# Microstates of an Einstein solid

1. Feb 4, 2015

### MostlyHarmless

My confusion isn't exactly with a homework problem, but more with an example that is key to understanding a homework problem. So I am posting here anyway.

1. The problem statement, all variables and given/known data

The example is of an Einstein solid, with N=3 oscillators. The book lists the multiplicity of each macrostate, with presumably each macrostate as the total energy units of the system. It says "There is just one microstate with total energy 0, while there are 3 microstates with one unit of energy, six with two units, and ten with three units."

Giving 20 total microstates, then it lists them all in a table.

2. Relevant equations
Finally it gives the formula:
$${\Omega}(N, q)= {{q+N-1}\choose{q}}=\frac{(q+N-1)!}{q!(N-1)!}$$ Where N is the number of oscillators and q is the energy units.

3. The attempt at a solution
My problem is, the book doesn't explicitly say what q is for this example, but I'm assuming it is 4? So using N=3 and q=4 for that equation gives me 15 microstates, using q=3 gives 10 microstates, q = 5 gives 21.

So I'm not sure what I'm missing. I'm going to step away from the problem for a minute and hopefully I can find my mistake, but in the meantime I'm posting it here. Thanks in advance.

2. Feb 4, 2015

### TSny

q is the number of energy units. Then $\Omega(q, N)$ is the number of ways of distributing q energy units among N oscillators. So, when N = 3 and q = 2, you get $\Omega(2, 3) = 6$, in agreement with what you stated for this case.

It looks like maybe you are trying to interpret $\Omega(q_0,N)$ as the sum of all microstates with different number of energy units q = 0, 1, 2, ...q0. But $\Omega(q_0,N)$ is just the number of microstates with fixed total energy corresponding to q0 energy units.

Last edited: Feb 4, 2015
3. Feb 4, 2015

### MostlyHarmless

Ok, so $\Omega$ only gives the multiplicity for a single macro state? I was misinterpreting what q meant. Thank you.

4. Feb 4, 2015

Yes.