How does time dilation affect the synchronization of clocks in
different gravitational potentials, as described by the general theory
of relativity?
Clocks at different gravitational potentials cannot be synchronised in the conventional sense used in SR using Einstein clock synchronisation convention. The clocks run at different rates, so in order to synchronise the clocks, one of the clocks has to have its output multiplied by a factor, so that it runs at the same rate as the other clock.
For example if we consider two clocks, one located at sea level and
another at a high altitude (e.g., on a mountain), how would the
difference in gravitational potential between the two locations
influence the rate at which each clock ticks
As you go up a mountain there are two opposing effects. Gravitational time dilation speeds up clocks due to being further away from the centre of the Earth. Velocity related time dilation slows down clocks due the higher tangential velocity that results from being at a greater radius from the centre of rotation of a rotating body. In broad terms, the gravitational effect dominates and clocks speed up as you go up a mountain.
The best way to calculate the exact result for time dilation in the gravitational field of a non-rotating spherically symmetric body, is to use the Schwarzschild metric:
$$\begin{equation}
\mathrm d\tau^2 = \left(1 - \frac{2GM}{rc^2}\right) \mathrm dt^2 - \left(1 - \frac{2GM}{rc^2}\right)^{-1} \frac{dr^2}{c^2} - \frac{r^2}{c^2} \left(\mathrm d\theta^2 + \sin^2 \theta \, \mathrm d\phi^2\right)\end{equation},$$
where $\tau$ is the proper time of a particle moving in the gravitational field and all other quantities are as measured by a reference observer at infinity. For constant radius $(r)$, $\mathrm dr=0$ and for circular motion in the equatorial plane, ($\theta = \pi/2$) and $(\mathrm d\phi = 0)$, the above equation simplifies to:
$$\begin{equation}
\frac{\mathrm d\tau^2}{\mathrm dt^2} = \left(1 - \frac{2GM}{rc^2}\right) - \frac{r^2}{c^2} \left(\frac{\mathrm d\theta^2}{\mathrm dt^2} \right)\end{equation}.$$
Since $\mathrm d\theta/\mathrm dt$ is equal to the angular velocity $\omega$, we can replace that expression with $\frac{\mathrm d\theta }{\mathrm dt} = \omega_{\infty} = \frac{v_{\infty}}{r}$ where $\omega_{\infty}$ and $v_{\infty}$ are the angular velocity and the horizontal tangential velocity respectively, both according to the Schwarzschild observer at $r = \infty$. We can now conveniently rewrite the metric as:
$$\rightarrow \mathrm d\tau= \mathrm dt \sqrt{1 - \frac{2GM}{rc^2} - \frac{v_{\infty}^2}{c^2} }.\tag{1}$$
We now have a nice expression of time dilation as a function of gravitational potential and velocity. Since the coordinate velocity $(v_{\infty})$ measured at infinity is slower than the local velocity $(v)$ measured by a local observer at $r$ that is stationary relative to the gravitational field, by a factor of $\sqrt{1-2GM/(rc^2)}$, the metric can be written in terms of local velocity as:
$$\rightarrow \mathrm d\tau= \mathrm dt \sqrt{1 - \frac{2GM}{rc^2} - \frac{v^2}{c^2}\left(1-\frac{2GM}{rc^2}\right)}. \tag{2}$$
When the local velocity of a particle is measured to be $c$, the above equation collapses to $\tau = 0$, which is the expected result for a photon, indicating we are on the right path.
The above equation can be factored to:
$$\rightarrow \mathrm d\tau= \mathrm dt\ \sqrt{\left(1 - \frac{2GM}{rc^2}\right)}\ \sqrt{\left( 1 - \frac{v^2}{c^2}\right)}. \tag{3}$$
And now, it can be seen that the total time dilation is simply the product of the gravitational time dilation factor and the velocity time dilation of special relativity, when we use local velocity.
For an inertial particle following a circular orbit, we can use the well-known equation
$$v_{\infty}=\sqrt{\frac{GM}{r}}\tag{4}.$$
Substituting this expression into equation $(1)$ gives us the fairly well-known equation:
$$\mathrm d\tau = \mathrm dt \sqrt{ 1- \frac{3GM}{rc^2}},\tag{5}$$
I have digressed a bit from the main question, but I wanted to show that the relatively little-known equation $(3)$ is exact and consistent with known results of the Schwarzschild metric.
So far, we have only considered horizontal tangential velocity.
Vertical velocity is subject to both gravitational length contraction and gravitational time dilation, so the coordinate vertical velocity $v_{\parallel \infty}$ in terms of local velocity is subject to the inverse gravitational gamma factor squared. When we express the vertical coordinate velocity ($dr/dt$) in terms of local velocity, we end up with equation (2) again, confirming that equation (3) is universal and independent of the direction of the velocity as long as we use local velocity.
So far, we have only compared the proper time of an arbitrary clock moving in a gravitational field to a static clock at infinity. To compare the proper time of a clock at radius $r_1$ with local velocity $v_1$ with another clock at radius $r_2$ with local velocity $v_2$, we use:
$$\frac{\mathrm d\tau_1}{\mathrm d\tau_2} = \frac{\sqrt{\left(1 - \frac{2GM}{r_1c^2}\right) \left( 1 - \frac{v_1^2}{c^2}\right)}}{\sqrt{\left(1 - \frac{2GM}{r_2c^2}\right) \left( 1 - \frac{v_2^2}{c^2}\right)}}\tag{6}$$
It is easy enough to plug in values for a GPS satellite and a clock on the surface of the Earth into the above equation to find the required time dilation adjustment factor for the GPS satellite so that it synchronises with the ground clock.
It can also be seen that if $r_1=r_2$, then the gravitational time dilation factors cancel out, and locally, the equations of special relativity apply even in a gravitational field.
All the above assumes the Earth is not rotating. If we want to be really fussy, we should probably use the Kerr metric for a rotating gravitational body, and, using the same procedures as used for the Schwarzschild metric, we end up with:
$$\mathrm d\tau= \mathrm dt \sqrt{1 - \frac{2GM}{rc^2} - \frac{v_{\infty}^2}{c^2} +\frac{4GJ_{\infty}}{rc^3} - \frac{J_{\infty}v_{\infty}}{M^2 r^2 c^4} }$$
which is basically the same as equation $(1)$ obtained from the Schwarzschild metric with a couple of extra terms to account for the rotation of the gravitational body. As far as I know, the Kerr metric is not used to calibrate the GPS satellite clocks. It may be that the additional terms are insignificant for the Earth, and some people would dispute whether the Kerr metric is actually valid for real-world cases of solid rotating bodies. If there is sufficient interest and when I have sufficient spare time, I might do some numerical calculations using real data to see what difference using the Kerr metric makes.
