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Chemical potential equilibrium cosmology

  1. Jun 6, 2015 #1
    In Dodelson's "Introduction to Modern Cosmology" at p. 61 he introduces a non- equilibrium number density
    $$n_i = g_i e^{\mu_i/T} \int \frac{d^3p}{(2\pi)^3} e^{-E_i/T}$$
    and an equilibrium number density
    $$n_i^{(0)} = g_i \int \frac{d^3p}{(2\pi)^3} e^{-E_i/T},$$
    from which it follows that the equilibrium number density, has the same form as number density of non-equilibrium just with a chemical potential of zero.

    Question: Why does the chemical potential vanish for the equilibrium case?
     
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  3. Jun 6, 2015 #2

    ChrisVer

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    ask yourself: What is the chemical potential? (that's a thermodynamics question).

    As a fast answer: that's because in equllibrium there is not preferred direction of the reaction. So by its definition the chemical potential will vanish.
     
    Last edited: Jun 6, 2015
  4. Jun 7, 2015 #3
    The chemical potential arises in the first law of thermodynamics as
    $$dE = TdS -pdV + \mu dN$$
    and hence it represents the energy added to the system when one changes the particles in the system at constant entropy and volume. I dont see why the chemical potential has to vanish in equilibrium: if two subsystems of a larger system is in equilibrium, their chemical potentials must equal, but not vanish.
     
  5. Jun 7, 2015 #4

    ChrisVer

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    The chemical potential is gives you the change of the free energy as you change the number of particles:
    [itex] \mu := \frac{\partial E}{\partial N}\Big|_{S,V}[/itex]
    Now in equlibrium, the free energy should be minimum (otherwise work can be produced along a reaction's direction- so you get a preferable arrow).
     
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