurement you are talking about? How exactly would you measure the magnitude of the thermal radiation from a hot surface to a cold surface?

But here's the thing, Tim. Backradiation has never not been a part of thermodynamics.
#221 Ollie Hughes 2013-09-14 19:31
Quoting Squid2112:
Mr. Shore,

Again, condescension! Mr. Vaughan posed a very good question. He quite humbly conveyed his thoughts in a succinct manner. You, on the other hand, replied with crassness and authoritarian narcissism. And to make matters worse, you did it twice more. For purporting to be such an intellect and subject matter expert, you sure are stupid. This was a very simple question and paradigm, for which I was eager to read a coherent and thoughtful response. I am disappointed. However, as always, I am learning new things. Today's lesson for me; Pay no attention to Mr. Shore's responses, he is an asshat.

Now, perhaps someone else out there can take a "thoughtful" stab at explaining what Mr. Vaughan was alluding to. I would very much like to understand this subject better, and I enjoy learning from those who ACTUALLY project intellect upon the subject matter. I would like to understand how something (matter, radiation, whatever) with a lower frequency can add energy to something (matter, radiation, whatever) possessing a higher frequency. Please forgive me Mr. Vaughan if I am not conveying your thoughts accurately, but I believe this IS the core gist of the question, and this DOES speak directly to the topic and subject matter at hand.

Thank you,
Squid

I don't think anyone can answer Mr.Vaughan" good question! The whole question of the scientific cause of temperature induced electromagnetic radiation has ever bee4n answered. The three professors here mumble about statistical mechanics, and the kinetic theory of heat, without ever considering the limitations of each. Dr. Brown seems the only one comfortable with idea that the two T^4 in the S-B equation are truly radiative potentials and that such electromagnetic radiative power transfer "only" is from the higher to lower potential. Still of all I know that have precisely measured this electromagnetic radiation, none can say "why it is".
#222 Joel Shore 2013-09-14 19:36
Let's try to look at this objectively. Who is more arrogant, a person who studies and works long and hard in a field and, as a result, feels that he has at least some expertise in the field, or someone who has (apparently) studied almost nothing about the field but nonetheless apparently feels that he is just so innately brilliant that he has a lot of expertise in the field nonetheless?
#223 Joel Shore 2013-09-14 19:53
"If you look at the energy budget as published by the IPCC (which IS intended to be a fairly accurate average), you see that the models actually call for the back-radiation to be absorbed BY THE SURFACE.

If on the other hand it were a matter back-radiation preventing the surface from cooling as the system moves toward equilibrium, ..."

This is a distinction without a difference. Just about any physics book you could find (certainly everyone I can find, which is quite a few) would explain to you that an object in cooler surroundings cools more slowly when its surroundings are warmer (or maybe we can say "less cool") than when they are cooler. And, this is because there is an exchange of radiation between the object and its surroundings, with a warmer object absorbing radiation emitted from its surroundings but (as long as it is warmer than its surrounding) emitting more radiation that is absorbed by its surroundings.

None of this is controversial whatsoever in physics circles. It is only in the strange Slayers world that people believe that thermodynamics applies at the microscopic scale and hence that a warmer object can't absorb any radiation from a cooler object (or it can't in any way affect it that it does or whatever).

Also, you are confusing two different senses of the word "cooling". When we say that back-radiation decrease the cooling, we are not only talking about some case where the object is actually cooling over time. We are also talking about the case where the object is in a steady-state. The point is that in the absence of the back-radiation, the Earth would be able to radiate all of the radiation it receives from the sun "more easily", i.e., at a lower steady-state temperature than in the presence of back-radiation.

It is just like your house if you insulate it better: It is less easily able to cool and hence it will be at a higher steady-state temperature when you run the furnace than it would be before you added the insulation.

Or, take a sink that has a partially stopped up drain: If you run the water at a certain rate, then you will get a certain level of water in the sink. If you block the drain even more, that water level will rise because now the pressure at the drain has to be higher to have the same rate of flow of water down the drain as is coming in from the faucet.
#224 Ollie Hughes 2013-09-14 20:02
Quoting Tim Folkerts:
There is a bit of 'academic arrogance' in Joel, but you are also rather arrogant here. You, on your own authority, proclaim that Joel is misusing statistical mechanics.

Rather than simply calling "bullshit" you should point out a specific error. You say you disagree -- so what is the statistical mechanical error in anything Joel said that you disagree with?

Perhaps I should, to someone I respect. so:
The idea of using statistical mechanics, and the kinetic theory of heat, to "declare" that all objects must radiate in all directions with a flux appropriate, and measured,to a surface at a temperature "near" zero Kelvin, with an emissivity of 100%, must be a gross error in all physical theories.
#225 Joel Shore 2013-09-14 20:04
Quoting GHEbreaker:

According to Robert Brown, if backradiation exists, then the tungsten filament should change its temperature from 3000K to 3100K, by receiving Thermal energy from the aluminum foil, and then the 3100K filament should increase still further the temperature of aluminum foil, up - say - 473K, and then the aluminum foil should "backradiate" energy to the filament increasing still further the temperature up to 3300K, and so on...like a perpetuum motion, or like an endless tennis game.

and that's simply CRAZY!!

Not crazy at all. Infinite series of this type occur quite often in physics. As I think you are well aware, not all infinite series diverge. For example, the infinite series 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... converges to two.
#226 Joel Shore 2013-09-14 20:11
"Backradiation , according to the AGW/GHE supporters, should mean that the core (tungsten filament) of the light bulb would increase its temperature, because it would receive further Thermal energy, and THAT'S CRAZY!!"

Why is that crazy? Is it crazy if I predict that I can increase the temperature of my house just by putting more insulation in the attic, even though I have not increased the rate of thermal energy produced by the furnace? No...It is not crazy, because the steady-state temperature of my house is not just determined by the rate of energy production in my furnace. It is also determined by how easy it is for the house to transfer away the energy that the furnace produces.
#227 Ollie Hughes 2013-09-14 20:25
Quoting Tim Folkerts:
Not being facetious -- could you tell me what sort of specific measurement you are talking about? How exactly would you measure the magnitude of the thermal radiation from a hot surface to a cold surface?

But here's the thing, Tim. Backradiation has never not been a part of thermodynamics.

Thermally induced electromagnet radiation when absorbed is generally converted and expressed as a sensible heat gain or as some endothermic chemical process. The chemical process can only be measured indirectly, Latent heat of evaporation, or the changes in biomass.the sensible heat part is easy. Measure the delta temperature across a thermal resistance that all the flux must crosd the results must be both the power of the flux and the direction of the thermal flux.
#228 Ollie Hughes 2013-09-14 20:58

Please, that is radiance "potential to radiate", a precise scientific word used to distinguish between what "can be", and radiation "that which happens". Radiance is but remotely related to what is radiated or transferred.,

See my construction of a thermal fluxmeter here in this thread. The measurement of rate of sensible heat transfer to/from is easy, to any accuracy recognised by the NIST.
#229 Ollie Hughes 2013-09-14 21:09
Tim,

One of the most difficult things with respect to conveying a scientific idea is dealing with the fact that your audience brings with it all kinds of baggage. Unless one is open to and skilled at dealing with other people's baggage you really can't make any progress in conveying a scientific idea. IMO.

That is true, audience with baggage. Many thinking folk have learned to distinguish "snake oil" from "truth". In the Nuevo science process what is presented as "truth" always becomes "snake oil"
#230 Joel Shore 2013-09-14 21:14
Quoting Ollie Hughes:
Can they at least give som time frame for the "invention" of the non-measurable "back-radiation"? Was the term used before the AGW mess? Why has GHE changed from being dependent on the missing "back radiation" to be just an claimed, but never demonstrated, reduction of the electromagnetic radiation from the warm earth to cold space? Can a time frame for this latest change in GHE be established?

Actually, the idea that radiation goes only from the warmer to colder objects is an invention of the Slayers. It appears nowhere in the physics literature. I don't know about the exact history of our understanding, but my physics textbook from 1983 (Serway, "Physics for Scientists and Engineers", after introducing the law P = sigma*A*e*T^4 says

"A body radiates and also absorbs electromagnetic radiation at rates given by Eq. 17.11. If this were not the case, a body would eventually radiate all of its internal energy and its temperature would reach absolute zero. The energy that the body absorbs comes from the surroundings, which also emit radiant energy. If the body is at a temperature T and its surroundings are at a temperature T_0, the net power gained (or lost) as a result of radiation is given by
P_net = sigma*A_*e*(T^4 - T_0^4) (17.12)
When a body is in equilibrium with its surroundings, it radiates and absorbs energy at the same rate and so its temperature remains constant. When a body is hotter than its surroundings, it radiates more energy than it absorbs, and so it cools..."

If, in your opinion, 1983 is already beyond the date where the people in the Black Helicopters corrupted the authors of physics textbooks, I am sure you can go further back. S. Chandrasekhar's book "RADIATIVE TRANSFER" published in 1960 probably doesn't have any statement quite as pithy, but I see nothing in it that prohibits radiation from going from warmer to cooler objects.

As for the specific term "back radiation", I don't think that term was used much at all simply because radiation is radiation. I don't particularly like the term, although I can see why with the geometry of interest, one could think of it as radiation coming back to the Earth. Slayers have used the term much more than it was ever used by non-Slayers, likely because they have found traction in making it sound like some novel new phenomenon. As I have shown you in but one of countless examples, physics textbooks simply talk about bodies both emitting and absorbing radiation from their surroundings; there is no concern whatsoever about whether the surroundings are cooler or warmer until such time as one wants to know what the net effect is (i.e., if it emits more than it absorbs or absorbs more than it emits).

"Why has GHE changed from being dependent on the missing "back radiation" to be just an claimed, but never demonstrated, reduction of the electromagnetic radiation from the warm earth to cold space?"

It is simply two ways of saying the same thing. One is focused on the net effect while the other is focused on the transfers that lead to the net effect, just as it is talked about in the quote that I provided you from Serway.
#231 Joel Shore 2013-09-14 21:16
I said: "but I see nothing in it that prohibits radiation from going from warmer to cooler objects."

