Suppose that we want to compute the total time dilation for a clock located in an orbiting satellite relative to the clock in our cell phone on the ground.

Consider two different approaches below.

  1. Use special relativity and compute time contraction due to the relative velocity. Use approximation of General Relativity in the Newtonian limit and compute time dilation due to lower gravity and then find the total time dilation.

  2. Don't use Special Relativity. Stick to the approximation of General Relativity based on the symmetry and find Schwarzschild metric and the geodesic for the Earth limits. Find the time dilation assuming a relative velocity in the metric.

The question is:

Which of them are more justified and provide a better approximation? Are they equivalent? What happens when the relative velocity of the satellite is zero?

How good is the approximation in either of the two approaches above.

When we pick the second approach and use the Schwarzschild metric we get this equation:

$$ dt' = \sqrt{1-\frac{3GM}{c^2r}}dt = \sqrt{1-\frac{3r_s}{2r}}dt $$

where $r_s$ is the Schwarzschild radius: $r_s = 2GM/c^2$.

Here we not only assume the asymptotic flat metric to measure $r$ but also switch to Newtonian gravity when we want to cancel $v$:

$$ v = \sqrt{\frac{GM}{r}} $$

So it appears that in the second approach there are many more approximation assumptions.

  • $\begingroup$ You'll find that these calculations have already been done on this site. You could compare both calculations and see how much difference there is (not much!). $\endgroup$ Sep 23, 2016 at 18:14
  • $\begingroup$ For calculation (2) see: What is the correct formula for gravitational time dilation for a satellite in a circular orbit?. Actually that contains enough info for you to do calculation (1) as well. $\endgroup$ Sep 23, 2016 at 18:17
  • $\begingroup$ @JohnRennie can you how the equation is derived from the metric in the circular orbit case? The $\frac{3}{2}$ factor in particular. $\endgroup$
    – user56963
    Sep 23, 2016 at 18:37
  • $\begingroup$ I derive the equation at the end of this answer $\endgroup$ Sep 23, 2016 at 19:24
  • 1
    $\begingroup$ @JohnRennie I mean how to cancel the velocity $v$ without using the Newtown gravity and only in GR. The factor $\frac{3}{2}$ is there because we combine Newton gravity law into Einstein's General Relativity. If we stick to the GR the velocity should remain. Yet again it is not a four velocity. $\endgroup$
    – user56963
    Sep 23, 2016 at 20:58

3 Answers 3


The gravitational time dilation for a static "shell" observer in the Schwarzschild metric observed from infinity is $$ d\tau_{\rm shell} = dt\left( 1 - \frac{r_s}{r}\right)^{1/2}\ .$$ We could label this as the "General Relativistic time dilation".

The speed of an object in a circular orbit around a central mass, as measured by the stationary shell observer at radius $r$, is $$ v_{\rm shell} = c\left(\frac{r_s}{2(r-r_s)}\right)^{1/2}\ .$$

The time dilation due to the orbital motion with respect to the local inertial frame of the shell observer is $$d\tau = \frac{dt_{\rm shell}}{\gamma_{\rm shell}}\ ,$$ where $\tau$ here is proper time measured by the orbiting observer. We could label this as the "Special Relativistic time dilation".

In a local inertial frame, $dt_{\rm shell} = d\tau_{\rm shell}$, so $$d \tau = \gamma_{\rm shell}^{-1}\ d\tau_{\rm shell} = \left( 1 - \frac{r_s}{r}\right)^{1/2} \gamma_{\rm shell}^{-1}\ dt\ ,$$ and we are effectively multiplying the "General Relativistic" and "Special Relativistic" time dilations to get$^{*}$ $$ d\tau = \left[ \left(1 - \frac{r_s}{r}\right)\left(1 - \frac{r_s}{2(r-r_s)}\right) \right]^{1/2} = \left(1 - \frac{3r_s}{2r}\right)^{1/2}\ .$$ No approximations have been made, this is an exact result.

EDIT: To address the last part of the question.

In the more "holistic" version of the calculation - directly using the Schwarzschild metric - the question arises as to why one uses the "Newtonian" value of $v = rd\phi/dt = (GM/r)^{1/2}$? Note that this is not an assumption or an approximation, it is exactly true for the Schwarzschild metric.

Apologies, but this is a lengthy derivation. To see this we have to go back to geodesics in the Schwarzschild metric, which are defined in terms of the two constants of motion that are often labelled $E/mc^2$ and $L/m$, the coordinate specific energy and coordinate specific angular momentum respectively.

In terms of these quantities, the rate of change of radial coordinate with respect to proper time can be written $$ \frac{dr}{d\tau} = \pm c\left[ \left(\frac{E}{mc^2}\right)^2 - \left(1 - \frac{r_s}{r}\right)\left(1 + \frac{L^2}{m^2r^2c^2}\right) \right]^{1/2}\ , \tag*{(1)} $$ where $$ \left(1 - \frac{r_s}{r}\right)\frac{dt}{d\tau} = \frac{E}{mc^2}\ , $$ $$ r^2 \frac{d\phi}{d\tau} = \frac{L}{m}\ . $$

