# Thermodynamics: Microstates

1. Sep 29, 2014

### teme92

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

Consider an isolated (fixed total energy) system of N atoms each of which may exist in three states of energies: −e, 0, e. Let us specify the macrostates of the system by N, E (the total energy), and n (the number of atoms in the zero energy state).

a) Identify explicitly and write out the microstates corresponding to the N = 3, E = 0, n = 1 and N = 3, E = 0, n =3 macrostates (use −, 0, + to denote the state of the atoms).

2. Relevant equations

E=ne

3. The attempt at a solution

I'm not sure exactly how to write the microstate but I understand what it is. I was thinking along the lines of the statistical weight Omega where:

Omega=Omega(N,E,n)

And then saying Omega(n) = N!/(n!(N-n)!), and then subbing in the N and n values given. But E isn't taken into account here so I'm confused a tad. Any help would be much appreciated.

2. Sep 29, 2014

### BvU

You mean $E=\Sigma_i n_i e_i$, right ?

But isn't this a much simpler exercise where all you have to do is come up with microstates that satisfy the macrostate properties ?

I suppose you can assume the atoms can be distinguished (as in a lattice), but it doesn't say so in your wording of the exercise. They may well be indistinguishably floating around.

N=3 is already a given, so all you have to do is for each case to think of ways to get E=0 with n=1 and n=3, respectively. In particular the latter seems rather straightforward.

3. Sep 29, 2014

### BvU

Re $\Omega$: that gives you the weight of the macrostate. Might coincide with the number of microstates you find ! And that is not really a coincidence: the statistical weight of a macrostate IS the number of microstates that can bring that macrostate about!
In your first case you have N = 3 and n=1, second case gives you $\Omega=$ ...

4. Sep 29, 2014

### teme92

Hey BvU, thanks for the reply. So are you saying I sub N=3 and n=1 into the formula I previously said of Omega=N!/(n!(N-n)!)?

If so then I get Omega = 3 and Omega = 1 for N=3, n=3

5. Sep 29, 2014

### BvU

That's right. But the exercise wants you to list all microstates explicitly.
Note that N!/(n!(N-n)!) only tells you in how many ways you can form a group of size n out of a total of N.
So perhaps the number of possible microstates in the first case is a bit more...

Do you now have an idea already on how to specify a microstate ?

6. Sep 30, 2014

### teme92

No I don't understand what the question wants me to do.

7. Sep 30, 2014

### BvU

and describe a microstate that has one atom in the zero energy state, whilst the two others cause a total energy of zero.
You might give the atoms names, like B, C, F. Now add +,-,0 to each of them in such a way that n=1 and E=0.

8. Sep 30, 2014

### teme92

So will there be a B+, B0 and B- aswell as a C+,C0, C-, F+,F0,F-?Or is it B+, C0 and F-

9. Sep 30, 2014

### BvU

There are only three atoms, so: the latter

Now invent a few more that have the same N, n and E ....

10. Sep 30, 2014

### teme92

I'm gonna call them n+, n0 and n- because in a follow on question it asks it in this form. If there are only three atoms, how do invent more? What equation do use to invent them?

11. Sep 30, 2014

### BvU

You use $\Sigma_i n_i = N$, $\Sigma_i n_i E_i = E$, and the n in the macrostate designation (N, n, E) is a given.

Very confusing that you now choose n+, n- and n0. This means you cannot distinguish between the atoms any more.

And here's me thinking you would easily come up with e.g. B0, C+ and F- as a second candidate for the first case!

12. Sep 30, 2014

### BvU

(N, n, E) = (3,1,0) can only be achieved with one atom in each of the three states, right ? So for the + you can pick one of the three, then for the 0 you can pick one of the two remaining and the other goes into the - state. If the atoms are distinguishable, that gives 3*2*1 or 3! = 6 ways to get that macrostate.

However, if the atoms are indistinguishable (which they probably are not in your exercise), you have to divide by the number of ways to line up the three atoms, also 3!=6. Leaves only 1(+), 1(0), 1(-), one single microstate !​

So now what for the second case ?

And - because I spoiled your first case - you might indulge me by listing off all microstates that correspond to the macrostate with N = 4 and E = e for the case of distinguishable atoms...

13. Oct 2, 2014

### teme92

Thanks for the help, I handed in the problem set and my questions were answered!