Of course, that should read "but I see nothing in it that prohibits radiation from going from cooler to warmer objects."
#232 Ollie Hughes 2013-09-14 21:27
None of your house insulation increases any energy. It can increase temperature (whatever that may be). Your fake "back radiation" must increase energy or it is truly "not".
#233 Joel Shore 2013-09-14 21:31
Quoting Ollie Hughes:
Can they at least give som time frame for the "invention" of the non-measurable "back-radiation"? Was the term used before the AGW mess?

Given the evidence that I have provided from but one of many physics textbooks, let's turn your questions around: Can you at least give some time frame for the invention of the idea that radiation from colder bodies is not emitted toward warmer bodies, or is not absorbed by warmer bodies (or, in violation of Conservation of Energy, it somehow has no effect)? Was this notion around before the Slayers invented it?
Quoting Ollie Hughes:
Tim,

One of the most difficult things with respect to conveying a scientific idea is dealing with the fact that your audience brings with it all kinds of baggage. Unless one is open to and skilled at dealing with other people's baggage you really can't make any progress in conveying a scientific idea. IMO.

That is true, audience with baggage. Many thinking folk have learned to distinguish "snake oil" from "truth". In the Nuevo science process what is presented as "truth" always becomes "snake oil"

Well, hopefully, your experiences with AGW will not have left you permanently jaded toward science and scientific progress. But, picking out the snake oil salesmen isn't hard, IMO. What's hard is breakthroughs/discoveries.
#235 Ollie Hughes 2013-09-14 21:39
Quoting Joel Shore:
I said: "but I see nothing in it that prohibits radiation from going from warmer to cooler objects."

Of course, that should read "but I see nothing in it that prohibits radiation from going from cooler to warmer objects."

Indeed, Make up your mind! That Joel Shore sees nothing or something does not affect
what "is" in this physical world
#236 Ollie Hughes 2013-09-14 21:49
#237 Tim Folkerts 2013-09-14 22:01
Quote:
But here's the thing, Tim. Backradiation has never not been a part of thermodynamics.
Quote:
Ok, now you have me REALLY confused! I better let you argue with yourself for a while.
Quoting Joel Shore:

Actually, the idea that radiation goes only from the warmer to colder objects is an invention of the Slayers.

Well, it's one of those kind of elementary and mundane things that a physicists might learn as a college freshman or sophmore. So it's easy to forget.
#239 Tim Folkerts 2013-09-14 22:13
On the flip side, unless students are open to learning AND have sufficient background, teaching will never be effective. Every field has assumed knowledge -- medicine, law, sports, cooking.

Trying to teach me the nuances of cricket is not going to work, because I have never played and don't understand the game. I would have to invest my OWN time until I knew the rules; I would have to watch some games; maybe even have a pitcher throw bowler bowls a few balls to me. THEN an expert could help me to learn the finer points.

Similarly, unless someone already knows a fair amount of thermodynamics and math, they will not be a good student (and a worse teacher) in these discussions.
Quoting Tim Folkerts:
Quote:
But here's the thing, Tim. Backradiation has never not been a part of thermodynamics.

Quote:
Ok, now you have me REALLY confused! I better let you argue with yourself for a while.
Jim McGinn:
Uh, I don't know what to say. Are you, maybe, not a native english speaker?

Jim McGinn
#241 Lonny Eachus 2013-09-14 22:22
I retract my comment about the temperature differential being different; it was late at night and I wasn't thinking clearly.

You need not have repeated your bit about back-radiation reducing cooling. I am not an idiot and I understood you the first time. But that has NOTHING whatever to do with the point I was making.

Quote:
Also, you are confusing two different senses of the word "cooling". When we say that back-radiation decrease the cooling, we are not only talking about some case where the object is actually cooling over time. We are also talking about the case where the object is in a steady-state.
I made no such error. I can see where you might have thought I did, though. I did mention cooling, but I only intended it to be an example, not an exclusive case. I could have worded it better.

As it is, though, I retract my comment about the temperature differential being different. I really don't want to get into that argument anyway, as it isn't important to my point.

Quote:
It is just like your house if you insulate it better: It is less easily able to cool and hence it will be at a higher steady-state temperature when you run the furnace than it would be before you added the insulation.
No, it isn't, and I explained this to you earlier. Typical thermal insulation prevents heat loss via conduction, not radiation. If the OUTSIDE of the insulation radiates less, it is only because there was less heat transfer to that surface from the inside VIA CONDUCTION. That is not the process under discussion here, and it is not even a valid analogy. On the contrary, it is such an elementary and obvious error that I have to wonder how anyone knowledgeable about the subject could make it.

Just so you don't misunderstand ME: I understand you analogies. They are crystal clear to me. But I am not convinced they are valid analogies at all. The insulation analogy is DEFINITELY not valid.

Please dispense with the analogies, which I already understand, and simply point me to an EXAMPLE of a system in which a spheroid, wrapped in a back-radiating blanket, radiates at its surface (inside said blanket) at a higher intensity than the total incoming radiation, per unit area.

That's all I'm asking for. Not arguments or analogies, but an example. As I stated earlier, I cannot accept Venus as an example because that would be assuming the effect in question.
#242 Tim Folkerts 2013-09-14 22:23
Quote:
Thermally induced electromagnet radiation when absorbed is generally converted and expressed as a sensible heat gain or as some endothermic chemical process.
Ah .. so you are talking about the NET radiation. For example, radiation from a 300 K object to a 300 K detector would read zero for absorbed energy. The incoming 300 K radiation cannot be converted into energy that makes the detector warmer. If the object gets a little warmer, then there will be a small gain in the detector. If the object gets a little cooler, then there will be a small loss in the detector.

P/A = εσ [ T_object^4 - T_detector^4 ]

So then, yes, there has never be a case where a measured NET energy has gone from colder to warmer.

Others find it handier to talk about εσ T_object^4 as the incoming radiation -- a value that depends ONLY on the temperature of the radiating object. Similarly εσ T_detector^4 as the outgoing radiation -- a value that depends ONLY on the temperature of the radiating detector.

This is just semantics. The two are mathematically equivalent.
#243 Ollie Hughes 2013-09-14 22:28
Well, hopefully, your experiences with AGW will not have left you permanently jaded toward science and scientific progress. But, picking out the snake oil salesmen isn't hard, IMO.
What's hard is breakthroughs/discoveries.

I am jaded toward science and scientific progress. I observe none at all. I observe many new things from those that can do. Nothing from those that cannot do, but only teach. It is the change from those that do, that admit no knowledge, but only the numbers with truth, to the opposite. The I know everything from professors that is the conversion from "truth" to "snake oil".
Truth is always wiggly and must change from moment to moment. Snake oil comes in the same old bottle.
#244 Lonny Eachus 2013-09-14 22:35
Quoting Joel Shore:
Actually, the idea that radiation goes only from the warmer to colder objects is an invention of the Slayers.

Did you never actually read LaTour, or are you just being an ass? I'm not accusing, just asking.

NOBODY (other than maybe one or two people in the comments, which don't represent anybody's "official" position) claimed that radiation "goes only from warmer to colder objects". Nobody.

That is a gross mis-characterization of their actual argument, which is about net, not "any".

This is the grossest example of a straw man I have seen yet. You are knocking down a claim that nobody has made.
Quoting Ollie Hughes:

Please, that is radiance "potential to radiate", a precise scientific word used to distinguish between what "can be", and radiation "that which happens". Radiance is but remotely related to what is radiated or transferred.,

See my construction of a thermal fluxmeter here in this thread. The measurement of rate of sensible heat transfer to/from is easy, to any accuracy recognised by the NIST.

I saw your instructions. Very interesting. Somewhat out of my wheelhouse. I don't even think I could identify a thermocoupler. I'm more of a theory guy.
#246 Lonny Eachus 2013-09-14 22:48
That *IS* interesting, but unfortunately I mis-stated the problem here earlier. In his rebuttal of Spencer, Dr. Latour actually made the point that it isn't the surface of the bulb that is actually in question, it is the filament.

In order for the experiment to actually be a valid example, it is the system SOURCE that would have to be warmer. The surface of the bulb is not the source.

So in order to be valid, you would have to measure the filament.

Another method might be to enclose the bulb inside another container, with little or no conductive communication between the bulb and the container. Then perhaps you could consider outside of the bulb to be the "source", and attach the thermistor to it. Then measure the temperature of the outside container, with and without the foil.

I'm not sure it would make any difference; still it would be a more valid comparison.

In any case, mea culpa. I had read the rebuttal some time ago but I wasn't thinking about it at the time.
Quoting Joel Shore:
Quoting Ollie Hughes:
Can they at least give som time frame for the "invention" of the non-measurable "back-radiation"? Was the term used before the AGW mess?

Given the evidence that I have provided from but one of many physics textbooks, let's turn your questions around: Can you at least give some time frame for the invention of the idea that radiation from colder bodies is not emitted toward warmer bodies, or is not absorbed by warmer bodies (or, in violation of Conservation of Energy, it somehow has no effect)? Was this notion around before the Slayers invented it?

Jim McGinn:
After an extensive investigation we have located the source of this clear violation of the laws of thermodynamics. And the answer to the question is seventh graders. All of them. So, were the slayers ever students in seventh grade physics classes? The evidence mounts.
#248 Ollie Hughes 2013-09-14 23:07
Quoting Tim Folkerts:
Thermally induced electromagnet radiation when absorbed is generally converted and expressed as a sensible heat gain or as some endothermic chemical process.

Ah .. so you are talking about the NET radiation. For example, radiation from a 300 K object to a 300 K detector would read zero for absorbed energy. The incoming 300 K radiation cannot be converted into energy that makes the detector warmer. If the object gets a little warmer, then there will be a small gain in the detector. If the object gets a little cooler, then there will be a small loss in the detector.

P/A = εσ [ T_object^4 - T_detector^4 ]

So then, yes, there has never be a case where a measured NET energy has gone from colder to warmer.

Others find it handier to talk about εσ T_object^4 as the incoming radiation -- a value that depends ONLY on the temperature of the radiating object. Similarly εσ T_detector^4 as the outgoing radiation -- a value that depends ONLY on the temperature of the radiating detector.

This is just semantics. The two are mathematically equivalent.

