# Energy flux versus temperature change

1. Apr 8, 2012

### kmarinas86

If I have two spheres of the same radius, they can still have different temperatures. However:

What if the colder sphere is closer to the ideal of being a blackbody than the hotter sphere?

http://en.wikipedia.org/wiki/Emissivity

In an open system without internal energy generation, both spheres would get colder. However, the colder sphere would lose energy faster than the hotter sphere. In other words, the temperature difference between the two spheres would increase.

However, if enough of the energy of one sphere were to be lost to the other sphere, and vice versa, then it would mean that the hotter sphere would receive more energy than it lost, while the colder sphere would lose more energy than it gained.

What is obvious is a net transfer of energy. What is not obvious is what exactly that would do to the hotter sphere. Would it get hotter? Or would contribute to a phase change in the hotter sphere?

2. Apr 8, 2012

### Staff: Mentor

Are you envisioning that these two spheres are only in thermal contact with each other and that the only means of heat transfer is via radiation?

3. Apr 8, 2012

Yes.

4. Apr 8, 2012

### Staff: Mentor

Then I think that the only effect of one body being a graybody is to slow the approach to thermal equilibrium. Any radiation emitted by the blackbody at one of the reflective frequencies of the graybody will simply be re-absorbed by the black body, slowing the heating of the blackbody. Any radiation emitted by the graybody will be absorbed better by the blackbody.

5. Apr 8, 2012

### Staff: Mentor

Then I would imagine it would go even slower. In the limit of a graybody where there was no absorption/emission at other than a single frequency and that frequency did not match then they would simply remain the original temperatures and not transfer any energy at all.

6. Apr 8, 2012

### kmarinas86

What if the colder sphere has a high emissivity at low-frequencies, the same frequency range where it peaks (thus helping it to emit heat), and a low emissivity at high-frequencies of which is scant inside the colder sphere (so negligible effect on limiting emission), the same high-frequencies which are at the same time being emitted by the hotter sphere (thus the heat of the hotter sphere is reflected away by the colder sphere), while the hotter sphere has a low emissivity in high-frequency range in which it would otherwise peak (thus reducing the emission of its own heat, providing yet another limitation on the transfer of heat from the hotter sphere to the colder sphere), while it has high emissivity at low-frequencies of which are scant inside the hotter sphere (so negligible effect on increasing emissions by the hotter sphere), permitting heat flux of low-frequencies from the colder sphere. Let's say that there exist super-low-frequencies which the colder sphere has a high emissivity for, while the hotter sphere has a low emissivity for them. So energy received by the hotter sphere from the colder sphere as low-frequencies get downgraded as super-low frequencies that get trapped by the hotter sphere. The result of all that is a net energy flux from the colder sphere to the hotter sphere. What does the energy affect when it is transferred like that?

Last edited: Apr 8, 2012
7. Apr 8, 2012

### Staff: Mentor

If the hot sphere has a low emissivity for them then it also has a low absorptivity. It will not absorb the radiation emitted by the colder sphere. This is approximately the situation that I was describing earlier where they simply do not interact thermally.

The result is never a net energy flux from the colder to the hotter sphere.

8. Apr 8, 2012

### kmarinas86

Imagine that we are talking about spherical shells filled with inert gases. A mirror-like low emissitivity shell limits the transfer of radiation in both directions, not just into the sphere. But this value depends on the frequency of the light. Light into a greenhouse experiences higher emissitivity going in, but due to scattering, a frequency reduction, and thus lower emissitivity going out. Thus it is easier to absorb it at first, but once it is in, it is not so easy for it to escape.

Last edited: Apr 8, 2012
9. Apr 8, 2012

### Staff: Mentor

Not only are they proportional, they are equal:

"a corollary of Kirchhoff's law is that for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity"

10. Apr 8, 2012

### kmarinas86

I have corrected this in my previous post. They are not constant with wavelength, even if they are proportional. As energy scatters inside the sphere, n*h increases and f declines, so wavelength increases. This can have negative effects on *subsequent* emission (via lower emissitivity of the sphere at longer wavelengths), while absorption is still as normal for the incoming light (prior to scattering inside).

11. Apr 8, 2012

### Staff: Mentor

I understand that we are talking about emissivity functions which are not constant wrt wavelength. That is what I have been describing for the last several posts. It doesn't matter.

Pick any given wavelength.

If the hot body and the cold body both have high emissivity or both have low emissivity then obviously energy at that wavelength goes from hot to cold.

If the hot body has high emissivity and the cold body has low emissivity then the hot body produces a lot of radiation at that wavelength and the cold body absorbs some small fraction and reflects the rest, with net energy from hot to cold.

If the cold body has high emissivity and the hot body has low emissivity then the cold body emits some modest amount of radiation which is almost entirely reflected by the hot body and the hot body emits a small amount of radiation which is almost entirely absorbed by the cold body, with net energy transfer from hot to cold.

Repeat at every other wavelength.

12. Apr 8, 2012

### kmarinas86

You're not poining out the fact that we have a changing wavelength when the light scatters. You are comparing different scenarios, each with a different wavelength. That's different than one scenario with separate before and after wavelengths.

