# Light emission from a hole in a cavity

1. Apr 16, 2013

### opaka

1. The problem statement, all variables and given/known data
Sethna 7.7

Assume that the hole of area A is is on the upper part of the cavity, perpendicular to the z axis. The vertical component of the velocity of each photon is therefore vz= c cos(θ), where θ is the angle between the photon velocity and the vertical. The photon distribution just inside the boundary of the hole is depleted of photons with v < 0 (very few photons come into the hole from the outside), but it is almost unaffected by the presence of the hole for photons with v > 0. The Planck distribution is isotropic, so the probability that a photon will be moving at an angle θ (and therefore velocity v) is the perimeter of the θ circle on the sphere divided by the area of the sphere, 2∏ sin θ dθ/ 4∏=1/2 sin θ dθ. Show that the probability density for a particular photon to have velocity v is independent of v in the range (-c,c), and thus is 1/2 c.

b) An upper bound on the energy emitted from a hold of area A is given by the energy in the box as a whole times the fraction A c dt/v of the volume within c dt of the hole. Show that the actual energy emitted is 1/4 of this upper bound. Hint: You will need to integrate $\int p(v) v dv$ from 0 to c.

2. Relevant equations

probablity density p(v)= annular shell/shell volume - which I thought would be 1/2 sin θ dθ

3. The attempt at a solution

Not much - I couldn't figure out part a at all, so I just plugged the answer of 1/2 c into the integral of part b, and changing the limits of integration to radians (0 to ∏ /2) I'm still confused - 1/2 c^3 $\int$ cos θ *sin θ dθ = 1/4 c^3 (sin θ)^2 = 1/4 (c^3).

Any ideas or tips?

Opaka

Last edited by a moderator: Apr 16, 2013