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I Newtonian limit of Schwarzschild metric

  1. Apr 20, 2016 #1
    If I am asked to show that the tt-component of the Einstein equation for the static metric
    ##ds^2 = (1-2\phi(r)) dt^2 - (1+2\phi(r)) dr^2 - r^2(d\theta^2 + sin^2(\theta) d\phi^2)##, where ##|\phi(r)| \ll1## reduces to the Newton's equation, what exactly am I supposed to prove?
     
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  3. Apr 20, 2016 #2

    PeterDonis

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    Compute the Einstein tensor for this metric and plug it into the Einstein Field Equation; take its 0-0 component; and show that the resulting equation reduces to Newton's equation.
     
  4. Apr 20, 2016 #3

    dextercioby

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    I think it leads to Poisson's equation for the gravitostatic potential. ## \nabla^2 \phi (\mathbb{r}) = -\rho ##
     
  5. Apr 21, 2016 #4
    I have got ##G_{tt} = - \frac{2(-1+2\phi)(\phi + 2 \phi^2 +r \phi')}{(r+2r\phi)^2}##
    How do I proceed from here? I am getting a first derivative of ##\phi## instead of second derivative.
     
  6. Apr 21, 2016 #5

    PeterDonis

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    Yes, this looks ok.

    Yes, but remember that the Einstein tensor has two pieces: ##G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R##, where ##R_{\mu \nu}## is the Ricci tensor, and ##R## is the Ricci scalar. So the EFE is really ##R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = 8 \pi T_{\mu \nu}##. You might try calculating the two pieces separately to see if there are second derivatives there.

    Also, you might take a look at Carroll's online lecture notes on GR, chapter 4, which has a discussion of this calculation.
     
  7. Apr 22, 2016 #6
    I don't understand how this would affect anything. Even if Ricci tensor and Ricci Scalar have second derivatives, what ultimately matters is this particular sum.
     
  8. Apr 22, 2016 #7

    PeterDonis

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    Not necessarily. Take a look at Carroll's notes. The short version: there is an alternate way of writing the EFE, which moves the trace term from the LHS to the RHS:

    $$
    R_{\mu \nu} = 8 \pi \left( T_{\mu \nu} - \frac{1}{2} g_{\mu \nu} T \right)
    $$

    For the case under discussion only the 0-0 component of this equation is significant, and only ##T_{00}## is significant in the trace ##T##.
     
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