Synchronizing Atomic Clocks in Inertial Motion: Planet or Black Hole?

[snailhunter]

Say there are two atomic clocks floating through space, both completely stationary with respect to each other. They are exactly synchronized, and not under the influence of any significant gravitating bodies. One is significantly farther ahead in the direction of motion than the other, but they both move at the same speed. Let's call the one farthest ahead in the direction of motion clock A, and the other clock that trails behind it clock B.
They eventually begin to approach a large planet (it's straight ahead of them, so they will fall straight to the ground). Since clock A is closer to the planet than clock B at any given time (by a decent amount), clock A begins to feel the effects of the planet's gravity before clock B. From clock B's point of view, clock A begins to accelerate away. Now from what I understand, any object in free fall isn't undergoing any proper acceleration, and is still considered to be in inertial motion (it's just inertially moving through curved spacetime). So even though clock B sees clock A move away at an accelerated rate, would they still remain synchronized? Can they still be treated as though they're still in the same inertial frame of reference, even though technically their relative velocities are changing over time?
Feel free to replace "planet" with "black hole" as well, if that makes it any more interesting.

 

[stevebd1]

The shell velocity (i.e. relative to a specific r) for an object that has fallen from rest at infinity radially to a non-rotating object of mass is (where c=1)-

[tex]v_{\text{(shell rain)}}=\sqrt{\frac{2M}{r}}[/tex]

The shell velocity for an object that is falling inward at speed v_far from a great distance to an object of mass is-

[tex]v_{\text{(shell hail)}}=\left[\frac{2M}{r}+v_{\text{far}}^2\left(1-\frac{2M}{r}\right)\right]^{1/2}[/tex]

which reduces to v_{(shell rain)} when v_far=0

The total time dilation of an object falling towards an object of mass is a product of the time dilation due to velocity and gravity-

[tex]d\tau_{\text{(total)}}=dt\sqrt{1-v_{\text{(shell hail)}}^2} \cdot \sqrt{1-\frac{2M}{r}}[/tex]

based on the above, while two objects traveling at the same speed in the same direction but at a distance from each other might be in sync in flat space, once the lead object begins to fall towards an object of mass, its time dilation will increase over the second object and remain that way, even when they both arrive at the surface of the object of mass, due to the lead object having spent longer in the object of mass gravitational field.

 

[snailhunter]

The shell velocity (i.e. relative to a specific r) for an object that has fallen from rest at infinity radially to a non-rotating object of mass is (where c=1)-

[tex]v_{\text{(shell rain)}}=\sqrt{\frac{2M}{r}}[/tex]

The shell velocity for an object that is falling inward at speed v_far from a great distance to an object of mass is-

[tex]v_{\text{(shell hail)}}=\left[\frac{2M}{r}+v_{\text{far}}^2\left(1-\frac{2M}{r}\right)\right]^{1/2}[/tex]

which reduces to v_{(shell rain)} when v_far=0

The total time dilation of an object falling towards an object of mass is a product of the time dilation due to velocity and gravity-

[tex]d\tau_{\text{(total)}}=dt\sqrt{1-v_{\text{(shell hail)}}^2} \cdot \sqrt{1-\frac{2M}{r}}[/tex]

based on the above, while two objects traveling at the same speed in the same direction but at a distance from each other might be in sync in flat space, once the lead object begins to fall towards an object of mass, its time dilation will increase over the second object and remain that way, even when they both arrive at the surface of the object of mass, due to the lead object having spent longer in the object of mass gravitational field.

Thank you. Just to make sure I understand this right, does that mean that before either object hits the ground, the only time dilation that occurs will be due to the change in velocity of clock A relative to clock B (which is a "coordinate" or "geometric" acceleration, as opposed to a proper acceleration)?

I think this question will help clarify what I'm really trying to understand:
Take the example with the two clocks and planet again. Clock A, which is farther ahead in the direction of motion than clock B begins to accelerate toward the planet completely due to the planet's gravity. Now replicate this exact scenario, except there's no planet, and it's completely in flat space. Now, clock A uses on-board boosters to geometrically accelerate away from clock B exactly the same as it did in the planet scenario, except now there is a proper acceleration. Will the time dilation affect be the same? Will clock B see clock A tick slower by the exact same amount as it would in the first example (ignoring the part after they hit the planet, that is)?

 

[Naty1]

"Just to make sure I understand this right, does that mean that before either object hits the ground, the only time dilation that occurs will be due to the change in velocity of clock A relative to clock B"

nope...also see last sentence post #2


Two phenomena affect time synchronization: relative velocity and gravity (gravitational potential). You can deduce whatever example results you wish from those two facts.

xamples...
If two clocks are stationary but within different gravitational potentials clock synchronization will be lost...If two clocks have different elative velocity with the same gravitational potential, clock synchornization will be lost...etc,etc

Relative to a given reference frame, clocks tick slower based only on their velocity, not acceleration. But the twin paradox demonstrates the frame-independent fact that if two clocks compare readings, move apart, and then later reunite and compare readings again, then if one clock moved inertially between meetings while the other accelerated at some point, the one that accelerated will have elapsed less time.

 

[PAllen]

"Just to make sure I understand this right, does that mean that before either object hits the ground, the only time dilation that occurs will be due to the change in velocity of clock A relative to clock B"

nope...also see last sentence post #2


Two phenomena affect time synchronization: relative velocity and gravity (gravitational potential). You can deduce whatever example results you wish from those two facts.

xamples...
If two clocks are stationary but within different gravitational potentials clock synchronization will be lost...If two clocks have different elative velocity with the same gravitational potential, clock synchornization will be lost...etc,etc

Relative to a given reference frame, clocks tick slower based only on their velocity, not acceleration. But the twin paradox demonstrates the frame-independent fact that if two clocks compare readings, move apart, and then later reunite and compare readings again, then if one clock moved inertially between meetings while the other accelerated at some point, the one that accelerated will have elapsed less time.

There is an ambiguity here based on the meaning of acceleration. Consider a observer in a space station orbiting a planet, and another observer in a rocket accelerating such that they remain stationary relative to the planet (under the simplifying assumption that the planet isn't rotating). They synchronize clocks, and wait for the space station to orbit coming back to the 'accelerating' stationary observer. It is, in fact, the the space station clock which will have fallen behind even though they followed an inertial path. (I calculated this directly using the Schwarzschild metric).

The resolution is that there are more than one geodesic paths between the 'leave' and 'meet' spacetime events. The geodesic with longest proper time (the one shooting directly away from the planet and back, just in time, in free fall the whole time) will have a clock ahead of all other clocks connecting these events; however the accelerating clock can show a longer time than other geodesics.

If you restrict your self to a small enough region in both space and time, there will only be one geodesic, and there will be no exceptions to the ordinary rule of 'inertial ages fastest'.

[EDIT: A way to see that something like this must be so: Given 2 or more geodesics between two events, with different proper time along them, then there must be non-inertial paths 'close' to the longer time geodesic that have longer proper time than the shorter time geodesic.]