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I understand that higher the gravitational potential, "slower" time runs. So a clock on international space station will run "faster" than a clock at sea level.

However does it matter where the gravity is coming from? (i know it shouldnt since the gravitation force formula has no direction, just mass and distance)

Here is the part that i want to understand:

say you have an object of mass $M$ at a distance of $r$ from the observer $O_1$

and you have $4$ smaller objects of mass $M/4$ each placed at the point of a compass with the observer $O_2$ at the center at a distance $r/4$

Will the two observers experience the same time dilation?

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    $\begingroup$ Hi zaftcoAgeiha and welcome to Physics.SE! Please see this help post to learn how to write your equations in a way nicer way i.e. in $\LaTeX$, in order to improve legibility. I did it this time for you but it'd be better if you would do it by yourself in the future. Thanks! $\endgroup$ – Gonenc Feb 11 '16 at 19:06
  • $\begingroup$ cool! did not know we can use latex here :) $\endgroup$ – WindowsMaker Feb 11 '16 at 19:37
  • $\begingroup$ I think I'm misunderstanding. In your example, isn't the second observer experiencing zero net force/zero net gravity? $\endgroup$ – barrycarter Feb 11 '16 at 21:10
  • $\begingroup$ i guess that is part of the question. The fact that you're experiencing 0 force, does that mean you are not experiencing any time dilation? say you have a clock on the international space station, you are experiencing time dilation. Now say the station starts to accelerate away from earth at exactly 1g. you are not experiencing any force anymore but you should still feel time dilation, no? $\endgroup$ – WindowsMaker Feb 11 '16 at 21:28
  • $\begingroup$ I don't think so. If you're experiencing zero force, you are in an inertial reference frame, meaning you are not accelerating and there is no time dilation. In other words, in an inertial reference frame, you age more than in any other frame (as a note, you'd have to accelerate at slightly less than 1g since gravity decreases with distance from Earth) $\endgroup$ – barrycarter Feb 12 '16 at 4:54
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The relationship between time dilation and gravitational potential that you mention is a weak field approximation i.e. it applies only when the curvature of spacetime is small. In this limit the time dilation is given by:

$$ \frac{dt_B}{dt_A} = \sqrt{1 - \frac{2(\Phi_A - \Phi_B)}{c^2}} \tag{1} $$

The quantity $\Phi_A - \Phi_B$ is the difference in the Newtonian gravitational potential energy between $A$ and $B$, and $dt_B/dt_A$ is the time dilation of $B$'s clock relative to $A$'s clock.

In this approximation it does not matter what distribution of masses causes the difference in gravitational potential, so in the example you give the time dilation due to your four small masses would be equal to that of the single large mass.

But remember that this is only an approximation. The time dilation is calculated from the metric that describes the spacetime, and the metric for four smaller masses is different from that of a single larger mass.

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