Light emission from a hole in a cavity

[opaka]

Homework Statement

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 v_z= 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.

Homework Equations

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

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

 

[mark726]

Hello Sethna 7.7,

Thank you for your forum post. I am a scientist and I would be happy to help you with your questions.

For part a, we can use the given equation for the probability density p(v) = 1/2 sin θ dθ to show that the probability density for a particular photon to have velocity v is independent of v in the range (-c,c). We can do this by integrating p(v) from 0 to c.

∫p(v)dv = ∫1/2 sin θ dθ = 1/2 ∫sin θ dθ = 1/2 (-cos θ) from 0 to c = 1/2 (-cos c + cos 0) = 1/2 (-cos c + 1)

Since cos c = -1, the final result is 1/2 (1 + 1) = 1, which means that the probability density for a particular photon to have velocity v is independent of v in the range (-c,c).

For part b, we can use the given equation for the upper bound of energy emitted from a hole of area A, which is given by E = (A c dt/v) (volume within c dt of the hole). We can then use the result from part a to show that the actual energy emitted is 1/4 of this upper bound.

∫p(v)v dv = ∫1/2 sin θ dθ * c cos θ dθ = 1/2 c ∫sin θ cos θ dθ = 1/2 c (-1/2 cos^2 θ) from 0 to ∏/2 = 1/4 c (1 - cos^2 ∏/2) = 1/4 c (1 - 0) = 1/4 c

This means that the actual energy emitted is 1/4 of the upper bound given by E = (A c dt/v) (volume within c dt of the hole).

I hope this helps. Let me know if you have any further questions or if you need any clarification.


Opaka