Planck formula and density of photons

[EmilyRuck]

Hello!
Let's consider again a system of atoms with only two permitted energy levels E_1 and E_2 > E_1. When electrons decay from E_2 level to E_1, they generate a photon of energy E_{21} = E_2 - E_1 = h \nu. The number of photons (per unit frequency, per unit volume) emitted by such a system in thermal equilibrium at a temperature T can be determined dividing its black body radiation \rho (\nu) by h \nu:

<br /> n_{ph} (E_{21}) = \rho (\nu) \displaystyle \frac{1}{h \nu} = \displaystyle \frac{8 \pi h \nu^3}{c^3 \left( e^{h \nu / (k_B T)} - 1 \right)} \frac{1}{h \nu} = \frac{8 \pi \nu^2}{c^3 \left( e^{h \nu / (k_B T)} - 1 \right)} = \frac{8 \pi E_{21}^2}{h^2 c^3 \left( e^{h \nu / (k_B T)} - 1 \right)}<br />

where h is the Planck constant, k_B is the Boltzmann constant, c is the speed of light. This computation is about spontaneous and stimulated emission in a LASER system.

In http://www.springer.com/cda/content/document/cda_downloaddocument/9784431551478-c2.pdf document (page 10, formula (2.13)), n_{ph} (E_{21}) is slightly different. It is

<br /> n_{ph} (E_{21}) = \displaystyle \frac{8 \pi n_r^3 E_{21}^2}{h^3 c^3 \left( e^{h \nu / (k_B T)} - 1 \right)}<br />

So:

1) If the above computation is correct, why in the book the denominator contains h^3 instead of h^2?

2) What can the refractive index be related to? Maybe is it due to the fact that the material is a semiconductor and not the vacuum?

Thank you anyway!

Emily

 

[TeethWhitener]

I don't know if I can give you definitive answers here, but it's clear that the equation from the document has the wrong units, while the equation you derive has the correct units. So maybe the one in the book is a typo?

What can the refractive index be related to? Maybe is it due to the fact that the material is a semiconductor and not the vacuum?

This was my thought, given that the refractive index is raised to the same power as the speed of light. The derivation of the Planck formula involves counting the number of photon modes in the interval $\nu +d\nu$. This assumes that these modes look something like $A \sin (kx + \omega t)$, which implicitly gives the speed $c =\frac {\omega }{k}$. For photons, this speed is generally assumed to be the speed of light, but if you're not in a vacuum, you need to correct the wave speed by the refractive index of the medium.

Caveat: this is my best guess. I didn't specifically go back through the whole derivation of Planck's law.

 

[Henryk]

EmilyRuck,
Both formula are correct, just different quantities.
You derived number of photons per volume per unit frequency. The formula in the book is number of photons per volume per unit energy, thence the difference by a Planck's constant.

The refractive index does have to be included in the formula. It changes the density of states for photons in the material. Yes, the book formula is derived for a solid material, not vacuum.

 

[EmilyRuck]

TeethWhitener, I agree with you, the refractive index is at the same power as c and in different media photons have different velocities.
Henryk, it was difficult because the book spoke about "density" without specifying anything else; so I thought it was per unit frequency.
Thank you both!