We can now express the circular velocity as seen by a distant observer, $v_{\rm circ} = r d\phi/dt$ as : $$ v_{\rm circ}^2 = r^2\left(\frac{d\phi}{dt}\right)^2 = r^2\left(\frac{d\phi}{d\tau}\right)^2 \left(\frac{d\tau}{dt}\right)^2 = \frac{L^2}{m^2r^2} \left(1 - \frac{r_s}{r}\right)^2 \left(\frac{mc^2}{E}\right)^2\, . $$ But for a circular orbit $dr/d\tau=0$ and eqn. (1) gives us a relation between $E$ and $L$ -- $$ \left(\frac{E^2}{mc^2}\right)^2 = \left(1 - \frac{r_s}{r}\right)\left(1 + \frac{L^2}{m^2r^2c^2}\right) \ ; $$ and substituting this into the equation for $v_{\rm circ}^2$, we get $$ v_{\rm circ}^2 = \frac{L^2}{m^2 r^2} \left(1 - \frac{r_s}{r}\right) \left(1 + \frac{L^2}{m^2 r^2 c^2}\right)^{-1} . \tag*{(2)} $$

The last part of the puzzle is to find an expression for $L/m$ in terms of the radial coordinate for a circular orbit. This is done by writing an expression for the effective potential in the Schwarzschild metric and finding where it is a minimum. Thus: $$ \frac{V_{\rm eff}}{mc^2} = -\frac{r_s}{2r} +\frac{L^2}{2m^2r^2c^2} - \frac{r_sL^2}{2m^2 r^3 c^2}\ . $$ Differentiating and equating to zero, we get $$ \left(\frac{L}{m}\right)^2 = \frac{c^2\,r^2\,r_s}{2r - 3r_s}\ . \tag*{(3)} $$

Replacing $L^2/m^2$ in eqn. (2) using eqn. (3), \begin{eqnarray} v_{\rm circ}^2 & = & c^2\left(\frac{2r_s}{2r- 3r_s}\right) \left(1 - \frac{r_s}{r}\right) \left(1 + \frac{r_s}{2r-3r_s}\right)^{-1} \nonumber \\ & = & c^2 \left(\frac{2r_s}{2r- 3r_s}\right) \left(\frac{r - r_s}{r}\right)\left(\frac{2r - 3r_s}{2r -2r_s}\right) \nonumber \\ & = & c^2\, \frac{r_s}{2r} = \frac{GM}{r}\, . \end{eqnarray} Thus the coordinate speed according to a distant observer is $\sqrt{GM/r}$, the same as the Newtonian result!


They are different effects. Special relativity (SR) will do the first part, just calculate from Newton the velocity and you get a slowdown at the clocks in the satellite. With GR you get a faster clock because of the gravitational time dilation. Both apply, subtract one from the other

In GPS orbits the GR effect is about 45 usec per day, and the SR effect is 7 usec per day. The net effect is 45-7= 38 usec per day. The GPS clocks go that much faster than earthbound clocks. If you were measuring frequencies at the GPS satellites you'd measure a redshift at the GPS satellites (faster clock, lower freq)

See https://en.wikipedia.org/wiki/Error_analysis_for_the_Global_Positioning_System

The clock reported times are adjusted for that.

If the relative velocity of the satellite was zero there would be no SR effect, it'd be 45 usec faster. And it would fall.


Following on from @ProfRob's method, instead of considering the SR and GR "time dilations" separately, we can work directly from the Schwarzchild metric.

(Using signature $+,-,-,-,-$)

$$\boxed{ds^2 = c^2\left(1-\frac{r_s}{r}\right)dt^2 - \left(1-\frac{r_s}{r}\right)^{-1}dr^2 - r^2\sin^2\theta\,d\phi^2 - r^2d\theta^2}$$

Circular orbit, so let $\theta=90^\circ \implies d\theta = 0, \sin\theta=1$. We also have $d\phi = \omega dt$, so $\displaystyle r^2d\phi^2 = v^2dt^2 = \frac{GM}{r}dt^2 = c^2\frac{r_s}{2} dt^2$.


$$ds^2 = c^2\left(1-\frac{3r_s}{2r}\right)\,dt^2$$

and comparing with an instantaneous rest frame (IRF) $ds^2 = c^2\,d\tau^2$, we have that:

$$d\tau = \sqrt{1-\frac{3r_s}{2r}}\,dt$$

Note we have not have to consider the Geodesic equation, or make any further approximations than those in deriving the Schwarzchild metric itself.

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    $\begingroup$ Right, but you have used $r^2(d \phi/dt)^2 = v^2 = GM/r$ without proof. There is no obvious reason why that Newtonian result should be true in GR. To prove this requires consideration of the orbits and geodesics. Note that it is $d\phi/dt$, not $d\theta/dt$, since $d\theta=0$. $\endgroup$
    – ProfRob
    Mar 20, 2023 at 12:18
  • $\begingroup$ @ProfRob Fixed the $\theta$... I agree that's an assumption made, but as you say we could prove this separately. I still think there's value in this derivation for working directly from the metric, rather than considering two separate effects as you have done. Do you agree? $\endgroup$
    – Joe Iddon
    Mar 21, 2023 at 9:16
  • $\begingroup$ As I say in my answer and link to a previous answer. "Note, it isn't really correct to label them as separate relativistic corrections. To see the calculation done holistically, see physics.stackexchange.com/a/110675/43351 " $\endgroup$
    – ProfRob
    Mar 21, 2023 at 19:12
  • $\begingroup$ @ProfRob I agree. Your calculation is not 'holistic' though, whereas mine is. If we assume the form of $v$ is Newtonian (your answer provides a good derivation) then my method is succinct and distinct from your method. $\endgroup$
    – Joe Iddon
    Mar 22, 2023 at 20:28
  • $\begingroup$ The OP knows how to derive the time dilation in the way you've done it. The question IS about whether the Newtonian $v$ is appropriate. $\endgroup$
    – ProfRob
    Mar 22, 2023 at 21:56

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