# I Is a clock on the equator time dilated?

1. Aug 12, 2018

### Grimble

Clocks in orbit are time dilated due to their orbital velocity.
Do we measure a similar effect for clocks on the equator when measured against clocks at the poles?

2. Aug 12, 2018

### FactChecker

I am sure that they would be. I'm not sure that anyone has ever proved it because the separation distance might be a problem. Also that larger diameter around the equator complicates things. The greater effect for clocks in orbiting satelites is the increased speed of the clocks due to the increased altitude and decreased gravitational pull. So the orbiting clocks in GPS satelites actually go faster than on Earth.

3. Aug 12, 2018

### Staff: Mentor

There are two competing effects here. There is the velocity that you already noticed and there is also the equatorial bulge. The velocity causes the equator clock to slow down, but the equatorial bulge causes the equator clock to speed up due to being higher in a gravitational field.

It turns out that for any pair of clocks on the geoid these two effects exactly cancel. So a clock on the geoid at the pole will run at the same rate as a clock on the geoid at the equator.

Last edited: Aug 12, 2018
4. Aug 12, 2018

### stevendaryl

Staff Emeritus
They exactly cancel? What's the argument for that?

Approximately, the time dilation at different locations is given by $\Delta \tau \approx \frac{1}{c^2} (\Delta V - \Delta T)$, where $\Delta V$ is the change in gravitational potential (per unit mass) and $\Delta T$ is the change in kinetic energy (per unit mass). For the time dilation to be zero between a clock at the North Pole and a clock at the equator, it must be that $\Delta V = \Delta T$.

$\Delta T = \frac{1}{2} v^2 = \frac{1}{2} (\frac{2 \pi R}{T})^2$

where $R$ is the radius of the Earth and $T$ is the time for one rotation.

$\Delta V \approx g \Delta R$

where $\Delta R$ is the change in radius from the North Pole to the equator, and $g$ is the gravitational acceleration.

So you're saying that

$g \Delta R \approx \frac{1}{2} (\frac{2 \pi R}{T})^2$

What's the argument for that?

5. Aug 12, 2018

### Staff: Mentor

We're talking about the geoid here, which is the hypothetical surface of the earth if it were a perfect fluid (or to very good approximation, all ocean) so the tangential speed has the right relationship to the potential in the weak field approximation. Google for "geoid time dilation" will find some derivations.

6. Aug 12, 2018

### FactChecker

I googled it but couldn't spot it. It might be hiding amungst all the GPS articles. Do you know of a specific reference? Is it theoretically exactly equal for some reason?

7. Aug 12, 2018

### stevendaryl

Staff Emeritus
An argument that I saw from my Googling research was this: Choose a rotating coordinate system in which the Earth is at "rest". In this coordinate system, the "gravitational field" includes both real gravity and centrifugal force. The surface of the ocean is an equipotential surface in this "gravitational field". But mathematically, the gravitational potential $V$ is related to the metric as follows: $1 + V = \sqrt{g_{00}}$. So equipotential implies equal value for $g_{00}$. And proper time is related to coordinate time through: $d\tau = \sqrt{g_{00}} dt$ (for a clock at "rest"). So clocks everywhere at sea level have the same proper time.

8. Aug 12, 2018

### Staff: Mentor

In the rotating frame of reference the centrifugal force points up and the gravitational force points down and the combined force forms a potential. The geoid then is an equipotential surface. Since time dilation is related to the change in potential this equipotential surface is also an “equidilation” surface.

Edit: I see you already found this!

Last edited: Aug 12, 2018