Yes but no, Mathematically the actual heat transfer between two thermal radiative potentials is correct and as measured is the same. The difference is that the three Professors claim a large energy transfer from the higher temperature to absolute zero kelvin and then a reverse transfer of energy from zero kelvin back to them lower temperature. The transfer from zero Kelvin to a higher temperature has never been demonstrated and the idea of net is false! the actual transfer is the "only" not the net. And the concept of back radiation as the only reason for Greenhouse effect is hereby demolished. Any questions?
#249 Ollie Hughes 2013-09-14 23:15
Quoting Tim Folkerts:
Actually, I just did this experiment yesterday. I glued a small thermistor to a light bulb. The glass warmed to ~ 55 C when uncovered. With Al foil, covering the bulb, the temperature of the thermistor shot up to over 100 C within a couple minutes.

So what does that mean with that forced convection from that temperature.
#250 Ollie Hughes 2013-09-14 23:29
Others find it handier to talk about εσ T_object^4 as the incoming radiation -- a value that depends ONLY on the temperature of the radiating object. Similarly εσ T_detector^4 as the outgoing radiation -- a value that depends ONLY on the temperature of the radiating detector.

Please demonstrate any of what you claim, mainly that the radiative exitance depends "only" on the temperature of the emitter
Quoting Ollie Hughes:
Truth is always wiggly and must change from moment to moment.

This quote is great.
#252 Tim Folkerts 2013-09-15 00:03
Quote:
The difference is that the three Professors claim a large energy transfer from the higher temperature to absolute zero kelvin and then a reverse transfer of energy from zero kelvin back
This is a new spin that I haven't heard before. The actual claim is that there is a large energy transfer from the higher temperature object (eg the earth's surface) to a lower temperature object (eg the atmosphere) which is NOT at absolute zero. There is a SMALLER energy transfer from the lower temperature temperature object back to the higher temperature object.

It is amazing how many of the objects are actually strawmen, incorrectly predicting what others think. Nothing is demolished; and, no, I have no questions for you.

PS If the cold object is indeed 0 K, then there can be no back-radiation.
#253 Ollie Hughes 2013-09-15 00:03
Very good Tim, pleas tell us of your misunderstanding of thermodynamics and math.
There are many many earthlings, on each or both,
that have learned by making and forgetting more mistakes than you will ever do in your lifetime.
You can learn, by carefully listening. All you need to do is ask! then disagree. This does not make friends!
#254 Tim Folkerts 2013-09-15 00:07
Quote:
So in order to be valid, you would have to measure the filament.
I plan to do that soon. One version of the experiment might use an actual light bulb, but other geometries should show the effect more dramatically.
#255 Ollie Hughes 2013-09-15 00:15
This is immaterial. A 10 um photon is a 10 um photon. It behaves the same way whether it came from a block of ice or from the sun.
What arrogance, Try to demonstrate what you claim
#256 Tim Folkerts 2013-09-15 00:31
Quote:
What arrogance
I'm arrogant??? I give the 'textbook' answer that is accepted by basically all scientists everywhere.
All electrons are identical (except for energy & momentum).
All protons are identical (except for energy & momentum)
All photons are identical (except for energy & momentum)

You arrogantly claim on your own authority that all textbooks and all scientist are wrong, without presenting any evidence. You are the one making an extraordinary claim, not me!
#257 Ollie Hughes 2013-09-15 01:25
Quoting Lonny Eachus:
That *IS* interesting, but unfortunately I mis-stated the problem here earlier. In his rebuttal of Spencer, Dr. Latour actually made the point that it isn't the surface of the bulb that is actually in question, it is the filament.

In order for the experiment to actually be a valid example, it is the system SOURCE that would have to be warmer. The surface of the bulb is not the source.

So in order to be valid, you would have to measure the filament.

Another method might be to enclose the bulb inside another container, with little or no conductive communication between the bulb and the container. Then perhaps you could consider outside of the bulb to be the "source", and attach the thermistor to it. Then measure the temperature of the outside container, with and without the foil.

I'm not sure it would make any difference; still it would be a more valid comparison.

In any case, mea culpa. I had read the rebuttal some time ago but I wasn't thinking about it at the time.

Lonny,
Most are thinking of very narrow scientific and mathematical answers. There is none.
the filament with almost no radiating surface
with an imaginary cross sectional area.can produce an luminescent radiating bulb the bulb that has a surface area that does provide the luminance and illumination of the lamp.
The professors do not notice that the 3600 Kelvin filament temperature and the 360 Kelvin bulb temperature produce a tremendous
convective heat transfer with the backfill argon gas. Your Professors are again promoting scientific fakery.
#258 David Cosserat 2013-09-15 12:08
Tim,

I like experiments. That's the only way forward to unlock the entrenched positions represented by commentators on this thread.

The question is why the temperature of your light bulb shoots up from 55degC to 100degC when you cover it with aluminium foil.

Is the temperature rise due to the reflection of radiant energy from the foil's inner surface? Is it due to absorption of radiant energy at the foil's inner surface? Is it due to conduction of thermal energy from the outer surface of the bulb glass to the foil's inner surface?

CONDUCTION?
Yes. The aluminium is an extra layer of thermal conductive material in addition to the glass etc. of the bulb. An extra layer of insulation (however low - aluminium is a good conductor) means more impedance to flow. Hence the interfaces will rebalance with higher temperatures all the way up the ambient-foil-glass-filament impedance path.

ABSORPTION?
Yes. To the extent that the aluminium (not being a perfect reflector) absorbs radiant energy coming from the filament, the resulting thermal energy has to flow through the aluminium towards its outer surface rather than simply being radiated or convected directly to ambient from the outer surface of the glass, as happened before the foil was introduced. So the introduction of the foil has caused an additional obstacle to flow.

REFLECTION?
Yes, for 2 reasons:

(1) Some proportion of the radiant energy reflected back from the inner surface of the aluminium foil will pass straight through the glass and be absorbed by the filament. Compared with the situation before the foil was introduced, this means that there is now an imbalance between the electrical power input to the filament (same as before) and the energy flowing through from the filament to ambient (lower than before). This will cause the filament temperature to rise to compensate for this additional obstacle to flow.

(2) Some proportion of the radiant energy reflected back from the inner surface of the aluminium foil will be absorbed by the glass. This means that there is now more energy flowing into the glass than before. This will cause the glass temperature to rise to compensate for this additional obstacle to flow.

[continued below...]
#259 David Cosserat 2013-09-15 12:09
[...continued from above]
So to summarise, although we don't know what proportions each of the above mechanisms will contribute to the overall temperature rise up the impedance path (at least not without doing detailed calculations based on all the relevant physical parameters!) we nevertheless do know with 100% certainty that the temperatures of the layers WILL rise until a new steady-state temperature profile is achieved. The new steady state temperature of the glass/foil interface in the particular case of your experiment being 100degC.

BUT SO WHAT?
Tim, having said all that, my problem is that I can't see how your experiment assists in disproving (or proving) the Slayer concept that radiant energy cannot flow from a cooler to a warmer body. If my analysis above is correct, nowhere is energy having to flow from a cooler to a warmer interface. (Reflection does not violate that rule when the radiation returns to its emitting surface.)
Quoting Tim Folkerts:
It is amazing how many of the objects are actually strawmen, incorrectly predicting what others think.

IOW, the whole discussion is the result of semantic confusion.

Quoting Tim Folkerts:
Nothing is demolished; . . .

Everything is demolished. We now realize both sides in this discussion have been talking past each other because both sides have not bothered to consider the epistemological POV of their opponents.

In terms of pure science, however, the physicists in this conversation are right (back radiation can be ignored because it is already accounted for at a fundamental level) and, therefore, Warmies are wrong because their accounting tactics mistakenly assume that backradiation is not already accounted for.
#261 Tim Folkerts 2013-09-15 14:19
Quote:
the physicists in this conversation are right (back radiation can be ignored because it is already accounted for at a fundamental level) and, therefore, Warmies are wrong ...
Except the physicist in this conversation ARE the "warmies". Our 'accounting tactics' include proper radiation (either explicitly as "back-radiation" or implicitly as "net radiation"). We are able to discuss it either way, and either way it shows a warming of the earth's surface as a direct result of IR radiation.
Quoting Tim Folkerts:
Quote:
the physicists in this conversation are right (back radiation can be ignored because it is already accounted for at a fundamental level) and, therefore, Warmies are wrong ...

Except the physicist in this conversation ARE the "warmies". Our 'accounting tactics' include proper radiation (either explicitly as "back-radiation" or implicitly as "net radiation"). We are able to discuss it either way, and either way it shows a warming of the earth's surface as a direct result of IR radiation.
Rest assured, I fully realize that. Moreover I know how you will do it:
The appearance of scientific validity of arguments from Warmies is 100 percent dependent on semantic ambiguity and the appearance will persist only for as long as the semantic ambiguity is concealed from their audience. Moreover, once a semantic ambiguity is revealed as scientfically invalid it is only a matter of time before the Warmy (or some other Warmy) will use some sort of semantic trick to reestablish the appearance of scientific validity.

" . . . like Mike's Nature trick."

Jim McGinn
What about the Vortex phase of water.
#263 Joel Shore 2013-09-15 15:28
If the claim is about net, then nobody disagrees. All models of the greenhouse effect, whether they be "toy" models like the steel greenhouse or full-fledged climate models have the net radiative transfer being from the warmer Earth to the colder atmosphere.
#264 Joel Shore 2013-09-15 15:30
Temperature is a measure of internal energy.
#265 Curt 2013-09-15 15:44
Tim, I did those measurements a few months ago. You need professional grade voltmeters and ammeters, but nothing too expensive. Wrapping foil tightly around a clear 40W bulb increases filament temperature 10-11K. It increases bulb surface temperature by over 200K (so be very careful of potential fire danger!)
Wrong. Temperature can he used to represent energy but at best it is a "proxy" for energy. It can also be used as a proxy for flux (when energy flows). Temperature measurements just give you a indication of how excited the molecules are at a certain point. Abstracted from a context, temperature tells you nothing directly at all about either energy or flux.

IMO, physicists (dragon slayers) know this innately, and therefore are not prepared to enunciate it. In sharp contrast, Warmies don't understand this much at all and when one attempts to educate them they become defensive and ultimately ignore this understanding deliberately.
#267 Ollie Hughes 2013-09-15 16:24
Quoting Joel Shore:
Temperature is a measure of internal energy.

