Gravitational time dilation and a race If there is a person far away from a gravitational source(higher gravitational potential) the faster time passes relative to a person closer to that gravitational field. Is this correct?(this is how I understand it)
So, my question is, if there is someone far from a source of gravity and a person closer to that gravity source then that means that person A's time will be running faster compared to B's clock. If A and B had a race and they were both going $10 \; \mathrm{m/s}$ and traveled a distance of $5 \; \mathrm{km}$ who would win the race?
I was thinking about this all day and, I think that from an outside observer that person A should finish first due to his time being faster than person B.  

 A: If the local velocity as measured by the particles themself is 10m/sec and the rulers measuring the distance show the distances in the frame of a far away observer then the one farther away from the source of gravity would win, since the particle which is nearer to the source of gravity would have his velocity shapirodelayed by a larger amount than the person which is farther away. You can also see this in this animation, where the light rays farther away from the mass overtake the ones which are nearer, although locally they all have a velocity of 1c:

A: $\let\D=\Delta$
Let me begin giving a precise statement of the problem.
We are in the neighbourhood of a spherical nonrotating star (or planet) of mass $M$. Two tracks have been built: the first at (Schwarzschild) radial
coordinate $r_1$, the other at $r_2>r_1$. Both tracks have length $d$, as
measured by local instruments (formulas would be $d=r_1\D\phi_1$ and
$d=r_2\D\phi_2$ for equatorial tracks).
Race times are measured by local stationary clocks. Since $d$ is the
same and the speeds $v$ are the same, measured times will also be the
same:
$$\D\tau_1 = \D\tau_2 = d/v.$$

who would win the race?

A clear-cut definition of "win" is needed. If times $\D\tau_1$ and
$\D\tau_2$ were compared (say by radio automatic transmission of clocks readings to a referee) the race would be a tie. But we can consider two other ways of comparing times:


*

*Sending two light signals from clocks in 1 to those in 2. The
signals start $\D\tau_1$ apart and arrive $\D\tau_1'$ apart (calculation follows). The reverse could also be done, and is treated in an analogous way.


Using Schwarzschild's metric, the well-known relation between
Schwarzschild time $\D t$ and proper time $\D\tau$ of a stationary
clock results:
$$\D\tau = \D t\,\sqrt{1 - 2M/r}\qquad(1)$$
(for sake of brevity I have chosen units where $c=1$, $G=1$). The only other thing one must know is that during light propagation $\D t$ is conserved. So, for the same $\D t$, we will have
$${\D\tau_1' \over \D\tau_1} = \sqrt{1 - 2M/r_2 \over 1 - 2M/r_1}$$
since both times refer to runner #1 but are measured on different clocks: $\D\tau_1$ at radius $r_1$, $\D\tau_1'$ at radius $r_2$. From $r_2>r_1$ we see that $\D\tau_1'>\D\tau_1=\D\tau_2$ and runner #2 is declared the winner.


*Sending two light signals from clocks in 1 to a faraway clock,
telling Schwarzschild time $t$. The same is done for clocks in 1. Let
$\D t_1$ and $\D t_2$ be the time intervals at the arrival of these
signals.


Again, from eq. 1:
$$\D\tau_1 = \D t_1\,\sqrt{1 - 2M/r_1}$$
$$\D\tau_2 = \D t_2\,\sqrt{1 - 2M/r_2}.$$
Recalling $\D\tau_1=\D\tau_2$ and dividing
$${\D t_1 \over \D t_2} = \sqrt{1 - 2M/r_2 \over 1 - 2M/r_1}$$
$$\D t_1>\D t_2 $$
and #2 is the winner.
