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Gravitational Waves and Potential

  1. Nov 9, 2008 #1
    What's the relationship between gravitational waves and (traditional/stationary) gravitational potential?

    Does the (traditional/stationary) potential become a "wave" under acceleration? Or do both exist under acceleration? Do both use energy from the source? I think I saw somewhere that GR predicts gravitational waves...true? Is this related in any way to the Unruh (acceleration) effect? Are these effects (waves, Unruh) evidence of the "absolute" nature of acceleration?

    (http://en.wikipedia.org/wiki/Bill_Unruh)

    Wiki says:
    http://en.wikipedia.org/wiki/Gravitational_Radiation

     
  2. jcsd
  3. Nov 9, 2008 #2

    tiny-tim

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    Hi Naty1! :smile:

    Here's another wiki link (about the decomposition of the Riemann curvature tensor into the Ricci tensor and the Weyl tensor) which may help: http://en.wikipedia.org/wiki/Ricci_decomposition#Physical_interpretation
    EDIT: and this link from John Baez's website:
     
    Last edited: Nov 9, 2008
  4. Nov 10, 2008 #3
    Tiny-tim...great descriptions to get me started...thanks..
     
  5. Nov 13, 2008 #4
    An electromagnetic wave that needn't have any source???

    How does one rationalize something like that?

    Regards,

    Bill
     
    Last edited: Nov 13, 2008
  6. Nov 13, 2008 #5
    will an accelerated observer also emits gravitational waves?
     
  7. Nov 14, 2008 #6
    It's a wave originating from other than the source description reflected in the Ricci tensor. Like an initial condition.
     
  8. Nov 14, 2008 #7
    Sure. Although extremely,extremely small as even massive bodies emit only small amounts of gravitational radiation (power).

    And an accelerating observer, in the absence of external gravity, also observes curved space time via the equivalence principle (I think). I think this is why special relativity (no gravity) requires inertial frames to keep space time observations flat.

    I still not quite sure whether an accelerating observer measures local light at "c". I believe I read there is a theory that basically says instantaneous readings while accelerating measure "c". These would apparently be inertial observers.
    But there are so many references which say "free falling observers" (non accelerating) in general relativity measure light at "c" that I am still suspicious...it may be a matter of context. I know Einstein came close to relying exclusively on free falling (local) frames and there had to be a good reason.
     
  9. Nov 14, 2008 #8

    tiny-tim

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    Weyl curvature tensor is not related to sources

    Yup! :smile:

    The source tensor (the stress-energy tensor) only has ten independent parameters,

    but the Riemann curvature tensor has twenty :eek:

    Einstein's field equations ("matter tells space how to curve") only relate the source tensor to the Ricci tensor, which of course also has ten parameters (natch :wink:) …

    the remaining ten parameters of the Riemann curvature tensor (in other words, the ten parameters of the Weyl tensor) are entirely independent of the source tensor, and therefore independent of all sources (assuming Einstein is right! :biggrin:)
     
  10. Nov 17, 2008 #9

    DrGreg

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    I'd just like to remind you of this reply I gave you in another thread. "Curvature of spacetime" is an intrinsic property of spacetime and does not depend on the observer. In the absence of gravity, spacetime is always "flat" whether you are an inertial observer or not. It's just that (to use the analogy of that other post) non-inertial observers draw a curved grid on flat graph paper.

    Non-inertial observers in the absence of gravity do indeed see some (but not all) of the effects that "stationary" observers see near black holes, but you shouldn't attribute those effects to "spacetime curvature" -- attribute them to a non-inertial frame instead.

    This is more a question of the established terminology than anything else.

    (For the benefit of other readers: as for your last paragraph, we are currently discussing that in another thread.)
     
  11. Nov 19, 2008 #10
    I just read your "this reply"...which was # 11 on the other thread .....don't believe I saw that ...excellent description and likely the source of my confusion.....

    I'll think about it and post tomorrow...
     
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