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Time dilation in an elliptical orbit

  1. Sep 21, 2015 #1
    The formula for time dilation in a circular orbit is readily available but the literature seems to indicate it would not be so simple in the case of an elliptical orbit, and no simple formula seems to be available.

    Given that time dilation in a circular orbit adds the velocity effect (GM/r) to the gravitational effect (2GM/r)

    t = sqrt(1 - (GM/r + 2GM/r)/c²)
    t = sqrt(1 - 3GM/rc²)

    And given that velocity at perihelion in an elliptical orbit is

    vp = sqrt(GM((2/r) - (1/a)))

    Can it be deduced that time dilation in an elliptical orbit at perihelion is

    t = sqrt(1 - (GM((2/r) - (1/a)) + 2GM/r)/c²)
    t = sqrt(1 - (2GM/r - GM/a + 2GM/r)/c²)
    t = sqrt(1 - (4GM/r - GM/a)/c²)
    t = sqrt(1 - GM((4/r) - (1/a))/c²)
     
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  3. Sep 21, 2015 #2

    bcrowell

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    You're more likely to get replies if you mark up your math properly in LaTeX. We have a howto here: https://www.physicsforums.com/help/latexhelp/ . It's very difficult for people to read math that isn't written like math.

    In general, the time elapsed on a path through spacetime is given by ##s=\int ds##, where ##ds^2=g_{ij}dx^idx^j##.

    In your #1, you seem to be flipping back and forth between talking about two different things: (a) the time dilation as an object moves past a particular point in a static spacetime, and (b) the time dilation of an object that moves along a path (a periodic orbit) in spacetime. If a is meaningful, then it can't depend on the rest of the path. It's not completely obvious to me, however, that a is meaningful. To define what this really means, you'd have to relate it to at least some hypothetical, possible measurement process, which would involve synchronizing clocks and then later reuniting and comparing them. In the case of a static spacetime, the implicit understanding about what we mean by gravitational time dilation is that we synchronize clocks at P, transport one clock to some other spatial point of interest Q, leave it there for a very long time, and then transport it back for comparison. Since the spacetime is static, a spatial point like Q has a well-defined identity, and you can leave the clock at Q for as long as you like, so that any time dilation incurred during transport to and from Q can be made negligible.

    Is this the kind of thing you have in mind, or are you thinking of something more like clock synchronization via radio signals? If the latter, then Doppler shifts are not going to be cleanly separable from time dilation.
     
    Last edited: Sep 21, 2015
  4. Sep 21, 2015 #3
    Thanks, here is the question with better wording and math:

    The formula for time dilation in a circular orbit is readily available but the literature seems to indicate it would not be so simple in the case of an elliptical orbit, and no simple formula seems to be available.

    Given that time dilation in a circular orbit adds the velocity effect (##v²##) to the gravitational effect (##2GM/r##)

    [tex]t = \sqrt{1 - \left( v^2 + 2GM/r \right) / c^2}[/tex]
    [tex]t = \sqrt{1 - \left( GM/r + 2GM/r \right) / c^2}[/tex]
    [tex]t = \sqrt{1 - 3GM / rc^2}[/tex]

    And given that velocity in an elliptical orbit is

    [tex]v = \sqrt{GM \left( \frac{2}{r} - \frac{1}{a} \right) }[/tex]

    Can it be deduced that time dilation in an elliptical orbit is

    [tex]t = \sqrt{1 - \left(GM \left( \frac{2}{r} - \frac{1}{a} \right) + 2GM/r \right) / c^2}[/tex]
    [tex]t = \sqrt{1 - GM \left( \frac{4}{r} - \frac{1}{a} \right) / c^2}[/tex]

    I have no specific measurement process in mind. Just wanted to know if the algebraic logic holds true in this case and if time dilation at any point in an elliptical orbit can be predicted using the last equation.

