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Question about gravity waves energy and momentum

  1. Jun 29, 2012 #1
    where in the Einstein field equations is the energy and momentum of the gravity waves themselves Rμγ- 1/2gμγ R = 0 doesn't mean the space is flat there are waves in the fabric of spacetime itself so how and where does the energy and momentum of these waves manifest itself Tμγ is the energy density or the matter density so the answer must lie on the right hand side of the equation but where?
  2. jcsd
  3. Jun 29, 2012 #2


    Staff: Mentor

    In so far as the energy and momentum of gravitational waves is reflected in the EFE, it is on the LHS of the Einstein Field Equation, not the RHS. Gravitational waves can be present in regions of spacetime where there is zero stress-energy, i.e., the RHS of the EFE is zero. GWs are really just dynamic fluctuations in the curvature of spacetime, and everything related to the curvature of spacetime is on the LHS of the EFE.

    The reason for the qualifier in my first sentence above is that (I believe) GWs are actually a form of Weyl curvature, not Ricci curvature, and only Ricci curvature appears on the LHS of the EFE. So it's possible (at least in principle) to have a solution of the EFE with gravitational waves present, where nothing relative to the energy and momentum carried by the waves appears on the RHS of the EFE, at least in some region of spacetime.

    The Wikipedia page on gravitational waves, here...


    ...and the Usenet Physics FAQ entry on energy conservation in GR, here...


    ...have good information that's relevant to this subject.
  4. Jun 29, 2012 #3
    Thank you very much I appreciate your response and information
  5. Jun 29, 2012 #4


    Staff: Mentor

    You're welcome. Also I just noticed this is your first thread here, welcome to PF!
  6. Jun 30, 2012 #5
    thank you very much....another question I know that for example that the Ricci tensor....is from the contracted Riemann Tensor via the metric...but what is the Weyl tensor and how do the Ricci and Weyl curvature differ? Thanks!
  7. Jun 30, 2012 #6
    The Weyl tensor contains the "other" information which is lost when contracting the Riemann tensor to form the Ricci tensor. The Ricci tensor contains the information about how the volume of a body changes under tidal forces, and the Weyl tensor contains the information about those tidal forces and how they change the shape of a body. The information in the Weyl tensor + the information in the Ricci tensor = the information in the Riemann tensor.

    Where the Ricci tensor is the trace part of the Riemann tensor, the Weyl tensor is the traceless part (so contraction on any of its indices yields zero).
  8. Jun 30, 2012 #7


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    Science Advisor

    Gravitational waves have an "effective" stress energy tensor. This appears if you linearize the highly nonlinear equations - it's rather similar to the shift in the (linearized) operating point that you get if you apply AC signals to a non-linear electric circuit.

    The actual stress-energy tensor in a region containing no matter is still zero, however, even when the effective tensor is non-zero due to the presence of gravitational waves.

    This is one of many and various issues that's related to the problem of defining energy in GR. When you talk about "the" energy density of gravitational waves, you are implicitly assuming that it's well defined. There are multiple definitions of energy in GR, so it's not so well defined. Furthermore, none of the definitions that we do have allow the energy of the gravitational field to be given a specific location.
  9. Jul 1, 2012 #8


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    Yes, and as John Baez says in his article The Ricci and Weyl Tensors the tidal stretching and squashing caused by gravitational waves would not change the volume as there is 'only' Weyl- but no Ricci-curvature.

    But, beeing not an expert I still have no good understanding, what Weyl-curvature really means and would appreciate any help.

    If I think of spacetime curvature I have effects like Shapiro-Delay/time dilation/the sum of light ray triangles etc. in my mind. But it seems that the Weyl curvature is not responsible for anything else than the tidal effects happening in the x-y-plane of the wave. Is that right? Perhaps it is sufficient to say, a plane is not curved. The MTW talkes about the plane-wave solution.

    Otherwise gravitational wave measurements should be obscured by Shapiro-Delay/time dilation to a certain extent.

    On the other side, as the gravitational wave propagates, it takes energy with it, a part of which can be transformed into work, if I remember correctly. How should I think of this energy, as it doesn't curve spacetime in the above mentioned way?
  10. Jul 1, 2012 #9
    thank you for your input I appreciate it
  11. Jul 1, 2012 #10
    thank you very much
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