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I have been working on building intuition and experience with solving problems related to the microcanonical ensemble, but I haven't been able to solve the following problem:

If you have $N$ independent oscillators, each having the same oscillation frequency $\omega$, given the energy $E=(N_E+\frac{N}{2})\hbar\omega$, find the number of states with energy E and calculate the entropy as a function of the energy.

What I tried/thought:

Since the oscillators don't interact and are said to have the same vibrational frequency, I don't see how the energy could be anything other than the sum of their vibrational energies, which would be a constant given that $\omega$ seem to be a constant. I'm also unsure how to interpret $N_E$, so I'm unable to move forward.

My gut tells me to do something like this to calculate the number of states,

$\frac{E}{N}=(\frac{1}{2}+\frac{N_E}{N})\hbar\omega$

from which, if I'm reading incorrectly that all having the same vibrational frequency means all vibrating at the same frequency and instead that if they vibrate they vibrate at the same frequency, I would extract that the fraction of particles vibrating to produce a system with a certain energy E would be given by $\frac{N_E}{N}=\frac{E}{N}-\frac{\hbar\omega}{2}$. The term subtracted from $\frac{E}{N}$ would represent a ground energy for each atom. The number of states that would satisfy this would be $N\choose N_E$, though I don't think this reduces to a nice number.

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The energy levels of a system of $N$ independent oscillators, all with the same frequency $\omega$, have the form $$ E(n_1,\dots,n_N) = \sum_{i=1}^N \hbar\omega\left(n_i+\frac12\right), $$ where $n_i$ is the number of quanta of energy in $i$-th oscillator. Hence, in your formula, $$ N_E = \sum_{i=1}^N n_i. $$ From here it is straightforward to obtain, that the number of different states of the system of N oscillators with energy corresponding to the number $N_E$ is equal to $$ \Gamma(N_E) = \sum_{n_1,\dots,n_N \geq 0} \Delta(n_1+\dots+n_N - N_E), $$ where $\Delta(x) = 1$ if $x=0$ and $\Delta(x) = 0$ if $x\neq0$. A simple explicit formula for the $\Gamma(N_E)$ can be derived in different ways.

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  • $\begingroup$ Thanks. Just to see if I understood $\Gamma(N_E)$, this is equivalent to the number of partitions of $N_E+N$ - not $N_E$ because $n_i$'s can be zero - into N parts, right? $\endgroup$ – Fhoenix Apr 15 at 17:21
  • $\begingroup$ @Fhoenix, I would rather say, that $\Gamma(N_E)$ is exactly the number of ways to write $N_E$ as a sum of $N$ nonnegative integers. $\endgroup$ – Gec Apr 15 at 17:50

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