Derive energy density proportional to emitted power per unit area

[Happiness]

The following derives the relation that for a blackbody radiation the energy density is proportional to the energy emitted per unit area over unit time.

The average energy density $d\psi$ is obtained by dividing the radiant energy $dE$ received by the surface $dB$ in 1 second by the cylindrical volume $dV$ occupied by the radiation in 1 second.

I don't understand why $d\psi$ is uniform anywhere inside the cylinder, in particular, at points near the surface $dS$. At a point far away from $dS$, the point is always the same distance $\rho$ away from every part of $dS$. But this is not true for a point near $dS$. The point would be closer to some parts of $dS$. Then the energy density $d\psi$ at the point should not be uniform but vary depending on how far the point is from the center of the surface $dS$.

 

[gtoguy97]

The energy density $d\psi$ at any point in the cylinder is proportional to the energy emitted per unit time and area from the surface $dS$, i.e., $d\psi \propto E/A \cdot t$.We can derive this relation by considering an element of surface area $dA$ of the surface $dS$ and the energy it emits in one second, $E/A \cdot t$. This energy spreads out in a spherical wave with an intensity $I$, where $I = E/A \cdot t$. The total energy emitted by the surface is equal to the energy of the wave at any distance $r$ from the surface, i.e., $E = 4 \pi r^2 I$.The energy density at any point in the cylinder is then given by the energy of the wave divided by the cylindrical volume, i.e., $$d\psi = \frac{4 \pi r^2 I}{\pi r^2 h} = \frac{4I}{h} $$where h is the height of the cylinder.Therefore, the energy density at any point in the cylinder is proportional to the energy emitted per unit time and area from the surface $dS$, i.e., $d\psi \propto E/A \cdot t$.