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A Assumptions of the Schwarzschild solution to the EFE

  1. Feb 17, 2016 #1
    Given that no assumption is of a point energy is necessary to derive the vacuum (Schwarzschild) solution to the EFE, why is the solution assumed to apply to spacetime surrounding a point energy?
     
  2. jcsd
  3. Feb 17, 2016 #2

    Dale

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    It doesn't have to be a point. It could be the vacuum outside any spherically symmetric distribution of matter.
     
  4. Feb 17, 2016 #3
    Sure, but point or otherwise, there is no need to assume ANY nearby energy to derive the vacuum solution. See below (from https://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution#Assumptions_and_notation)

    1. A spherically symmetric spacetime is one that is invariant under rotations and taking the mirror image.
    2. A static spacetime is one in which all metric components are independent of the time coordinate e358efa489f58062f10dd7316b65649e.png (so that 0f3b24a6ef8dee4954525b7379840c17.png ) and the geometry of the spacetime is unchanged under a time-reversal 77922a3584dc13ac8d79c428dd968c14.png .
    3. A vacuum solution is one that satisfies the equation 1c79950933f8776e47df1985aae41dad.png . From the Einstein field equations (with zero cosmological constant), this implies that 039270e362e40fa42563913afd67fa9c.png since contracting fb04e5bf20a01ac94b18db795e9cf41a.png yields bbac866fef4adae62451a4d889a21c4f.png .
    4. Metric signature used here is (+,+,+,−).
    Dispensing with the static assumption - Birkhoff's theorem[edit]
    In deriving the Schwarzschild metric, it was assumed that the metric was vacuum, spherically symmetric and static. In fact, the static assumption is stronger than required, as Birkhoff's theorem states that any spherically symmetric vacuum solution of Einstein's field equations is stationary; then one obtains the Schwarzschild solution. Birkhoff's theorem has the consequence that any pulsating star which remains spherically symmetric cannot generate gravitational waves (as the region exterior to the star must remain static).
     
  5. Feb 17, 2016 #4

    Dale

    Staff: Mentor

    None is assumed. You can have M=0.

    All you assume is spherical symmetry and vacuum. Those assumptions lead to a 1 parameter family of solutions. Those range from flat to varying degrees of curvature. The parameter has units of mass and is interpreted as the total mass inside the vacuum region and can be 0.

    It isn't an assumption, it is a derived result.
     
    Last edited: Feb 17, 2016
  6. Feb 17, 2016 #5

    PeterDonis

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    It isn't. See below.

    That's right; but this "vacuum solution" doesn't have to describe an entire spacetime. It can also describe a portion of a spacetime. For example, as Dale said, it could describe a vacuum region outside a spherically symmetric massive object. The full spacetime would consist of two regions, described by different solutions of the EFE: the Schwarzschild vacuum solution for the vacuum region, and a spherically symmetric non-vacuum solution for the matter region. As long as the proper conditions are met at the boundary between the regions (these are called "junction conditions" in GR), the combined solution is perfectly valid.
     
  7. Feb 19, 2016 #6
    On the RHS of the EFE, energy is rigorously defined as a rank 2 Tensor in 4D for all locations of the spacetime manifold. Then, in the vacuum solution, energy is used as a scalar that serves to make sense of a constant of integration in the derivation of the vacuum solution, 1/S.

    af13c98093b646d1865f5563c60c5615.png

    What is the relationship between the energy tensor and the energy scalar? What is the transformation?
     
  8. Feb 19, 2016 #7

    Dale

    Staff: Mentor

    More correctly that is called the stress energy tensor. All of its components are identically 0 in a vacuum solution, by definition.

    There isn't such a transformation. The stress energy tensor is 0 regardless of the value of S.
     
  9. Feb 24, 2016 #8
    By whatever name one uses--energy, stress-energy, stress-energy-momentum or energy-momentum--the tensor is zero in all its components in the vacuum solution. However, a scalar energy term is non-zero in the vacuum solution (associated as noted above the 1/S constant of integration term). Either there is a transformation that relates the stress-energy tensor to this energy scalar or the stress-energy tensor fails to capture the information embodied in this energy scalar term. If the former is true, then what is the transformation relating the stress-energy tensor to this energy scalar? If the latter is true, it implies an incompleteness to the stress-energy tensor and thus the EFE.