Wrong What is the internal energy of an isotherm?
#268 Tim Folkerts 2013-09-15 16:24
Glad to hear that my basic results are confirmed. I didn't wrap the foil especially tight-- I was going for a rough first approximation. Also, my thermistors are only rated to 150 C, so I stopped before it got too closer to these temperatures.

I am also glad you could do the next step and actually determine the temperature of the filament. As you say, with a good meters, it should not be too tough to measure the resistance = (V/I) of the filament and watch it increase as the foil is added around the bulb.

In one fell swoop, our experiments refute the analysis of the top post and refute those who are "sure" the filament won't warm up.
#269 Ollie Hughes 2013-09-15 16:46
Quoting Joel Shore:
Quoting Ollie Hughes:
Can they at least give som time frame for the "invention" of the non-measurable "back-radiation"? Was the term used before the AGW mess?

Given the evidence that I have provided from but one of many physics textbooks, let's turn your questions around: Can you at least give some time frame for the invention of the idea that radiation from colder bodies is not emitted toward warmer bodies, or is not absorbed by warmer bodies (or, in violation of Conservation of Energy, it somehow has no effect)? Was this notion around before the Slayers invented it?

Rumford (early 1800's)stated that heat cannot transfer transfer on opposite directions simultaneously. Clausius 1854 stated "the second law of thermodynamics" This law has never been falsified. Show any use of the term "back radiation" prior to 1985. It is a warmist invention that is false
#270 Joel Shore 2013-09-15 17:16
Then you ought to be able to find a physics textbook that makes this claim. I haven't been able to find any, but I have found plenty that directly contradict it.
#271 Joel Shore 2013-09-15 17:25
And yet, I have shown you an example of a physics textbook (and could easily show you a dozen more) that talks about radiation from colder surroundings being absorbed by a warmer body. In fact, you won't be able to find a textbook that agrees with you.

And, the problem is that you don't understand what "heat" is. Heat and radiation are not synonymous. Heat is the NET MACROSCOPIC energy flow. You need look no further than Wikipedia to find this out.

As I have already explained, whether the term "back radiation" is used or not is irrelevant. That is not a scientific term; it is just a term used sometimes in this case given the specific geometry. And, it is a term used much more by Slayers than anybody else.

If you understand that radiation goes from colder to warmer bodies as well as warmer to colder but want to not use the term "back radiation", I am fine with that. The more I see the term, the more I dislike it because it feeds into the Slayer sophistry that this is some sort of novel effect.
#272 Ollie Hughes 2013-09-15 17:35
Quoting Tim Folkerts:
Quote:
What arrogance

I'm arrogant??? I give the 'textbook' answer that is accepted by basically all scientists everywhere.
All electrons are identical (except for energy & momentum).
All protons are identical (except for energy & momentum)
All photons are identical (except for energy & momentum)

You arrogantly claim on your own authority that all textbooks and all scientist are wrong, without presenting any evidence. You are the one making an extraordinary claim, not me!
Only other professors accept your textbook answers. Each scientist has his own answer, which if not different from your textbook, will soon be different.
You claim CO2 molecules vibrate at a frequency
2E13 Hz -- which corresponds to a 15 um IR photon. So CO2 can absorb and emit 15 um photons. Three extraordinary claims in one.
Textbooks have theory not science. No on has observed a CO2 molecule at that frequency.
We can observe absorption an emission electromagnetic energy by CO2 molecules at that wavelength. It is only the fake kinetic theory of heat that has folk actually believing that there is some physical motion at that rate. You claim photons "have" energy
while this is just another fake theory. Show any of your fake "15 micron photons" being emitted form a block of ice toward a pot of boiling water?. You seem to claim manny things while arrogantly claiming that the textbooks you teach from, are some sort of truth in science. They are theories useful in helping students think and learn. To claim the theories are valid science is true arrogance!

eaving
#273 Joel Shore 2013-09-15 17:38
Slayers need to either put up or shut up in regards to their silly claim that the Second Law forbids radiative transfer from a colder body to a warmer body. Show us ***ANY*** physics textbook that agrees with you on that...ANY one.

Every physics textbook that I have ever seen in fact interprets the second term in the Stefan-Boltzmann Equation as being radiation from the colder body (or surroundings) absorbed by the warmer body.

The heat is the net macroscopic flow of radiation which, clearly from the form of the S-B Equation, is from the hotter to the colder body. However, this net effect is arrived out through an exchange of radiation between the bodies.

You have to either directly show that your interpretation of the Second Law has any justification in the physics literature OR admit that you are making up laws of physics to suit your prejudices.

You guys are talking nonsense that NO recognized physics authority would support and it is time that you provide evidence (which, of course, you won't be able to) to support your bogus interpretation of the laws of physics.
#274 David Cosserat 2013-09-15 17:46
Tim,

You say: i]In one fell swoop, our experiments refute the analysis of the top post and refute those who are "sure" the filament won't warm up.

But how do you respond to my analysis at principia-scientific.org/latest-news/301-robert-brown-backradiation-and-mathematics.html#comment-1564 where I argue that the only filament heating (very minor because of the tiny view factor) is due to reflection, not back radiation, and that most of the warming is due to increased thermal impedance?
#275 David Cosserat 2013-09-15 17:49
#276 Ollie Hughes 2013-09-15 17:54
Quoting Tim Folkerts:
Quote:
I think if they did rigorously test this notion and actually quantify the GHE characteristics of molecules, these result would reveal that there is little difference in the GHE characteristics of GHEs (CO2, methane) and the normal constituents of the atmosphere N2, O2.

Simply stating an opinion ("I think") is not very scientific of you! The data is readily available; the quantum mechanical theory is well-founded. This is not something we need to have opinions about.

Try looking at Spectralcalc www.spectralcalc.com/spectral_browser/db_intensity.php
For "Species" select H2O, CO2, O2, & N2.
For "Spectral Range" try somerthing like 1-40 microns.
Under "Options" select 'Scale by Atmospheric Abundance'

Then plot the results. The GHGs are clearly orders of magnitude stronger absorbers. AND they absorb in the important wavelengths (around 10 um)

Quote:
Sir, Lir? What in the world are you talking about?

Joel was simply following the abbreviations that John Marshall had introduced. From the context, it is pretty clear these mean "Solar infrared radiation" (0.7-4 um) and "Longwave infrared radiation" (4-100 um)

Quote:
Specralcalc does some of this. MODTRAN is a common calculator for such things. You can play with a user-friendly interface here:
forecast.uchicago.edu/Projects/modtran.html
Lowtran, Modtran, and Hitran were never developed to do any "radiative transfer calculations." That code and database was developed to do estimates on atmospheric "seeing". The use for radiative heat transfer is another Warmest fallacy. Check out the development by the Air Force Cambridge Laboratory circa 1970.
#277 Ollie Hughes 2013-09-15 18:28
Quoting Tim Folkerts:
The difference is that the three Professors claim a large energy transfer from the higher temperature to absolute zero kelvin and then a reverse transfer of energy from zero kelvin back

This is a new spin that I haven't heard before. The actual claim is that there is a large energy transfer from the higher temperature object (eg the earth's surface) to a lower temperature object (eg the atmosphere) which is NOT at absolute zero. There is a SMALLER energy transfer from the lower temperature temperature object back to the higher temperature object.

This is bold faced deception. From your own comments here you have looked at the charts
of radiative budget from the IPCC In those charts the flux from the surface has a value such that the receptor is near zero kelvin, The flux from the atmosphere also has a value such that the receptor is near zero kelvin. This was done by spliting your S_B equation into to two parts To deceptively fool the public into thinking that such downwelling flux called "back radiation" actually exists and warms the earth surface. Your S-B equation correctly predicts the maximum possible only flux from the surface, but only if both the surface and the atmosphere were lambertian surfaces with 100-% emissivity.
Every part of the GHE theory is made up and fake nonsense and is demolished.

PS If the cold object is indeed 0 K, then there can be no back-radiation.
Another nonsense claim. Ddemonstrate a zero kelvin object with 100% emissivity!
There is never any back radiation.
#278 Tim Folkerts 2013-09-15 18:57
Quote:
The flux from the atmosphere also has a value such that the receptor is near zero kelvin.
Again this is semantics. I see better now what you are thinking, but it is a very unusual interpretation.

The equation that describes the radiant heat from a hot object to a cold object is
P/A = εσ(T_hot)^4 - εσ(T_cold)^4 ]
(give or take a few geometrical factors that we can ignore for now). This is well accepted by physicists and engineers around the world. It agrees with both theory AND practical application.

Usually this is interpreted either as
1) The whole thing represents the net radiant energy flow (ie the "radiant heat").
2) Each half represents the radiant energy flow from each surface separately.

You seem to want to interpret this as
P/A = εσ[(T_hot)^4 - (0 K)^4] - εσ[(T_cold)^4 - (0 K)^4]
Mathematically it is the same. But I really don't see what it gains.
#279 David Cosserat 2013-09-15 19:05
Joel Shore says: Slayers need to either put up or shut up in regards to their silly claim that the Second Law forbids radiative transfer from a colder body to a warmer body.

Joel,

I agree. All I see here is juvenile behaviour and name calling and worse.

Where are the heavyweight Slayers, Siddons, Postma, Latour, Schreuder, Hertzberg, etc? Don't they care that their colleagues here are making fools of themselves?

I am still waiting for an official Slayer explanation of how they can, on the one hand, resolutely maintain that energy cannot possibly be transferred from a colder heat sink at a constant temperature Tv to a warmer constantly-powered body at steady-state temperature Tc ... yet on the other hand can equally resolutely maintain that varying the value of Tv will cause Tc to vary.

All they need to do is to explain clearly and succinctly their proposed mechanism (and supporting math) explaining how varying Tv can possibly vary Tc without any back radiation. Then we can devise a mutually agreed experiment to decide empirically between Warmist physics (back radiation) and Slayer physics (still unknown).

It is such a simple and obvious question that I find it incredible that they cannot answer it succinctly and clearly.

After several months of work on this, always keeping an open mind, I am very disappointed with the lack of progress. It makes me all the more resolute about getting to the bottom of what is going on here. Are the Slayers in self-destruct? Do they even exist? Are they just playing games?
#280 David Cosserat 2013-09-15 19:07
Joel Shore says: Slayers need to either put up or shut up in regards to their silly claim that the Second Law forbids radiative transfer from a colder body to a warmer body.

Joel,

I agree. All I see here is juvenile behaviour and name calling and worse.

