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Metric outside a spherically symmetric source.

  1. Apr 22, 2010 #1
    I've been learning GR in my sparetime, and occationally I run into a conceptual problem that stalls my progress. Here is a question that has come up. I expect that this is a stupid question, but it's really bugging me, and an explanation will help me move forward more efficiently.

    If we wish to find the metric in the region outside a spherically symmetric source, the approach is to note that the energy-momentum tensor vanishes in the exterior and the Einstein field equation becomes:

    [tex]R_{\mu \nu}=0[/tex]

    So, I understand that a vanishing Ricci tensor does not imply a vanishing Rieman tensor ([tex]R_{\mu \nu \alpha \beta}[/tex]), and the exterior space need not be flat. However, it's not clear to me how the Einstein field equation can "know" about the source mass and generate the correct metric. It seems to me that the metric would be different between a nonrotating neutron star vs. a nonrotating planet, for example.

    So, my question is, what is the correct way for me to interpret this situation? If I solve for the metric with an equation that makes no reference to the source, how do I get the correct metric for a massive object that warps space, vs. a small object that barely warps space. My instinct says that the correct approach would be to match an exterior solution to the interior solution with boundary conditions, but I see no mention of this in the books I'm studying. I do see where a Newtonian limit is taken to bring the source mass back into the fold, so perhaps this is just an alternative way to get the right answer? Is this the correct interpretation, or am I missing something important here?
     
  2. jcsd
  3. Apr 22, 2010 #2
    you should consider the weak-field limit ...

    g00=1+2phi/(c^2)+...

    where phi is the Newtonian gravitational potential ...

    when M(mass) goes to zero then g00 goes to one and the metric would be Minkowski metric ....
     
  4. Apr 22, 2010 #3

    Mentz114

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    It has been done. You can splice a Schwarzschild interior and exterior spacetime. I have it in Stephani's 1984 book, and it must be covered in other books. Stephani remarks,

    "In interpreting the constant M, one must note that it is a measure of the total effective mass of the star ... and is in fact proportional to the coordinate volume, not the true three dimensional volume ... "
     
  5. Apr 22, 2010 #4
    which M do you mean ? ... the one that appears in Schwarzschild metric (1-2m/r ,geometric mass ) or the one that appears in the Newtonian gravitational potential ( the active gravitational mass ) ?....
     
  6. Apr 22, 2010 #5

    Mentz114

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    M is the Schwarzschild radius, 2m in your post. But that quote doesn't mean much out of context.
     
  7. Apr 22, 2010 #6
    Thank you both of your input. This is helpful indeed.

    I think the problem I'm having is that the use of the Newtonian limit, which brings in the critical source information, seems like it needs some justification when presented to someone first learning the subject. In hindsight (and I mean about 1 meter of hindsight), I can see the validity since this appears to be a static case, and the Newtonian theory of gravity is a static theory. However, the presentations I've read seem to make some implied assumptions without justification or even identification, and thus leave open questions.

    First, what is the general method that would work in all (or at least many) cases? Is the idea of matching boundary conditions, and/or solving in all regions, the general approach, and would always work in principle?

    Second, is the use of the Newtonian limit only valid for this static case? I would think so, but it would be good to know for sure as I try to move forward. For example, would a rotating source mass invalidate this method and require an approach which solves Einstein's field equation everywhere, not just exterior to the source?
     
  8. Apr 22, 2010 #7

    Mentz114

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    The only semi-rigorous derivation of 'linearized GR' I've seen is in Stephani and it's not straight forward at all. Some sources just let all the metric components go to 1 except g00 and then use 1-g00 as a potential. That is dodging the issue, I agree.

    Most people who look for solutions prefer vacuum solutions. There is even a class of spacetimes called Weyl vacuums, where the metric is defined in terms of two unspecified functions of the coords, and the conditions for the Einstein tensor to disappear is that the Laplacian of the function must be zero. In this case, one finds a vacuum solution by specifying crucial information about the source.

    The Kerr metric describes the exterior of a rotating spherical source. So the answer to your final question is 'no'. I understand your misgivings and had some robust discussions on this subject. The fact is that vacuum solutions exist and give good agreement with observations.

    Incidentally, the rotating source is still considered 'static' but not stationary. None of the metric coefficients depending on time is what makes a solution static. Compare with FLRW which is not static.
     
    Last edited: Apr 22, 2010
  9. Apr 22, 2010 #8

    nicksauce

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    Other way around. To quote directly from Carroll's GR book, "rotating systems that keep rotating in the same way at all times will be described by metrics that are stationary but not static".
     
  10. Apr 22, 2010 #9
    I consulted Carroll's book at the library this evening. It's a very good book and I will buy it for sure. Yes, he clearly states that the constantly rotating case is not static, but is stationary. Even though the metric does not have direct time dependence, apparently the nonzero position-time cross terms (i.e. the dt-dphi portion of the metric in this case) imply non-static behavior. I guess this is related to the frame-dragging that occurs.

    Anyway, I'm comfortable enough with the answers I been given to continue studying. I'm not ready to study the Kerr solution yet, and I'll have to wait to see if the Newtonian limit (in the vacuum solution) is used there. My gut says that the non-static case will not suit a Newtonian limit, but perhaps the frame dragging falls off at large distance and it still applies. There is still much for me to learn here, and I can revisit these concepts and questions in the future.

    Thank you to all for your help.
     
    Last edited: Apr 22, 2010
  11. Apr 23, 2010 #10

    bcrowell

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  12. Apr 23, 2010 #11
    This very much confuses me as well.

    The reason is this:
    Setup the lagrangian with the Einstein-Hilbert GR action term plus a non-interacting matter term (well, it still interacts via gravity, but I mean only that).
    It is my understanding that with varying the action we will find:
    1) Einstein's field equations
    2) The geodesic equation of motion for a particle

    Now, we can take the Newtonian limit to figure out what the constant in the Einstein-Hilbert action is. After that, the theory should be fully specified... no?

    Instead, when we solve Einstein's field equations (EFE), which are necessarily differential equations, and thus will have integration constants in the solutions, how do we determine what those constants should be?

    It seems wrong to be forced to turn to the Newtonian limit again to fix the solution. If this is required then that means the EFE + the geodesic equations are not enough!. What is missing?

    If instead there is another way to set those constants, and the Newtonian limit is just a convenient (equivalent and simpler) means to do this, then so be it. But I'd love to know how, if just in principle, one could set these constants using just the EFE. Since those constants of integration (in this vacuum case) don't contribute to the Ricci curvature ... and the EFE only constrain the Ricci curvature, then it seems (perhaps incorrectly) that the EFE wouldn't be able to say anything about those terms. Set the terms to anything we want and R_uv = 0.

    So, to echo elect_eng's comments. Yes, the need to use the Newtonian limit (after it has already been used to fix the "coupling constant" in the theory) very much bothers me too. Could someone more knowledgeable than I please help satisfy my curiosity here?
     
  13. Apr 23, 2010 #12
    As I read further in to this theorem, I have to say you are correct. This theorem is precisely the answer to one of my questions.

    This theorem was actually the next step in my study plan, and it is, in fact, in the books I'm reading. Hence, my statement "... the presentations I've read seem to make some implied assumptions without justification or even identification, and thus leave open questions" was unfair. I just did not recognize the significance of this theorem and how it related to one of my questions. Once again, thank you!
     
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