Originally posted by @joe-shmo
Here is a PhD's quick answer.
https://www.quora.com/Does-time-run-slower-on-the-equator-than-at-the-poles
He says the effects cancel.
I was also wondering about whether the centrifugal force felt in our rotating frame effects the clocks as far as relativity is concerned, it does...
One question:
He goes on to say, "And the earth, being o ...[text shortened]... ugh it's quite small, of order 1.4E-11, it's well within the precision of modern atomic clocks."
I think he's wrong. Sorry to be slow about this, but I had to get around to actually doing it, look up all the numbers and then feed them into a spread sheet. The result is that a clock at the pole will gain about 3 nanoseconds per second relative to one at the equator (assuming I haven't goofed up somewhere).
The way to calculate this is to start with the Schwartzschild metric (I'll check the Kerr Metric as well, but I really doubt it will make any difference).
The metric is:
c²dτ² = (1 - rₛ/r)c²dt² - dr² / (1 - rₛ/r) - r² (dθ² + sin²θ dφ² )
t is the elapsed time as measured by an asymptotic observer (one far away from the Earth and stationary with respect to the Earth's centre).
r, θ, and φ are the normal polar coordinates.
dx where x is one of t, r, θ, or φ refers to a small change in one of these coordinates along the worldline of either of the two twins.
τ is the elapsed proper time along the worldline of one or other of the twins. In other words the time shown on their clocks.
rₛ = GM/c² is the Schwartzschild radius.
Neither twins world lines involve r or θ changing so we have that dr = dθ = 0. At the pole θ = 0 and sin(0) = 0. So the change in proper time dτ for the twin at the pole is given by:
c²dτₚ² = (1 - rₛ/r)c²dt²
and we have
dτₚ/dt = sqrt(1 - rₛ/r)
This is the rate that the twin at the poles clock ticks relative to the asymptotic observers.
For the twin at the equator θ = π/2 and sin(π/2) = 1. So in this case we have:
c²dτₑ² = (1 - rₛ/r)c²dt² - r²dφ²
dφ/dt is just the angular velocity ω and the linear velocity v = ωr. So we can write:
dτₑ/dt = sqrt( 1 - rₛ/r - ωr²/c² ) = sqrt (1 - rₛ/r - v²/c² )
So, were the Earth a perfect sphere the observer at the pole would have the faster clock. When I took the differing radii into account and put the numbers into a spreadsheet I got:
dτₚ/dt = 1 - 3.448E-10 for the twin at the pole and
dτₑ/dt = 1 - 6.977E-10 for the twin at the equator.
Dividing these two gives us the rate at which the polar twin's clock gains on the equatorial twin's clock:
dτₚ/dτₑ = 1 + 3.489E-10 > 1
However this calculation does not take the effect of rotating frames into account (I'd need to use the Kerr Metric and have some sort of estimate to the moment of inertia of the Earth, or at least its angular momentum). It might be that my results converge with your PhD's answer if I did that (which would be interesting as I wouldn't have expected the Kerr metric to be needed for this).