Where are the heavyweight Slayers, Siddons, Postma, Latour, Schreuder, Hertzberg, etc? Don't they care that their colleagues here are making fools of themselves?

I am still waiting for an official Slayer explanation of how they can, on the one hand, resolutely maintain that energy cannot possibly be transferred from a colder heat sink at a constant temperature Tv to a warmer constantly-powered body at steady-state temperature Tc ... yet on the other hand can equally resolutely maintain that varying the value of Tv will cause Tc to vary.

All they need to do is to explain clearly and succinctly their proposed mechanism (and supporting math) explaining how varying Tv can possibly vary Tc without any back radiation. Then we can devise a mutually agreed experiment to decide empirically between Warmist physics (back radiation) and Slayer physics (still unknown).

It is such a simple and obvious question that I find it incredible that they cannot answer it succinctly and clearly.

After several months of work on this, always keeping an open mind, I am very disappointed with the lack of progress. It makes me all the more resolute about getting to the bottom of what is going on here. Are the Slayers in self-destruct? Do they even exist? Are they just playing games?
#281 Ollie Hughes 2013-09-15 19:08
Quoting Joel Shore:
I said: "but I see nothing in it that prohibits radiation from going from warmer to cooler objects."

Of course, that should read "but I see nothing in it that prohibits radiation from going from cooler to warmer objects."

Except the second law of thermodynamics,
and the fact that you can only claim never demonstrate "back radiation"

Chandrasekhar's book "RADIATIVE TRANSFER" published in 1960 probably doesn't have any statement quite as pithy, but I see nothing in it that prohibits radiation from going from warmer to cooler objects.

That book is correct because "back radiation"
was not invented yet. Nowhere in that book is anything that says an object thermally radiates a flux "only" determined by its own temperature. All thermal radiation is determined by a difference in radiative potential not by the potential of itself.

It is simply two ways of saying the same thing. One is focused on the net effect while the other is focused on the transfers that lead to the net effect, just as it is talked about in the quote that I provided you from Serway.

No the whole idea of GHE and "back radiation"
was created to deliberately deceive the public an governments. You talk of a reduction but that is still false. Such has never been demonstrated. Thermal radiation from the surface to the atmosphere is insignificant compared to the thermal radiation of the atmosphere itself to cold space. Nno thermal radiation exists from the atmosphere to the warmer surface.
#282 Ollie Hughes 2013-09-15 19:35
I am not a slayer (whatever that is) buy it is easy to demonstrate the lack of such radiation.
Demonstrate with a real experiment that "all"
the power in your experiment goes in a single direction, "away from the source" to do this requires the ability to measure thermal flux independent of temperature. Such devices are available or can be constricted. measureing absolute temperatures will not demonstrate that there is no power available to transfer colder to warmer. by any means including thermal radiation. This I think the only claim your slayers are making. It is the warmests and professors that claim thermal radiation is different. Why not ask them to demonstrate their claims! I can demonstrate that the warmest idea is false. Where is their experiment to prove their claims.
#283 Curt 2013-09-15 19:41
David - I did I set of experiments earlier this year that should address your concerns. I described them at WUWT here:

wattsupwiththat.com/2013/05/28/slaying-the-slayers-with-watts-part-2/

I compared the effect of transparent, reflective, and absorptive/re-emittant "shells" over the bulb (but not it contact with it). The transparent glass shell, while it had a resistance to thermal conduction around a thousand times higher than the metal shells, left the bulb the coolest.

Replacing the glass shell with either of the metal shells and watching the bulb warm up to a new steady-state temperature, the only conclusion is that radiative energy from outside the bulb, even though from an object colder than the bulb, increased the internal energy and therefore temperature of the bulb. The electrical power into the bulb did not increase, and in fact decreased a tiny bit.
Howse about a direct response to my questions.
#285 Ollie Hughes 2013-09-15 19:50
If you look carefully at Roy Spencers steel greenhouse thought experiment demonmstrates
that "all" power from the inner powered sphere to the shell via thermal radiation is always simultaneously radiated to a colder sink.there simply is no power available to return to the powered sphere.
#286 Ollie Hughes 2013-09-15 20:20
Quoting Curt:
Tim, I did those measurements a few months ago. You need professional grade voltmeters and ammeters, but nothing too expensive. Wrapping foil tightly around a clear 40W bulb increases filament temperature 10-11K. It increases bulb surface temperature by over 200K (so be very careful of potential fire danger!)

How ever did you determine that the temperature of the filament only went up by 11 Celsius? Tungsten's thermal coefficient of resistance is very non linear at those temperatures. What were the numbers? If the resistance went up the power to be dissipated went down. If you got the bulb temperature to increase by 200 Celsius, you were seriously interfering with the Convective cooling of the glass. How did you measure that temperature?
How do you know that you were measuring a temperature increase in the glass rather than that of the temperature sensor?
What "claim?" And what is "it?"
Quoting Ollie Hughes:
Quoting Joel Shore:

I see nothing in it that prohibits radiation from going from cooler to warmer objects."

Except the second law of thermodynamics, . . .

Ollie, if you're saying the actual units of energy (electro magnetic engergy packets) don't actually move both ways between hotter and colder entities then you are wrong.

Take a deep breath and read the whole thread before you respond.
#290 Tim Folkerts 2013-09-15 20:34
Using P/A = εσ(T_hot)^4 - εσ(T_cold)^4 ], please explain how power is radiated from the inner shell to the outer shell if both are at the same temperature? The power from the inner to the outer shell would have to be
P/A = εσ(T_inner)^4 - εσ(T_outer)^4 = 0 W/m^2
#291 Ollie Hughes 2013-09-15 21:49
Quoting Tim Folkerts:
Using P/A = εσ(T_hot)^4 - εσ(T_cold)^4 ], please explain how power is radiated from the inner shell to the outer shell if both are at the same temperature? The power from the inner to the outer shell would have to be
P/A = εσ(T_inner)^4 - εσ(T_outer)^4 = 0 W/m^2

I did not say they were at the same temperature
I sad that they both radiated the same power always outward to a colder sink. Call this total of all power power P1. The temperature of the cold sink Tcold is an isotherm capable of providing or dissipating any power required to maintain that temperature.
The inner powered sphere radiating alone to the Tcold sink with establish equilibrium temperature Tsphere1
Introducing a shell between the sphere and the coldest sink seriously interferes with such radiation. the effect on the sphere is the same as reducing its emissivity to 1/2 its true value. Thermodynamically the temperature of the sphere must icrease to radiate the same power to Tcold at with 1/2 the emissivity, as per the Whole S-B equation. Call this new temperature Tsphere2. The introduction of the shell neat the sphere with high emissivity changes the entire geometry the sphere can no longer radiate to Tcold, it must radiate to Tshell At the new temperature Tsphere2 will radiate all of P1 to the shell which spontaneously reaches temperature tshell which is just high enough to radiate power P1 to Tcold with a larger surface area than the sphere. the details of Tshell depend on the surface area increase. All power P1 from the sphere passes through the shell unchanged and none is returned. In thermodynamic terms the shell reaches a temperature to radiate Ts P1 to Tcold. The sphere then reaches the temperature required to radiate P1 to Tshell.
All temperatures and fluxes are in thermodynamic equilibrium and all are accounted fore with absolutely no "back radiation" from the radiance of the inner surface of the shell. There is no power from anywhere for the nonsense radiation.
#292 Joel Shore 2013-09-15 22:04
Well, what any physics textbook would tell you is that the interpretation of the second term in the S-B equation describing the net radiative transfer [Power = sigma*A*(T_hot^4 - T_cold^4)] is due to radiation from the colder to the hotter object (and the first term is the larger amount of radiation from the hotter to the colder object).

However, if you want to believe, against all evidence, that this is not correct, I guess it really doesn't matter: You are not disputing the greenhouse effect, which arises from the mathematics and not from the physical interpretation of the terms.
#293 Ollie Hughes 2013-09-15 22:07
Ollie, if you're saying the actual units of energy (electro magnetic engergy packets) don't actually move both ways between hotter and colder entities then you are wrong.

Take a deep breath and read the whole thread before you respond.

I have but that is quite a task. The warmests
If the energy goes both ways with different temperatures it must also go both ways with no temperature difference. This is both physical and conceptual nonsense. Thermal radiation goes one way only. the warmests here insist that the slayers demand an infinite temperature increase with back radiation this is true only if there "is" back radiation.
there is no back radiation. There is only the correct radiation from hot to cold as per the whole S-B equation with two not one thermal radiative potentials. Now the professors chime in with there "net", which is the "only"
physical transfer ever observed. The rest is a professors dream only.
#294 Joel Shore 2013-09-15 22:22
(1) The overlap between physicists and Dragon Slayer is vanishingly small. Even among physicists who are skeptics in regards to the seriousness of AGW, such as Robert Brown, Fred Singer, ..., there is acceptance of the basic physics of the greenhouse effect and general embarrassment about Slayers (as expressed by Robert above).

(2) Like I said, temperature is a measure of internal energy. Yes, the relationship is not always very simple. It is in an ideal gas, which most gases not too close in density to their liquid state are good approximations thereof, where in fact temperature and internal energy are directly proportional to each other. In liquids and solids, the relationship is more complicated...but still the internal energy increases as temperature increases. (Only at phase transitions can you get the situation where internal energy increases with no increase in temperature.)
#295 Tim Folkerts 2013-09-15 23:03
So now you are arbitrarily changing the emissivity of the inner shell to try to get the numbers to work?

Just to be clear -- do you agree that the inner shell will be 2^(0.25) times higher than the outer shell?
#296 Ollie Hughes 2013-09-15 23:36
Quoting Tim Folkerts:
So now you are arbitrarily changing the emissivity of the inner shell to try to get the numbers to work?

Just to be clear -- do you agree that the inner shell will be 2^(0.25) times higher than the outer shell?

I sad only that the temperature of the spnere
with 1/2 the emissivity and no shell as the temperature of the sphere and its whole emissivity "and the shell in place.

