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What's the curvature of spacetime describe?

  1. Feb 23, 2009 #1
    Does the curvature of spacetime describe gravity or tidal force?

     
  2. jcsd
  3. Feb 23, 2009 #2

    atyy

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  4. Feb 24, 2009 #3

    A.T.

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    - Tidal force implies curvature of spacetime.
    - Zero tidal force implies flat spacetime.

    But with gravity it is more complicated:

    - Uniform gravity for example doesn't imply curved spacetime.
    - Zero gravity in the center of the earth doesn't imply flat spacetime.

    So you are only safe to say: The metric of spacetime describes both, gravity and tidal forces.
     
  5. Feb 24, 2009 #4
    Both....or neither... As AT posted:
    ....

    but I don't believe we have such in our universe....Things like planets/stars etc have both.....a uniform gravitational field is what Einstein imagined for his "equivalence" principle...say from an infinite plane of mass...but mass in our unviverse is like point masses, hence tidal effects.

    For a slightly different view:
    http://en.wikipedia.org/wiki/De_Sitter_relativity

     
    Last edited: Feb 24, 2009
  6. Feb 24, 2009 #5

    atyy

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    A.T. states the exceptions to the rule of thumb.

    In case you are interested, a specific cases of this point is the metric of a uniform spherical thin shell: http://arxiv.org/abs/gr-qc/0008065 (Eqn 1 only, rest of article not relevant).

    I guess an example of this might be metric of the Tolman-Oppenhemier-Volkoff equation, but I'm guessing, haven't actually checked what the curvature is at the centre: http://en.wikipedia.org/wiki/Tolman-Oppenheimer-Volkoff_equation.
     
  7. Feb 24, 2009 #6

    A.T.

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    A good rule of thumb might be this (please someone bright verify this):

    If the have a metric (like the http://en.wikipedia.org/wiki/Schwarzschild_metric" [Broken]), and you look at the derivates of the temporal component in regards to spatial coordinates then:
    1st derivate is related to gravity
    2nd derivate is related to curvature & tidal forces

    Obviously one can find metrics, where at certain space coordinates only one of the derivates is zero.
     
    Last edited by a moderator: May 4, 2017
  8. Feb 24, 2009 #7

    atyy

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    This isn't the rule of thumb, and I don't think it helps with the exceptions either.

    This is the rule of thumb that gravity=curvature=geodesic deviation=tidal forces. See A.T.'s post for exceptions to the rule of thumb.

    The usual way of motivating it is to compare the "local" and "non-local" lift experiments. In Carroll's words, we distinguish between experiments done in "small enough regions of spacetime" versus experiments "in a very big box". It is the non-local, "in a very big box" experiments that show up curvature and geodesic deviation of "true" gravity. http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll4.html

    Mathematically this corresponds to curvature and geodesic deviation being characterised by second derivatives, in the sense that higher order derivatives are more non-local than lower order derivatives (I think this is not too cheating a way to think about it, even though strictly speaking, all derivatives are local, since their limits at a point are well-defined). The relevant equations are Eq. 3.21 where the Christoffel symbol is defined in terms of the metric; Eq 3.67 where the curvature is defined in terms of derivatives of Christoffel symbols; and Eq 3.113 which gives the tidal forces in terms of the curvature. http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html
     
  9. Feb 25, 2009 #8

    Ich

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    That's true as long as gtt acts as the gravitational potential. Generally, you can extract curvature from the metric components, and "gavity" is not even well defined.
     
  10. Feb 26, 2009 #9
    If uniform gravity doesn't imply curved spacetime, how about Newtonian gravity?
    Dose Newtonian gravity imply curved spacetime or just curved time?
     
  11. Feb 26, 2009 #10

    atyy

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    Newtonian gravity itself can be reformulated "essentially exactly" as curved spacetime (Newton-Cartan theory). You can find more detail in Malament's http://arxiv.org/abs/gr-qc/0506065. I don't know whether "essentially exactly" is exactly enough that totally uniform gravity is possible in Newton-Cartan theory.

    Newtonian gravity is an excellent approximate description of certain regimes of GR. In those regimes, one can choose coordinates such that Newtonian gravity is the curvature of time. In GR, there is no such thing as totally uniform gravity. There are situations in which gravity is present, and large regions of spacetime are flat, but these flat regions are not infinite in size, and there must be curvature somewhere.
     
  12. Feb 27, 2009 #11

    A.T.

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    "Uniform gravity" is a theoretical special case of Newtonian gravity, where the Newtonian field is constant for each point of space. It exist only approximately in small regions of space.

    You cannot have only one dimension of a manifold intrinsically curved, so you cannot have "just curved time". But you can have a metric, where the distances along the time dimension vary depending on the space coordinate. This does not imply intrinsic curvature yet, but it implies gravity (and gravitational time dilation). http://www.physics.ucla.edu/demoweb...lence_and_general_relativity/curved_time.gif" Note that the cone doesn't have intrinsic curvature, an therefore the title is put in quotation marks.
     
    Last edited by a moderator: Apr 24, 2017
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