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Homework Help: Grand Canonical Partition function question.

  1. Jun 7, 2008 #1

    MathematicalPhysicist

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    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 [tex]\epsilon[/tex].
    (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:
    [tex]Z_G=1+e^{\beta *\mu}+e^{\beta *(2\mu -\epsilon)}[/tex]
    (because the particles are indisitiguishable, the partition function equals: [tex]Z_G^{N}/N![/tex]
    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: [tex]-NmB[/tex]
    Now I'm using the next equation:
    [tex]U=(\frac{\mu}{\beta}\frac{d}{d\mu}-\frac{d}{d\beta})log(Z_G)[/tex]
    After some manipulations, I get to the next equation:
    [tex]\frac{e^{2\beta *\mu -\beta *\epsilon}\epsilon}{1+e^{\beta *\mu}+e^{2\beta * \mu-\beta *\epsilon}}=mB[/tex]
    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 :
    [tex]\mu =-\tau (\frac{d\sigma}{dN})[/tex]
    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. jcsd
  3. Jun 8, 2008 #2

    MathematicalPhysicist

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    So another question of mine get unanswered, great site.
     
  4. Jun 8, 2008 #3

    Physics Monkey

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    You can think of the system as a collection of N 4-level sites, but your [tex] Z_G [/tex] (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 [tex] n! [/tex]?

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

    Hope this helps.
     
    Last edited: Jun 8, 2008
  5. Jun 8, 2008 #4

    Mapes

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    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

    [tex]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)[/tex]

    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

    [tex]U=TS-PV+\mu N+BN_2m[/tex]

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

    [tex]d\Lambda=-S\,dT-P\,dV+\mu\,dN-N_2m\,dB[/tex]

    where [itex]\Lambda=U-TS-N_2mB[/tex] is the new characteristic function. Then

    [tex]\Lambda=-kT\ln Z_G[/itex]

    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}[/tex]

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

    Thoughts on this? I'm not sure why your energy is [itex]-NmB[/itex] instead of [itex]N_2mB[/itex], and I'm not immediately seeing where your U equation comes from.
     
  6. Jun 8, 2008 #5

    MathematicalPhysicist

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    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:
    [tex]Z_G=(1+e^{\beta \mu}+\frac{e^{2\beta \mu - 2\beta \epsilon}}{2!})^N[/tex]
    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
    [tex]U=(\frac{\mu}{\beta}\frac{d}{d\mu}-\frac{d}{d\beta})log(Z_G)[/tex]
    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) [tex](e^{-\beta \epsilon })^2/2![/tex]
    What do you think, thanks for the point you made they made things much clearer.
     
  7. Jun 8, 2008 #6

    MathematicalPhysicist

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    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:
    [tex]\mu = \frac{log(n/n_Q)}{\beta}[/tex], but I don't have the mass of the system in order to calculate n_Q among other parameters, so What do I miss?
     
  8. Jun 10, 2008 #7

    Physics Monkey

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    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.
     
  9. Jun 12, 2008 #8

    MathematicalPhysicist

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    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:
    [tex]<N>=\lambda \frac{d}{d\lambda}log(Z)[/tex]
    where:
    [tex]\lambda =exp(\frac{\mu}{\tau})[/tex]
    did you refer to this equation?
     
  10. Jun 12, 2008 #9

    Physics Monkey

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    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 [tex] \mu [/tex], and by setting this equation equal to the number you want, N, you will be able to determine [tex] \mu [/tex].

    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.
     
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