The temperatures of both the sphere and shell will change with the increase in area of the shell, both of them.
The temperature of the sphere will go to your value only if the shell and sphere sizes are the same. This is impossible. Yet all of the power P1 goes through the shell in all cases, there is no power to return to the sphere ever. There is no back radiation. I gave you the means for calculating the temperatures of both the sphere and shell with the shell at any larger size. All classical thermodynamics with no photons and no "back radiation", only the "correct" solution to the whole S-B equation.
#297 Curt 2013-09-15 23:39
Ollie - I used a quadratic relationship for tungsten resistivity vs temperature from the CRC Handbook of Chemistry and Physics. I measured the filament resistance at room temperature (23C) at 26.2 ohms. Tungsten resistivity at 23C is calculated as 123.66 nano-ohm meters.

Applying voltage to the bulb through a variac (variable transformer), I took measurements at 30, 60, 90, and 120VAC settings, using professionally calibrated "true RMS" voltmeters and ammeters, both on an open bulb and the same bulb wrapped in aluminum foil.

The ratio of measured voltage to current in each case gave me the filament resistance. I could see this increase with voltage in both cases, and the resistance at 120VAC was almost 15 times the room temperature resistance.

At every voltage setting, the resistance with foil on was significantly higher than without the foil - well past any uncertainty in the measurements. From the calculated resistance, I used the ratio of measure filament resistance to tungsten resistivity at room temperature to calculate the resistivity of the metal in that case. Then I solved the quadratic equation for absolute temperature.

A few of the measurements:

Bare bulb:
90.0Vrms 0.2878Arms 312.71ohms 1475.96nohm-m 3019.8K
120.0Vrms 0.3318Arms 361.66ohms 1706.98nohm-m 3332.4K

Foil-covered bulb:
90.1Vrms 0.2862Arms 314.81ohms 1485.86nohm-m 3033.6K
119.9Vrms 0.3302Arms 363.11ohms
1713.82nohm-m 3341.5K

The thermal capacitance of the bulb's glass surface is thousands of times higher than the thermal capacitance of the thermocouple junction, so I am not unduly worried about what temperature I am really measuring.

Aluminum foil is about the last thing you would want to use to suppress conductive/convective heat transfer, but it's probably the first thing you would want to use to suppress radiative heat transfer. While it probably had a bit of the former effect, other tests I had done controlling for conductive/convective losses showed that "back radiation" from a foil shell not in contact with the bulb could lead to a bulb surface temperature 70K higher than with a transparent shell.

wattsupwiththat.com/2013/05/28/slaying-the-slayers-with-watts-part-2/
#298 Ollie Hughes 2013-09-15 23:53
Quoting Joel Shore:
Well, what any physics textbook would tell you is that the interpretation of the second term in the S-B equation describing the net radiative transfer [Power = sigma*A*(T_hot^4 - T_cold^4)] is due to radiation from the colder to the hotter object (and the first term is the larger amount of radiation from the hotter to the colder object).

However, if you want to believe, against all evidence, that this is not correct, I guess it really doesn't matter: You are not disputing the greenhouse effect, which arises from the mathematics and not from the physical interpretation of the terms.

I do dispute your alleged greenhouse effect as it depends on your fake concept of "back radiation". you insist on changing the concept of GHE from back radiation to nothing.
You do not know if the transmissivity of the atmosphere has changed with the higher leavel of CO2 You have no way of knowing how much energy is absorbed or radiated by the earth's surface. Such has never been measured.
You simply have nothing to go on.
#299 Ollie Hughes 2013-09-16 00:17
Using P/A = εσ(T_hot)^4 - εσ(T_cold)^4 ],
I have no Idea where you came up with such an equation T^he S-B equation reads
P/A(max) = εσ(T_hot^4 - T_cold^4 )
where ε is the product of the emissivities of each surface. The Two radiative potentials inside the pasens() must be evaluated first thus resulting in the maximum one way power transfer between the two surfaces. And yes
at zero potential difference no power at all the is transferred in either direction quite independent of the radiance of the surfaces.
Why such sloppy science?
#300 David Cosserat 2013-09-16 02:55
Ollie,

1. The two equations you quote are algebraically identical, whatever physical significance you do (or do not) ascribe to each of its two terms.

2. Where did you get the idea that your epsilon is "the product of the emissivity of each surface". I have never heard anybody else saying that except you. I thought it was meant to be the emissivity of the warmer surface on the grounds that it is used (obviously) in the first term as the emissivity of the hotter surface and in the second term as the absorptivity (same value as emissivity for an opaque body) of the hotter surface, the presumption being (whether justified or not) that the emissivity of the cooler body is 1.

By the way I am still waiting for a reply to my straightforward challenge to you at principia-scientific.org/latest-news/301-robert-brown-backradiation-and-mathematics.html#comment-1687.

But I am not holding my breath.
#301 Greg House 2013-09-16 04:24
Quoting Curt:
David - I did I set of experiments earlier this year that should address your concerns. I described them at WUWT here: wattsupwiththat.com/2013/05/28/slaying-the-slayers-with-watts-part-2/

Your experiment seems very much to be a fake, because you apparently faked one important detail.

I commented on it on that WUWT thread, but my comment was deleted by Watts or a moderator, so I published it on the Joe Postma blog: climateofsophistry.com/2013/05/13/slayers-putting-up-not-shutting-up/#comment-2235
#302 David Cosserat 2013-09-16 04:44
Curt,

Yes I read your article very carefully. And also Postma's rebuttal. I would very much like to correspond with you in detail over this. It can't be done in 'sound bites' on a blog trail. So please do contact me at cosserat@gmail .com and join me in what I hope will become a growing group dedicated purely to the experimental approach. Between us (even where we differ) we will crack this barmy intellectual logjam.
#303 Kelvin Vaughan 2013-09-16 05:05
Sorry for being thick but,

If I plunge my hand into liquid nitrogen it will freeze instantly.

Now the radiated heat from my hand is relative to its temperature. As it cools it looses less and less heat, but it radiates at the same initial rate as before I plunged it in. The rate of radiation from my hand didn't suddenly increase.

So why did it freeze almost instantly?
Quoting Ollie Hughes:
Ollie, if you're saying the actual units of energy (electro magnetic engergy packets) don't actually move both ways between hotter and colder entities then you are wrong.

Take a deep breath and read the whole thread before you respond.

I have but that is quite a task. The warmests
If the energy goes both ways with different temperatures it must also go both ways with no temperature difference.

Yes. In fact that only time any entity does not emit electro magnetic energy (EME) is when its temperature is at absolute zero.

Quoting Ollie Hughes:

This is both physical and conceptual nonsense.

Speaking of physical and conceptual nonsense, can you explain how (in the context of your model/understanding) the cooler entity knows that it is the cooler entity and, apparently, turns itself off (stops emitting EME)?
Quoting Ollie Hughes:

Thermal radiation goes one way only.

Really? How did you determine this? Are you going to write up your findings or do we all have to just take your word on this amazing discovery?
Quoting Ollie Hughes:

The warmests here insist that the slayers demand an infinite temperature increase with back radiation this is true only if there "is" back radiation.

I agree that their (Warmies') absurd interpretation is only possible if back radiation (as they define it) does exists. But the fact that they have misinterpreted its significance doesn't mean that backradiation (as they define it) doesn't exist.
Quoting Ollie Hughes:

there is no back radiation. There is only the correct radiation from hot to cold as per the whole S-B equation with two not one thermal radiative potentials. Now the professors chime in with there "net", which is the "only"
physical transfer ever observed. The rest is a professors dream only.

Right, the net is the only thing measured/measurable, but that doesn't mean that the flow of energy does not travel in all directions (including from colder to warmer) at all times.

Right? If not then I think it is incumbent on you to offer some kind of explanation as to how colder objects are able to sense they are in the vicinity of warmer objects and are able to turn off their outflow of EME?
#305 Curt 2013-09-16 12:39
Greg - You'd get a lot further if you'd learn to phrase things like "I don't see how this could possibly be true" rather than flinging libelous (yes, libelous) accusations about whenever you see you don't understand.

You'd also get a lot further if you really tried to learn anything about how the real world works. If you had ever worked with real heat transfer systems, you would no that putting a vertical plane many inches to the side of a heat generating body has no detectable effect on the convective cooling of the body. The air heated by the body rises up (and this is not blocked at all) creating a lower pressure near the body. With the other dimensions still wide open, there is no limitation on providing air to keep the cooling stream wide open. Repeat the experiment yourself - even if you constrain things a lot more, you cannot detect any air speed of replacement air coming toward the bulb. And until this speed would become significant, there is no meaningful inhibition of radiative cooling. If you had ever designed a cooling system, you would know this.

And in the case of the glass shell, you persist in thinking that standard glass is not transparent to the near infrared that the bulb outputs. As has been pointed out to you numerous times, standard glass is overwhelmingly transparent to wavelengths as long as 2500 nm, which covers the vast majority of both solar and incandescent light output.
#306 David Cosserat 2013-09-16 12:42

Well said. Your very pointed exposure of Ollie's obfuscations is spot on.

I don't know why you bother, though. He is just a re-incarnation of Ewiljan who was banned for being rude and obstructive. Now he comes back a reformed character in that respect but is still just as incoherent.
#307 David Cosserat 2013-09-16 12:53
Ollie,

What is it with you that you always get the wrong end of the stick?

In the Steel Greenhouse the shell has two surfaces each of area A whilst the core has one surface of area A. So it is able to radiate twice the energy, half to space and half back to the core. Slayers don't argue with the math which is impeccable if you are using Warmist physics (which says back radiation from the shell adds energy to the core, so the core is radiating twice as much towards the shell than it would do if the shell were removed.

So you have used the wrong argument and thereby shot yourself in the foot.
#308 Tim Folkerts 2013-09-16 15:29
Quote:
If I plunge my hand into liquid nitrogen it will freeze instantly.
I have plunged my hand into liquid nitrogen! It did not freeze in the least!

Granted, I only did it for a second because I am not stupid, but for brief times, you can indeed do this. The vaporized N2 forms a thin insulating gas layer around your hand which greatly inhibits conductive loss. The loss from radiation is small enough that your hand will only feel a mild cool sensation.

*******************************

Back to the main point... why do you say "the radiated heat from my hand is relative to its temperature" but then say "as it cools ... it radiates at the same initial rate". As it cools, it must necessarily radiate less.

