# Derive energy density proportional to emitted power per unit area

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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##.

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##.