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Force in General Relativity

  1. Sep 3, 2006 #1
    "Force" in General Relativity...

    Is there a way to define via Geommetry or other concept the quantities we call "Force" and "potential"?..in the same that as it happens with Newtonian mechanics..

    [tex] \dot \pi_{ab}= -\nabla U_{ab} [/tex] ?

    or even simpler [tex] Force= \frac{\delta L}{\delta g_{ab}} [/tex]ç

    And What would be the meaning of "Force"?..is perhaps what makes the two close geodesic to diverge or to "twist" the Space-time?...
  2. jcsd
  3. Sep 3, 2006 #2


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    The geometric concept of force is the 4-force, sometimes known as the Minkowski force. You should be able to read about it in the Wikipeda or any book on SR that uses 4-vectors. (You might want to read about 4-vectors, too).

    Basically, the 4-force is the derivative of the energy-momentum 4-vector with respect to proper time.

    This is a SR concept. It imports to GR, though, as do most SR concepts. Basically, worldines that follow geodesics have zero 4-accelerations. 4 acceleration can be defined as the covariant derivative of the 4-velocity. The total force on a system is its mass times its four-acceleration. Strictly speaking this import applies only to zero volume systems (point particles) due to concerns about where and when the force is acting, but this shows up as a minor observer-dependence in the value for "mass" in the above equation, which can usually be ignored for reasonble forces and small volumes (i.e. anything that acts remotely like a rigid body).

    Electromagnetic fields have a 4-potential, the wiki article on it is pretty good assuming it hasn't been vandalized recently


    There isn't any similar concept of potential in general relativity (some people like to call the metric coefficents the 4-potential)
  4. Sep 4, 2006 #3
    - But if you can "define" a Force then you could define a "potential2 and hence a "potential" Energy and a "Kinetic" energy or using "Poisson Bracket" with momenta:

    [tex] \dot \pi_{ab} =[ \pi_{ab} , H] [/tex]
  5. Sep 4, 2006 #4


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    Not really. The situation is this.

    Given a particular worldline (of zero volume), one can define a force that an observer following that worldline would measure with an acceleromater.

    That is not sufficient to be ble to unambiguously define gravity as a "force" in general. For slowly moving objects there is an interpretation available of gravity as a force. This runs into difficulties with as simple a situation as defining the force on a rapidly (i.e. relativistically) moving object, however. While one can compute the 4-accleration for any particular wordline, interpreting gravity as a force requires that one pick out particular moving worldlines as being "straight lines of constant velocity". This turns out to be ambiguous and dependent on the coordinates used. The worldlines that follow geodesics experience no force at all, so they don't define the notion of "straight" that is wanted here.
  6. Sep 4, 2006 #5
    There are two types of forces: Inertial 3-forces and 4-forces. In the absense of 4-forces the inertial 3-force on a particle = f = dp/dt. In the absense of inertial 3-forces the 4-force is given by F = dP/dt.

    The inertial forces are defined by a set of 10 quantities which are known as the gravitational potential. The inertial 3-forces can be defined in terms the velocity of the particle and a combination of the second derivatives of the gravitational potential.

    Geodesic deviation is defined by the tidal gravitational 4-acceleration.

  7. Sep 5, 2006 #6


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    Pete's is one of those people who like to call the metric the "gravitational potential". The 10 quantites he is talking about are just the metric coefficients. The metric is a symmetric 4x4 tensor, because of symmetry it has 10 rather than 16 unique coefficients.
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