The rapid cooling occurs shortly after plunging an object into the N2. Once the surface cools, then conduction can rapidly cool the object.
#309 Tim Folkerts 2013-09-16 15:55
David, both emissivities matter. The relationship is not simple if neither surface is a blackbody. You can find a quick intro here: en.wikipedia.org/wiki/Radiative_heat_transfer#Radiative_heat_transfer

Basically, unless you want to go into GREAT detail, it is much simpler to start with the approximation that one surface (eg the earth's surface) is indeed a blackbody.
#310 Ollie Hughes 2013-09-16 20:40
Quoting David Cosserat:
Ollie,

What is it with you that you always get the wrong end of the stick?

David: What is it with you that you always get the wrong end of the stick?
You are the one that wants folk to agree!
IMHV back radiation does not exist! The interior of the shell radiates nothing Such is an invention of the idiot Roy Spenser.
Do the whole STEEL SHELL thing with no "back radiation".,It is easy, you get perfict results with all power accountedfor , no extra power "created" and all temperatures exact!

In the Steel Greenhouse the shell has two surfaces each of area A whilst the core has one surface of area A. So it is able to radiate twice the energy, half to space and half back to the core. Slayers don't argue with the math which is impeccable if you are using Warmist physics (which says back radiation from the shell adds energy to the core, so the core is radiating twice as much towards the shell than it would do if the shell were removed.

Not at all, the core (sphere) is radiating
all of the power that exists in the system.
(P1). That is all it has to radiate. You claim that it receives an equal amount from the shell thus must radiate twice then amount. this is false. The new higher temperature could radiate that much if it were still radiating to Tcold. It is not.
It is radiating to Tshell exactly P1 to the shell at Tshell, and the shell radiates exactly P1 at temperature Tshell to the isotherm at Tcold. If this were not true, by the laws of thermodynamics, the temperatures would spontaniously adjust to reach such an equilibrium. All is consistant with the proper use of the S-B equation, which only describes the max flux between two different radiative potentials.
No Physics text prior to 1973 claimed anything else. however none of these texts "deny" (back radiation) as is correctly claimed by all three professor, because none of the authors,then, had no idea that anyone would be so stupid as to claim such.
After 1973 was created by an idiot professor the idea that everything thermally radiates independently of any possible absorber at a lower temperature of such thermal radiation.
Come along in 1985 and we have Climate Clowns
being very cleverly, "that stupid" to promote their scam.
David, you have used the wrong argument and thereby shot yourself in the foot.
You also have no real understanding of how greatly and professionally you were scammed,
You are indeed showing off your colors as a luke-warmer!
#311 Ollie Hughes 2013-09-16 21:14
Quoting David Cosserat:
Ollie,

1. The two equations you quote are algebraically identical, whatever physical significance you do (or do not) ascribe to each of its two terms.
Mathematically they are very different.
The real S-B equations had the difference of radiative potential in parentheses (). this is precise and means that the difference in potential "must" be done first with the sign of the result producing the the direction of the only measurable radiant flux. More importantly it indicates that neither potential can be zero as thr radiative potential of zero Kelvin is a mathematicaly 'not a number".

2. Where did you get the idea that your epsilon is "the product of the emissivity of each surface". I have never heard anybody else saying that except you. I thought it was meant to be the emissivity of the warmer surface on the grounds that it is used (obviously) in the first term as the emissivity of the hotter surface and in the second term as the absorptivity (same value as emissivity for an opaque body) of the hotter surface, the presumption being (whether justified or not) that the emissivity of the cooler body is 1.
The wrong equation has the emissivities subtracting. Both emissivities are a fraction that proportionate the emission and absorbtion independently, thus the must be multiplicative. This does not come from textbooks, it comes from making measurements.
that point out "your" mistale.

By the way I am still waiting for a reply to my straightforward challenge to you at principia-scientific.org/latest-news/301-robert-brown-backradiation-and-mathematics.html#comment-1687.

But I am not holding my breath.

I have been checking your posts on Tallbloke.
I am not impressed!
#312 Tim Folkerts 2013-09-16 22:06
Ollie, just for completeness, what would you predict for the "steel greenhouse"?

Let's make it concrete:
* all surfaces are blackbodies
* the shell has negligible mass and in a small distance above the inner sphere.
* the inner sphere is heated with a 240 W/m^2 heater.

I claim the shell will radiate 240 W/m^2 to space and be 255 K
Furthermore, I claim the inner sphere will be 303 K.

Do you agree or disagree?
#313 Ollie Hughes 2013-09-16 22:22
Quoting David Cosserat:

Well said. Your very pointed exposure of Ollie's obfuscations is spot on.

I don't know why you bother, though. He is just a re-incarnation of Ewiljan who was banned for being rude and obstructive. Now he comes back a reformed character in that respect but is still just as incoherent.

Indeed I ollie just wish to sit back with a beer and watch. ewiljan sees the same but somehow thinks he can fix.

Good Bye Will J.
#314 Ollie Hughes 2013-09-16 23:22
Quoting Tim Folkerts:
Ollie, just for completeness, what would you predict for the "steel greenhouse"?

Let's make it concrete:
* all surfaces are blackbodies
* the shell has negligible mass and in a small distance above the inner sphere.
* the inner sphere is heated with a 240 W/m^2 heater.

I claim the shell will radiate 240 W/m^2 to space and be 255 K
Furthermore, I claim the inner sphere will be 303 K.

Do you agree or disagree?

Your claims are correct for a non-physical thought problem. Note all power from the shell is tradiated to a cold sink there is no power available to "back radiate".

Let me destroy your "thought problem" in
physical terms, because you are supposed to
be instructing in physics not thought!

* all surfaces are blackbodies
No surface is a blackbody. A blackbody is a conceptual term only. Never a physical instantiation of such.

* the shell has negligible mass and in a small distance above the inner sphere.

I have no idea of what you mean by "small"
At a seperation of more thn two wavelengths at the lowest frequency of radiant heat transfer. which is at 300 Kelvin, about 6 thousants of an inch. in a vacuum all heat transfer would be radiative
* the inner sphere is heated with a 240 W/m^2 heater.
Ok whatever!! wahat real power incase your sphere collapsees. increasing the temperature.

I claim the shell will radiate 240 W/m^2 to space and be 255 K
Furthermore, I claim the inner sphere will be 303 K.
Your temperatures are close if space is less than 10 Kelvin. If space is 77 Kelvin both your numbers are Bull Shit.
All thermal radiative flux depends on the temperature and absorbtivity of the absorber. Yout textbooks are nonsense. Your notion that thermal radiation must transfer to zero Kelvin is also Bull Shit.
Try to learn how to do radiative heat transfer corectly!!
#315 Ollie Hughes 2013-09-16 23:51
Quoting Joel Shore:
Well, what any physics textbook would tell you is that the interpretation of the second term in the S-B equation describing the net radiative transfer [Power = sigma*A*(T_hot^4 - T_cold^4)] is due to radiation from the colder to the hotter object (and the first term is the larger amount of radiation from the hotter to the colder object).

However, if you want to believe, against all evidence, that this is not correct, I guess it really doesn't matter: You are not disputing the greenhouse effect, which arises from the mathematics and not from the physical interpretation of the terms.

No radiative flux from cold to warm has ever
been observed. Your claimed theoretical nonsense is only theoretical nonsense.
#316 Ollie Hughes 2013-09-16 23:58
Quoting Ollie Hughes:
Quoting Joel Shore:

I see nothing in it that prohibits radiation from going from cooler to warmer objects."

Except the second law of thermodynamics, . . .

Perhap you have misinterpreted it,
Clausius statement has never ben falsified.
All from stastical mechanics and the Kinetic
theory of heat. has never even been accepted.
Quoting Joel Shore:
If the claim is about net, then nobody disagrees. All models of the greenhouse effect, whether they be "toy" models like the steel greenhouse or full-fledged climate models have the net radiative transfer being from the warmer Earth to the colder atmosphere.

All that arguing, over one word: "net".
#318 Ollie Hughes 2013-09-17 01:08
Quoting Joel Shore:
If the claim is about net, then nobody disagrees. All models of the greenhouse effect, whether they be "toy" models like the steel greenhouse or full-fledged climate models have the net radiative transfer being from the warmer Earth to the colder atmosphere.

All that arguing, over one word: "net".

Indeed, a culture difference that must be defended. You can call it "net", I call it "only". If you think that I also must call it "net", You die a horrable death.
#319 Ollie Hughes 2013-09-17 02:27
Right? If not then I think it is incumbent on you to offer some kind of explanation as to how colder objects are able to sense they are in the vicinity of warmer objects and are able to turn off their outflow of EME?

It is your mistaken claim of EME from
only one temperature. Such has never been observes. I need no explanation of what others claim. Go observe what physically is.
If you are confused, I may be able to offer a different way of looking at that, that might make sense! What is Is nomatter what the professors say!
#320 David Cosserat 2013-09-17 04:08
Ollie,

It is a pleasure to read a rare response from you that does not contain typos and misplaced words, etc., that sometimes make it very difficult to understand the points you are making. Thanks for that.

I said: "Slayers don't argue with the math which is impeccable if you are using Warmist physics."

The point I was making is that you are wasting your time arguing with Warmist math that is based on Warmist physics, because the math that Warmists use is utterly consistent with Warmist physics.

That doesn't mean the Warmist physics is necessarily correct. I was simply criticising you for wasting time arguing with their math rather than with their physics, as you were attempting to do in the case of the Steel Greenhouse math. But I would defend to the last your right to argue with their physics.

The same of course goes for Warmists. It is pointless them criticising Slayer math if it is based on Slayer physics. The trouble in this second case is that, when challenged, Slayers are not too hot at providing the math that supports their physics...

When I previously asked you at at principia-scientific.org/lates t-news/301-robert-brown-backra diation-and-mathematics.html#c omment-1687 to come up with the math (as it would apply to my experiment) to support your version of the physics, I was hoping to receive from you math that I could use to predict the experimental outcome, in the event that your version of the physics was correct. You did not reply. That is prevarication.

If you want your Slayer (or, if you prefer, non-Warmist) physics to be taken seriously, you need to respond to my challenge to come up with the precise math as requested in my link above. It seems that you can't or won't do that. And that leaves you in possession of a proto-scientific theory.

More's the pity! I (and I hope every other decent person on this planet) would love you to be right. If my simple experiment can show that your physics is correct and that the Warmist physics is wrong, it really would Slay the Dragon and consign GW alarmism to the dustbin of history, which is where I personally believe (but cannot yet prove) it belongs.

