Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

General Relativity Gravitation Calculations

  1. Sep 20, 2011 #1
    Question on calculating gravitation per General Relativity and the field equations:

    I understand that Einstein's field equations describe spacetime. But I’m not sure of how they are used when evaluating gravitation. For example, the equations describe how spacetime is “curved” by a large object of mass. But how does it say how a smaller object will be influenced by that curvature.

    If it could be broken down for me:

    The equations are hard to make sense of. Say, if Earth (A) is the gravitational body with such and such mass, and the acceleration of a ball (B) with such and such a mass, in the air some (X) meters above the Earth’s surface, is effected by the curvature of spacetime (C?). Could you identify these characters in the field equations for calculating the force of gravity on a body?
  2. jcsd
  3. Sep 20, 2011 #2
    To solve for the trajectory of a test particle in GR, one calculates the 'geodesic' along the proper manifold for the local-geometry.
    The gravity determines the geometry of the manifold (local space-time), and the trajectory is simply the multidimensional equivalent of a straight-line (a 'geodesic'). Calculating that 'straight line' gives you the trajectory.
    See: http://en.wikipedia.org/wiki/Geodesic_equation#Affine_geodesics
  4. Sep 21, 2011 #3
    Thanks for the reply.

    Is there something more along the lines of calculating the weight of an object?
  5. Sep 21, 2011 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    If you have a stationary object and a static geometry, the force required to stay in place can be written as [itex]\nabla_{\alpha} u{\alpha}[/itex] where [itex]u^{\alpha}[/itex] is the four-velocity of the object in question.

    (I'm not sure if this will be directly helpful to you or not, because I don't know what level of knowledge of the math baackground you have).

    This is well defined because its easy to define a co-located "stationary" reference object when you have a static geometry.

    You can then measure the acceleration of your test object relative to your rest object, and call this the "weight" of the object.

    In a non-static geometry, it becomes difficult if not impossible to find some reference observer to measure the relative acceleration to to measure any sort of traditional "force". What you can do, instead, is measure the tidal force, which is the relative acceleration of two objects a certain distance apart.

    If your objects are following geodesics, the relative acceleration (and hence the loal derivative of the force) is given exactly by the geodesic deviation equation

    acceleration = [tex]R^{\alpha}{}_{\beta}{\gamma}{\delta}\,\, u^{\beta} u^{\delta} \,\, d^{\gamma}/tex]

    Here [itex]u^{\alpha}[/itex] is the four-velocity, as before, and [itex]d^{\alpha}[/itex] is the separation vector.

    Unfortunately, while in flat space-time you can integrate the derivative of the force without worrying about the specific path, in curved space-time this integral may become path dependent.

    If the objects are NOT following geodesics, there are some interpretational issues which will affect the results of the above calculation. Howeer, it would be premature to go into the details, especially as under normal circumstances the errors are very small. For all practical purposes, if you have an object that's accelerating at modest rates (say less than 10,0000 g's), you won't notice the difference.
    Last edited: Sep 21, 2011
  6. Sep 21, 2011 #5
    That is not trivial but doable if you are willing to make a few simplifications.

    1. We assume the Earth (A) is not rotating and is a perfect ball.
    2. We assume the mass of the ball (B) is negligible.

    Then we can use the Schwarzschild solution (although for a mass the size of the Earth hardly useful as the results will be very close to Newton's solutions), this solution is derived from the field equations and includes the assumption that Newton's law of gravitation is correct at infinity and for c->infinity.

    If you are interested how the Schwarzschild solution is derived you can check: http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution
  7. Sep 21, 2011 #6


    User Avatar
    Science Advisor
    Gold Member

  8. Sep 21, 2011 #7
    Thanks guys. I'm wondering if I describe in words what GR says about weight, comparing it to Newtonian gravity.

    The weight of an object = the energy content of that object (it's mass) multiplied by the local spacetime curvature.

    Would this be incorrect?
  9. Sep 27, 2011 #8
    Relevant to my last post, does it make sense to imagine that the field equations define/describe the curvature of spacetime around a gravitational body?

    Say we have a gravitational body of a known mass, e.g. the Earth. Could we say that at a place X distance from the center of the body (assuming beyond the surface of the object), there will be a calculated degree of spacetime curvature (also assuming there are no other large bodies to be concerned with)? So the equations would tell us the spacetime curvature at X, and therefore the gravitational acceleration of anything at that distance. Then, putting an object of known mass at that distance, e.g. a 1kg ball, we could predict the gravitational force on that object.

    Or is this not how the field equations work?
  10. Oct 10, 2011 #9
  11. Oct 10, 2011 #10


    User Avatar
    Science Advisor

    Well, if you consider any static, spherically symmetric object then the geometry of space - time outside the source would be described by the schwarzchild metric and from that metric one can determine the independent components of the Riemann tensor [itex]R^{\alpha }_{\beta \mu \nu }[/itex] or scalar curvatures like the Ricci scalar or even [itex]R^{\alpha \beta \gamma \delta }R_{\alpha \beta \gamma \delta } = \frac{48M^{2}}{r^{6}}[/itex] and this will give you a scalar curvature at each r for which it is defined but note that this is not giving a curvature for a distance r from the center of the body because for the schwarzchild metric the r coordinate is not the distance from the center of the spherically symmetric, static object (it is not the same r as that in euclidean space). In order to figure out how much an object would "accelerate" for this space - time you would have to form a separation vector that connects the geodesic the particle is traveling on in the schwarzchild space- time to a nearby geodesic and see how the separation vector changes. You can write this as [itex]\frac{D^{2}\xi ^{\alpha }}{D\tau ^{2}} = R^{\alpha }_{\beta \gamma \delta }U^{\beta }U^{\gamma }\xi ^{\delta }[/itex] where [itex]\boldsymbol{\xi }, \mathbf{U}[/itex] are the separation vector and 4 - velocity of the particle respectively and the components of [itex]R^{\alpha }_{\beta \mu \nu }[/itex] would of course come from the schwarzchild metric but remember that this describes gravitational tidal forces.
  12. Oct 10, 2011 #11


    User Avatar
    Science Advisor
    Gold Member

    Spacetime curvature relates to gravitational tides rather than "acceleration". It relates to the rate of change of "gravitational acceleration".

    Imaging one object close to the Earth and another object much further away from Jupiter, such that both measure the same local "acceleration due to gravity". The one nearer the Earth would be in a region of greater spacetime curvature, because gravity "changes more rapidly" near the Earth compared with a long way from Jupiter.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook