Do clocks tick faster when gravitational forces are weaker?

Question

A professor last year taught us that "gravity slows clocks," when teaching about the relationship between gravity and time. This led me to think about places, such as intergalactic space, where gravitational forces are weaker and if clocks there would tick relatively faster. "Gravity slows clocks", seems to imply that when there is more gravity, then clocks tick slower. So, does less gravity mean clocks tick faster? Rephrased more technically, do clocks tick faster when gravitational forces are weaker?

Answer 1

The (relative) rate of clocks depends on the gravitational potential not force. For example, a clock at the center of the Earth would experience no acceleration due to gravity, but would tick slower than one at the Earth's surface. As others have mentioned, this gravitational time dilation was predicted by General Relativity and has been confirmed by many, many experiments (including the operation of the GPS satellites).

Answer 2

Pop sci (e.g., YouTube) regularly says gravitational fields slow clocks, and then the comments are flooded with Flat Space Timers claiming acceleration changes how clocks function, not time itself. Ug. It's maddening.

As pointed out by @Eric Smith, it's not force (which is local), the difference between two clocks depends on the potential difference between them (at least in weak fields, nothing I say here would necessarily be true near an event horizon).

In fact, you would be hard pressed to turn an acceleration, $\vec g$, into a time delay without multiplying it by a distance (Which gives you a potential energy per unit mass, which has dimension of $c^2$).

The time dilation formula go like:

$$ \gamma = \frac{1}{\sqrt{1-\frac{2\alpha}{c^2}}}$$

The kinematic Lorentz factor has:

$$ \alpha = \frac 1 2 v^2 $$

(kinetic energy per unit mass), while gravity is:

$$ \alpha = \int_{r_1}^{r_2}\vec g(r')d\vec r' \rightarrow \frac{MG}{r} $$

or the difference in gravitational potential per unit mass.

So for a reference, the difference in clock rate between two points in gravitational field is the same as the Lorentz factor from the velocity a fiducial particle would pick up falling from the higher point to the lower point.

Also: if you apply it to the Twin Paradox, the space twin moving at $v$ and turning around at a distance $D$ at $t'=D/v/\gamma$, the simultaneous time back on Earth ($x'=-D/\gamma\equiv D'$)is given by the Lorentz transform:

$$ t_{\pm} = \gamma(t' + \frac{vx'}{c^2})\rightarrow \frac D v \pm vD $$

where $\pm$ refers to the outbound and inbound velocity, respectively.

So the time change defining space-twin's "now" back on Earth is:

$$ \Delta t = 2vD = \Delta v D $$

Note that $\Delta v$ is an acceleration integrated over the turnaround time (which can be zero in though experiment), so the formula looks like:

$$ \Delta t = Dg $$

where $g$ is a uniform universal gravitational field in the accelerating rocket frame.

This is why some people say you need gravity to resolve the Twin's Paradox, but you don't..it's just a coordinate transform from the outbound frame to the inbound frame, but it looks just like gravity.

Answer 3

Yes, this is a well-known effect referred to as gravitational time dilation: https://en.wikipedia.org/wiki/Gravitational_time_dilation