General Relativity – Does Gravitational Time Dilation Occur Due to Height or Field Strength?

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The reason why you have to tune differently the atomic clocks in GPS is because the GPS is higher or because there is less gravity there, or both? In other words in a constant gravitational field which doesn't differ with height, will time dilation still occur?

They say that the reason why people on the first floor age slower than people on higher floors is because as you get further from the earth gravity weakens. Is that true? Does the difference in the field cause that or just the distance?

If pure distance doesn't matter then why do we say that for a spaceship accelerating forward clocks at the front tick faster than clocks at the back since both are accelerating at the same rate?

Best Answer

The time dilation is due to a difference in the gravitational potential energy, so it is due to the difference in height. It doesn't matter whether the strength of the gravitational field varies, or how much it varies, all that matters is that the two observers comparing their clocks have a different gravitational potential energy.

To be more precise about this, when the gravitational fields are relatively weak (which basically means everywhere well away from a black hole) we can use an approximation to general relativity called the weak field limit. In this case the relative time dilation of two observers $A$ and $B$ is given by:

$$ \frac{dt_A}{dt_B} \approx \sqrt{ 1 + \frac{2\Delta\phi_{AB}}{c^2}} \approx 1 + \frac{\Delta\phi_{AB}}{c^2} \tag{1}$$

where $\Delta\phi_{AB}$ is the difference in the gravitational potential energy per unit mass between $A$ and $B$.

Suppose the distance in height between the two observers is $h$, then in a constant gravitatioinal field with acceleration $g$ we'd have:

$$ \Delta\phi_{AB} = gh $$

If this was on the Earth then taking into account the change in the gravitational potential energy with height we'd have:

$$ \Delta\phi_{AB} = \frac{GM}{r_A} - \frac{GM}{r_B} $$

where $r_A$ and $r_B$ are the distances of $A$ and $B$ from the centre of the earth and $M$ is the mass of the Earth. Either way when we substitute our value of $\Delta\phi_{AB}$ into equation (1) we're going to get a time dilation.

As for the accelerating rocket: the shortcut is to appeal to the equivalence principle. If acceleration is equivalent to a gravitational field then it must also cause a time dilation in the same way that a gravitational field does.

Alternatively we can do the calculation rigorously. The spacetime geometry of an accelerating frame is described by the Rindler metric, and we can use this to calculate the time dilation. The Rindler metric for an acceleration $g$ in the $x$ direction is:

$$ c^2d\tau^2 = \left(1 + \frac{gx}{c^2}\right)^2c^2dt^2 - dx^2 - dy^2 - dz^2 $$

We get the time dilation by setting $dx=dy=dz=0$ to give:

$$ c^2d\tau^2 = \left(1 + \frac{gx}{c^2}\right)^2c^2dt^2 $$

and on rearranging this gives:

$$ \frac{d\tau}{dt} = 1 + \frac{gx}{c^2} $$

which is just the equation (1) that we started with.

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