# Einstein's formula for specific heat

1. Jun 15, 2010

### Piano man

I'm working through a derivation of Einstein's formula for specific heat and I'm stuck.

So far I've been working off Planck's assumption of quantised energy $$E=n\hbar\omega$$ and the energy probability $$P(E)= e^{\frac{-E}{k_b T}}$$, using the fact that the mean expectation energy is $$\langle E \rangle= \frac{\sum_n E P(E)}{\sum_n P(E)}$$ to get total energy $$U=3N\langle E \rangle=\frac{3N\sum_n n\hbar\omega e^{-n\hbar\omega/k_b T}}{\sum_n e^{-n\hbar\omega/k_b T}}$$

The next step is where my problem is. The derivation I am studying says the above expression is equal to $$3Nk_b T\left[\frac{\hbar\omega/k_b T}{e^{\hbar\omega/k_bT}-1}\right]$$, which when differentiated wrt T gives the Einstein formula, but I don't see how that step is made.

Any ideas?
Thanks.

2. Jun 15, 2010

### hikaru1221

I haven't found how to derive the result from your way yet, but here is an idea:
The numerator is somewhat like $$A = e^{-1} + 2e^{-2} + ...$$ and the denominator is somewhat like $$B = 1 + e^{-1} + e^{-2} + ...$$
Notice that: $$A = (B-1) + (B-1)e^{-1} + (B-1)e^{-2} + ...$$ and that B is a geometric series and easily computed.