By Paul Hellings

2014-12-23 08:54:03 8 Comments

In the Wikipedia article on Gravitational Time Dilation I see two formulae for the gravitational time dilation. One contains $\sqrt{1 - r_0/r}$ but the other $\sqrt{1 - 1.5 r_0/r}$. What formula should I use to calculate the time dilation due to gravity inside a satellite? The speed induced time dilation is of course just special relativity.

With the first of these two formulas I can perfectly reproduce the graphs on time dilation as a function of $r$, I find that the two contributions to time dilation cancel at 1.5 Earth radii, all numbers are OK... But why not the second formula?


@John Rennie 2014-12-23 10:19:43

To separate time dilation into a gravitational time dilation and a special relativistic time dilation is a somewhat artificial distinction. To calculate the time dilation you calculate the length of the target object's world line, and this is affected both by the curvature of spacetime and the relative motion. Your end result might be split into separate bits depending on the context, but the calculation cannot be split into a gravity bit and a special relativity bit.

A satellite hovering stationary above the earth and one in a circular orbit obviously have very different worldlines, and it shouldn't surprise you to find their time dilation is different. Note that to calculate the time dilation for the orbiting satellite you can't just start with the time dilation for a stationary satellite and just multiply it by some special relativity derived factor. In both cases you have to do a full GR calculation to calculate the satellites proper time.

So if you're describing a satellite in a stable circular orbit calculate the total time dilation using the equation:

$$ t_0 = t_f\sqrt{1 - \frac{3}{2}\frac{r_0}{r}} $$

or for a satellite hovering use:

$$ t_0 = t_f\sqrt{1 - \frac{r_0}{r}} $$

A few extra interesting points (since we're all nerds here :-)

You've spotted that for the orbiting satellite the time dilation becomes infinite at $r = \tfrac{3}{2}r_0$. This radius is known as the photon sphere because it's the distance at which a light ray will orbit a black hole (though note this is an unstable orbit). Since the time dilation becomes infinite as the velocity approaches $c$, or more precisely as the world line approaches a null geodesic, it shouldn't surprise you that your equation gives an infinite time dilation.

Note also that no massive particle can have a stable circular orbit at $r < 3r_0$, despite what the time dilation formula says. Any massive particle attempting an orbit for $r < 3r_0$ will inevitably be drawn into the black hole unless some assistance, e.g. rocket motors, is used to sustain the orbit.

@Hammar 2017-02-20 08:43:46

“if one twin spent 79 years living at an altitude 1 foot higher than her sister, the first twin would end up approximately 90 billionths of a second older” Is it possible to calculate how fast should the elevated twin moves/orbit so her time dilation equal her sister time dilation on the ground ?

@mmeent 2020-05-18 17:06:07

"Any massive particle attempting an orbit for r<3r0 will inevitably be drawn into the black hole unless some assistance, e.g. rocket motors, is used to sustain the orbit." This incorrect. The unstability of unstable circular orbits can work both ways. Either plunging into the black hole, or sending the particle back out on either an eccentric orbit (for $3r_0>r> 2r_0$) or out to infinity (for $2r_0>r> 3/2r_0$).

@Ehryk 2014-12-23 09:47:00

Use $\sqrt{1 - 1.5 r_0/r}$

It's under the heading 'circular orbits'. The other equation,

$\sqrt{1 - r_0/r}$

Is for the time dilation near non rotating bodies.

@Hypnosifl 2014-12-23 15:49:10

The issue is not whether the body itself is rotating or non-rotating, but whether the clock is orbiting the body or if it's just "hovering" (using rockets, say) at a constant radius and angular coordinate in Schwarzschild coordinates.

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