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Resultant time dilation from both gravity and motion

  1. Jun 2, 2010 #1
    When a frame is moving in relation to an observer in his rest frame at infinity, and the frame is in a gravitational well, is the resultant time dilation simply the sum of the motional and gravitational dilation, e.g.


    Where [tex]\tau[/tex] is proper time and [tex]t[/tex] is measured by the observer?

    If, not what is the correct expression?
  2. jcsd
  3. Jun 2, 2010 #2
    There is no reason why it would be the sum , you can calculate the expression easily from the Schwarzschild metric:

    [tex](cd\tau)^2=(1-r_s/r)(cdt)^2-(1-r_s/r)^{-1}(dr)^2-(rd\theta)^2-(rd\phi sin\theta)^2[/tex]

    Make [tex]d\theta=dr=0[/tex]
    Last edited: Jun 3, 2010
  4. Jun 2, 2010 #3


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    For an object in a circular orbit, the total time dilation is a product of gravitational and velocity-based time dilation--see kev's post #8 on this thread and post #10 here. But cases other than a circular orbit would probably be more complicated.
  5. Jun 2, 2010 #4

    Jonathan Scott

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    In non-relativistic situations, you can simply fall back on Newtonian theory:

    The fractional time dilation (that is, the difference in time rate divided by the original time rate) due to velocity is equal to the ratio of kinetic energy to rest energy.

    The fractional time dilation due to gravity is equal to the ratio of potential energy to rest energy.

    The combined effect simply adds the fractions together to give the overall fraction (which is equivalent to multiplying the time dilation factors for each of the two effects).

    For free fall (including any shape of orbit around a static mass), the sum of kinetic energy and potential energy is constant, so the time dilation is constant (and so is the total energy, as in Newtonian theory).

    The relative time rates for different orbits can be compared using Newtonian potential theory.
  6. Jun 2, 2010 #5
    Thank you very much. All replies were very useful.
  7. Jun 3, 2010 #6
    Hi Jesse,

    I don't think the expressions put down by kev in that post are correct. The correct result is derived from the Schwarzschild metric, the periods of two clocks situated at radiuses [tex]r_1[/tex] and [tex]r_2[/tex] respectively is expressed by the ratio:


    where [tex]r_s[/tex] is the Schwarzschild radius.The above is valid for a uniform density sphere.
    Start with the Schwarzschild metric:

    [tex](cd\tau)^2=(1-r_s/r)(cdt)^2-(1-r_s/r)^{-1}(dr)^2-(rd\theta)^2-(rd\phi sin\theta)^2[/tex]

    and make [tex]d\theta=dr=0[/tex] for an object orbiting at [tex]r=constant[/tex].

    If [tex]d\theta=d\phi=0[/tex] we get the expression for an object moving radially, which is still different from kev's expressions. In kev's notation:


    where [tex]v=\frac{dr}{dt}[/tex]
    Last edited: Jun 3, 2010
  8. Jun 3, 2010 #7


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    kev wasn't talking about an object moving radially, as I said before he was dealing with the scenario of an object in circular orbit. pervect also found that for this case, the total time dilation was "almost" a product of SR and GR time dilations here...I think the difference was just because pervect was using coordinate velocity in Schwarzschild coordinates in the part of the equation that looked "almost" like SR time dilation, whereas kev was using the local velocity as seen in a freefalling frame for an observer whose coordinate velocity in Schwarzschild coordinates is zero at the moment the orbiting object passes it.
  9. Jun 3, 2010 #8


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    I believe that the equation

    \frac{dt}{d\tau} = \frac{1}{\sqrt{1-v^2/c^2}\sqrt{1 - 2GM/rc^2}}

    always applies (for radial, tangential or any other motion) where v is speed relative to a local hovering observer using local proper distance and local proper time.

