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Does a Gravitational Field Contribute to itself

  1. Jan 1, 2007 #1

    -Job-

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    I think i came across this before. Does the size of an object's gravitational field contribute to its gravitational field?
     
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  3. Jan 1, 2007 #2

    pervect

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    The short answer is yes - the long answer involves a lot of cautions about what , if anything, the term "gravitational field" actually means.
     
  4. Jan 1, 2007 #3

    Chris Hillman

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    For the cognoscenti, I point out that an arXiv eprint by Costa and Herdeiro, "A gravito-electric analogy based on tidal tensors", offers a thoughtful critical review of some often mentioned mathematical analogies between gtr and EM (in particular, the GEM formalism). This is one of the few eprints I have seen which distinguish clearly between the electro(magneto)gravitic tensor (i.e. the electric(magnetic) parts of the Riemann tensor) and the electric(magnetic) parts of the Weyl tensor, and give some reasons why those in the know prefer the former.

    Pervect, you should be interested in their (8), which compares the EM Gauss law with a gtr analog, in which the trace of the electrogravitic tensor (taken wrt some timelike congruence)

    [tex]{R^a}_{man} \, U^m \, U^n = {E[\vec{U}]^a}_{a} = 4 \pi \, \left( \rho + 3 p \right) [/tex]

    aka the Raychaudhuri scalar, appears as the analog of charge density. Compare the expository eprint by Baez and Bunn, "The Meaning of the EFE".

    One of the many interesting points implied by this eprint concerns the question of which variable (metric, connection, curvature) should be taken as "the potential", "the field", etc., in analogy with EM. The answer of course depends upon how you are thinking about this. From the POV of this eprint, the answer would be "the metric" and "the connection", respectively. But these are a matrix-valued 0-form and 1-form, whereas from a more abstract POV we would expect 1-form and 2-form respectively, and thus would be impelled to answer "the connection" and "the curvature". Overall, this is one the clearest eprints I've seen making the larger point that in gtr, generally speaking, the short answer to most questions is "it depends".

    In todays pickings, gr-qc/0612189 will probably also be of interest. In the past year there has been a noticeable surge of interest in repairing the lack of a truly practical general theory treating elastic deformations in gtr. (Several have been offered over the years, but to date none have really caught on in the open literature, as far as I can tell.)
     
    Last edited: Jan 1, 2007
  5. Jan 1, 2007 #4

    pervect

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    I should probably explain my very terse answer a little more.

    If you look at Eisntein's field equations

    G_uv = 8 Pi T_uv

    the right hand side of the equations, T_uv, the "stress-energy tensor" contains only terms due to NON-GRAVITATIONAL fields, i.e it contains terms to to matter, and even electromagnetic fields, but that is all.

    How then, does the idea that "gravity causes gravity" make sense? It makes sense because the left hand side of the equation, G_uv, can be regarded as a non-linear differential equation (whose variables are the metric coefficients and various partial derivatives of the metric coefficients).

    It is the non-linear part of this equation that basically causes the appearance of "gravity" causing "gravity" when viewed from a linearized (weak field) perspective.
     
  6. Jan 1, 2007 #5

    pervect

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    Thanks Chris - I don't have time to read it at the moment, but http://arxiv.org/abs/gr-qc/0612140 looks very interesting.
     
  7. Jan 1, 2007 #6

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    So this is just an interpretation of the non-linearity of the left-hand side, and not like a static object's gravitational field grows over time due to this, right?
     
  8. Jan 1, 2007 #7

    pervect

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    Right - a static object's gravitational field does not grow over time. However, there is (to borrow a phrase from one of my textbooks) there is a "pit in the potential" (or effective potential) for the gravitational field of a single body represented by the Schwarzschild metric in GR. This can also be interpreted as "gravity gravitating".

    For instance, if we imagine a charged black hole that is charged just enough so that a test charge with some particluar charge/mass ratio experiences no force at infinity, when we bring that same test charge closer to the black hole, the test charge will experience a net attraction - the gravitational attraction will increase faster than the electrostatic repulsion.

    (Area * electric field (both measured locally) is constant for the electrostatic case, while area * force at infinty is constant for the static gravitational case.)
     
  9. Jan 2, 2007 #8

    Haelfix

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    Thats one way to look at it Pervect.. The other is to look at the right hand side, and note that even though it contains only matter fields, these fields do depend on the metric (or the coordinates).

    So gravity seems to creep its way into the right hand side as well, even if its naively hidden at first.

    Thats why when you solve the field equations you always have to run consistency checks as the field equations do possess a sort of back reaction or recursive character inside them.

    The whole thing is clearer if you look at the formalism with an Einstein Hilbert action. Its quite apparent that upon varying everything in sight, you will end up with pure gravitational terms.
     
    Last edited: Jan 2, 2007
  10. Jan 2, 2007 #9

    pervect

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    There's lots of ways of looking at it :-).

    It's clear that in GR, pressure causes gravity (pressure is part of the stress-energy tensor) and it's quite arguable that gravity causes pressure. Though one could also take the position that I took earlier as well, that pressure, being part of the RHS of the equation, isn't directly due to gravity but rather due to matter's attempt to fight gravity.

    If one does view gravity as causing pressure, then it's easy to see the feedback loop, and incidentally to get some feel for why gravitational collapse can't be resisted.
     
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