# Gray body vs Black body, a few questions

1. Oct 6, 2011

### fluidistic

I have 2 questions.
1)Can a gray body irradiate more than a black body for a given wavelength if both bodies are at the same temperature?
2)If the answer to the previous question is yes, do you know any material that irradiate enough in the visible spectrum that we can see it with our eyes for a room temperature (about 300 K)?

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M thoughts:
1)I think the answer is yes. It would explain the graph of http://en.wikipedia.org/wiki/File:EffectiveTemperature_300dpi_e.png, i.e. that the Sun irradiates more than a black body in most part of the visible spectra, even though both the Sun and the black body have the same temperature.
2)I have no idea but I find this interesting. I'd love to see some material emitting enough in the visible spectra for us to see it and touch it (low temperature).

2. Oct 6, 2011

### Redbelly98

Staff Emeritus
Hi there, hope you are well!

1) No. A blackbody emits the maximum amount of radiation at any wavelength.

3. Oct 6, 2011

### fluidistic

Hmm I see so this implies that the Sun's surface temperature is noticeably greater than 5777 K. Do you know its real surface temperature?
By the way, how do you know/deduce this?
P.S.:I am well, thank you. Hope you're well too :)

4. Oct 6, 2011

### Ken G

The Sun doesn't have a "real" surface temperature, because it has a range of temperatures, both over the surface, and with depth below the surface. So all you can do is find a kind of effective temperature, based either on its spectrum or its luminosity (the latter is what it looks like they've done in your plot). When you do that, you get something in the neighborhood of your 5777 K number, but since that's not the only temperature there, there's no guarantee you won't find wavelengths that are brighter than a 5777 K blackbody. The problem is that there is a lot of line absorption in the UV, which causes 'line blanketing" and makes the actual local temperature a bit higher than the effective temperature you need to get the luminosity right. So regions of the spectrum that are not line blanketed reflect a slightly higher temperature than the effective average of 5777 K.

Last edited: Oct 6, 2011
5. Oct 6, 2011

### fluidistic

Ok thanks I see. So the fact that I see wavelengths that are brigther than a black body at 5777 K does imply that there are regions over the Sun's surface that are hotter than 5777 K.
Thus the answer of Redbelly98 is right. But is there a way to mathematically deduce that a black body would radiate more than a gray body for any wavelength?

6. Oct 7, 2011

### xts

For every body its emission and absorption coefficient must be equal - otherwise 3rd law of thermodynamics would be violated.

Ideal black body is such, that absorbs all (its absorption coeff=1), so its emission coeff is also 1.
Gray body reflects some light, so its absorption coeff is smaller than 1, so its emission coeff is also smaller than 1.

7. Oct 7, 2011

### fluidistic

Okay so basically what you mean is related to Stefan-Boltzmann's law which states that the power radiated (which I think I can assignate to the irradiance) will be greater for a black body than for a gray body. However this isn't so simple when I talk about single wavelengths.
I know that $P=A\varepsilon \sigma T^4$ where $\varepsilon$ is worth 1 for a black body and less than 1 for any gray body. But this does not imply that for any wavelength a gray body will emit $1/ \varepsilon$ times less than a black body. The law means that the area under the curve of Planck's law for the black and for the gray body will differ by a factor $\varepsilon$ of the gray body. Not that the function "Planck's law" is greater by a factor $1/\varepsilon$ for the black body vs the gray body, for all wavelengths. (Please correct me if I said something wrong)

8. Oct 7, 2011

### xts

It implies! ε may be a function of wavelength, but for every wavelength absorption and amission coeffs must be equal.

Try the following gedankenexperiment:
Container made of ideally reflecting mirrors, divided by halves with a dichroic filter (transparent for very narrow range of wavelengths, reflective for others). Black body in one part, gray body in other.
Now the thermal equilibrium depends not on Stefan's integrated radiation, but on a narrow slice of Planck's radiation.

Dichroic filters are physically real, e.g. Fabry–Pérot ones.

Last edited: Oct 7, 2011
9. Oct 7, 2011

### Ken G

Another way to make xts's argument that doesn't require any filters (though having them is a useful empirical device) is to note that in thermodynamic equilibrium, every single process you can name must balance its inverse process. So if the surface of the Sun really only had one temperature associated with it, then you could surround the Sun with a thick shell at that temperature, and you'd have to have thermodynamic equilibrium between the Sun and that shell, as everthing that is in contact is at T. But since every single process would then balance its inverse process, you could choose any wavelength you like, and indeed any point on the surface of the Sun and any ray radiating from that point, and the intensity of the radiation in both directions along that ray would have to take on its thermodynamic (Planck function) value. But if the incident light along that ray is partially reflected by the Sun, then the Sun itself must emit less light along that ray, to compensate for what was reflected.

10. Oct 9, 2011

### fluidistic

Ah I see guys! Well I'm glad to have been corrected.
However I still have a small doubt as the reason why the absorption coefficient and the emissivity must be equal for any wavelength for any body (not only the black body).

11. Oct 10, 2011

### xts

Take again my previous post. Let P be power emitted by black body in some temperature in an infitesimal frequency (or wavelength) range df.
Consider two bodies in the same temeperature: black and your non-black one, separated by idealised dichroic filter transparent for a range df around f.
In order to fulfill 3rd law of thermodynamics the total energy flux through the filter must be equal to 0. As the filter is reflective for all other frequencies (so the flux for them is 0 in any direction), we consider only frequency f.
Let absorption coeff of your body is a, and emission coeff is e.
Flux from black to gray is P.
Flux from gray to black is (1-a)P + eP: (1-a)P is a reflected back part of the light coming from black body, eP - emitted.
So you have (1-a)P+eP=P, thus e=a.
QED

12. Nov 15, 2011

### fluidistic

Hmm I just thought about it, but what about fluorescent materials? Can't they radiate more than a black body in some part of the EM spectrum (especially in the visible spectrum) compared to a black body?

13. Nov 15, 2011

### Matterwave

Fluorescent materials are not in equilibrium. They shine for a while and then die away.

All of these arguments being made have assumed thermodynamic equilibrium (at least local thermodynamic equilibrium).