    I derived this in posts #9 and #7 of the thread "Speed in general relativity" (and repeated in post #46).
  10. Jun 3, 2010 #9
    kev's expression for radial motion is not correct (see post #6 above). It is very easy to obtain the correct expressions.
    Last edited: Jun 3, 2010
  11. Jun 3, 2010 #10
    Yes, this is correct, provided "v" in your case is defined as:


    or as:

    [tex]\frac{r*sin\theta* d\phi/dt}{\sqrt{1-r_s/r}}=\frac{\omega rsin\theta}{\sqrt{1-r_s/r}}[/tex]


    (see post 6)
    Last edited: Jun 3, 2010
  12. Jun 3, 2010 #11
    So the correct expression is


    , right?

    Do you then define [tex]r\frac{\text{d}\phi}{\text{d}t}[/tex] as coordinate velocity?
  13. Jun 3, 2010 #12


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    Your post #6 seems to be addressing a different question than pervect and kev, since you are finding the ratio of ticking rates of two clocks orbiting at finite radius, while pervect and kev were deriving time dilation of an orbiting clock relative to a stationary clock at infinity (as in the commonly-used equation for gravitational time dilation). I suppose your expression would probably have a well-defined limit as r2 approaches infinity though. Anyway, it might be easier to deal with pervect's derivation rather than kev's, since pervect's equation is expressed entirely in Schwarzschild coordinates rather than including a non-Schwarzschild notion of "velocity". Do you disagree with pervect's conclusions here? If so, where's the first line you would dispute?
  14. Jun 3, 2010 #13
    Precisely. It addresses the question in the OP. (post 1). That is, what is the difference in rates for atomic clocks on the geoid.

    No, the first formula in post 6 is derived from :




    where [tex]\frac{d\tau}{dt}[/tex] is derived straight from the metric:

    [tex](cd\tau)^2=(1-r_s/r)(cdt)^2-(1-r_s/r)^{-1}(dr)^2-(rd\theta)^2-(rd\phi sin\theta)^2[/tex]

    Make [tex]d\theta=d\phi=0[/tex]:


    Pervect's formula in the post you linked is identical to mine. So, no dispute.
    Last edited: Jun 3, 2010
  15. Jun 3, 2010 #14

    I don't define anything.
  16. Jun 3, 2010 #15
    How do I interpret it then?
  17. Jun 3, 2010 #16
    But it looks from the Schwartzschild metric that it would be


  18. Jun 3, 2010 #17
    The Schwartzschild metric for constant r and [tex]\theta=\frac{\pi}{2}[/tex] gives us

    [tex]c^2\left(\frac{\text{d}\tau}{\text{d}t}\right)^2=c^2\left(1-\frac{r_s}{r}\right) - r^2\left(\frac{\text{d}\phi}{\text{d}t}\right)^2[/tex]

    If we divide both sides with c2 we get

    [tex]\left(\frac{\text{d}\tau}{\text{d}t}\right)^2=\left(1-\frac{r_s}{r}\right) - \left(\frac{r\frac{\text{d}\phi}{\text{d}t}}{c}\right)^2[/tex]

    "Factoring out" [itex]1-\frac{r_s}{r}[/itex] on the right side gives

    [tex]\left(\frac{\text{d}\tau}{\text{d}t}\right)^2=\left(1-\frac{r_s}{r}\right)\left(1 - \frac{1}{1-\frac{r_s}{r}}\left(\frac{r\frac{\text{d}\phi}{\text{d}t}}{c}\right)^2\right)[/tex]

    then, taking the square root gives the result in #16;


    I don't see where the mistake is. Would you please point it out to me?
  19. Jun 3, 2010 #18
    yes, fine
  20. Jun 3, 2010 #19
    I was using a notion of local velocity (v' = dr'/dt') as measured by a stationary observer at r.

    Since [tex]v' = dr'/dt' = (dr/dt)/(1-r_s/r)[/tex]

    the value of v' can be directly substituted into your expression to obtain:


    The two forms are numerically the same and in agreement with #8 by DrGReg here:
  21. Jun 3, 2010 #20
    There is no mention of any such convention in this post. Actually, there is no derivation, the expression is put in by hand, you simply multiplied the kinematic factor by the gravitational factor.
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