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 Newtown limits and compute time expansion due to the less 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 Newtown 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!). $\endgroup$ – John Rennie Sep 23 '16 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. $\endgroup$ – John Rennie Sep 23 '16 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 '16 at 18:37
  • $\begingroup$ I derive the equation at the end of this answer $\endgroup$ – John Rennie Sep 23 '16 at 19:24
  • $\begingroup$ @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 '16 at 20:58

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.

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The gravitational time dilation of a static object in the Schwarzschild metric observed from infinity is $$ dt'_{\rm GR} = dt\left( 1 - \frac{r_s}{r}\right)^{1/2}\ .$$

The speed of an object in a circular orbit around a central mass is $$ v = c\left(\frac{r_s}{2(r-r_s)}\right)^{1/2}\ .$$

The time dilation due to the orbital motion is $$dt'_{\rm SR} = \frac{dt}{\gamma}\ .$$

Multiplying the "General Relativistic$^*$" and "Special Relativistic" time dilations we get $$ dt'_{\rm total} = \left[ \left(1 - \frac{r_s}{r}\right)\left(1 - \frac{r_s}{2(r-r_s)}\right) \right]^{1/2} = \left(1 - \frac{3r_s}{2r}\right)^{1/2}\ .$$

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