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I Schwarzschild Metric

  1. Jun 18, 2017 #1
    Is there a less boring way of deriving the Schwarzschild solution? The derivation itself is easy to going with; what I don't like is computing all the Christoffel symbols and Ricci tensor components --there are so many possible combinations of indices. I know that by using some constraint conditions one don't need to consider some of the Christoffel symbols, but still one end up with many equations.

    (I wonder how many time Schwarzschild himself took to derive that solution for the first time....)
    BTW, I've read that he was at a hospital at that time; good place to do calculus if you can.
     
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  3. Jun 18, 2017 #2

    Vanadium 50

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    "Less boring" seems awfully subjective.
     
  4. Jun 18, 2017 #3
    GR is one of the more computational fields of physics. Unless you have some physical intuition, this is what you'll be spending your time doing (or just use an algebraic package to do it for you...)
     
  5. Jun 18, 2017 #4

    PeterDonis

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    By "less boring" I assume, from your further statements about having to compute so many possible combinations of indices, that you mean something like "requires less computational drudgery". The best way I know of to do that is to first derive Schwarzschild coordinates, i.e., to demonstrate that any spherically symmetric spacetime can be described by coordinates ##(t, r, \theta, \phi)## in which the metric looks like this:

    $$
    ds^2 =j(t, r) dt^2 + k(t, r) dr^2 + r^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)
    $$

    Notice that we have already considerably reduced the possible combinations of indices, since the angular part of the metric is just the standard metric on a 2-sphere with radius ##r##, and as far as computing the two unknown functions ##j## and ##k## goes, you don't even have to worry about the angular part at all; in other words, the problem is effectively reduced to only two coordinates instead of four. Also, since there is no ##dt dr## cross term in the metric, the possible index combinations are further reduced. MTW takes this approach in their derivation; I think Carroll's notes also go into it.

    Once you have the metric as above, computing its Einstein tensor (you only need to compute the ##tt##, ##tr##, and ##rr## components, as noted above) and setting each component equal to zero (because you want a vacuum solution, with vanishing stress-energy tensor), enables you to find the two unknown functions, and to show that they turn out to be functions of ##r## only, not ##t## (the latter requires rescaling the ##t## coordinate). I wrote up this part in an Insights article some time ago (you can search on "A Short Proof of Birkhoff's Theorem" in the Insights section), but I think it's worth working through it yourself.
     
  6. Jun 19, 2017 #5

    vanhees71

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    Well, as Leibniz has put it, it's insulting the human mind to do "boring" purely technical calculations. Nowadays it's easy to get good computer-algebra programs (even for free) that do the job for you. I use Mathematica, but deriving the Christoffel symbols and the Einstein tensor from the usual ansatz (I prefer @PeterDonis 's in #4, because it's then also shown that a spherically symmetric pseudometric implies that it's static) involves only taking derivatives and sums and thus should be possible with any CA, and as I said, there are free ones (e.g., maxima).
     
  7. Jun 19, 2017 #6

    martinbn

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    If you want to see a different way, although you didn't say which way you found boring, you can look at Geroch lecture notes.

    If you want to compute curvature, then it is better to do the computations the Cartan way.
     
  8. Jun 19, 2017 #7
    vanhees71 good way of avoiding calculus at hand. By the way, I have tried to compute the Ricci tensor on Mathematica (for the Schwarzschild metric) and I got zero all times. Is it a coincidence for this particular metric that that tensor vanishes? It's not obvious at first that it has to be zero for the Einstein's equations to be satisfied, because there seems to be possible to get zero from the combination ##R_{\mu \nu} - \frac{1}{2}g_{\mu \nu}R## other than every component ##R_{\mu \nu}## (and consequently ##R##) being equal to zero.

    PeterDonis I think this is the most beautiful way of learning something, because as you pointed out, it's doing the calculations that one can discover more and more things that otherwise would be hidden in a software.

    I will try to check out these two. Thanks.
     
  9. Jun 19, 2017 #8

    PeterDonis

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    It isn't. The easiest way to see that is to take the trace of the equation.
     
