I must warn you that this is probably not the way that your instructor would like you to learn. On the other hand I think it cuts away all the clutter surrounding the derivation. There isn't anything very different from what you might expect but it's a little backward compared to other derivations I've seen. Let me explain.

Back in the day, when I was walking to school in 5 ft snowdrifts, Statistical Mechanics was a new sort of way to do calculations and was not really tested or trusted. Good results were obtained but it was not really trusted. I'm going to use it to derive the average energy that is coming off a blackbody.

So, say we have a blackbody. Statmech says that we have, for the average energy of the radiation coming off the blackbody is

$\displaystyle \overline{E} = \dfrac{ \int _0 ^{ \infty } E ~ P(E) ~ dE }{ \int_0 ^{ \infty } P(E) ~ dE }$

where P(E) is the probability a wave emitted from the blackbody has an energy given by the Boltzmann distribution $\displaystyle P(E) = \dfrac{e^{-E/(kT)}}{kT}$.

If we follow down this derivation we will get the Rayleigh-Jeans law which is incorrect. Planck's genious (or stubbornness) was to assume that the energy coming off the blackbody was quantised in little packets $\displaystyle \Delta E = nh \nu$, where n is the number of oscillators emitting the radiation. This means that we use a sum instead of an integral in the average energy:

$\displaystyle \overline{E} = \dfrac{ \sum_{n = 0} ^{ \infty } E ~ e^{-E/(kT)}/(kT) }{ \sum_{n = 0}^{ \infty} e^{-E/(kT)}}$

This is our starting point for the Mathematics.

Now, Planck postulated that energy of a wave from the blackbody takes the form $\displaystyle E = n h \nu $, so we put this into the summations:

$\displaystyle \overline{E} = \dfrac{ \sum_{n = 0}^{ \infty } \dfrac{n h \nu }{kT} e^{-(n h \nu )/(kT) }}{ \sum_{n = 0}^{ \infty } \dfrac{1}{kT} e^{-(n h \nu )/(kT)}}$

I'm going to neaten this up a bit. Let's set $\displaystyle \alpha = \dfrac{h \nu }{kT}$:

$\displaystyle \overline{E} = kT ~ \dfrac{ \sum_{n = 0}^{\infty} \alpha ~ n e^{-n \alpha }}{ \sum_{n = 0}^{ \infty } e^{-n \alpha }}$

Doing these sums is a bit of a task. You might have run across the method before: we are going to take a derivative and end up with a simpler form that can be summed. Notice that:

$\displaystyle - \alpha ~ \dfrac{d}{d \alpha} ln \left ( \sum_{n = 0}^{ \infty } e^{- n ~ \alpha } \right ) = \dfrac{ - \alpha \dfrac{d}{d \alpha } ~

\sum_{n = 0}^{ \infty } e^{-n \alpha }}{ \sum_{n = 0} ^{ \infty } e^{- n \alpha }}$

Now to do switch the sum and the derivative. (Unlike Mathematicians Physicists can do this without checking if it is legal.

)

$\displaystyle - \alpha ~ \dfrac{d}{d \alpha} ln \left ( \sum_{n = 0}^{ \infty } e^{- n ~ \alpha } \right ) = \dfrac{ - \alpha \dfrac{d}{d \alpha } \sum_{n = 0}^{ \infty } e^{-n \alpha }}{ \sum_{n = 0} ^{ \infty } e^{- n \alpha }}$ $\displaystyle ~ = \dfrac{ - \sum _{n = 0}^{ \infty } \alpha \dfrac{d}{d \alpha } e^{-n ~ \alpha } }{ \sum_{n = 0}^{ \infty } e^{ -n ~ \alpha }} = \dfrac{ \sum_{n = 0}^{ \infty } n ~ \alpha ~ e^{-n \alpha } }{ \sum_{n = 0}^{ \infty } e^{-n ~ \alpha }}$

(I highly doubt anyone in their right mind would expect you to come up with this. I had to look it up myself!)

Putting a bit of this together:

$\displaystyle \left ( - \alpha ~ \dfrac{d}{d \alpha} ln \left ( \sum_{n = 0}^{ \infty } e^{- n ~ \alpha } \right ) \right ) \cdot \left ( \sum_{n = 0}^{ \infty } e^{-n ~ \alpha } \right ) = \sum_{n = 0}^{ \infty } n ~ \alpha ~ e^{-n \alpha }$

Going back a bit to remind us where we are:

$\displaystyle \overline{E} = kT ~ \dfrac{ \sum_{n = 0}^{\infty} \alpha ~ n e^{-n \alpha }}{ \sum_{n = 0}^{ \infty } e^{-n \alpha }} =$ $\displaystyle kT~ \dfrac{ - \alpha ~ \dfrac{d}{d \alpha} ln \left ( \sum_{n = 0}^{ \infty } e^{- n ~ \alpha } \right ) \cdot \left ( \sum_{n = 0}^{ \infty } e^{-n ~ \alpha } \right ) }{ \sum_{n = 0}^{ \infty } e^{-n ~ \alpha }} =$ $\displaystyle -kT \alpha ~ \dfrac{d}{d \alpha} ln \left ( \sum_{n = 0}^{ \infty } e^{- n ~ \alpha } \right ) $

Now, the sum is simply an infinite geometric series with $\displaystyle r = e^{- \alpha }$ so

$\displaystyle \sum _{n = 0}^{ \infty } e^{-n ~ \alpha } = \dfrac{1}{1 - e^{ - \alpha }}$

Thus

$\displaystyle \overline{E} = -kT \alpha \dfrac{d}{d \alpha } \ln \left ( \dfrac{1}{1 - e^{ - \alpha }} \right )$

(Almost done!)

Taking the derivative and noting that $\displaystyle kT \alpha = h \nu$

$\displaystyle \overline{E} = h \nu \dfrac{1}{ \left ( 1 - e^{- \alpha } \right ) ^{-1} } \cdot (-1) \left ( 1 - e^{ - \alpha } \right ) ^{-2} e^{- \alpha }$

$\displaystyle \overline{E} = \dfrac{h \nu e^{- \alpha }}{1 - e^{- \alpha }} = \dfrac{ h \nu }{e^{ \alpha } - 1}$

and finally:

$\displaystyle \overline{E} = \dfrac{h \nu }{ e^{(h \nu)/(kT)} - 1}$

As it happens we have actually derived the Bose-Einstein distribution (as opposed to the Maxwell distribution that we started with.) However Planck didn't have access to this trick. It would have made the derivation much simpler.

I think I'm going to take a break here and get to the other part of the problem later.

-Dan

Addendum: I should mention that much of the above derivation was borrowed liberally from "Quantum Mechanics of Atoms, Molecules, Solids, Nuclei, and Particles," (second edition) by Robert Eisberg and Robert Resnick.