1. Dec 7, 2006

### romeo6

Hey folks, I really need help setting up a canonical ensemble.

I am using the einstein model and have an energy:

$$E_\nu=A + \sum_{i=1}^{2N} \hbar\omega n_i+\sum_{j=1}^{N} \hbarh\omega n_j$$

where A is some constant.

Now, I need to build the partition function which I 'think' looks like

$$Q=\sum_{i=1}^{2N} e^{-\beta \hbar n_i}+\sum_{j=1}^N e^{-\beta \hbar n_j}$$

...but I'm not too confident. Can anyone help me with the partition function please???

2. Dec 8, 2006

### alfredblase

nevermind :/ sorry

3. Dec 9, 2006

anyone? :)

4. Dec 9, 2006

### Physics Monkey

Hi romeo6,

The definition of the partition function in the canonical ensemble is simply this: $$Q = \sum_{states} e^{- \beta E}.$$

The sum is over all possible states of your system. Now your system is composed of many oscillators, so how do you specify the state of your system? This tells you what to sum over.

The energy E which appears in the exponent is the total energy of your system in a given state.

Note that the sum is over states of the whole system, not just states of a single oscillator. Similarly, the energy which goes in the exponent is the energy of your whole system, not just the energy of a single oscillator.

Hope this helps.

Last edited: Dec 9, 2006
5. Dec 9, 2006

### romeo6

Thanks physics monkey. How does this look?

$$Q=\sum_{i=1}^{2N} e^{-\beta \hbar n_i+A}+\sum_{j=1}^N e^{-\beta \hbar n_j}$$

thanks again.

6. Dec 9, 2006

### Physics Monkey

romeo6, be careful, your expression isn't quite right. See my comment just above the "hope this helps". Right now you're exponentiating each oscillator energy separately and then adding, but what you need to do is add up the energies of all the oscillators in a given state and then exponentiate.

Last edited: Dec 9, 2006
7. Dec 9, 2006

### romeo6

Hmmmm....is it something more like:

$$Q = \sum e^{- \beta (A + \sum_{i=1}^{2N} e^{-\beta \hbar n_i+A}+\sum_{j=1}^N e^{-\beta \hbar n_j})}$$

Thats a little weird looking...

8. Dec 9, 2006

### Physics Monkey

Why are you exponentiating inside the exponential? Is $$A + \sum_{i=1}^{2N} e^{-\beta \hbar n_i+A}+\sum_{j=1}^N e^{-\beta \hbar n_j}$$ the energy of your system? Those exponentials dont even have the units of energy. You already wrote the energy of your system in your first post, so I know you know what it is. Put that in the exponential and then figure out what the states of your system are.

9. Dec 9, 2006

### romeo6

Hmmmmm...throw a thousand darts...lol, one hits the bull...

$$Q=\sum_{i,j=1}^{2N} e^{-\beta( \hbar n_i+A +\hbar n_j)}$$

10. Dec 9, 2006

### Physics Monkey

Still not there. Is $$\hbar n_i+A +\hbar n_j$$ the energy of your whole system? (you're at least missing some omegas). Again, you wrote the energy of your system in your first post. I'll repeat it here for you:
$$E = A + \sum_{i=1}^{2N} \hbar \omega n_i + \sum^N_{j=1} \hbar \omega n_j$$. You should put the energy of your whole system in the exponential. The state of your whole system is specified by giving the integers $$n_i$$ for each oscillator and you must sum over all states.

11. Dec 9, 2006

### romeo6

Sorry, I missed off the omega by mistake.

$$Q=\sum_{i,j=1}^{2N} e^{-\beta\omega( \hbar n_i+A +\hbar n_j)}$$

I think that what is confusing me is how to do the sum up to say N and 2N, I mean it would look something like:

$$Q=\sum_{i,j=1}^{2N} e^{-\beta\omega( \hbar (1+2+3+...(2N-1)+2N)+A +\hbar(1+2+3+...N))}$$

But this looks far too messy. Of course I could put the summation in the exponential, but I already tried that earlier and that wasn't correct either.

I'm sorry, I hope this isn't sounding dum!

Last edited: Dec 9, 2006
12. Dec 9, 2006

### Physics Monkey

No, it doesn't sound dumb. In my book not knowing what to do is fine so long as you try hard to learn what to do.

The sum you have written in your exponential isn't right. You don't sum just the numbers 1 to 2N, you sum the numbers n_1 to n_2N. Now what these numbers are depend on the state of your system. For example, one possible state is to have n_1 = 5, n_2 = 6, n_3 = 2, and all other n's zero. This state has an energy $$A + \hbar \omega 5 + \hbar \omega 6 + \hbar \omega 2$$ and so one term in the sum which is your partition function will be $$e^{-\beta(A+13\hbar \omega)}$$. In other words, for each set of numbers $$\{n_i\}$$ you should add one term to the partition function which is the exponential of (-beta times) the energy of the system corresponding to that set of n's.

If you're still having trouble, try setting N to be something small and writing a few terms out. You'll get the hang of it.

Last edited: Dec 9, 2006
13. Dec 10, 2006

### romeo6

Thanks again physics monkey.

Ok, so both oscillators the sum goes from 1 to N (hence the 2 in the exponential below and the first expansion) and one of the oscillators goes up to 2N (hence the second term in the exponent).

$$Q= e^{-\beta\omega (2\hbar (n_1+n_2+...n_n)+\hbar(n_{n+1}+n_{n+2}+...n_{2n})+A)}$$

Does this look a litte better?

Last edited: Dec 10, 2006
14. Dec 11, 2006

### dextercioby

$$Z=\sum_{n} \mbox{exp}\left(\frac{1}{kT}\left(A + \sum_{i=1}^{2N} \hbar\omega n_i+\sum_{j=1}^{N} \hbar\omega n_j\right)\right)$$

Daniel.

15. Dec 11, 2006

### romeo6

Thanks Dexter. :)

I now want to derive the equation of state.

In a classical gas this is just PV=nRT

For the case above do I use the partition function to find expressions for P,V and T and just plug them in to the classical case?

16. Dec 12, 2006

### dextercioby

What do you mean "plug them" in the classical case ? Which "classical case" ?

Daniel.

17. Dec 12, 2006

### romeo6

Well, my question is how do I find the equation of state for this system.

Do I simply duplicate PV=nRT, however calculate P, V and T using the above partition function?

18. Dec 12, 2006

### dextercioby

You have the partition function, all you need to do in order to derive the statistical thermostatics is apply the theory. That is, obtain the free energy.

Daniel.