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A GR index gymnastics -- Have I misunderstood something or typo?

  1. Jun 5, 2017 #1
    Hello there,

    I am learning GR and in the cosmology chapter, we are using the metric
    $$
    ds^2 = - dt^2 + a^2(t) \left[ \frac{dr^2}{1 - \kappa r^2} + r^2 d \Omega \right].
    $$

    Suppose now that ##U^\mu = (1,0,0,0)## and the energy momentum tensor is
    $$
    T_{\mu \nu} = (\rho + p)U_\mu U_\nu + p g_{\mu \nu}.
    $$

    My professors says that this implies ##T^{\mu \nu} = \mathrm{diag}(\rho, p,p,p)##. Surely this is a typo, or is it just me that dont understand how to raise indecies? Thanks
     
  2. jcsd
  3. Jun 5, 2017 #2

    PeterDonis

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    What do you get when you try to raise the indexes on ##T_{\mu \nu}##?
     
  4. Jun 5, 2017 #3
    $$
    T^{\mu \nu} = g^{\mu \rho}g^{\nu \sigma} T_{\rho \sigma} = g^{\mu \rho}g^{\nu \sigma}\left[(\rho + p)U_\rho U_\sigma + pg_{\rho \sigma} \right] = (\rho + p)U^\mu U^\nu + p g^{\mu \nu} =
    \begin{bmatrix}
    \rho + p &0&0&0\\
    0&0&0&0\\
    0&0&0&0\\
    0&0&0&0
    \end{bmatrix}^{\mu \nu} +
    \begin{bmatrix}
    -p &0&0&0\\
    0&p\frac{1 - \kappa r^2}{a^2}&0&0\\
    0&0&p \frac{1}{a^2r^2}&0\\
    0&0&0&p \frac{1}{a^2r^2\sin^2\theta}
    \end{bmatrix}^{\mu \nu}
    =
    \begin{bmatrix}
    \rho &0&0&0\\
    0&p\frac{1 - \kappa r^2}{a^2}&0&0\\
    0&0&p \frac{1}{a^2r^2}&0\\
    0&0&0&p \frac{1}{a^2r^2\sin^2\theta}
    \end{bmatrix}^{\mu \nu} \neq \mathrm{diag}(\rho,p,p,p)^{\mu \nu}
    $$
    Is this correct, or is my professor right and I wrong?
     
  5. Jun 5, 2017 #4

    PeterDonis

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    You are right if you use the metric you wrote down. I suspect, though, that your professor was implicitly using a different metric when he made his claim--one that is only valid in a small patch of spacetime around a given event. Are you familiar with the concept of local inertial coordinates in GR?
     
  6. Jun 5, 2017 #5
    Inertial coordinates are free fall coordinates, the coordinates of a freely falling observer?

    This is in the context of deriblant the Friedmann equations, so surely means comoving coordinates?
     
  7. Jun 5, 2017 #6

    PeterDonis

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    Local inertial coordinates are coordinates in which a chosen free-fall observer is at rest, and in which the metric is the Minkowski metric. But in a curved spacetime, such coordinates can only cover a small patch of spacetime centered on some particular event. They are not the same as global comoving coordinates.
     
  8. Jun 5, 2017 #7
    This is in the context of deriving the Friedmann equations, so surely we are using comoving coordinates?
     
  9. Jun 5, 2017 #8

    PeterDonis

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    Can you give more specific references, such as the textbook you are using?
     
  10. Jun 5, 2017 #9

    PeterDonis

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    As a general comment, the Friedmann equations are usually derived by looking at the 0-0 component of the Einstein Field Equation (which doesn't bring in the terms involving ##p## that you are finding to differ from your professor's claim), and the trace of the Einstein Field Equation--which makes all of the extra factors involving the metric coefficients cancel out. So it's also possible that your professor was implicitly referring to taking the trace when he talked about ##\mathrm{diag}(\rho, p, p, p)##.
     
  11. Jun 5, 2017 #10
    Carroll -- Spacetime and geometry. Page 333
     
  12. Jun 5, 2017 #11

    PeterDonis

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    I don't have the actual book, but I'm familiar with the online lecture notes of his that the book was based on. Equation (8.18) in Chapter 8 of those notes says:

    $$
    T^\mu{}_\nu = \mathrm{diag} \left( - \rho, p, p, p \right)
    $$

    Notice two key differences: only one index is raised, and the sign of ##\rho## is flipped. Could this have been what your professor was actually saying?
     
  13. Jun 5, 2017 #12
    No, then I wouldn't have needed to ask. Thank you for your help :)
     
  14. Jun 5, 2017 #13

    Orodruin

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    It is possible your professor is implicitly using a tetrad, i.e., an orthonormal basis, instead of a coordinate basis.

    Edit: Also, note that comoving coordinates are not free fall coordinates. Comoving observers in a local inertial coordinate frame will move relative to each other at a velocity proportional to the (comoving) distance between them, i.e., Hubble's law.
     
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