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Gravity & geodesics

  1. Feb 24, 2006 #1
    How is the earth on a geodesic? Forgive me if this is basic. I had associated the time dilation effect with acceleration & now it turns out gravity isn't a "force" even though GR in the equivalence principle models it as an acceleration.

    To rephrase, the earth is constantly accelerating & yet maximizing its proper time because...
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  3. Feb 24, 2006 #2


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    ...because the zero-variation-in-action path through the four-dimensional pseudo-Riemannian non-euclidean geometry near a gravitating source is a geodesic in spacetime. Note the involvement of time. Earth's worldline is approximately an "elliptical helix" in spacetime.
  4. Feb 24, 2006 #3
    Ok. So why the equivalence principle then? The theory seems to imply that it's both inertial motion & accelerated motion. I'm guessing that's due to a change of frames?
  5. Feb 24, 2006 #4


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    Probably your mistake was to associate time dilation with acceleration in the first place. It's hard to be sure because "associated with" is a very vague term.

    But if you look at the formula for time dilation due to motion, time dilation is a function of velocity, not by acceleration. (This is a SR effect, which causes the so-called twin paradox).

    If you look at the formula for time dilation due to gravity (gravitational red shift, a GR effect), it is a function of potential energy, and (again) not acceleration.

    If you want the mathematical details, any time a body follows a path that minimizes or maximizes an intergal, the body must obey certain differential equations called the Euler-Lagrange differential equations.

    These are usually taught in physics well before GR, in Lagrangian mechanics. I don't know much about your background, I'm going to assume from your question that you are not familiar with them yet. If I'm right, the following brief treatment will probably bo too rapid, but there's no way to compress a semester physics course into a post.

    One way of writing down the geodesic equations for a body is to write down the intergal for its proper time, which then gives the associated Euler-Lagrange differential equations that describe how it moves. When you do this for the Earth, you find that it orbits the sun, the resulting differential equations are almost exactly the same as the Newtonian differential equations.

    For more on the principle of least action and the Euler-Lagrange differential equations, try the wikipedia article


    EF taylor's website would probably be another good resource as well:


    Writing down the intergal for proper time is easy. In SR, we simply solve for [itex]d\tau[/itex] given

    [tex]d\tau^2 = dt^2 - dx^2[/tex]

    (I've made c=1 for simplicity)

    In GR, we add in the metric coefficients

    [tex] d\tau^2 = g_{00} dt^2 + 2 g_{01} dxdt + g_{11} dx^2[/tex]
  6. Feb 25, 2006 #5
    The earth is on a geodesic in spacetime (not just space).
    How and why?
    Gravity is a force. Its called an inertial force. There are two types of forces in nature, inertial forces and non-inertial forces (non-zero 4-force).
    You're neglecting to mention the frame of reference in your questions. This is where the problem is. First off the earth has a non-zero spatial acceleration with respect to the rest frame of the sun, but it has zero spatial acceleration in its own (freely-falling) frame. Also, proper time is not always maximized in GR. All that is required is that it has a stationary value.

  7. Mar 18, 2006 #6
    Could you explain what you mean by stationary value? I'm encountering that a lot.
  8. Mar 22, 2006 #7
    One of the main and most difficult problem in general relativity is to know how to go from a point A to a point B in the neighborough of A because within this theory there is no long distance homogeneity. The result is that coordinates system for A is not the coordinates system for B and that one have to imagine a stratagem to join both systems in a coherent way.

    Concerning the question about the geodesics. The equivalence principle states that the coordinates of B (they are depending on the coordinates of A) must vary in such a way that B has no acceleration if B is not under the influence of any force.
  9. Mar 22, 2006 #8
    I suppose it means that the ratio : the variation of time for B relatively to the variation of time for A is zero.
  10. Mar 23, 2006 #9
    It means that the functional (i.e. integral of a parameter(s)) has a "first derivative" (whatever that means) of zero. Notice the typical idea from calculus y = f(x). y'(x=a) = 0 does not mean that there is a local minumum at x = a but only that the tangent to the curve is parallel to the x=axis. For all the gory details please see


  11. Mar 23, 2006 #10
    This brings up a question I've wanted to ask for a while.

    The vacuum metric "moves" so that
    [tex]\int \sqrt{-g}\mathcal{L}\quad d^4x[/tex]
    is stationary.

    And particles move so that
    [tex]\int d\tau.[/tex]
    is stationary.

    Does this summarise GR?
  12. Mar 23, 2006 #11
    What meaning are you attaching to "metric moves"?


  13. Mar 23, 2006 #12
    The components changing with respect to each other is what I mean.

    EDIT: I suppose I actually mean its degrees of freedom changing with respect to each other, since component implies we must have a frame, but the metric exists independent of one.
  14. Mar 23, 2006 #13


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    Of course, you'll have to specify the meanings of the various symbols.
    In a variational principle, you have to specify the type of variations under consideration (vary the metric? the connection? something else?)... as well as some boundary conditions. By "moves", you probably mean "is a classical solution to the equations of motion". By "particle", you probably mean "inertial particle".
  15. Mar 23, 2006 #14
    Yes that's a good point robphy. The [itex]\mathcal{L}[/itex] should be the Ricci scalar [itex]R[/itex]. And the first integral is varied with respect to the metric.

    I'm not entirely sure if I do. What difference would that make? Also, could you explain to me what kind of boundary conditions you mean?
    Last edited: Mar 23, 2006
  16. Mar 23, 2006 #15
    Note - In the present case it is not the metric that is being varied. It is the pathlength which is being varied. The metric merely defines "path length". So there is no variation of the metric here. The components of the metric are functions of position and the position of the "test path" is varied. But one does not say that the metric has changed. Does that clear it up a bit?

    Last edited: Mar 23, 2006
  17. Mar 23, 2006 #16
    I see no reason to assume that [itex]\mathcal{L}[/itex] is the Ricci scalar. In those places where there is no matter the Ricci tensor and hence the Ricci scalar are both zero. This would give a zero path length for any variation in empty space (e.g. inbetween planets). And what I just said would hold in all possible spacetimes.

  18. Mar 23, 2006 #17


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    For geodesic motion, your particle is travelling inertially (i.e. unaffected by non-gravitational influences).

    Concerning boundary conditions: for example,... you might require that your path variations don't change the endpoints.

    My point was that one is obligated to specify what is being varied. My reference to the metric and the connection concerned the first problem posed... not the second.
  19. Mar 23, 2006 #18


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    To say "In those places where there is no matter the Ricci tensor and hence the Ricci scalar are both zero." already uses the the equations of motion, which is what you are trying to find.

    Admittedly, the entire problem hasn't been sufficiently well-posed.
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