I try to understand the proper meaning of the relation between the energy and the entropy carried by a thermal radiation. I usually find the formula $$ j_S=\frac{4}{3}\frac{j_E}{T} \,\,\,\,\, (1), $$ which I understand as "whenever a blackbody at temperature T emits or absorb an energy $Q$, he generates an amount of entropy $Q/3T$ which is either transmitted to the radiation (during emission) or kept in the body (during absorption)."

Now, a paper of P. Wurfel entitled Generation of entropy by the emission of light (1988) reads

Nowhere along the path of the photons emitted into free space is any entropy generated. [...] We recognize that it is not the emission of photons into a vacuum, but rather the lack of absorption from the vacuum, that generates the entropy.

which is also quite close to what I understand from here.

  1. How could absorbing radiation decrease the generated entropy ? How does absorption affect the entropy $S_c$ created during the emission ?

  2. I thought $j_s$ represented the entropy acquired by a body absorbing the radiation, corresponding to an entropy creation of $j_E /3T$. Isn't that contradictory with the idea of absorption reducing the entropy creation ?

  3. What does eq(1) represent in the end ? Is what I wrote above correct ?


1 Answer 1


For the record, here is -I think- one way to answer this question.

Let's consider a black body at temperature $T_{BB}$, absorbing a radiation at $T_{rad}$ and emitting a radiation at $T_{BB}$. The exchange surface is $S$ and we perform a balance over a duration $dt$.

From an energy perspective, the black body receives an amount $Q_{in}=\sigma T_{rad}^4 Scdt$ of heat and emits an amount $Q_{out}=\sigma T_{BB}^4 Scdt$.

From an entropy perspective, let us consider the balance of the total system $\{ \rm{Blackbody + radiation} \}$: $$ dS_{BB} = \frac{Q_{in}}{T_{BB}} - \frac{Q_{out}}{T_{BB}}\\ dS_{rad} = \frac{4}{3} \sigma T_{BB}^3 cdSdt - \frac{4}{3} \sigma T_{rad}^3cdSdt\\ dS= dS_{rad}+dS_{BB} = \sigma \frac{T_{rad}^4}{T_{BB}} \left(1 +\frac{1}{3}u^4 - \frac{4}{3}u \right)cdSdt $$ where $u=\frac{T_{BB}}{T_{rad}}$.

Few remarks about this expression :

  • Since the system is isolated, $dS$ is the created entropy during the absorption + emission proccess
  • The created entropy is positive or null. It equals $0$ if and only if $u=1$.
  • If the black body does not absorb any radiation (ie $T_{rad} = 0$), the created entropy is $\frac{1}{3}\sigma T^3$

So we recover here classical results, and notably the fact that the amount of created entropy depends on whether the black body absorbs radiation at the same time it emits radiations.

One thing remains unclear to me, though.

When calculating the maximal efficiency for a solar energy conversion device, Lansberg considers that the amount of entropy transmitted to the absorber is not $dS_{BB}^{in}=Q_{in}/T_{BB}$, but $dS_{BB}^{in}=\frac{4}{3}\sigma T_{rad}^3$. In the picture developed here, this makes sense only if $dS=0$, in which case the conversion efficiency is trivially $0$...


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