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Quick one-line on lowering indices

  1. Mar 1, 2015 #1
    tensorlowering.png


    At low speeds and assuming pressure ##P=0##,
    [tex]T^{\alpha \beta} = \rho U^\alpha U^\beta [/tex]
    [tex] g_{\alpha \mu} g_{\gamma \beta} T^{\alpha \beta} = \rho g_{\alpha \mu} g_{\gamma \beta} U^\alpha U^\beta [/tex]
    [tex]T_{\gamma \mu} = \rho U_\mu U^\beta g_{\gamma \beta} [/tex]

    Setting ##\gamma = \mu = 0##:

    [tex] T_{00} = \rho U_0 U^\beta g_{0 \beta} [/tex]

    Since ##g_{0 \beta} \backsimeq \eta_{0 \beta} ## and the only non-zero term is ##\eta_{00} = -1##, combined with ##U_\alpha U^\alpha = -c^2##:

    [tex] T_{00} = \rho U_0 U^0 g_{00} = \rho c^2 [/tex]

    I'm still learning tensor calculus, would that be considered a proper derivation?
    Also, is ##g_{ij} \backsimeq \eta_{ij}## the reason why ##T_{ij} \backsimeq 0##?
     
  2. jcsd
  3. Mar 1, 2015 #2

    PeterDonis

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    It's correct as far as it goes, but you left out two things.

    First: why do you assume ##P = 0##? The place you got this from only said a time-independent source at low speeds; it didn't say zero pressure. What eliminates the pressure terms?

    Second: how do you know that ##U_0 U^0 = U_{\alpha} U^{\alpha}## for this case (which is what lets you substitute ##- c^2## for ##U_0 U^0##)? In general, ##U_{\alpha} U^{\alpha} = U_0 U^0 + U_1 U^1 + U_2 U^2 + U_3 U^3##. What eliminates the other three terms?

    No. The correct reason is connected to the questions I asked above.
     
  4. Mar 2, 2015 #3
    Kinetic energy ##\propto k_BT## is much less than rest mass energy

    [tex] P \backsimeq n k_B T << nMc^2 \backsimeq \rho c^2 [/tex]
    So ##P \backsimeq 0##.

    ##U_\alpha U^\alpha = -c^2## because of invariance when you evalue ##U = (c,0)## in the rest frame.
     
  5. Mar 2, 2015 #4

    ChrisVer

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    I also have a question, why do you take [itex]g_{\mu \nu} \approx \eta_{\mu \nu} [/itex]?
     
  6. Mar 2, 2015 #5
    In the weak field limit of gravity, ## g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} ## where ##|h_{\mu \nu} << 1##.
     
  7. Mar 2, 2015 #6

    ChrisVer

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    aha, so you take the weak field limit...because I didn't see that in written in the text you attached. I would just say that in the low speed limit:
    [itex] U^a = \begin{pmatrix} c \\ u_x \\ u_y \\ u_z \end{pmatrix} \approx c \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} [/itex]
    and this gives immediately the components of [itex]T^{ab}[/itex]
     
  8. Mar 2, 2015 #7

    PeterDonis

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    Not in all cases. I was asking what makes it true in this case.

    I wasn't asking why ##U_\alpha U^\alpha = -c^2##. I was asking why ##U_0 U^0 = - c^2##, i.e., why ##U_0 U^0 = U_\alpha U^\alpha##. I think ChrisVer gave the answer to that, though.

    Small nitpick: in the low speed limit, and in the rest frame of the fluid...
     
  9. Mar 2, 2015 #8

    ChrisVer

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    wouldn't you agree that in the low speed limit this approximation holds (I just used [itex]u_i/c \approx 0[/itex])? The thing is that the text says "taking the low speed limit" (together with time independence of the source)
     
  10. Mar 2, 2015 #9

    PeterDonis

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    Sure, but "low speed" requires a choice of frame; "speed" is relative. I was only pointing out that the implicit choice of frame being made is the rest frame of the fluid. I said it was a small nitpick. :wink:
     
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