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While learning about ways in which heat can be transferred I came to the section on Radiation. I noticed one thing peculiar which wasn't that obvious.

A body at a temperature $T_b$ emits thermal energy via electromagnetic radiation. The rate of heat transfer is given by the following formula:

$$H_{\text {rad}} = \varepsilon_{rad} \sigma A T_b^4$$

A body also absorbs radiation from surrounding which is at a temperature $T_{s}$. The rate of absorption of heat is given by:

$$H_{\text {abs}} = \varepsilon_{abs} \sigma A T_s^4$$

And for the fact $\varepsilon_{\text {abs}} = \varepsilon_{\text {rad}} = \varepsilon$.

-Why is this so? I mean why is it necessary for these two to be one and the same?

One reason that I can think of is that if that wasn't the case then it's would violate the Zeroth law of Thermodynamics. But

Since $\varepsilon_{\text {abs}} $ and $\varepsilon_{\text {rad}} $ are properties of materials. Therefore I think that there should be an explanation if we look at it from this angle.

  • So what it is?
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    $\begingroup$ I suggest you read up on Kirchhoff’s law of radiation for an explanation $\endgroup$
    – Bob D
    Feb 4, 2020 at 15:44
  • $\begingroup$ is there anything more fundamental than the 0th law? $\endgroup$
    – hyportnex
    Feb 4, 2020 at 16:04
  • $\begingroup$ @hyportnex sorry I used the wrong word. Actually I want a theoretical explanation rather than invoking a law. $\endgroup$
    – user249968
    Feb 4, 2020 at 16:06
  • $\begingroup$ If you want a mechanism, there are the Einstein coefficients. $\endgroup$
    – user137289
    Feb 4, 2020 at 16:16
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    $\begingroup$ Time reversal symmetry is sufficient. $\endgroup$
    – Jon Custer
    Feb 4, 2020 at 17:43

2 Answers 2

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Assume a body in equilibrium. You shine light on it, let it heat up, and at some point it will reach a constant temperature. At that point input and output have to be the exact same. This can be argued at every temperature. So whenever the first equation is valid, the second must be too.

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    $\begingroup$ It can also argued at every wavelength too. $\endgroup$
    – ProfRob
    Feb 5, 2020 at 6:42
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I think that I got the answer, but I would like to know of what others think of this. So here it is.

As soon as I saw the comment by Jon Custer, it became crystal clear to me why these two quantities might be one and the same. As he points out in his comment :

Time reversal symmetry is sufficient.

So since the process of emission would look the same like absorption if we reverse the time. Therefore $\varepsilon_{\text {abs}} = \varepsilon_{\text {rad}}$.

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    $\begingroup$ Hm, something to think about: some processes are not symmetric under time. Why is this one? $\endgroup$
    – pwatts
    Feb 4, 2020 at 18:45
  • $\begingroup$ Hm, what does this have to do with Kirchhoff's law of thermal radiation? $\endgroup$
    – Bob D
    Feb 4, 2020 at 21:24

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