    NB, r is radius, a is semi-major axis, G is gravitational constant, M is mass of orbited body, c is the velocity of light
     
  5. Sep 21, 2015 #4

    bcrowell

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    Is this an orbit that's at velocities small compared to c? If not, then your expression for v will not be accurate.

    Time dilation doesn't give a formula for the time on a clock. In fact, if you look at your formulas, you'll see that they don't have units of time.
     
  6. Sep 21, 2015 #5
    I realize that. I am not familiar with the correct way of writing "rate of change in time with respect to an immobile observer" but that is what I mean by "t".
    I am looking for the formula that defines rate of change in time for small OR large velocities, ie orbitting a small OR large mass M. The question, if I may repeat it is, will the following formula allow the calculation of time dilation for an observer in an elliptical orbit?

    [tex]t = \sqrt{1 - GM \left( \frac{4}{r} - \frac{1}{a} \right) / c^2}[/tex]
     
  7. Sep 21, 2015 #6

    bcrowell

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    You could write it as a ratio like ##t_1/t_2##.


    This is a lot more complicated. You assumed Newtonian conservation of energy, which doesn't apply. See https://en.wikipedia.org/wiki/Schwarzschild_geodesics .
     
  8. Sep 21, 2015 #7
  9. Sep 21, 2015 #8

    bcrowell

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    The answer is no.
     
  10. Sep 21, 2015 #9
    So what is the equation for calculating time dilation in an elliptical orbit?
     
  11. Sep 21, 2015 #10

    bcrowell

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    One thing you'll find out if you read the WP article I linked to is that there are no elliptical orbits.

    You need to do some work in carefully defining and researching your question. Asking people for "the equation" isn't a substitute for that.
     
  12. Sep 21, 2015 #11
    Your position would be correct if I was a student; but I really am just looking for an answer.
    According to the answer to the following post, the equation, simplified, would be at perihelion:

    [tex]\delta t = \sqrt{1 - GM \left( 1 - \frac{a}{r} \right) \left( 1 - e^2 \right) /rc^2}[/tex]
    https://www.quora.com/Is-the-subjective-relativistic-time-experienced-by-a-body-in-circular-orbit-more-less-or-the-same-as-a-body-moving-in-an-elliptical-orbit [Broken]

    Many thanks for your help.
     
    Last edited by a moderator: May 7, 2017
  13. Sep 22, 2015 #12

    PeterDonis

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    A question that is not well-defined doesn't have a well-defined answer.

    Which also isn't well-defined. What elliptical orbit (or trajectory that is approximately an elliptical orbit--as bcrowell pointed out, in GR there are no closed elliptical orbits) is being compared with a given circular orbit? In other words, how am I supposed to pick out the (trajectory that is approximately an) elliptical orbit that is "comparable" to a given circular orbit? You don't say, and neither does the quora post. (The answer you are referring to assumes that the "comparable" circular orbit is one whose radius is equal to the semi-major axis of the elliptical orbit; but the question itself doesn't say that, and it's not clear that the questioner even understands that that's a key issue.)

    This doesn't look right. The equation in the quora post (which is, btw, already a Newtonian approximation, so it won't be valid for large velocities or strong fields, and you said you were looking for an answer valid in those regimes) is

    $$
    d\tau = dt \sqrt{1 - \frac{2m}{r} - \frac{ma}{r^2} (1 - e^2) \left( 1 + \frac{e^2 \sin^2 \theta}{(1 - 2m/r)(1 - e \cos \theta)^2} \right)}
    $$

    They use ##m## to mean ##GM / c^2##, so it looks to me like this would simplify, for the case of an orbit in the equatorial plane for which ##\theta = \pi / 2##, to

    $$
    d\tau = dt \sqrt{1 - GM \left( 2 - \frac{a}{r - 2 GM / c^2} \left[ 1 - \frac{2GM}{c^2r} (1 - e^2) - e^4 \right] \right) / r c^2}
    $$
     
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