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Multiplicity of a Einstein Solid, Low Temperature Limit

  1. Feb 4, 2009 #1

    TFM

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

    (a)

    The formula for the multiplicity of an Einstein solid in the “high-temperature” limit,
    q >> N, was derived in one of the lectures. Use the same methods to show that the multiplicity of an Einstein solid in the “low-temperature” limit, q << N, is

    Ω(N,q)=(eN/q)^q (When q≪N)


    (b)

    Find a formula for the temperature of an Einstein solid in the limit q << N. Solve for the energy as a function of temperature to obtain U=Nϵe^(-ϵ/(k_B T)), where ε is the size of an energy unit.


    2. Relevant equations

    N/A

    3. The attempt at a solution

    Okay, I have started (a):

    [tex] \Omega (N,q) = (\frac{(N - 1 + q)!}{(n - 1)!q!}) [/tex]

    N large:

    (N - 1)! approx = N!

    [tex] \Omega (N,q) = (\frac{(N + q)!}{N! q!}) [/tex]

    Take logs

    [tex] ln(\Omega (N,q)) = ln(N + q)! - ln(N!) - ln(q!) [/tex]

    Use Stirling aprrox:

    [tex] ln N! \approx N ln(N) - N [/tex]

    [tex] ln(\Omega (N,q)) = (N + q)ln(N + q) - (N + q) - (Nln(N) - N) - (qln(q) - q) [/tex]

    Cancels down to:

    [tex] ln(\Omega (N,q)) = (N + q)ln(N + q) - Nln(N) - qln(q) [/tex]

    Now I have to use the Taylor Expnasion for q << N, but I got slightly confused here.

    Could anyone please offer some assistane what I need to do from here?

    Many thanks in advance,

    TFM
     
  2. jcsd
  3. Feb 4, 2009 #2
    You need to write the (N+q) log like this (below) and then just copy the lecture derivation, I think. You can show how you use the taylor expansion, but I would have said it's not neccessary

    [tex]ln(N+q)=ln(N(1+q/N))[/tex]
     
  4. Feb 5, 2009 #3

    TFM

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    Okay, so now:

    [tex] ln(\Omega (N,q)) = (N + q)ln(N + q) - Nln(N) - qln(q) [/tex]

    Use Taylor Expansion:

    [tex] ln(N+q)=ln(N(1+q/N)) [/tex]

    [tex] ln(N+q)=ln(N) + ln(1+q/N)) [/tex]

    now the notes say:

    [tex] ln(N+q) = ln(N) + q/N [/tex]

    Althought I am not sure why the last ln has dissapeared?

    Now plug into omega to give:

    [tex] ln(\Omega (N,q)) = (N + q)(ln(N) + q/N)- Nln(N) - qln(q) [/tex]

    Multiply out:

    [tex] ln(\Omega (N,q)) = NlnN + q + q^2/N + qlnN - Nln(N) - qln(q) [/tex]

    Cancel down:

    [tex] ln(\Omega (N,q)) = q + q^2/N + qlnN - qln(q) [/tex]

    Now the notes have that:

    because q >>N,

    N^2/q approx 0

    So for

    because q <<N,

    This should still be

    q^2/N approx 0

    thus:

    [tex] ln(\Omega (N,q)) = q + qlnN - qln(q) [/tex]

    take exponentials

    [tex] \Omega (N,q) = e^q + (N/q)^q [/tex]

    Thus

    [tex] \Omega (N,q) = (eN/q)^q [/tex]

    Does this look okay?
     
  5. Feb 5, 2009 #4
    This is just rearranging (not using a Taylor series)

    You should say, [tex] ln(N+q)=ln(N(1+q/N))=ln(N) + ln(1+q/N)[/tex]

    You can then take the Taylor expansion of the last term in the above, namely
    [tex]ln(1+q/N)[/tex], provided you assume q<<N so that q/N<<1. If you look up the Taylor expansion for ln(1+x) you will find that it says this is approximately equal to x in the limit x<<1.
    Here there is an error. When you exponentiate an equation the entire right side is raised to the same exponential. Also [tex]e^x+e^y \neq e^{x+y}[/tex]
     
  6. Feb 5, 2009 #5

    TFM

    User Avatar

    Okay so:

    [tex] ln(\Omega (N,q)) = q + qlnN - qln(q) [/tex]

    so I have to take exponentials,


    [tex] \Omega (N,q) = e^q + e^qlnN - e^qln(q) [/tex]

    [tex] \Omega (N,q) = e^q + Ne^q - qe^q [/tex]

    So now do we use the log/exponential rules to give?

    [tex] \Omega (N,q) = (eN/q)^q [/tex]

    Also,

    Find a formula for the temperature of an Einstein solid in the limit q << N. Solve for the energy as a function of temperature to obtain U=Nϵe^(-ϵ/(k_B T)), where ε is the size of an energy unit.

    Where woul be aq good place to start for here?
     
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