D.27 Ra­di­a­tion from a hole

To find how much black­body ra­di­a­tion is emit­ted from a small hole in a box, first imag­ine that all pho­tons move in the di­rec­tion nor­mal to the hole with the speed of light $c$. In that case, in a time in­ter­val ${\rm d}{t}$, a cylin­der of pho­tons of vol­ume $Ac{\rm d}{t}$ would leave through the hole, where $A$ is the hole area. To get the elec­tro­mag­netic en­ergy in that cylin­der, sim­ply mul­ti­ply by Planck’s black­body spec­trum $\rho$. That gives the sur­face ra­di­a­tion for­mula ex­cept for an ad­di­tional fac­tor ${\textstyle\frac{1}{4}}$. Half of that fac­tor is due to the fact that on av­er­age only half of the pho­tons will have a ve­loc­ity com­po­nent in the di­rec­tion nor­mal to the hole that is to­wards the hole. The other half will have a ve­loc­ity com­po­nent in that di­rec­tion that is away from the hole. In ad­di­tion, be­cause the pho­tons move in all di­rec­tions, the av­er­age ve­loc­ity com­po­nent of the pho­tons that move to­wards the hole is only half the speed of light.

More rig­or­ously, as­sume that the hole is large com­pared to $c{\rm d}{t}$. The frac­tion of pho­tons with ve­loc­ity di­rec­tions within in a spher­i­cal el­e­ment $\sin\theta{\rm d}\theta{\rm d}\phi$ will be $\sin\theta{\rm d}\theta{\rm d}\phi$$\raisebox{.5pt}{$/$}$​$4\pi$. The amount of these pho­tons that ex­its will be those in a skewed cylin­der of vol­ume $Ac\cos\theta{\rm d}{t}$. To get the en­ergy in­volved mul­ti­ply by $\rho$. So the en­ergy leav­ing in this small range of ve­loc­ity di­rec­tions is

$$\rho Ac {\rm d}t \cos\theta \frac{\sin\theta{\rm d}\theta{\rm d}\phi}{4\pi}$$

In­te­grate over all $\phi$ and $\theta$ up to 90 de­grees to get $\frac14{\rho}Ac{\rm d}{t}$ for the to­tal en­ergy that ex­its.

Note also from the above ex­pres­sion that the amount of en­ergy leav­ing per unit time, unit area, and unit solid an­gle is

$$\frac{\rho c}{4\pi} \cos\theta$$

where $\theta$ is the an­gle from the nor­mal to the hole.