I came across a seemingly simple derivation of Kirchhoff's Law:

Consider two bodies A and B placed in a constant temperature enclosure, and B is a perfect black body. After some time both of them will attain thermal equilibrium and will settle down at the same temperature.
Assume that some energy Q is incident on both of them, and since they must maintain thermal equilibrium, they must emit as much heat as they absorb. One could write for A: $E_a=aQ$, where $a$ is the absorptivity of A. and for B: $E_b=Q$. Some algebra gives $\frac{E_a}{E_b} = a$. However, by definition, $\frac{E_a}{E_b} = e$ so $a=e$.

I have a couple of questions to ask here:

  1. Shouldn't one account for the reflection of radiation by A? After all, Kirchhoff stated this for A made of 'any arbitrary material'. And once this is accounted for, I do not think $a=e$ would hold. Besides, if we take a perfect reflector, $r=1$ and therefore $a=0$. We'll actually end up dividing by zero.
  2. The exact mathematical statement was that the spectral emissive power and absorptivity are in a ratio of a function of only temperature and wavelength. Here, we are considering that equal amounts of radiation irradiates $A$ and $B$. But if the bodies are made of different materials, the rise in temperatures would be different, and so would the final temperature. And therefore we must end up with something like $\frac{E_a}{a} = \frac{E_b}{1}= f(\lambda, T)$ and this is clearly not equal to $Q$, which is supposed to be a constant.

P.S. It would be great if someone could share Kirchhoff's original derivation to this law, I believe it would really help. I couldn't find it anywhere on the internet.

  • $\begingroup$ If the body is in tehrmal equilibrium, it emits as much heat as it absorbs, regradless of how much radiation is reflected. I am however puzzled by the phrase Assume that some energy $Q$ is incident on both of them, since I see now reason why they should remain in tehrmal equilibrium... $\endgroup$
    – Roger V.
    Commented Jan 12, 2022 at 8:40
  • $\begingroup$ In the case $r=1,a=0$, you do not divide by zero as $E_a$ is in the denominator. Also what is the problem of $\epsilon=a$ in that case? $\endgroup$
    – Mauricio
    Commented Jan 12, 2022 at 11:07

1 Answer 1


You're confusing situations here. Reflectivity plays a part if these bodies were not insulated, but a constant-temperature box forces them to be at the same temperature. If $A$ or $B$ were cooler than the constant temperature, they would radiate less energy and absorb more until they heated up; the excess energy would be supplied by the box. So no, we don't have to factor in $r$, which only affects how fast $A$ would reach thermal equilibrium ($r=1$ would take an infinite amount of time).

But if the bodies are made of different materials, the rise in temperatures would be different, and so would the final temperature

Again, this only applies without insulation, but the box takes care of this.

Kirchoff's law was the precursor to Planck's law, as Wikipedia says,

Kirchhoff's great insight was to recognize the universality and uniqueness of the function that describes the black body emissive power. But he did not know the precise form or character of that universal function... The correct form of the law was found by Max Planck in 1900...

So I think more importantly, emissivity = absorbance in equilibrium holds at any wavelength. Maybe the derivation you posted is incomplete.


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