The following argument is from my textbook, An Introduction to Thermal Physics by Daniel Schroeder. If you are familiar with the derivation of Stefan's Law from the energy density of a photon gas, you can scroll down to the bottom where I state my question.
I am looking for a more simple explanation for the result of the following derivation.
Using Bose-Einstein statistics, we have derived that the energy density of a photon gas is $$\frac{U}{V} = \frac{8\pi^5}{15} \frac{(kT)^4}{(hc)^3}$$ We would like to now understand how photons are emitted from an idealized black body.
Consider a box filled with a photon gas with a hole of area $A$ on one side. We are concerned with computing the amount of energy that escapes per unit time.
The photons that escape the hole during a time interval $dt$ were once pointed at the hole from somewhere within a hemispherical shell centered at the hole. Say the shell has radius $R$. The thickness of the shell is $c$ $dt$.
Below is a diagram of a cross-section of the described situation.
The shaded chunk of the shell has volume $$\text{volume of chunk} = R \, d\theta \times r \sin \theta \, d\phi \times c \, dt$$ where $\phi$ is the azimuthal angle not shown (it ranges from $0$ to $2 \pi$, and can be thought of revolving the semi-circle around its central axis).
Assuming a uniform density of photons in the box, the energy in the chunk is just the energy density above times the volume of this chunk: $$\text{energy in chunk} = \frac{U}{V} c dt R^2 \sin \theta d\theta d\phi$$ However, not all of the energy in this chunk will escape through the hole. Assume the photons in the chunk are have random directions. The probability of any single photon escaping through the hole is equal to the apparent area of the hole (as viewed from the chunk's location) divided by the total area of a sphere of radius $R$ centered at the chunk (i.e. the total area of the places the photon could end up). $$\text{probability of escape} = \frac{A \cos \theta}{4 \pi R^2}$$ Therefore, the energy that escapes from this chunk is $$\text{energy escaping from chunk} = \frac{A \cos \theta}{4 \pi} \frac{U}{V} c dt \sin \theta d\theta d\phi$$ We can find the total energy escaping by integrating over all hemispherical angles $$\text{total energy escaping} = \int_0^{2\pi} d\phi \int_0^{\pi/2} \frac{A \cos \theta}{4 \pi} \frac{U}{V} c \sin \theta d\theta = \frac{A}{4} \frac{U}{V} c dt$$
This final result gives us Stefan's law: $$\text{power per unit area} = \frac{c}{4} \frac{U}{V} = \frac{2 \pi^5}{15} \frac{(kT)^4}{h^3 c^2} = \sigma T^4$$
We went through all of this trouble just to find that the power emitted per unit area is equal to the energy density times the speed of light, divided by four. The factor $c$ makes sense: the photons (which are carrying the energy) are traveling at that speed, so the rate of energy emission should be proportional to $c$ times the energy density. So, the only non-obvious thing the geometric derivation provided is the factor $\frac{1}{4}$.
My question is...
..is there a simpler explanation for this factor of $\frac{1}{4}$? Is there some sort of symmetry argument? Or do you have to deal with integrating over all these angles?
Here is another way to explain the factor of $1\over 4$.
Let the outward unit vector normal to the hole be $\hat n$. Consider the photons in a thin layer inside the hole moving in the direction of the unit vector $\hat u$. If $\hat n\cdot \hat u<0$, none of these photons will escape the hole. If $\hat n\cdot \hat u>0$, a layer of these photons of thickness $\hat n\cdot \hat u\,c\,dt$ will escape in time $dt$. If the energy density of the photons is $\rho$, the flux per area measured at the hole will therefore be $f\, \cal Ec \rho $, where $f$ is the fraction of unit vectors $\hat u$ with $\hat n\cdot \hat u>0$ and $\cal E$ is the expected value of $\hat n\cdot \hat u$ when it is averaged over the unit vectors $\hat u$ with $\hat n\cdot \hat u>0$.
By symmetry, $f={1\over 2}$. To find $\cal E$, take a flat, horizontal, transparent circular disc of radius $1$ and place a transparent hemisphere of radius $1$ on top of it. Shine a laser beam which has flux per area equal to $1$ vertically upwards through the horizontal disc. $\cal E$ will equal the average flux per area as measured at the exit hemisphere. Since the total fluxes in and out of the system must be equal, this quantity must equal the ratio of the area of the disc to the area of the hemisphere, which is ${\pi\over{2\pi}}={1\over 2}$.