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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?

WindowsMaker
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1 Answers1

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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.

John Rennie
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