13. Apr 8, 2012

### Staff: Mentor

Scattering doesn't change the wavelength, except via Doppler shift. I was assuming that the bodies were at rest wrt each other.

14. Apr 8, 2012

### kmarinas86

You don't even need the Doppler effect. Phosphorescent glow under UV light is a case in point.

15. Apr 8, 2012

### Staff: Mentor

That isn't scattering. That is absorption and emission. It always goes from hot to cold as I described above.

16. Apr 8, 2012

### kmarinas86

The point is that "absorption and emission", to not use the word "scattering", mind you, can change the wavelength.

You just assumed "that the bodies were at rest wrt each other" to argue that the wavelength does not change. The wavelength does change. So your assumption is wrong.

So that assumption must be unnecessary for your earlier argument to be valid. This assumption however, IS required for your reasoning to work:

...obvious lack of a plural.

For an energy in-of-itself to go from "hot to cold" implies a wavelength increase. A wavelength increase affects its relationship with the material that it may interact with. Thus, a material may be better or worse at emitting it. This relationship is by no means monotonic:

http://en.wikipedia.org/wiki/Optical_window

Thus, as energy can be downgraded in frequency it may lead to increases or decreases in the energy's ability to be absorbed and re-emitted by the atmosphere. The "optical window" as the picture depicts, represents a region of the EM spectrum in which the Earth's atmosphere has a low emissivity. Those regions largely ignore the atmosphere and pass right through.

This is where you get it completely wrong. A greybody (whether hot or cold) cannot be reduced to simplistic notions as having "high" or "low" emissivity. A simple grasp of the nature of the optical window proves that such a simplistic viewpoint is wrong. These variations in emissivity are VERY significant, and they cannot be described using simplistic formulas because the actual values depend on the atmospheric composition, consisting of a various mixture of chemicals.

A notable amount of radiation is generated due to re-emission of energy at different wavelengths. The amount of visible rays absorbed by the Earth is greater than the amount of visible rays emitted by the Earth because the visible rays get re-emitted at lower frequencies, while UV part of the spectrum is past the Sun's peak color, so it doesn't quite make up for that drop in visible light.

Your analysis (along together with your prior statement that "not only are they proportional, they are equal"), in contrast, projects that the amount of light at a given wavelength absorbed by a body will equal the amount of light at that wavelength emitted by a body. That is simply not right in the case of anything capable of emitting or absorbing a broken visible and/or infrared spectrum. Such are deviations from Maxwell-Boltzmann statistics, and thus they are automatically non-equilibrium in nature.

See above.

Last edited: Apr 8, 2012
17. Apr 9, 2012

### Staff: Mentor

You are interested in the process of energy transfer from one body to the other. Absorption and emission do not change the wavelength during energy transfer. A given wavelength that is emitted by one body is either absorbed or reflected by the other. It is simply not possible for the wavelength emitted by one to be different from the wavelength absorbed by the other in this scenario.

Once the wavelength is absorbed the energy and temperature of the absorbing body increases. It then emits the energy according to its own characteristic emission spectrum, but that is a new emission-absorption transfer of energy. In each emission-absorption transfer the wavelength does not change.

All of my comments above apply.

I never said that it did. I made the big deal of picking a wavelength in the beginning so that I wouldn't have to keep on writing "at that wavelength" throughout the description. And at a given wavelength a greybody can be reduced to having a "high" or "low" emissivity in this scenario, which can change at a different wavelength. Which is what I said.

I certainly never said that, nor does it follow from my analysis. I never discussed energy absorbed and emitted by the same body at all. I only discussed energy emitted by one body and absorbed by the other, which is the only mechanism of energy transfer in this scenario. Wavelength does not change in that process, which is the energy transfer process.

Last edited: Apr 9, 2012
18. Apr 9, 2012

### kmarinas86

http://en.wikipedia.org/wiki/File:Fluorescent_minerals_hg.jpg - Collection of various fluorescent minerals under ultraviolet UV-A, UV-B and UV-C light. Chemicals in the rocks absorb the ultraviolet light and emit visible light of various colors, a process called fluorescence.

The wavelength changes in the above picture "scenario".

It seems like energy transfer goes both ways. We have a chain of energy transfers, one after the other. The wavelength cannot be conserved, and it is not realistic to think that it is. Don't you think?

Last edited: Apr 9, 2012
19. Apr 9, 2012

### Staff: Mentor

Sure, but the "above picture 'scenario'" is not relevant to your scenario. The "above picture 'scenario'" is fluorecense. That is energy which is absorbed by some object, A, at one wavelength and emitted by that same object, A, at a different wavelength.

In your scenario you are interested in energy which is emitted by A and then absorbed by some other object, B. The wavelength of energy emitted by A is always the same as the wavelength of energy absorbed by B, absent any Doppler or relativistic effects. Do you disagree with that?

Last edited: Apr 9, 2012
20. Apr 9, 2012

### kmarinas86

Do you even know what the scenario is?

It is not "my" scenario. The topic is the system to be described, including all absorption, re-emission etc. not mentioned in the opening post. The OP is what, apparently, you think establishes the full context of the scenario. It does not.