    As to the argument that the vacuum solution is local and that "non-local" energy might affect local curvature, presumably a complete stress-energy tensor would encode the "local" effects of "non-local" energy.
     
  10. Feb 24, 2016 #9

    PAllen

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    You can search these forums for many discussions of the following points:

    - the total energy of universe (including gravitational) is only well defined for asymptotically flat spacetime. The Schwarzschild solution is asymptotically flat, so it has a well defined total energy that can be associated with one free parameter of the solution. Cosmological solutions are not asymptotically flat and it is accepted that (pick your formulation, I prefer the first) : total energy is not defined; total energy is not conserved.
    - a tensorial quantity defining local energy including gravitational energy cannot be defined. This is actually mathematically proven in the sense that if you list the minimum properties you would expect for localization of energy, it can be proven that no such tensorial object can exist in GR. Some people define a non-tensorial, coordinate dependent object (pseudo-tensor) representing energy. The majority of GR practitioners don't find this useful, precisely because it is non-tensorial, and (as expected from this) has very odd properties except in special coordinates (e.g. harmonic coordinates).
     
  11. Feb 24, 2016 #10

    Dale

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    It fails to capture it. There is no relationship between the two, they are completely different things.

    I don't know how to be more clear about that.
     
  12. Feb 24, 2016 #11

    Dale

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    Nonsense. It just means that like any differential equation you have to include the boundary conditions. If you have ever used any differential equation then this should be a familiar concept.
     
  13. Feb 24, 2016 #12
    Can you give a good reference on your second point?
     
  14. Feb 24, 2016 #13

    PAllen

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    Landau and Lifshitz, Classical Theory of Fields is a good starting point on why you must use a pseudo-tensor, and what their properties are.
     
  15. Feb 24, 2016 #14
    Given that both the stress-energy tensor (obviously via the EFE) and the energy scalar--as I call it--(via the vacuum solution as a constant of integration, 1/S) both associate with the same thing, namely, spacetime curvature, it seems problematic to call them totally different things. Rather, they seem highly similar.

    A primary difference appears precisely in the boundary conditions of each implied by the assumptions of the vacuum solution. Specifically, the vacuum solution sets a boundary condition of zero for the stress-energy tensor and no boundary condition for the energy scalar.

    In this context, it seems more accurate to say that the vacuum solution to the EFE associates with two, not one free parameter, namely r and 1/S (the energy scalar).

    Again, all of this begs the question: given that both affect local spacetime curvature, what is the relationship between the stress-energy tensor and the energy scalar? Clearly, in the current formulation of GR, they are independent. However, on deeper consideration, I don't think the answer is at all obvious.
     
  16. Feb 24, 2016 #15

    PAllen

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    You have been given the answer several times the the SET of the SC solution is identically zero everywhere. How do you claim that connects to the mass parameter of the solution (which is NOT curvature scalar of any type). When the SET is zero, the type of curvature allowed is Weyl curvature, and there are curvature scalars that are non-zero for Weyl curvature (e.g. the Kreschmann invariant). However, Weyl curvature is not constant in the SC solution, so the Kretshcmann invariant is a function of position, not a single number.

    The way to recover the mass parameter from curvature (but NOT from the SET) in the SC solution is to do particular measures of aggregate curvature measured at infinity. Two of these are the ADM mass (computed at spatial infintiy) and Bondi mass (computed at null infinity). In the SC solution you have that ADM mass = Bondi Mass = mass parameter of solution. Note that ADM mass = Bondi mass implies the solution has no radiation. Also, that either of these being defined at all means asymptotic flatness. All of these are true for the SC solution but not, e.g. for any cosmological solution. None of these total curvature measures are defined at all for cosmological solutions.
     
  17. Feb 24, 2016 #16

    Dale

    Staff: Mentor

    I agree

    @redtree in the future you should probably not mark your threads as "Advanced" if you don't know the difference between a boundary condition and a source term.
     
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