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Questions on Weinberg Cosmology Book

  1. Jun 15, 2013 #1
    I am following up Weinberg Cosmology book, but I have one question.

    In chapter 3.1, we have Eq (3.1.3) and (3.1.4)

    [itex] s(T) = \frac{\rho(T) + p(T)}{T} [/itex]
    [itex] T\frac{dp(T)}{dT} = \rho(T) + p(T) [/itex]

    In Eq (3.1.5), we have the Fermi-Dirac or Bose-Einstein distributions.

    [itex] n(p, T) = \frac{4 \pi g p^2}{(2 \pi \hbar)^3} \frac{1}{exp(\sqrt{p^2 + m^2} / k_B T) \pm 1} [/itex].

    From using this number distribution, the author said we have the energy density and pressure of a particle mass m are given by Eq (3.1.6) and (3.1.7).

    [itex] \rho(T) = \int n(p, T) dp \sqrt{p^2 + m^2} [/itex]
    [itex] p(T) = \int n(p, T) dp \frac{p^2}{3\sqrt{p^2 + m^2}} [/itex]

    Here, energy density is straightforward by the definition of number density.
    But, for pressure, the author said it can be derived from Eq(3.1.4), the second equation on this post.

    However, I cannot derive this pressure equation using Eq(3.1.4). Can somebody help me do this?

    Thank you.
     
  2. jcsd
  3. Jun 22, 2013 #2

    Bill_K

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    Science Advisor

    My Weinberg seems a lot different from your Weinberg! But tracing through it, here's where the p(T) equation comes from. For a particle (using c = 1), the energy is E = γm and the momentum is p = γmv. Thus p/E = v (Eq.1)

    For a system of particles, the energy-momentum tensor is Tμν = ∑n pnμ dxnν/dt δ3(x-xn), or using Eq.1, Tμν = ∑n pnμ pnν/En δ3(x-xn

    For a perfect fluid, the spatial components of Tμν are related to the pressure p by Tij = p δij.

    Thus p = (1/3) Ʃj Tjj = (1/3) Ʃn pn2/En δ3(x-xn).

    This gets you the p2 in the numerator, the E in the denominator, and the 1/3 out front.
     
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