Questions on Weinberg Cosmology Book

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SUMMARY

The discussion centers on deriving the pressure equation from Weinberg's Cosmology book, specifically from Eq (3.1.4). The equations provided include the Fermi-Dirac or Bose-Einstein distributions, which are essential for calculating energy density and pressure. The user seeks clarification on how to derive the pressure equation, p(T), using the energy-momentum tensor and the definitions provided in the book. The response outlines the relationship between energy and momentum for a system of particles, ultimately leading to the derivation of the pressure equation.

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  • Familiarity with energy-momentum tensors in relativistic physics
  • Knowledge of perfect fluid dynamics in cosmology
  • Proficiency in mathematical derivations involving integrals and equations of state
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Students and researchers in theoretical physics, particularly those focused on cosmology, particle physics, and thermodynamics. This discussion is beneficial for anyone looking to deepen their understanding of the concepts presented in Weinberg's Cosmology.

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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)

s(T) = \frac{\rho(T) + p(T)}{T}
T\frac{dp(T)}{dT} = \rho(T) + p(T)

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

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}.

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).

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

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.
 
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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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