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From summation to integral

  1. Jan 22, 2014 #1

    DrClaude

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    Staff: Mentor

    I'm teaching a course using D. V. Schroeder, An Introduction to Thermal Physics, and there is a "derivation" in the book that is making me cringe a bit. I would like the opinion of mathematicians on the subject.

    Take a (continuous) degree of freedom ##q## from which you can get the energy ##E(q) = cq^2##. Then artificially take ##q## to be discrete, with equally-spaced intervals ##\Delta q##. This allows you to write the partition function as the sum
    $$
    Z = \sum_q e^{-\beta E(q)} = \sum_q e^{-\beta c q^2},
    $$
    which you multiply by ##\Delta q / \Delta q##,
    $$
    Z = \frac{1}{\Delta q} \sum_q e^{-\beta c q^2} \Delta q.
    $$
    Then take ##\Delta q## to be small, such that
    $$
    Z = \frac{1}{\Delta q} \int_{-\infty}^{\infty} e^{-\beta c q^2} \, dq.
    $$

    It is this last step that I don't like at all. I don't like how ##\Delta q## is made infinitesimally small at one place but not the other.

    Also, since ##q## is continuous, I would have taken directly the partition function as
    $$
    Z = \int_{-\infty}^{\infty} e^{-\beta c q^2} \, dq,
    $$
    i.e., without the ##1 / \Delta q##. In the end, that factor cancels out, so the answer is the same.

    Any pointers on how to correctly transform a summation into an integral will be appreciated.
     
  2. jcsd
  3. Jan 22, 2014 #2
    What's your definition of an integral? If you're fine with Riemann integration, then it's already a shorthand for a limit of discrete summations.
     
  4. Jan 23, 2014 #3

    DrClaude

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    Staff: Mentor

    I have no problem with a sum becoming an integral. I am only unsure about the proper way of doing it.

    With a Riemann sum, there is already a factor ##\Delta x## that becomes ##dx## when going to the Riemann integral. What we have here is a normal sum, which is taken over a continuous variable. Do you simply replace the sum with an integral and the ##dx## that comes with it, or do you have to introduce it in a contorted way as above?
     
  5. Jan 23, 2014 #4

    Stephen Tashi

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    What would that mean, mathematically? If [itex] q [/itex] is a variable then how can the expected value of [itex] q [/itex] be a function of [itex] q [/itex]? If the derivation is to make mathematical sense, shouldn't this be the first thing to straighten out? Or is [itex] E(q) =c q^2 [/itex] intended only as the definition of the function [itex] E() [/itex] ?
     
  6. Jan 23, 2014 #5

    Office_Shredder

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    Gold Member

    It is going to be true that if [itex] \delta q [/itex] is very small, then
    [tex] \frac{1}{\delta q} \sum_{q} e^{-c\ beta q^2} \delta q \approx \frac{1}{\delta q} \int_{-\infty}^{\infty} e^{-c\beta q^2} dq [/tex]
    where the approximation is saying that the relative error is going to zero, i.e. the ratio of the two is going to 1. Since you are dividing/multiplying this by something else that has a [itex]\delta q[/itex] in it, then you are only really interested in the ratio going to 1, not that they are absolutely converging to each other.

    Alternatively, you can simply not take the limit until after you have canceled the extra [itex] \delta q[/itex] factors.

    Your third choice is to start off by not writing down something which goes to infinity as [itex] \delta q[/itex] goes to zero, but that might be harder to motivate (I don't know too much about the subject).
     
  7. Jan 27, 2014 #6

    FactChecker

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    When 1/Δq and Δq are introduced, it is not immediately clear that the summation is well behaved. Soon it is clear that the summation with the Δq factor is approaching a limit, L, (the integral value). Then, if you are only interested in the order of growth of the function as Δq goes to zero, you are free to go ahead and replace the summation with it's limit value, L. So we end up with a good representation of the order of growth of the function when Δq goes to zero. Apparently that is what they need.
     
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