  10. Jun 19, 2017 #9
    So that means that whenever ##T_{\mu \nu} = 0##, ##R_{\mu \nu} = 0 = R## as well?
     
  11. Jun 19, 2017 #10

    PeterDonis

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    Yes; again, you can see this for yourself by taking the trace of the vacuum field equation.
     
  12. Jun 19, 2017 #11

    George Jones

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    And then use the trace to find another useful form of the Einstein field equation,
    $$R_{\mu\nu} = 8\pi \left( T_{\mu \nu} - \frac{1}{2}T^\alpha_\alpha g_{\mu\nu} \right).$$
    The equation above shows that ##T_{\mu \nu} = 0## implies that ##R_{\mu \nu} = 0##, and the usual form of the field equation shows ##R_{\mu \nu} = 0## implies that ##T_{\mu \nu} = 0##. i.e., a vacuum if and only if ##R_{\mu \nu} = 0##.
     
  13. Jun 19, 2017 #12
    Can you show how to obtain the trace? I'm not getting it.
     
  14. Jun 19, 2017 #13

    George Jones

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    What do you get when you multiply both sides of
    $$R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = 8\pi T_{\mu\nu}$$
    by ##g^{\mu \nu}##?
     
  15. Jun 19, 2017 #14
    ##\frac{1}{2}R = 8 \pi T## ??
     
  16. Jun 19, 2017 #15

    George Jones

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    This is not what I get.

    What does ##g_{\mu \nu} g^{\mu \nu}## equal?
     
  17. Jun 19, 2017 #16
    For a diagonal metric in ##n## dimensional space, would it be ##g_{\mu \nu} \ g^{\mu \nu} = n##?

    Maybe then $$R = - 8 \pi T \ (1) \\ R_{\mu \nu} = 8 \pi\bigg(T_{\mu \nu} - \frac{1}{2}T^\alpha{}_\alpha \ g_{\mu \nu}\bigg) \ (2)$$ for then if we operate on ##(2)## with ##g^{\mu \nu}## we get ##(1)## again.
     
  18. Jun 19, 2017 #17

    George Jones

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    For the coordinate components of any (semi)Riemannian metric (not just for coordinate systems in which the components are diagonal),
    $$g_{\mu \nu} g^{\alpha \nu} = \delta_\mu^\alpha .$$
    Taking ##\alpha = \mu## gives
    $$g_{\mu \nu} g^{\mu \nu} = \delta_\mu^\mu = 4$$ (for spacetime.)

    Yes,
    $$\begin{align}
    \left( R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} \right) g^{\mu \nu} &= 8\pi T_{\mu\nu} g^{\mu \nu}\\
    R_{\mu\nu} g^{\mu \nu} - \frac{1}{2}Rg_{\mu\nu} g^{\mu \nu} &= 8\pi T\\
    R - \frac{1}{2}4R &= 8\pi T\\
    -R &= 8\pi T.
    \end{align}$$

    Substituting this for the scalar ##R## in
    $$R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = 8\pi T_{\mu\nu}$$
    gives the result.
     
  19. Jun 20, 2017 #18
    Thanks. The form of the equation ##-R = 8 \pi \ T## allow us to see that when ##T_{\mu \nu} = 0## and consequently ##T = 0##, ##R = 0## and consequently ##R_{\mu \nu} = 0##.
     
  20. Jun 20, 2017 #19

    George Jones

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    I am having a little difficulty understanding what you mean here, particularly the last part

    If all the components ##T_{\mu \nu}## are zero, then the right side of
    $$R_{\mu\nu} = 8\pi \left( T_{\mu \nu} - \frac{1}{2}T^\alpha_\alpha g_{\mu\nu} \right)$$
    is zero, and consequently the the left side, ##R_{\mu \nu}##, must also be zero.

    Similarly, if all the components ##R_{\mu \nu}## are zero, then the left side of
    $$R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = 8\pi T_{\mu\nu}$$
    is zero, and consequently the the right side, ##T_{\mu \nu}##, must also be zero.
     
  21. Jun 20, 2017 #20
    That's what I was trying to say.

    Now how do one shows that ##T - 2 T^{\alpha}{}_{\alpha}## is the same as ##T - 2T##? That is, why are we free to clear out the indices ##\alpha## appearing in ##T##?
     
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