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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.

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  • $\begingroup$ I think it is a $h^3$ in the denominator. $\endgroup$
    – Trimok
    Sep 17, 2013 at 17:02

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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 : $\dfrac{j^*}{E} = \dfrac{I'}{I} = \dfrac{c}{4}\tag{5}$

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  • $\begingroup$ " 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 (θ), and we have to consider that only one hemisphere is concerned " The flat infinitesimal surface is the surface of hemisphere?you are doing surface integral? Why the direction of radiation is not r but (θ)? I wonder why must consider only hemisphere? $\endgroup$
    – Outrageous
    Dec 10, 2013 at 22:56
  • $\begingroup$ @Outrageous : The intinitesimal surface is flat, no doubt, and the word "hemisphere" is maybe not the best choice... The idea, is that the outgoing photons coming from the little hole correspond to half of the directions (because half of the photons, the ingoing photons, have a direction such that they stay inside the black body, and you must not be taken in account these ones). The second problem, is that the outgoing photons have different directions, so you have to take in account the angle between the direction and the normal to the (flat) surface. $\endgroup$
    – Trimok
    Dec 11, 2013 at 10:50
  • $\begingroup$ @Trimok to correct dimensions, why does it have to be $c$ instead of, say $2c$? $\endgroup$
    – M. Zeng
    Feb 4, 2015 at 10:22

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