1
$\begingroup$

Consider a geostationary satellite at some arbitrary distance above the Earth's equator and consider a person directly below it, standing on the Earth's equator. Both satellite and person exist on the same radial line.

General relativity tells us that the observer on the Earth will experience time dilation due to being further inside the gravitational well. This will slow the person's clock relative to the satellites clock.

However, would it be correct to say that, since the satellite has greater linear velocity (it's further out and rotating with the same angular velocity as the person on Earth), we should be taking into account special relativity, which would cause a time dilation of the satellites clock relative to the person's clock? Therefore there would be two competing time dilations, that of the satellites clock and that of the person's clock, and the actual time difference between the clocks must take both effects into account.

I've had counter suggestions, saying that since both observers are stationary relative to each other in the rotating frame, special relativity doesn't play a role when comparing the satellites clock to that of the person.

$\endgroup$
2
$\begingroup$

Yes, both special relativity and general relativity have to be taken into account. The total time dilation is given by $$ \frac{\text{d}\tau}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{rc^2}\right) - \left(1 - \frac{2GM}{rc^2}\right)^{-1}\frac{v^2}{c^2}}, $$ where $\text{d}\tau$ is the time measured by a moving clock at radius $r$, and $\text{d}t$ is the coordinate time measured by a hypothetical stationary clock infinitely far from the gravitational field.

For a person standing on the equator, we have $r_\text{eq}=6378\,\text{km}$ and $v_\text{eq}=0.465\,\text{km/s}$, and for a geostationary satellite $r_\text{s} = 42164\,\text{km}$ and $v_\text{s} = 3.074\,\text{km/s}$. This enables you to calculate $\text{d}\tau_\text{eq}/\text{d}t$ and $\text{d}\tau_\text{s}/\text{d}t$, and finally the ratio $\text{d}\tau_\text{eq}/\text{d}\tau_\text{s}$.

See also this post for more details: https://physics.stackexchange.com/a/90764/24142

$\endgroup$
  • 2
    $\begingroup$ Also see Phil Fraundorf's picture showing SR time dilation and GR speedup, and the net effect for various orbits. $\endgroup$ – John Duffield Apr 7 '15 at 16:03
  • $\begingroup$ Pulsar: "both special relativity and general relativity have to be taken into account." -- That's certainly preferable instead of saying "SR and/or GR predict". However, your formulation gives the impression as if SR should be taken into account in addition to GR. "The total time dilation is given by [...]" -- I wonder (especially) about the meaning of $v$ in your formula. (The OP wrote of "linear velocity" ... this question makes my doubt more precise). "This enables you to calculate [...]" -- Consider giving a value of $\frac{G~M}{c^2}$. $\endgroup$ – user12262 Apr 7 '15 at 21:32
0
$\begingroup$

There is a slight error in the answer by Pulsar that I want to point out. If you are moving in a circular, pure non-radial orbit in a spherically symmetric gravitational field the time dilation is actually:

$$ \frac{\text{d}\tau}{\text{d}t} = \sqrt{ 1 - \frac{2GM}{rc^2} - \frac{v^2}{c^2}}, $$

However if you are moving in pure radial direction, then as Pulsar writes:

$$ \frac{\text{d}\tau}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{rc^2}\right) - \left(1 - \frac{2GM}{rc^2}\right)^{-1}\frac{v^2}{c^2}}, $$

The difference is in second order so in actual calculations it would not matter much which expression you use in the weak field of our solar system.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.