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Scalar field pressure and energy density

  1. Oct 3, 2006 #1
    Hi all,

    I'm hoping someone can help me out as I'm really stuck.

    With reference to the top of page 7 at http://faculty.washington.edu/mrdepies/Survey_of_Dark_Energy2.pdf

    I'd like to know how to get the quoted energy density and pressure of phi from the stress-energy tensor. Im very new to tensors and the notation involved. There are times I think I understand what is going on, but then I find I can't do simple problems, like get the pressure from the stress energy tensor.

    The way I'd get the energy density is by setting all indices in the stress-energy tensor to 0, but I'm not sure if that's correct?

    What would help me out massivly is a step by step way to get these answers (or point me to a site that explains how to get them, I've yet to find one). Once I understand this, I suspect a lot of other stuff I've been reading about will fall into place.

    Thanks in advance to anyone who can help :smile:
     
  2. jcsd
  3. Oct 4, 2006 #2

    hellfire

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    In the lecture notes it is applied the usual procedure to calculate the energy-momentum tensor: start with the Lagrangian of the scalar field and apply Noether's theorem to get its energy-momentum tensor.

    Having the expression for [itex]T_{\mu \nu}[/tex] (second formula in page 7) you will get the density as [itex]\rho = T_{00}[/itex] and the pressure as [itex]p = T_{11} = T_{22} = T_{33}[/itex]. You can assume that it is a perfect fluid, homogeneous and isotropic, and therefore [itex]\partial_1\phi = \partial_2\phi = \partial_3\phi = 0[/itex].
     
    Last edited: Oct 4, 2006
  4. Oct 4, 2006 #3
    I see,

    say i want to work out the pressure. On that second formula on page 7, do all the indices run from 0 to 3, or from 1 to 3. Or do alpha and beta run over a different number of indices from mu and nu?

    (edit) actually looking over it, I feel I'm missing something fundamental from this. If the metric is g = diag(-1, 1, 1, 1) I get: P = T(11) = T(22) = T(33) = -0.5*(d phi/dt)^2 - V
     
    Last edited: Oct 4, 2006
  5. Oct 4, 2006 #4

    hellfire

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    I think you just have to apply with care the second formula in page 7. Note that [itex]\alpha[/itex] and [itex]\beta[/itex] are the indices of the energy-momentum tensor and [itex]\mu[/itex] and [itex]\nu[/itex] are dummy indices that are summed over. Remember the condition of homogeneity and isotropy and also note that [itex]g^{\mu \nu} = g_{\mu \nu}[/itex] for g the Minkowski metric.

    [tex]T_{\alpha \beta} = \partial_{\alpha} \phi \partial_{\beta} \phi - g_{\alpha \beta} \left(\frac{1}{2} g^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi + V \right)[/tex]

    Start with the 00 term:

    [tex]T_{00} = \partial_{0} \phi \partial_{0} \phi - g_{00} \left(\frac{1}{2} g^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi + V \right)[/tex]

    In the sum over [itex]\mu[/itex] and [itex]\nu[/itex], only the 00 term is different from zero:

    [tex]T_{00} = (\partial_{0} \phi)^2 + \left(- \frac{1}{2} (\partial_{0} \phi)^2 + V \right)[/tex]
    [tex]T_{00} = \frac{1}{2}(\partial_{0} \phi)^2 + V[/tex]
    [tex]\rho = \frac{1}{2}(\partial_{0} \phi)^2 + V[/tex]

    The 11, 22 and 33 terms in the same way:

    [tex]T_{11} = \partial_{1} \phi \partial_{1} \phi - g_{11} \left(\frac{1}{2} g^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi + V \right)[/tex]
    [tex]T_{11} = - \left(- \frac{1}{2} (\partial_{0} \phi)^2 + V \right)[/tex]
    [tex]p = \frac{1}{2} (\partial_{0} \phi)^2 - V \right)[/tex]
     
    Last edited: Oct 4, 2006
  6. Oct 4, 2006 #5
    ahhhh, I get it now. Thank you so much!
     
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