I Reconciling smooth spacetime with variable, rotating, earth gravity

[John Helly]

TL;DR Summary

How do the gravity measurements from satellite square with smooth spacetime?

Aloha.

I'm studying Sean Carroll's Spacetime text and just beginning to understand the differential approximations to spacetime curvature. I'm trying to learn general relativity.

However, as an oceanographer, I am familiar with the gravimetry satellite measurements and wonder how to reconcile the variable distribution of gravity measurements across the rotating earth with the concept of spacetime curvature inducing apparent accelerations corresponding to Newtonian forces.

I can't get past the Newtonian concept of gravity associated with mass distribution as reflected in the satellite measurements. The only interpretation I can imagine is a curved spacetime field rotating like a vortex with the earth rotation. In which case, how could the gravity at a point remain fixedly attached to, say, a seamount (unerwater volcano)?

So, I would appreciate any thoughts on this question.

J.

 

[Dale]

Essentially you are correct. If you think of spacetime then a circular orbit is a helix. So the helical path of matter makes helical curves in spacetime.

 

[John Helly]

Mahalo. That helps greatly.
J.

 

[Ibix]

The only interpretation I can imagine is a curved spacetime field rotating like a vortex with the earth rotation.

There are a couple of points to note here. First is that Carroll will only cover analytically tractable spacetimes - prmarily the Schwarzschild spacetime, which is the spacetime around a perfectly spherically symmetric non-rotating mass. For most purposes that's an excellent approximation to the Earth, but obviously in your profession you are interested in the deviation from that approximation. Carroll won't cover that because he's teaching relativity at the equivalent level to studying exact elliptical orbits around point masses in Newtonian gravity.

Second, spacetime doesn't rotate or change in any way - it's the thing that provides a notion of time that lets you think about things changing. What you can do is imagine slicing 4d spacetime into a stack of 3d slices (the technical term for this is "foliating" spacetime) where each slice is what you call space at one instant of time. In a simple spacetime like Schwarzschild there's only one sensible way of doing this (although you can pick others) and the extreme uniformity of the mass distribution means that every "slice" is identical and spherically symmetric, which is how the metric ends up with no time or angular dependence.

If you modelled the Earth as a rigidly rotating non-uniform body you might expect to be able to find a foliation that has identical spatial slices with a complicated angular dependence and rotated with respect to each other. I suspect that this will be an extremely good approximation, but not quite accurate. An accurate model will emit tiny amounts of gravitational radiation and slow its spin (and I do mean tiny - I suspect that this process would take many multiples of the present age of the universe to have a measurable effect on the Earth's rotation rate), which would make each spatial slice very slightly different from its predecessor. Of course that difference is utterly negligible on any human scale, but it will instantly make the maths intractable without some approximation. Which is why Carroll is sticking to simple cases.

 

[PeterDonis]

how to reconcile the variable distribution of gravity measurements across the rotating earth with the concept of spacetime curvature

This is simple: the spacetime curvature is linked to the distribution of matter (and energy and stress and anything else that goes into the stress-energy tensor) by the Einstein Field Equation. As @Ibix noted in post #4, the solutions you see in textbooks are for very simple distributions of matter that have a high degree of symmetry--because those are the only cases for which we can actually solve the equations analytically. For real world distributions of matter, like that of the actual Earth, we have to solve the equations numerically instead. Detailed measurements of the actual spacetime curvature around the Earth due to its actual matter distribution would be analyzed this way. But the physics is still the same, and is described by the first sentence of this paragraph.

inducing apparent accelerations corresponding to Newtonian forces

No, this is not correct. Spacetime curvature is tidal gravity, and that is not "apparent accelerations corresponding to Newtonian forces". The latter are due to being in a non-inertial frame; when you're sitting or standing at rest on the Earth's surface, the Earth is pushing up on you, so you're not in free fall; you feel weight. If you drop a rock, it is in free fall, so to you it appears to accelerate downward. But from the point of view of the rock's rest frame, which is an inertial frame (a local one), you are the one who is accelerating--upwards. And if we attach accelerometers to both you and the rock, the one attached to the rock will read zero, while the one attached to you will read 1 g. So as far as GR is concerned, you are the one who is actually accelerating.

Spacetime curvature comes into play when you have multiple rocks, all in free fall near each other, that start out at rest relative to each other, but don't remain at rest relative to each other. That's an example of tidal gravity.