Do orbiting satellites experience gravitational time dilation?

[PAllen]

To emphasize the importance of the metric, I will show just how clearly and simply it explains the orbital versus static temporal relationship for this case.

First, the relevant metric for this case is the Schwarzschild metric, which you can find here:

http://en.wikipedia.org/wiki/Schwarzschild_metric

However, if we restrict ourselves to constant r, and one plane through the center of a planet, this reduces to:

d$\tau$^2 = g00 dt^2 - r^2 d$\theta$^2

where g00 may be treated as a positive constant.

Note, that as Peter explained, the notion of constant positions (r,$\theta$) in these coordinates has invariant meaning (provided by lack of motion relative to the massive body) - even though the particular coordinate values and units are purely conventional.

You can see that for a static clock ($\theta$ not changing), the rate of proper time to coordinated time is √g00. For any circular motion, geodesic (orbit) or forced, you have the following ratio of proper time to coordinate time.

√(g00 - (r θ')^2) , where θ' is the angular speed.

No matter what the functional form of θ'(t), integrating this over t such that a circle is traversed, will produce a smaller result than √g00 times the coordinate time of the circular path. Thus, the circular moving clock will elapse less time, no matter what the nature of its motion relative to static clocks.

 

[WannabeNewton]

Just to add on to PAllen, we can easily calculate the ratios of proper times of the static observer and circularly orbiting observer between two events coinciding on both their worldlines. Imagine for example that two observers $O$ and $O'$ are in a ship together that is in circular orbit at some allowed radius $R > 3M$ around a static spherically symmetric star (the allowed radii can be determined by using conserved quantities and the definition of stable/unstable circular orbits as the minima/maxima of the effective potential). At some event $p$, observer $O$ steps out with a rocket and hovers in place while observer $O'$remains in the circular orbit. The two meet again at event $q$ after $O'$ makes a complete orbit starting from $p$, at which point they compare their clock readings (which were synchronized at $p$).

The 4-velocity of $O$ between $p$ and $q$ is just $u = (1 - 2M/R)^{-1/2}\partial_t = \gamma \partial_{t}$. The 4-velocity $\tilde{u}$ of $O'$ between said events takes a bit more work. The angular frequency of the orbit as measured by an observer at infinity will be $\omega := \frac{\mathrm{d} \phi}{\mathrm{d} t} = \sqrt{M/R^3}$ so $\tilde{u} = \tilde{\gamma} (\partial_t + \omega \partial_{\phi})$ where $\tilde{\gamma}$ is yet to be determined. Using the fact that $g(\tilde{u},\tilde{u}) = -1$ we have that $\tilde{\gamma} = (1 - 2M/R - R^2\omega^2)^{-1/2}$. This is basically a combination of the gravitational time dilation factor and the kinematical time dilation factor when boosting from the local Lorentz frame of a static observer at $R$ to the local Lorentz frame of a circularly orbiting observer at $R$.

The proper time $\Delta \tau_{pq}$ as measured by $O$ between $p$ and $q$ is then $\Delta \tau _{pq} = \int _{p}^{q}\gamma^{-1}dt = (1 - 2M/R)^{1/2}\Delta t_{pq}$ whereas the proper time $\Delta \tau'_{pq}$ as measured by $O'$ between said events is $\Delta \tau' _{pq} = \int _{p}^{q}\tilde{\gamma}^{-1}dt = (1 - 3M/R )^{1/2}\Delta t_{pq}$ where $\Delta t_{pq} = \frac{2\pi}{\omega}$ is just the period of the circular orbit as measured by an observer at infinity. Thus $\frac{\Delta \tau _{pq}}{\Delta \tau' _{pq}} = (\frac{1 - 2M/R }{1 - 3M/R })^{1/2}$; you can see that $\Delta \tau _{pq} > \Delta \tau' _{pq}$.

You may object to this because $O'$ is freely falling whereas $O$ is accelerating so it would seem that $O'$s clock should read more proper time between $p$ and $q$ than that recorded by $O$s clock since, as noted, $O'$ is freely falling. The thing is that between two events in space-time, a free fall worldline only maximizes proper time amongst worldlines in a sufficiently small neighborhood of this free fall worldline in the function space of all possible worldlines between the two events. In our case the circular orbit corresponds to a free fall worldline that is not close in the function space to the worldline of the static observer.

 

[Dale]

You may object to this because $O'$ is freely falling whereas $O$ is accelerating so it would seem that $O'$s clock should read more proper time between $p$ and $q$ than that recorded by $O$s clock since, as noted, $O'$ is freely falling. The thing is that between two events in space-time, a free fall worldline only maximizes proper time amongst worldlines in a sufficiently small neighborhood of this free fall worldline in the function space of all possible worldlines between the two events. In our case the circular orbit corresponds to a free fall worldline that is not close in the function space to the worldline of the static observer.

Which is exactly why there is no general substitute for integrating the metric along the worldline, and one of the many reasons why you simply cannot understand GR without understanding the metric.