So put up the math for me, Ollie!!
#321 David Cosserat 2013-09-17 04:56
Ollie,

Your response to Tim Folkerts' very clear question is pure and simple sophistry.

You accept that the Warmist math is the same as your non-Warmist physics would also predict in the case of a theoretical thought experiment and then seek to cast that astonishing admission aside on the grounds that the thought experiment cannot be realised physically.

All right. So tell us how does the situation change when we modify Tim's request to make it physically possible and eminently testable in a laboratory.

Given the following independent variables...

The power input to the core is set at P
The surfaces all have emissivity e
The shell has mass M.
The shell is made of material having thermal conductivity k.
The core has surface area Ac
The shell's inner surface has area As1
The shell's outer surface has area As2
The environment is a heat sink at constant temperature Tenv

Now, assuming your physics (not Warmist physics), give us the algebra that predicts what the following three dependent variables will be once the system reaches steady state:

The core surface temperature Tc
The shell's inner surface temperature Ts1
The shell's outer surface temperature Ts2.

Give us your non-Warmist math, Ollie!!
#322 David Cosserat 2013-09-17 05:08
More sophistry and prevarication.

Where's the math, Ollie!!
#323 Tim Folkerts 2013-09-17 08:24
Quote:
Your claims are correct for a non-physical thought problem.
That is a start. Most "slayers" can't seem to agree even on the results for this thought problem.

Quote:
Note all power from the shell is tradiated to a cold sink there is no power available to "back radiate".
Again, this is semantics. The radiation from the sphere to the shell is
P/A(max) = εσ(T_sphere^4 - T_shell^4 )
This is what matters. Whether one chooses to think of this as net radiation or one chooses to think of this as photons flying both directions, the results are the same.

Quote:
I have no idea of what you mean by "small"
"Small" is understood in such situations as "much smaller than other relevant sizes". In this case, that (r_shell - r_sphere)
#324 Tim Folkerts 2013-09-17 08:40
1) space is NOT 77 K.
2) even if it were, the correction is small (about 0.5 K for the outer shell and less of the inner sphere).

Either get into the spirit of thought experiments and reasonable approximations, or give us the exact answer as outlined in David's comment so we can 'learn how to do radiative transfer correctly'.
Quoting Ollie Hughes:
Right? If not then I think it is incumbent on you to offer some kind of explanation as to how colder objects are able to sense they are in the vicinity of warmer objects and are able to turn off their outflow of EME?

It is your mistaken claim of EME from
only one temperature. Such has never been observes. I need no explanation of what others claim. Go observe what physically is.
If you are confused, I may be able to offer a different way of looking at that, that might make sense! What is Is nomatter what the professors say!

Believers believe. And when their beliefs are challenged they become defensive, stubborn, and evasive. Scientists--real scientists--are more concerned about their own thinking being wrong than they are about the appearance that they are competent/knowledgeable. So, real scientists always make their assumptions explicit, accessible.

They have an agenda. And they thrive on confusion and ambiguity. By being stubborn you are only feeding the paranoia of AGW proponents that their opponents are confused, evasive and incompetent.
#326 David Cosserat 2013-09-17 12:45

You have expressed in a nutshell exactly what's wrong with this thread and with the Slayer's really odd approach to debate generally.

It is a pity because I still think their theory is worth investigating deeply. But how can one do so when there are no Slayers to talk to who don't either run away or get abusive as soon as they are challenged?

Meanwhile I shall continue with my experimental approach. If you wish to help, please do contact me at .

This invitation extends also to anybody else, Warmisy or otherwise, who favours the empirical approach to resolving a scientific dispute rather than just expressing irritation and anger.
#327 Kelvin Vaughan 2013-09-17 14:52
Thanks for the reply Tim. Very interesting. I didn't know you could do that.

When I asked the question my brain was stuck in radiation mode. It wasn't until later I thought of conduction. My poor old brain is a bit slow now.
#328 Ollie Hughes 2013-09-17 18:35
Quoting David Cosserat:
Ollie,

Your response to Tim Folkerts' very clear question is pure and simple sophistry.

That is correct That is all a "thought problem" ever is.

You accept that the Warmist math is the same as your non-Warmist physics would also predict in the case of a theoretical thought experiment and then seek to cast that astonishing admission aside on the grounds that the thought experiment cannot be realised physically.

The warmisat math and physics with the mistaken concept from statustical mechanics that temperature must radiate blah in all directions I specifically deny!!

All right. So tell us how does the situation change when we modify Tim's request to make it physically possible and eminently testable in a laboratory.

Given the following independent variables...

The power input to the core is set at P
The surfaces all have emissivity e
The shell has mass M.
The shell is made of material having thermal conductivity k.
The core has surface area Ac
The shell's inner surface has area As1
The shell's outer surface has area As2
The environment is a heat sink at constant temperature Tenv.

Now, assuming your physics (not Warmist physics), give us the algebra that predicts what the following three dependent variables will be once the system reaches steady state:

The core surface temperature Tc
The shell's inner surface temperature Ts1
The shell's outer surface temperature Ts2.

Give us your non-Warmist math, Ollie!!
Ok but I refuse to do that in symbolic terms. requirement:
All values must be realizable in a laboratory somewhere. You can use symbols,
but each symbol that is a given "must" be given a value. Basic units like a spheres radius rather than surface area is best. ( you may give both) Terms like your thermal conductivity are vague because no uniform thickness is given. Just state the temperature difference between inner and outer for power P in degrees Celcius.
If the spheres are concentric, say so. The mass of the shell is noy important at thermodynic equilibrium. Is the size and shape of whatever (Tenv) large enough and complete so as to have no "view factor" effects on the spheres. If all thermal effects except radiation identically zero so state. Emissivity can be anywere between 5% and 95% and demonstratable at all anghes and wavelengths.
Please give a value for negligible or "close enough for government work" in percentage of
the calculated or measured value.

I think that will be enough. I will do the math in a step by step process asigning the result to the variable in the process.
#329 Ollie Hughes 2013-09-17 18:42
Quoting Tim Folkerts:
1) space is NOT 77 K.
2) even if it were, the correction is small (about 0.5 K for the outer shell and less of the inner sphere).

Either get into the spirit of thought experiments and reasonable approximations, or give us the exact answer as outlined in David's comment so we can 'learn how to do radiative transfer correctly'.

Thank you! Areal warmist that agrees that the temperature of the cold sink affects the temperature of all powered radiative heat transfers ( actually it is closer to 0.3 degree Celsius.
I will not get into thought problems as they are always ill defoned thus give false results from assumtions. Only a real experiment can give true results and point out the bad reasoning.
See my response to Davids post.
I tale it that you have not done a Thermpdynamic equiibrium with multiple surfaces and one isotherm.l
#330 Tim Folkerts 2013-09-17 21:04
Quote:
A real warmist that agrees that the temperature of the cold sink affects the temperature of all powered radiative heat transfers ...
Of course the temperature of the cold object affects the radiative heat transfer. The equation has been "textbook physics" for decades:
P/A(max) = εσ(T_sphere^4 - T_shell^4 )

Your statement applies very directly to the earth: 'that the temperature of the cold sink (atmosphere) affects the temperature of all powered radiative heat transfers (heat transfers from the surface (powered by the sun) to the cold atmosphere)'. The temperature of the earth is affected by the radiative transfers between earth & atmosphere.
#331 David Cosserat 2013-09-18 16:46
Ollie,

You latest reply is a bit disappointing. I really don't know why you don't want to produce formulae. Anyway, here are my answers to your queries:

1. You say: "All values must be realizable in a laboratory somewhere. You can use symbols, but each symbol that is a given "must" be given a value."

We must eventually derive a general algebraic formula that applies for all (sensible) values of the independent variables. Anyway, if you can't do that choose whatever sensible values you like for them appropriate for a laboratory experiment and work out the corresponding values for Tc and Tc by whatever numerical method you plan.

2. You say: "Basic units like a spheres radius rather than surface area is best." Use them instead if you insist. Let us call them Rs and Rc.

3. I agree that neither mass M nor thermal conductivity k are required. I only put them in to stop you arguing that my model was not realistic because the shell had no thermal mass and no thickness (as you had said before about the thought experiment). Seems I can't win... Actually I would be perfectly happy to assume negligible shell thickness and perfect conductivity through the shell surface, in which case you need only one shell area As(or if you prefer radius Rs). You also then don't need to worry about a temperature drop between the inner and out surfaces so you can have just one shell temperature Ts. As with mass and thermal conductivity, I only defined two different shell surfaces to ward off complaints from you that I was being unrealistic.

4. You say: "If the spheres are concentric, say so." I don't believe it makes any difference to the view factors between them but, yes, that is how this kind of experiment is generally described.

5. You ask: "Is the size and shape of whatever (Tenv) large enough and complete so as to have no "view factor" effects on the spheres."

Sorry, I omitted to define the environment properly. Assume a temperature-controlled vacuum chamber of concave walls and total inner surface area area Av. I called the environment temperature Tenv but I suggest we now rename it Tv for notational consistency.

The view factor Fsv from shell to vac is unity (providing the shell has no concave parts to its surface which, obviously it hasn't). The view factor Fvs from vac to shell is Fsv*As/Av, by the reciprocity rule. (Ah! Now you see why I prefer to work in surface area rather than radius!)

[continued below...]
#332 David Cosserat 2013-09-18 16:47
[...continued from above]

6. You say: "Emissivity can be anywhere between 5% and 95% and demonstratable at all anghes and wavelengths."

Well its up to you to choose the emissivity values, so anywhere between 95% and 5% is fine. I would suggest 95% since it is quite easy to achieve this in practice.

7. You say: "If all thermal effects except radiation identically zero so state."

Assume no conduction/convection between core and shell or between shell and vacuum chamber wall.

8. You say: "Please give a value for negligible or "close enough for government work" in percentage of the calculated or measured value."

I think plus or minus 1degC accuracy for the output variables Tc and Ts would be more than accurate enough to distinguish clearly between non-Warmist and Warmist computations which are likely to be tens of degrees apart.
#333 Ollie Hughes 2013-09-18 19:10
Not interested in helping a warmist promote fake science
#334 David Cosserat 2013-09-19 03:41
Ollie,

I knew you would cop out in the end. The only fake is you. I am a hard line skeptic. Goodbye.