What's the relationship between the energy density of a black-body and its radiant exitance?

Question

Through a bit calculation we can derive that in a cavity, the energy density $$u(f,T)=\overline{E(f)}\times G(f)=\frac{8\pi h}{c^3}\frac{f^3}{e^{h\nu /kT}-1}$$ If we take the integral over all frequency, we get $$U(T)=\frac{8πh}{c^3}\frac{(kT)^4}{h^3}{\frac{π^4}{15}}=C_{onst}T^4$$ And Stefan-Boltzmann Law claims that for a perfect black-body $$j^*=\sigma T^4$$ where $j^*$ is the radiant exitance, which is defined as the total energy radiated per unit surface area of a black body across all wavelengths per unit time.

And it just so happens that $\frac{\sigma}{C_{onst}}=\frac{c}{4}$, why is that?

P.S. The professor told me to refer to some thermodynamics book, where a more general case is discussed. But we don't have that book in our library and the professor's now out of town xD.

Answer 1

The energy density could be written :

$$U \sim I =\int d^3 \vec k \dfrac{|\vec k|}{e^{\beta \hbar c |\vec k|} - 1} \tag{1}$$ where $\vec k$ is the wave vector. From now on, we will use the notation $k = |\vec k|$.

With $d^3 \vec k = k^2~ \sin \theta ~d\phi ~d\theta ~d k$, $\phi$ varying from $0$ to $2\pi$, and $\theta$ varying from $0$ to $\pi$, we have :

$$U \sim I = \int_0^{2 \pi} d\phi \int_0^\pi \sin \theta ~d \theta \int dk ~k^2\dfrac{k}{e^{\beta \hbar c k} - 1} = 4 \pi\int dk \dfrac{k^3}{e^{\beta \hbar c k} - 1} \tag{2}$$

If we are interested by the power radiated by unit area ($j^*$), we can consider a closed box, with a little hole, and considering that the light of the photons is $c$, so at first glance, we may think that we have $j^* A t = U A (ct)$, so $j^* = cU$, but in fact, here we consider the flux through a flat infinitesimal surface, we have to take care about the angle between the infinitesimal surface and the direction of the radiation ($\theta$), and we have to consider that only one hemisphere is concerned ($\theta$ is varying only between $0$ and $\pi/2$) . so, the correct calculus is (now including the "obvious " $c$ factor for correct dimensionless) :

$$j^* \sim I' = c \int_0^{2\pi} d\phi \int_0^{\pi / 2} d \theta \sin \theta~ \int dk ~k^2 \dfrac{\vec k.\vec n}{e^{\beta \hbar c k} - 1} \tag{3}$$ where $\vec n$ is the normal to the infinitesimal surface,corresponding to the direction $\theta = 0$, so that $\vec k.\vec n = ~k\cos \theta$

So, we have :

$$j^* \sim I' = c \int_0^{2\pi} d\phi \int_0^{\pi / 2} d \theta \sin \theta~ \cos \theta~ \int dk \dfrac{k^3}{e^{\beta \hbar c k} - 1} = c\pi\int dk\dfrac{k^3}{e^{\beta \hbar c k} - 1}\tag{4}$$

So, obviously, we have : $\dfrac{j^*}{E} = \dfrac{I'}{I} = \dfrac{c}{4}\tag{5}$