Grand Canonical Partition function question.

1. Jun 7, 2008

MathematicalPhysicist

The question:
A system consists of N sites and N particles with magnetic moment m.
each site can be in one of the three situations:
1. empty with energy zero.
2. occupied with one particle and zero energy (when there isn't magnetic field around).
3. occupied with two particles with anti parallel moments with energy $$\epsilon$$.
(particles in the same site are indistinguishable).
the system is under the influence of external magnetic field,B, which acts on the particles.
the question is to: 1. find the chemical potential of the system. 2. find the entropy of the system. 3. find the energy of the system in limit of high tempratures and low.

my solution:
ok first i need to find the grand canonical partition function, i think this should be:
$$Z_G=1+e^{\beta *\mu}+e^{\beta *(2\mu -\epsilon)}$$
(because the particles are indisitiguishable, the partition function equals: $$Z_G^{N}/N!$$
now, what I think to do is to calculate the energy with this function and then equate it with the energy given which is i think is: $$-NmB$$
Now I'm using the next equation:
$$U=(\frac{\mu}{\beta}\frac{d}{d\mu}-\frac{d}{d\beta})log(Z_G)$$
After some manipulations, I get to the next equation:
$$\frac{e^{2\beta *\mu -\beta *\epsilon}\epsilon}{1+e^{\beta *\mu}+e^{2\beta * \mu-\beta *\epsilon}}=mB$$
after that I need to solve a quadratic equation wrt e^(beta*mu) and choose the positive solution, is this seems plausible?

with finding the entropy I can use the thermodynamic identity, that :
$$\mu =-\tau (\frac{d\sigma}{dN})$$
and integrate what i find for the chemical potential which is mu ofcourse (we are using the notation in kittel's which i think is used worldwide).

anyway, what do you think of my attempt at solution?

thanks in advance for any help.

2. Jun 8, 2008

MathematicalPhysicist

So another question of mine get unanswered, great site.

3. Jun 8, 2008

Physics Monkey

You can think of the system as a collection of N 4-level sites, but your $$Z_G$$ (the partition function of a single site, I presume) isn't quite right. In particular, doesn't the energy depend on the magnetic field?

Also, the sites are distinguishable, so are you sure you want to divide by $$n!$$?

One more thing, why do you think the total energy is $$-NmB$$? You can calculate the energy from $$Z$$ without any additional information in terms of $$\mu$$. What you need to do is find $$\mu$$ which is associated with the number of particles.

Hope this helps.

Last edited: Jun 8, 2008
4. Jun 8, 2008

Mapes

The sites presumably are distinguishable, so wouldn't there be a configurational term in the partition function representing possible arrangements of 0-, 1-, and 2-particle sites? Distinguishable sites would also suggest against dividing by N! The partition function might be something like

$$Z_G=\frac{N!}{N_2! N_3!(N-N_2-N_3)!}\left(1+e^{\beta N_2 mB}}+e^{\beta N_3 \epsilon}\right)$$

where the subscripts correspond to the three situations you described. But I haven't looked further, this may give the same results as your approach.

It looks like the energy is

$$U=TS-PV+\mu N+BN_2m$$

If we take T, V, N, and B as constant, we can Legendre transform this into

$$d\Lambda=-S\,dT-P\,dV+\mu\,dN-N_2m\,dB$$

where $\Lambda=U-TS-N_2mB[/tex] is the new characteristic function. Then $$\Lambda=-kT\ln Z_G$ and the chemical potential and entropy are [tex]S=-\left(\frac{d\Lambda}{dT}\right)_{\mathrm{V,N,B}}=k\ln Z_G+kT\frac{d\,\ln Z_G}{dT}$$

$$\mu=\left(\frac{d\Lambda}{dN}\right)_{\mathrm{T,V,B}}=-kT\frac{d\,\ln Z_G}{dN}$$

Thoughts on this? I'm not sure why your energy is $-NmB$ instead of $N_2mB$, and I'm not immediately seeing where your U equation comes from.

5. Jun 8, 2008

MathematicalPhysicist

Well as I see it in the simple partition problem if we had indistiguishable particles then we would take Z^N/N!, here we are given that two particles in the same site are indistiguishable, I don't see how I take into acount unless I do the next calculation:
$$Z_G=(1+e^{\beta \mu}+\frac{e^{2\beta \mu - 2\beta \epsilon}}{2!})^N$$
I think this is more sound cause I've taken into account the option of two particles in the same site indistigushable from each other.
Now from here I think that I may find the chemical potential, by using the fact that NmB equals the energy, and we may calculate the energy from the next equation
$$U=(\frac{\mu}{\beta}\frac{d}{d\mu}-\frac{d}{d\beta})log(Z_G)$$
Is that a sound argument or not as for the two particles in the last site, the simple partition function is (because they are indistiguishable) $$(e^{-\beta \epsilon })^2/2!$$
What do you think, thanks for the point you made they made things much clearer.

6. Jun 8, 2008

MathematicalPhysicist

Physics Monkey, Im not sure how to find the chemical potential without using the equation for thermal energy.
I mean perhpas I should use the fact what was the chemical potential before the magnetic field and after.
I'm not sure how to calculate though.
another equation involving the chemical potential is:
$$\mu = \frac{log(n/n_Q)}{\beta}$$, but I don't have the mass of the system in order to calculate n_Q among other parameters, so What do I miss?

7. Jun 10, 2008

Physics Monkey

Here is a hint. In the grand canonical ensemble you control the chemical potential rather than the particle number directly. So if you want your system to contain N particles on average then you had better adjust the chemical potential to a certain value. Is there a way to know how many particles your system contains just by using the partition function?

p.s. You're trying to use too specific of formulas.

8. Jun 12, 2008

MathematicalPhysicist

Well I can use the partition function in order to calculate the average number of particles which should be equal to the number of particles for a system of N>>1, is this ok to use it here, I don't see here an assumption that N>>1.

The equation I'm thinking of is:
$$<N>=\lambda \frac{d}{d\lambda}log(Z)$$
where:
$$\lambda =exp(\frac{\mu}{\tau})$$
did you refer to this equation?

9. Jun 12, 2008

Physics Monkey

Great, that is precisely the equation I had in mind. You can use it to determine the total number of particles as a function of $$\mu$$, and by setting this equation equal to the number you want, N, you will be able to determine $$\mu$$.

In statistical physics there is usually a standing assumption that N is large, otherwise fluctuations become important exactly as you said. Remarkably, the theory can often be useful even when N is not so large, but one has to be a more careful about things.