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Energy-mom tensor does not determine curvature tensor uniquely ?

  1. Jun 1, 2007 #1
    Energy-mom tensor does not determine curvature tensor uniquely ???

    If the energy momentum tensor is known, that fixes the Einstein tensor uniquely from the Einstein eqs. Einstein tensor is built from Riemann contractions so it doesn't fix Riemann uniquely.

    Does that mean a single energy momentum tensor can generate spacetimes with different curvature tensors?
     
    Last edited: Jun 1, 2007
  2. jcsd
  3. Jun 1, 2007 #2
    A particular spacetime is the "sum" of all energy-momentum tensors.

    An interesting question would be: can we demonstrate or reason that an energy-momentum tensor pertains to a point in spacetime in general relativity.
     
    Last edited: Jun 1, 2007
  4. Jun 1, 2007 #3

    robphy

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    It's probably not a good idea to think of one side of Einstein's Equation "generating" the other side. Indeed, there are many vacuum solutions. In many situations, the metric tensor appears on both sides of the equation. There's more to say on this point. But I don't have the time right now.
     
  5. Jun 1, 2007 #4

    pervect

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    I think the electromagnetic analogy is useful. Suppose you specify the location and direction of motion of all charges, by specifying the current density J.

    Have you completely specified a unique solution of Maxwell's equations? The answer is no, because you haven't specified the boundary conditions.

    Physically, one can see this by thinking about adding or subtracting an electromagnetic wave from the solution. It doesn't change the charge density at all, but generates a different solution, i.e. a different set of elecltric and magnetic fields, a different Faraday tensor.

    So giving the energy momentum tensor to Einstein's equations is not enough in a similar manner to the way that giving the charge density and current density to Maxwell's equations is not enough - one also needs boundary conditions.
     
  6. Jun 1, 2007 #5
    I have a conceptual problem with the EM analogy. Boundary conditions are necessary if you want to specify the influence of the outside world on the boundary of your system. If your system is the whole universe and we accept that EM waves can't just pop out of nothing but require accelerating charges to create them, then specifying the position of all charges in the universe at all times should produce an unique solution for the EM field. You are not allowed to add extra waves anymore without the physical charges creating those waves present.

    That sounds like a paradox between physics (can't add waves without charges creating them) and math (can add waves without charges).

    In the same way I don't think that fixing the curvature tensor should require anything more than specifying the energy-momentum tensor at all spacetime points.

    It is true that there are many vacuum solutions of Einstein eq. corresponding to the same Tmunu = 0 like Schwarzschild and gravitational waves but all these must have sources and their Tmunu is not zero over the whole of spacetime. Mathematically you can have gravitational wave without any source or a Schwarzschild around a singular point but physically that doesn't make sense. Again the same paradox as in the EM case.
     
    Last edited: Jun 1, 2007
  7. Jun 1, 2007 #6

    pervect

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    Mathematically, your physical statement about waves is handled by the boundary conditions, however.

    Maxwell's equations have a number of different solutions - the way we prevent EM waves from "popping out of nothing" is by specifying the boundary conditions.

    We use the same mathematically tool in GR to solve the same issue - gravitational waves "popping out of nothing" is prevented by the tool of applying the boundary conditions.
     
  8. Jun 1, 2007 #7
    What exactly is boundary conditions in GR? Specification of the metric and first derivatives on the 'boundary' of the manifold?
     
  9. Jun 1, 2007 #8

    pervect

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    Wald (pg 252) say that an initial value formulation of GR can be specified with a spacelike hypersurface [itex]\Sigma[/itex], a 3-metric on [itex]\Sigma[/itex], and a symmetric tensor field [itex]K_{ab}[/itex] whih represents the extrinsic curvature on [itex]\Sigma[/itex].

    There are probably other ways of doing things, for instance you can apparently specify a Klein-Gordon field by [itex]\Phi[/itex] and [itex]\frac{\partial \Phi}{\partial t}[/itex] on some spacelike hypersurface.
     
  10. Jun 1, 2007 #9
    So where is the initial value formulation ala Wald in finding the Schwarzschild solution? The spacelike hypersurface must be one of the spheres. The 3 metric on it is the spherical metric in polar coordinates. What plays the role of K?
     
  11. Jun 2, 2007 #10
    The Schwarzschild solution is an exact solution, therefore there are no boundary conditions or initial values (except of course for the mass). But note that the complete Schwarzschild solution connects two asymptotically flat regions and that there are multiple topologies possible.

    Interesting here is that we can connect two or more Schwarzschild solutions (at infinity) at both of these asymptotically flat regions.
     
    Last edited: Jun 2, 2007
  12. Jun 2, 2007 #11
    This thread is about exact solutions. The boundary conditions have to select somehow between all the possible vacuum solutions and pick up Schwarzschild.
     
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