When reading simplified explanations of time dilation experienced by satellites, such as those used for the GPS and other satellite navigation systems, the time dilation is often presented as having two contributions, one from special relativity (SR) caused by the difference in velocity, and another from general relativity (GR) caused by the difference in gravity ($\approx$ acceleration due to gravity).
But does the GR effect only arise because one of the clocks stands "heavy" on the surface of the Earth, while the other clock is "weightless"?
In the above diagram, $M$ is a heavy object (like the Earth). $A$ is a freely falling clock (of negligible mass) in circular orbit. And $B$ is another freely falling (massless) clock, in the perigee of some eccentric trajectory.
Let us say I have picked the situation in a way such that the current velocities of $A$ and $B$ are identical, both with respect to magnitude (speed) and direction. (I know it may not make sense in Riemannian geometry to say the velocity vectors in different locations are parallel, but I am doing it anyway). Then the SR time dilation should be zero in this instant.
Which of the following two contradictory statements is (more) correct:
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At the instant shown, there is no time dilation between the clocks $A$ and $B$, since both are falling freely. The free fall cancels the gravity of $M$ (appeal to equivalence principle).
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Because $A$ is deeper down in the gravitational "well" than is $B$, the clock $A$ runs slower than the clock $B$. This is the gravitational time dilation.
Best Answer
Option (2) is correct.
The time dilation, both the "GR" and "SR" components, arises from applying the Schwarzschild spacetime interval to determine the proper time interval for one observer compared to a fiducial Schwarzschild coordinate time interval $dt$. Then doing the same for the other observer and then taking the ratio to determine the relative time dilation.
Given the usual equation for the proper time interval $$c^2 d\tau^2 = c^2 \left(1 - \frac{r_s}{r}\right) dt^2 - \left(1-\frac{r_s}{r}\right)^{-1}dr^2 - r^2d\theta^2 - r^2\sin^2\theta d\phi^2\ ,$$ then in the situation you hypothesise, we could say that everything happens in the plane $\theta = \pi/2$ for both clocks and at the points mentioned the velocities are entirely tangential. Thus $dr = d\theta =0$.
Dividing through by $c^2 dt^2$, we have for both observers $$\frac{d\tau}{dt} = \left[ 1 - \frac{r_s}{r} - \frac{v_\phi^2}{c^2}\right]^{1/2}\ ,$$ where $v_\phi$ is the tangential velocity.
The relative time dilation between the two clocks is given by $$\frac{d\tau_A}{d\tau_B}= \left[\frac{1 - r_s/r_A - v_\phi^2/c^2}{1- r_s/r_B - v_\phi^2/c^2}\right]^{1/2} $$ and if $r_A < r_B$ then clock A runs slow relative to clock B.
Note that if, as is often the case, $r_A, r_B \gg r_s$ and $v_\phi^2 \ll c^2$ then a binomial expansion yields $$\frac{d\tau_A}{d\tau_B} \simeq 1 -\frac{1}{2}\left[ \frac{r_s}{r_A}-\frac{r_s}{r_B}\right] $$ and the $v_\phi$ terms disappear. If on the other hand you were to fix A to the surface of the mass, then $v_\phi$ in the numerator would be the surface rotation speed of the mass.