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Deriving the Metric from the Energy-Momentum Tensor

  1. Dec 13, 2012 #1
    Say we were given an expression for the energy-momentum tensor (also assuming a perfect fluid), without getting into an expression with multiple derivatives of the metric, are there any cases where it would be possible to deduce the form of the metric?
     
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  3. Dec 13, 2012 #2

    PAllen

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    I assume you mean stress energy tensor(?)

    I don't see how. Even the case of stress-energy tensor = 0 => vacuum, produces non-trivial system of second order partial differential equations for the metric. The only case, even for vacuum, that is simple to derive, is if you add spherical symmetry. Axial symmetry is already much harder, even for vacuum.
     
  4. Dec 13, 2012 #3

    Bill_K

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    :confused: All you have to do is solve Laplace's Equation.
     
  5. Dec 13, 2012 #4

    K^2

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    Laplace equation has nothing to do with gravity. Einstein Field Equations are much more difficult to solve.

    You can find a numerical solution in some cases. Even then, you almost always have to rely on symmetry of some sort, simply because of dimensionality. Say you want 100 grid points per dimension. This would already require you to solve the equations numerically over 108 grid points. If your stress-energy tensor is time-independent, and you consider only steady-state solutions, then you can assume metric to be time-independent, and that reduces it to 106 points. Much more reasonable, but still extremely difficult to work with due to non-linearity. And how many points are sufficient to get a convergent and useful solution will depend on specific distribution of stress-energy density, and again, since it's non-linear, it might be hard to predict in advance.
     
  6. Dec 13, 2012 #5

    PAllen

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    The simplest axial symmetric vacuum case I know of is the Kerr metric. The derivations of I've seen are far more complex than for spherical symmetry.
     
  7. Dec 13, 2012 #6

    PAllen

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    Right, and I was addressing: "without getting into an expression with multiple derivatives of the metric", for which I still can't think of any answer other than: No.
     
  8. Dec 13, 2012 #7

    K^2

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    Have to agree on that one. Trying to solve differential equation without taking derivatives seems like a fool's quest under best of circumstances.
     
  9. Dec 14, 2012 #8
    outside a given source,i.e. if energy momentum tensor is zero for a region with some specified symmetry(say spherical).one can guess the form of metric but with given energy momentum tensor one has to go with what others have said.
     
  10. Dec 14, 2012 #9

    Bill_K

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    The general nonrotating time-independent axially symmetric vacuum solution to Einstein's Equation is given very simply by a solution of Laplace's Equation. This is a standard topic covered in any introductory course in General Relativity. For example, see here.
     
  11. Dec 14, 2012 #10

    PAllen

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    I was thinking only of the general case - nothing assumed but symmetry.. In the spherical symmetry case, you get static for free. In the axial case you don't. However, if you add staticity + (axial) symmetry, that obviously simplifies it a lot.
     
  12. Dec 14, 2012 #11

    Bill_K

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    The Weyl family of exact solutions were among the earliest found. Just as for Schwarzschild, in the derivation of the Weyl solutions there is nothing assumed but symmetry, namely the two Killing vectors that generate axial rotations and time translations.

    Even for the stationary case there are a great many exact solutions known. Kerr turns out to be just the simplest member of an infinite family of solutions.
     
  13. Dec 14, 2012 #12

    PAllen

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    But time like killing vector is the extra symmetry. For spherical symmetry you don't need to assume it, you discover it without assuming it. For the axial case, that is not so. You have to assume it, and it greatly simplifies the solution.
     
    Last edited: Dec 14, 2012
  14. Dec 14, 2012 #13

    George Jones

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    Is the interior of a Schwarzschild black hole spherically symmetric? Does this spacetime have a timelike Killing vector?
     
  15. Dec 14, 2012 #14

    PAllen

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    No, but it has an extra killing vector which you don't have to assume; it can be derived only assuming spherical symmetry. In the interior, the extra killing vector happens to be spacelike.
     
  16. Dec 15, 2012 #15
    I have a question about the form of the metric shown on page two of the link. It appears that the time dilation there is set equal to the tangent length contraction as inferred by a distant observer as a coordinate choice for the metric form. But if that is the case, then upon referring back to Schwarzschild coordinates, where the invariants should remain the same, with limits working toward a finite surface area, zero time dilation, and infinite local acceleration for a static observer very near the event horizon, with the Weyl solution, it would instead work toward limits of infinite surface area at the place with zero time dilation at any r>0 and with zero local acceleration at any r>0 also. The only remedy that I can see so far, at least there is one, is if with the Weyl solution, the event horizon can only be at r = 0 always, so never with any mapped interior coordinates. Only then can the invariants work toward a finite surface area and infinite acceleration there. Is that right?
     
    Last edited: Dec 15, 2012
  17. Dec 15, 2012 #16
    When one assumes spherical symmetry, what are they really assuming? Is that inferred by a distant observer's coordinate system? If so, then it seems to me that we can always assume spherical symmetry in a relativistic universe just by taking all of the points around a mass (although the mass would have to be spherically symmetric initially and collapse uniformly) that have the identical invariants and map them out to some spherical r according to the distant observer. Otherwise, the universe would be "lopsided" in some way by some influence external to the mass that we do not account for with Relativity, much for the same reason that two inertial observers in a purely relativistic universe, both applying the same synchronization of their clocks, will measure the same speed 'v' of each other, correct?
     
  18. Dec 15, 2012 #17

    PAllen

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    We are speaking here of pure vacuum solutions. There are coordinate independent ways to state spherical symmetry, but it does suffice to use coordinates. For a give 4-manifold with everywhere vanishing Ricci curvature but non-vanishing Weyl curvature, either it is possible to introduce coordinates such that the metric expressed in those coordinates takes a certain form or it is not. It turns out that there is exactly one maximally extended pseudo-Riemannian 4-manifold with vanishing Ricci curavature and non-vanishing Weyl curvature that admits a coordinate system manifesting spherical symmetry (up to a single parameter, that can be interpreted as mass). For axial symmetric vacuum, there are both static and non-static solutions, and even the static solutions cannot be characterized by a simple parameter choice.
     
  19. Dec 16, 2012 #18

    atyy

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    Doesn't the energy-momentum tensor contain the metric?
     
  20. Dec 16, 2012 #19

    Nugatory

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    No. They are related by the Einstein Field Equation, which is a set of differential equations for the metric in terms of the stress-energy tensor, but the EFE alone does not contain enough information to completely specify the metric. Like any reasonable differential equation, the EFE gives you a family of solutions; the boundary conditions are required to know the specific member of that family that describes the physical situation at hand.
     
  21. Dec 16, 2012 #20

    atyy

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    How about Eq 8.15 of http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll8.html? Isn't that the metric in the stress-energy tensor?
     
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