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Vacuum solutions, and Explicit source terms

  1. May 9, 2010 #1
    This is a question about solving Einstein's Field Equations for some simple cases, but I'm not sure how to explain this without analogy. So bear with me for a brief discussion.

    Consider the electrostatic equation:
    [tex] \nabla^2 V = \rho / \epsilon_0[/tex] which show explicitly how the charges source the scalar field.
    Or we can solve for the vacuum solutions easier, which are of the form
    [tex] \nabla^2 V = 0[/tex].

    Now, if for some reason we knew how the fields should fall of towards infinity, one can use this "boundary" condition plus our vacuum solution to get a full solution without even dealing with the source term. But I wouldn't find this very satisfactory without knowing how to solve for the equations with the source term explicitly at least for some simple cases.


    Turning now to GR, we can start with a simple form of the Hilbert action depending on the matter terms + the Ricci curvature scalar times some constant. Vary with respect to the metric and we get Einstein's field equations. To fix the constant, we can take the Newtonian limit and relate it to G. Now that we've fixed that constant, there are no "free" parameters left in the field equations. We shouldn't ever need to touch Newton's law again to solve these equations.

    However, in all derivations I've seen of the metric outside a static stationary spherical mass, to fix some integration constants, they again use the Newtonian limit. More-over, it is difficult to even define what is meant by mass in GR, and one of the more useful definitions (Komar mass) again refers to the vacuum region. So most of the simple solutions never refer to the actual source term in GR! They solve / define those integration constants some other way.

    I'm really having trouble with this. It almost seems like the source term doesn't matter. I mean, if there was a theory of gravity which only gave vacuum solutions of the form [itex]R_{\mu\nu} = 0[/itex] just like GR (without a cosmological constant), could we even experimentally distinguish it from GR? I really feel the answer should be yes, but since the means of fixing the integration constants above never even refer to the source term ... it seems to point to no. So what is going on here? Since we can't even say (in general) what the mass content in a specific spacetime region is in GR, does this make GR's equivalent of the Poisson equation no more useful than its equivalent of the Laplace equation? This seems to me like if in EM we couldn't tell the difference between [itex] \nabla^2 V = \rho / \epsilon_0[/itex] or [itex] \nabla^2 V = a(\rho / \epsilon_0)^3[/itex].

    ---------------
    EDIT:
    Alright, said more succinctly in one example. How do I know this isn't a solution to Einstein's Field Equations? (units such that c=1)
    [tex]{d \tau}^{2} =
    \left(1 - \frac{G M}{r} \right) dt^2 - \frac{dr^2}{\displaystyle{1-\frac{G M}{r}}} - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)[/tex]
    It satisfies the vacuum equations. But it claims an event horizon doesn't form till r=GM instead of r=2GM.
     
  2. jcsd
  3. May 9, 2010 #2

    Jonathan Scott

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    As you said before, as far as I know, the boundary conditions are determined by matching Newtonian theory (which the above alternative form clearly does not).

    That's not very satisfactory, and as you have probably noticed before on these forums, I'm not even sure whether that has been done correctly (in that Schwarzschild and Hilbert apparently assumed physically different boundary conditions, both reducing to Newtonian at a distance, as pointed out by people such as Leonard S Abrams and Salvatore Antoci).
     
  4. May 9, 2010 #3

    atyy

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    Isn't it done by using Birkhoff's theorem, which says you have the Schwarzschild form outside the mass? Then to see how the Schwarzschild solution "takes over" from the non-vacuum solution inside the mass, one matches boundary conditions at the edge of the mass.

    Say something like chapter 10 of Schutz?
     
    Last edited: May 9, 2010
  5. May 9, 2010 #4

    bcrowell

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    I think atyy hit the nail on the head about Birkhoff's theorem.

    Cosmological solutions are directly testable by observation, and they depend on the source terms.
     
  6. May 9, 2010 #5

    Stingray

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    That is a solution to Einstein's equation. The point of looking at Newtonian theory (a second time) in examples like this is to find a physical interpretation for the integration constant M. You've written down Schwarzchild's spacetime with mass M/2, but the word "mass" is not necessary to use the solution.

    If you have a star with a smooth interior, the parameter appearing in the exterior vacuum solution is determined by an integral of the interior's stress-energy tensor. It is still helpful (though not necessary) to go to the Newtonian limit to see what the resulting formula reduces to.

    More generally, there is a lot that can be determined purely from the vacuum equation plus falloff assumptions. Unlike in electrostatics, you can even learn a lot about the equations of motion.

    Despite this, there are a lot of things that do require the source terms. Most of the examples that come to mind involve stellar structure: It helps you find the mass-radius relation of a star with given composition. It is also required to find various internal instabilities (some of which may have strong gravitational wave signatures when triggered). You can ask how much a star bulges for a particular angular momentum, or even what the near-field vacuum solution should be near a rapidly-rotating star (it isn't Kerr!). You can find how much a star bulges if another comes near it. Also, how efficient is tidal locking in this scenario?

    GR turns out to predict that the internal structure of a star matters very little for its overall motion. Most alternatives that have been proposed are much more sensitive.
     
  7. May 18, 2010 #6
    Unless I'm misunderstanding, Birkhoff's theorem actually just shows the most general form of the solution. To interpret the constants, one still needs to turn to something else.

    I looked up the solution in chap 10 of Schutz as you suggested, but it just defines the mass to be so. (The definition is presented in eq 10.28). Right after presenting it, he too makes arguments in reference of Newtonian gravity.

    And again the solution in no way referred to the "coupling constant" G from the source term, until just defining what the "integration constants" of the solution should be. Is there any place that actually works this out directly using the source term? Or maybe someone already answered my question and I'm missing something important here?


    To make a more concrete example:
    Consider Einstein-Cartan gravity. It has the same vacuum constraint as normal GR, that the Ricci curvature tensor is zero. But the source term is different. How then, does this affect the vacuum solutions? Since the GR solutions seem to avoid the source term and fix the "integration constants" by other means, I don't know how to even approach this.

    Can you help me see how to do this?
    What I'd like to study are line sources. I've seen these done with minkowski background with angle deficits around one axis. They then state that the angle deficit is linearly proportional to the mass or energy per unit length of the line source. How do I actually work this out? If someone could show me how to actually solve the source equations instead of just making assumptions about integration terms in vacuum solutions using Newtonian limits ... it would be much appreciated.
     
    Last edited: May 18, 2010
  8. May 18, 2010 #7

    Stingray

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    This is done for spherical stars in most textbooks. See, e.g., section 6.2 of Wald or chapter 23 of MTW.

    There's been a lot of work on the interpretation of various parameters in the cylindrically symmetric vacuum solutions to Einstein's equation. This is often done by providing a matter source that has the appropriate exterior solution. One example seems to be http://iopscience.iop.org/0264-9381/18/18/305".
     
    Last edited by a moderator: Apr 25, 2017
  9. May 18, 2010 #8
  10. May 18, 2010 #9

    atyy

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    I'm thinking of Eq. 10.41, which defines the "mass parameter" of the Schwarzschild solution in terms of matter properties, ie. take naming M the "mass parameter" to be unjustified until Eq 10.41.
     
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