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Sum with bose-states

  1. Jun 10, 2013 #1
    My book wants to sum the number of particles in each energy states for a gas of bosons, that it calculate the infinity sum:


    now if the E's are narrowly spaced it says we can approximate this with an integral.


    Now can anyone tell me using the definition of the Riemann-integral, how this integral approximates the sum? - because it doesn't really make sense to me.
    I have had a similar question about the approximation of the partition function over space, but I think this is a bit different.
  2. jcsd
  3. Jun 10, 2013 #2
    What is [itex]\rho(E)[/itex]? Is that the density of states?

    The density of states is defined as
    [tex]g(E) = \sum_i \delta(E-E_i)[/tex]
    (I use g in case it is different from what your [itex]\rho[/itex] is.)

    If your [itex]\rho(E)[/itex] is the density of states, then you can see that the integral would just reduce to the sum. This in itself is not an approximation; it is exact, however one can use approximations to compute the density of states (such as giving the delta function a finite width).
  4. Jun 11, 2013 #3
    I think my book means the density of states yes. But my book doesn't use delta functions, rather something like √(E) (depends on the physical circumstances but for Bose states it is apparantly proportional to the square root of E).

    I just don't see how using the Riemann integral:

    limΔx→0Ʃf(nx)Δx you can approximate a sum Ʃf(x) with an integral which contains a density. I belive you showed me some months ago that:

    Ʃf(nx) ≈1/Δx ∫f(x) dx, which I understood but I don't see how a density fits in to all this.
  5. Jun 12, 2013 #4


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    The idea is that you would have a sum over energy states indexed by some discrete label, say ##i##. Your sum is then

    $$S = \sum_{i=0}^{N} n(E_i),$$
    where ##E_i## is the energy of the ith state and ##N## is the maximum number of states (could be infinite). Now, for a large number of finely spaced energy states, we can approximate this sum as an integral over ##i##, which we now treat as though it were continuous:

    $$S = \sum_{i=0}^{N} n(E_i) \approx \int_0^N di~n(E(i));$$
    however, the energy is really the more natural variable to do the integral in, so we would have to change variables, with ##di = \frac{di}{dE}dE##. We identify ##di/dE## as the density of states ##g(E)##. Thus,

    $$S \approx \int_{E_0}^{E_N} dE~\frac{di}{dE} n(E) = \int_{E_0}^{E_N} dE~g(E) n(E).$$

    (You may wonder how we justify approximately the sum over i as an integral even though the i's are evenly spaced. See this Wikipedia article on Euler-Maclaurin summation, which shows that sums can be approximated as integrals (plus higher order correction terms).
  6. Jun 13, 2013 #5
    Okay what troubled me is the first approximation. But it simply is true for many wellbehaved functions that the sum is well approximated by the integral even though the i's have a finite space of 1 between them.
    What requirements is there for f for the approximation to work well? I could imagine that if it was slowly varying it would be good since the approximation if perfect for a constant curve.
  7. Jun 13, 2013 #6


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    The wikipedia article on Euler-Maclaurin series specifies that the summed function, let's call it f(x), must be "a smooth (meaning: sufficiently often differentiable), analytic function of exponential type ##< 2\pi## defined for all real numbers x in the interval" which the sum covers. This means you have to be able to smoothly extend the summand to continuous values.

    The article gives a second sum which it terms as an asymptotic series. I'm not sure if the above requirements need to strictly hold for the asymptotic series. I suspect the book on asymptotic methods by Bender and Orszag would comment on it.
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