How does the average energy from equation (41.15) of Feynman's lectures approach $kT$ as omega goes to zero?

Question

Feynman said the following equation should approach $kT$ as $\omega$ goes to zero, or $T$ goes to infinity.

$$\langle E \rangle = \frac{\hbar \omega}{e^{\hbar \omega/kT}-1}$$

Does anybody know how to prove this?

Answer 1

Get use of L'Hopital's rule where ' stands for the derivative

$$\lim_{\omega \to 0} \langle E\rangle= \lim_{\omega \to 0}\frac{(\hbar \omega)'}{(e^{\frac{\hbar \omega}{kT}}-1)'} = \lim_{\omega \to 0} kT e^{-{\frac{\hbar \omega}{kT}}}=kT$$

Answer 2

The expression for $\langle E \rangle$ is not defined at $\omega = 0$ as it would be ${0 \over 0}$. So we need to take the limit $\omega \to 0$ i.e.

$$\lim_{\omega\to 0} \langle E \rangle = ?$$

To solve this, very simply, use the fact that the Taylor expansion of the exponential function is

$$e^x\approx 1+x+{1\over 2}x^2 + ...$$ which in your case is $$e^{\hbar\omega \over k_B T}\approx 1 + {\hbar\omega \over k_B T}+{1\over 2}\left({\hbar\omega \over k_B T}\right)^2+...$$

so $$\langle E \rangle \approx {\hbar\omega \over 1 + {\hbar\omega \over k_B T}+{1\over 2}\left({\hbar\omega \over k_B T}\right)^2+... -1}$$ which becomes

$$\langle E \rangle \approx {\hbar\omega \over {\hbar\omega \over k_B T}+{1\over 2}\left({\hbar\omega \over k_B T}\right)^2} = {\hbar\omega\over\hbar\omega} {k_BT \over 1+{1\over 2}{\hbar\omega \over k_B T}} $$ and finally

$$\langle E \rangle \approx {k_BT \over 1+{1\over 2}{\hbar\omega \over k_B T}} $$ This expression is valid if $\omega\approx 0$ and describes the behavior of $\langle E \rangle$ at low frequencies.

From this, just take the limit $\omega\to 0$, and it should be clear that $$\lim_{\omega\to 0} \langle E \rangle = \lim_{\omega\to 0} \langle E \rangle {k_BT \over 1+{1\over 2}{\hbar\omega \over k_B T}} \to k_B T $$

Answer 3

Use the Taylor series for the exponential function,

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots.$$

In addition to giving you the $\omega\to 0$ limit, it will give you the low-frequency corrections.