# Double internal energy, how much will the entropy change?

Approaching the following question

Imagine we have a collection of $N = 100$ simple harmonic oscillators. The total internal energy is $U = q \epsilon$ where $q$ is the number of energy quanta. Assume $q >> N >> 1$. If we double the internal energy, by how much will the entropy of the collection change?

The question was originally found here under Question 5.

The answer is $\sigma_{final} - \sigma_{initial} = 69.3$

How is this determined?

I know $S = k_b ln \Omega$, where $S$ is entropy, $\Omega$ is the number of micro-states, $\sigma = \frac{ 1}{ \Omega}$, $k_b$ is the Boltzmann constant.

I know $<energy> = 1/2 k T$ per quadradic term.

I know $\frac{ 1}{ T} \equiv \frac{ \partial S}{\partial U}$

But how does doubling energy effect entropy? Specifically, what equation am I looking for to determine this?

You have $$U=N\langle E \rangle=N\ (\langle E_\text{kin} \rangle+\langle E_\text{pot} \rangle)=N k\ T,$$ $$\tfrac{\partial S(U)}{\partial U}=\tfrac{1}{k\ T}=\tfrac{N}{U}\ \ \Longrightarrow\ \ S(U)=N \ln(U)+S_0,$$ $$\Delta S=S(x\ U)-S(U)=N\ln(x\ U)-N\ln(U)=N \ln(x).$$ And $$100 \ln(2)=69.315...$$
• @rudolph9: The thread you posted an hour ago doesn't contain the question "What is the temperature?" but only the phrase "Why is the answer $T= \dots\$ ?" and here the $\sigma$ is mentioned but not involved in the formulation of the question. Commented Jul 22, 2012 at 22:05
• Also on the note of the $sigma$, the answer was given as $\sigma_{final} - \sigma_{initial} = 69.3$ Commented Jul 22, 2012 at 22:12