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How did Einstein derive the meaning of Schwarzchild's metric

  1. Dec 1, 2014 #1
    I have been recently working with Schwarzchild's solution:

    ds2= - (1- (2GM/rc2))c2dt2 + dr2/(1-2GM/rc2) + r2(dθ2 + sin2(θ)d∅2)

    Now, when deriving the various general relativistic tensors for this metric such as the Ricci tensor, I found the calculations to be painfully tedious and monstrous (but not impossible). As I did these calculations, I thought about the fact that I was essentially working to derive a stress energy momentum tensor that I (with my current interpretation of the tensor) would probably not get as much information from as I should.

    After all, I remembered how I derived the stress energy momentum tensor for the Morris-Thorne wormhole and then didn't fully know what that tensor implied.

    I know the supposed meanings of the individual elements of the tensor. For instance, I know that T00 is energy density while the rest of this row and column are energy flux. Every other element is momentum flux.
    As you can see, I know this part.

    However, there is more to it than this. For example:

    We know that the Schwarzchild metric describes the space-time geometry around a spherical, non-electrically charged, non-rotating body. The stress energy momentum tensor however, does not have an element that tells you the shape of the body or system that generates the space-time curvature, nor does it tell you that it is static or non-rotating or non-electrically charged (although that last one can be inferred based on the fact that there are no electromagnetic terms in the derivation of this metric).

    How exactly did Einstein get this information? How does one know from a metric what shape the body is or whether or not it is rotating (and other information like this)?

    I have one hypothesis about this, and that is the fact that this metric is in a spherical basis. If that is not it, then please tell me where one gets this information.
     
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  3. Dec 1, 2014 #2

    Dale

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    Einstein didn't do it, Schwarzschild did.

    He assumed spherical symmetry and no charge or rotation in the derivation as well as assuming a vacuum solution. Those weren't things that were recognized afterwards, but deliberately put in there to begin with.
     
  4. Dec 1, 2014 #3

    ShayanJ

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    Schwarzschild calculated this metric for a spherical distribution of mass and informed Einstein about it in a letter. So actually Schwarzschild calculated the stress-energy tensor for such a distribution and solved EFEs for it and so all along the way, he knew that what he's going to get, is the metric of a space-time in the presence of a spherical distribution of mass. Einstein understood this when Schwarzschild told him about it in the letter. In fact Einstein was surprised by Schwarzschild's work!

    From here!
     
  5. Dec 1, 2014 #4
    I see, but how exactly did Schwarzchild know that this particular space-time interval embodied those assumptions?
    In other words, how did he know that in making his assumptions, that he didn't inadvertently put some rotation in the system?

    In essence, what I am asking is, how does one properly come up with an assumption of specific conditions for a metric?
     
  6. Dec 1, 2014 #5

    Nugatory

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    The good news is that when you find the stress-energy tensor for the Schwarzschild metric, you won't find it difficult to interpret. The bad news is that you will find it disappointing, except at ##r=0## where it is undefined.
     
  7. Dec 1, 2014 #6

    Matterwave

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    I am a little confused about your methods. The Schwarzschild metric is a vacuum solution of the EFE's. This means that both the stress-energy tensor and the Ricci tensor which you say in your OP that you calculated, are actually 0... The derivation of the Schwarzschild metric in fact starts from these facts.

    Why spend time calculating a 0?
     
  8. Dec 1, 2014 #7

    Dale

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    So he started with the assumptions of spherical symmetry, 0 stress energy tensor, and time independence (static). The spherical symmetry automatically excludes rotation since rotation has at most axisymmetey not spherical symmetry.

    The fact that the stress energy tensor is 0 precludes anything with a charge since that would have an E field and therefore a non zero stress energy tensor.

    Combining the above with time independence yields a family of solutions with a single parameter, so they can only differ by one quantity. This parameter is a characteristic length which is proportional to mass.
     
  9. Dec 2, 2014 #8

    PeterDonis

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    Just a note: Schwarzschild didn't realize it, but the time independence actually doesn't have to be assumed separately; Birkhoff's Theorem, which was proved in 1923, shows that spherical symmetry + vacuum alone is sufficient to specify the same family of solutions; the time independence comes "for free", so to speak, as a consequence of the theorem.
     
  10. Dec 2, 2014 #9

    Dale

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    Yes, this is correct. I was listing the assumptions used by Schwarzschild (to my understanding), not the minimal set of assumptions needed.
     
  11. Dec 2, 2014 #10

    pervect

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    A suggestion - finding out how to compute geodesics (i.e. orbits) in the Schwarzschild space-time from the metric might be a useful thing to look at. It would also not require the knowledge of vector spaces, covectors, and tensors that seems to be (best guess on my part) to be an obstacle for the OP.

    I'm not sure of the best approach, I believe there are undergraduate approaches based on the principle of maximal aging. Some of the algebra is a bit intensive, but I think it requires less differential geometry than trying to understand parallel transport and geodesics.
     
  12. Dec 2, 2014 #11

    Matterwave

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    Well, it's not so hard to set up the problem. You basically want to maximize the integral $$\tau=\int d\tau=\int \sqrt{g_{\mu\nu}\frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}d\tau$$ which will give you the parallel transport equation with the Christoffel symbols: $$\frac{dx^\mu}{d\tau}+\frac{dx^\rho}{d\tau}\Gamma^\mu_{\rho\sigma}\frac{dx^\rho}{d\tau}=0$$ From there I start forgetting the details, but you basically have a series of ODE's and you can use some conserved quantities (e.g. magnitude of four-velocity is constant) to make them a little more manageable. And then with some initial 4-velocity, you will be able to get the orbits.

    It's actually just a ton of plug and chug at many points of the calculation... I wish I remembered more details in the calculation so I could actually be of help if the OP were to try this though lol.
     
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