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Lagrange multiplier for bose- einstein stats

  1. May 9, 2008 #1

    Why is -BEi used instead of +BEi as the lagrange multiplier for indistinguishable particles? How is it justified?

    I've been reading a book about statistical mechanics and it introduces lagrange multipliers first for distinguishable particles- it has ln(ni) + a + BEi = 0.

    (where a is alpha, B is beta, Ei is the energy)

    The thing is, for the indistinguishable case, it has -BEi instead of BEi, so that it looks like ln () + a -BEi = 0. it goes on to show how this becomes the boltzmann distribution is the number of levels,g, is much greater than the number of particles, n. But it seems that this only works if you have -BE rather than +BE.

    So it looks like some sort of ad hoc measure to make things work. dont get it, is there a reason that its - instead of +????????

  2. jcsd
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