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I Energy density and pressure of perfect fluid

  1. Jul 30, 2017 #1
    is there any way to derive p=rho/3 from 4 vector or stress-energy tensor?
  2. jcsd
  3. Jul 30, 2017 #2


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    This is true for (free) massless particles only and derives from the invariance of the action under scaling transformations. Take, as an example, a massless scalar field (in for simplicity special relativity). The Lagrangian is
    $$\mathcal{L}=\frac{1}{2} (\partial_{\mu} \phi) (\partial^{\mu} \phi).$$
    The action is obviously invariant under the scaling transformation
    $$x \rightarrow \lambda x, \quad \phi \rightarrow \frac{1}{\lambda} \phi,$$
    because then
    $$\mathcal{L} \rightarrow \lambda^{-4} \mathcal{L}, \quad \mathrm{d}^4 x=\lambda^4 \mathrm{d}^4 x.$$
    Using Noether's theorem you can show that the corresponding Noether current leads to
    i.e., the vanishing of the (covariant) trace of the canonical energy-momentum tensor.

    Another way to directly verify this is to calculate pressure and energy density of an ideal gas of massless particles. The energy-momentum tensor is given by
    $$\Theta^{\mu \nu}(x) = g\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi \hbar)^3} \frac{p^{\mu} p^{\mu}}{p^0} f_{\text{B/F}}(p^0), \quad p^0=\sqrt{m^2+\vec{p}^2}.$$
    Here, ##f_{\text{B/F}}## is the Bose or Fermi distribution, and ##g## some degeneracy factor counting intrinsic discrete quantum numbers like spin, isospin, flavor, color and the like.

    Since in this integral the particles are always on their mass shell, for ##m=0## you get indeed ##{\Theta^{\mu}}_{\nu}=0## because of ##p_{\mu} p^{\mu}=m^2=0##.
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