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Statistical mechanics: Sums of exponentials with sums.

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm working through an example from class and the textbook, but I'm confused about how the steps progress mathematically.

    The example involves the Gibb's partition for a paramagnet.

    [tex] \sum_{s} exp(\beta \mu B \sum_{i}^{N} s) [/tex]

    Where s = -a,-a+1...a for each spin.

    Am I right in thinking the sum in the exponent is a sum over this particles in the sample (N). Since a sum in an exponent is a product of the exponentials we get:

    [tex] [\sum_{s} exp(\beta \mu B s)]^{N} [/tex]

    However, I'm not sure what do with the second sum. I can't see a neat way proceed. Can anyone point me in the right direction?
     
  2. jcsd
  3. Sep 26, 2011 #2

    G01

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    The second sum is a sum over spin states. If you're dealing with quantum systems with a discrete number of spin states, then it's probably best to write the sum explicitly and re-express the result in a more simple form. HINT: use some hyperbolic trig identities.

    If you are treating the spin classically and the value of s can take on any value [itex]s_o cos( \theta )[/itex] (where theta is the angle from the z axis), then your best best will be to convert the sum to an integral in theta.
     
  4. Sep 27, 2011 #3
    Thanks for the response. I'm treating the spins as quantum. I managed to solve the taking out the factor of a and treating the rest as a geometric series i.e.

    [tex][e^{-\beta\mu B a} \sum_{s=1}^{(2a+1)} exp(\beta \mu B s)]^{N} [/tex]

    Which, using the formula for the sum of a finite geometric series gives:

    [tex][e^{-\beta\mu B a} \frac{1-e^{(2a+1)\beta\mu B}}{1-e^{\beta \mu B}}]^{N}[/tex]

    Problem, is I can't seem to resolve that down to a hyperbolic function and I'm sure it should. Any ideas?
     
  5. Sep 27, 2011 #4

    G01

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    Oh ok. So, your dealing with a quantum spin system of arbitrary integer or half integer spin, not just spin 1/2. That makes it slightly more complicated but, not by too much.

    The sum of the finite geometric series is the way to go.

    I don't think the partition function can be simplified directly to a single hyperbolic trig function as it stands, but for large values of a, your expectation values for the Energy, Magnetization, Specific heat, etc. definitely can be.

    Try this: Keep the partition function in exponential form, but simplify and multiply through the exponential,[itex]e^{-\beta \mu B a}[/itex]. Take [itex]\frac{\partial Z}{\partial \beta}[/itex] with everything in exponential form. Then you can use that, along with Z to find all the state variable expectation values. Then for large a, these quantities will be expressible in terms of only hyperbolic trig functions.
     
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