PatrickPowers said:
I have read that at sea level gravity is the same everywhere on the surface of the Earth. I have also read that motionless clocks at sea level will stay synchronized.
The second sentence is true (at least to a very good approximation; see below for some further comments). The first is not, at least not if by "gravity is the same" you mean the "acceleration due to gravity" is the same. The acceleration due to gravity is slightly smaller at the Earth's equator (on the surface) than at the poles.
The difference is due to the fact that, speaking somewhat loosely, the clock rate depends on the potential, and the acceleration depends on the spatial gradient of the potential. The Earth's surface is an equipotential surface--again, at least to a very good approximation. If the Earth were a perfect fluid, its surface would be a perfect equipotential surface. But the spatial gradient of the potential is not constant over its surface, because of rotation. See below.
http://arxiv.org/abs/gr-qc/0411060It seems to me that this should be a consequence of general relativity. If one instead uses Newtonian gravitation then gravity would be less at the equator because both the rotation of the Earth and the shape of the Earth as an oblate spheroid would decrease gravity.[/QUOTE]
This is true in GR as well, as the page Q-reeus linked to makes clear. As I noted above, the acceleration due to gravity is slightly smaller at the equator than at the poles, basically for the reasons you give. That page also gives Newton's reasoning for why the surface of the Earth is an equipotential surface, at least to a very good approximation: if the Earth's surface were not an equipotential surface, then the oceans would slosh around to fill up the lower places (i.e., lower potential) from the higher places (i.e., higher potential) to *make* the surface equipotential. The fact that the Earth's surface is not entirely ocean means that it can't adjust itself completely in this way; but over long enough time scales, even the solid crust of the Earth is not completely rigid, so it too will adjust itself to make the surface equipotential to a very good approximation.
This reasoning is still valid in GR; but GR gives us an additional tool, the spacetime metric, with which to specify more precisely exactly what the shape of the equipotential surface will look like. The problem is, as the page Q-reeus linked to notes, we don't have an exact solution for the metric of a rotating oblate spheroid. However, we can make a very good approximation as follows:
(1) The Earth's radius is much, much larger than its mass in geometric units (about a million times larger). (The "mass in geometric units" is GM / c^2, where M is the mass in conventional units--i.e., it is half the Schwarzschild radius of a black hole with the same mass.) This implies that the metric at the Earth's surface is very close to flat; i.e., the metric is the Minkowski metric plus small correction terms.
(2) Also, the Earth is rotating very slowly (i.e., the speed of a point on its surface is much much less than the speed of light). This implies that the metric is very close to the Schwarzschild metric, with small correction terms for the rotational velocity.
(3) The key metric coefficient we're interested in is [itex]g_{tt}[/itex], since that's the one that governs the "rate of time flow"; also, to a very good approximation, its spatial gradient is what governs the "acceleration due to gravity". (In other words, [itex]g_{tt}[/itex] is basically the "potential" I referred to above.)
Given all the above, we can write:
[tex]\frac{d \tau}{dt} = \sqrt{g_{tt}} = \sqrt{1 - \frac{2M}{r} - v^{2}}[/tex]
where I have used units where G = c = 1. M is the mass of the Earth, r is the radius at a given point on the surface, and v is the velocity of that point due to the Earth's rotation.
On the equator, we can write [itex]v = \omega r[/itex], where [itex]\omega[/itex] is the rotational frequency of the Earth, and since the metric is very close to flat (all the terms under the square root other than 1 are << 1), we can expand out the square root to obtain, to first order, the equation near the end of the mathpages page Q-reeus linked to:
[tex]\frac{d \tau}{dt} = 1 - \frac{M}{r} - \frac{1}{2} \omega^{2} r^{2}[/tex]
Taking the derivative of this with respect to r, as the mathpages page notes, gives the net acceleration due to gravity, and you can use these formulas to run the numbers and see that (a) the potential [itex]d \tau / dt[/itex] is the same, to a very good approximation, at the equator and at the poles, but (b) the acceleration due to gravity is slightly smaller at the equator than at the poles.