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 ?

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 $u=\frac{T_{BB}}{T_{rad}}$.

• 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 $\frac{1}{3}\sigma T^3$
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 $0$...