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Gravitational effects on particles orbit in a box

  1. Jul 6, 2011 #1
    Hi all,

    If one were to consider a classical particle in a box of length L in Minkowski space then clearly the time taken to go between the walls of the box is L/v with v velocity.

    In the GR case with some metric coefficients A(r) and B(r) (spherically symmetric system, say) it's clear that the distance traveled is longer and so the period should be longer as well, my question is how to calculate this? Let's say m = 0 so v = c.

    Do I take dt = 0 and integrate the line element ds from 0 to L over dr? I thought of trying to take the already computed flat space period and applying the time dilation method to it but that contains a radial coordinate component (obviously as it must be position dependent) but i'm unsure as to how to 'get rid' of this part, i.e. would r here simply represent the end points or again is some integral/averaging needed?

    I know this is probably kind of a trivial question but still it's doing my head in a bit =P
  2. jcsd
  3. Jul 7, 2011 #2


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    Could you explain your notation and give the metric explicitly?

    Longer than what? Longer why?
  4. Jul 7, 2011 #3
    Sure, sorry.

    Lets say i have a metric
    ds^2 = -A(r) dt^2 + B(r) dr^2 +r^2 d\Omega^2

    I would expect that if i were to designate length L as the size of the box in Minkowski space then the geodesic path length traveled to go between ends of the box in a curved space, say that for an attractive gravitational potential, would be longer.
  5. Jul 7, 2011 #4


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    So, you want to find a radial geodesic path for the above metric? The easy way to do is to note that none of the coordinates depends on t. So our system has a time-like Killing vector. And the dot product of the time-like killing vector with the four velocity of the particle remains constant

    In your example, if we assume that the particle has some 4-velocity r(tau), t(tau), then we can write

    g_{ij} [(dt/dtau), (dr/dtau) ] dot [1,0] = constant,which can be simplified to

    -A(r) dt/dtau = constant

    This constant can be thought of as the energy of the particle.

    Add one more constraint, that the magnitude of the four-velocity is always -1 (with your sign convention), and you've got enough equations to solve for a radial geodesic.

    IT's not that much harder to do a non-radial path. You could also write out the Christoffel symbols and use the geodesic equation, but that's MUCH harder - it's a lot easier if you can find the Killing vectors. If you're interested in the concpets, though, it's worth working it out with the Killiing vector approach first, then confirming that your solution satisfies the geodesic equations.

    [add]...the geodesic equations are well known, but I'll jot them down anyway

    \frac {d^2 x^i}{d \tau^2} + \Gamma^{i}{}_{jk} \: \frac{dx^j}{d\tau} \: \frac{dx^k}{d\tau} = 0

    Maybe you want a non-radial geodesic. That's a little harder, but not much. You can assume that the motion occurs in the equatorial plane (theta=0), and then you have another Killing vector (a spacelike one) due to the fact that the metric doesn't depend on phi, giving you the other equation you need. This space-like killing vector corrseponds to a conserved angular momentum, because position invariance -> a conserved momentum, angular momentum because your coordinate is an angular coordinate.

    Most textbooks should go through this in more detial if my run-through has been too fast.
    Last edited: Jul 7, 2011
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