# The gravitational field of a moving, rotating globe

1. Dec 3, 2014

### James_Harford

Consider a solid globe of mass M and of of uniform density. My understanding is that its external gravitational field, in the absence of any other forces, will apply an instantaneous acceleration to any small test mass directly towards the center of the globe.
1. Is that still true in the frame of the test mass when the globe is moving?
2. If so, is it true if the moving globe is also rotating about an axis perpendicular to its velocity vector? Or will it accelerate towards the center of mass of the globe (now distinct from its center) instead? Or will its instantaneous acceleration be in some other direction?
The *motivation* underlying these questions is to know whether world line of the local center of mass of a rotating system (e.g. two orbiting masses) has little or no physical significance outside of the system (the meaning of this statement is possibly vague, hence the formulation as the above two questions).

2. Dec 3, 2014

### bcrowell

Staff Emeritus
I assume this is a question about general relativity, not Newtonian gravity.

GR doesn't describe gravity as producing accelerations of test masses. The acceleration of a test mass is zero by definition.

The spacetime surrounding a spinning body is not the same as the spacetime surrounding a non-spinning body. You get effects like frame dragging, which were verified by Gravity Probe B. If you really want an exact general-relativistic description of a material body in static equilibrium, I believe it's somewhat complicated, since pressure acts as a source of gravity, but realistically for a body like the earth these are not big effects.

In practice, effects like frame dragging are orders of magnitude weaker than the effects of the non-sphericity of the rotating body. Although you said you wanted to assume a "globe," which I assume means a perfect sphere, that isn't physically possible for any realistic form of matter with a field strong enough to have appreciable general-relativistic effects.

I don't see the connection between these parts of the question and the issue of whether the c.m. is well defined. The c.m. is in general not well defined in GR for a distribution of mass on a curved spacetime background. This is why we get things like relativistic "gliders" (Harte, "Extended-body effects in cosmological spacetimes," http://arxiv.org/abs/0706.2909v2[/PLAIN] [Broken] ). This kind of thing happens because we define the c.m. using vector addition, but in curved spacetime, vectors have to be parallel-transported to the same place before you can add them, and parallel transport is path-dependent.

However, if you have something like the earth-moon system, surrounded by an asymptotically flat spacetime, then I think it certainly does have a well-defined c.m., since the far field asymptotically approaches that of a spherically symmetric body.

When you ask about the case where a system is moving, the first thing to realize is that GR doesn't have global frames of reference, so in a general spacetime this question doesn't really mean anything. However, in an asymptotically flat spacetime, the system looks Newtonian from a distance, so there are no big surprises, you can talk about frames of reference, and the c.m. of the system behaves as you would expect from special relativity.

Last edited by a moderator: May 7, 2017
3. Dec 5, 2014

### James_Harford

Thank your helpful observations. I attempted to reformulate the question in a way that avoids the many pitfalls of the first version that you pointed out, including removing all references to CM.

Here is attempt #2 -- I hope it is not too verbose or unclear -- and thanks again for your careful analysis.

Consider an extended mass (EM) of mass m with a well-defined boundary, and a mass density function that, within its rest frame, is both stable and symmetric about both an axis of revolution and an equatorial plane. My understanding is that if it is not rotating, its external gravitational field will apply a tidal effect (a 3-vector) in any small test mass in a direction that, at any particular moment in time, is "pointed" directly towards the center (midpoint of the axis) of the EM. By this, I mean that a vector that is "pointed" in this direction can be parallel transported from the test mass along a path that is tangent both to the test mass's space slice of that moment and to the vector, and which terminates at the center of the EM.
1. Is this still true in the frame of the test mass when the EM is moving (relative to the test mass)?
2. If so, is it true if the EM is rotating about its axis and moving tangent to its equatorial plane? Or will a tidal vector point elsewhere instead? If elsewhere, will all tidal vectors in a given space slice "point" at the same point?
The stability of such an EM is simply assumed.

The *motivation* underlying these questions is to know whether the world line of the well-defined axis midpoint of such an EM has little or no physical significance outside of the system.

4. Dec 5, 2014

### James_Harford

Correction: the word "my understanding" in the 3rd paragraph is not correct.as I do not know this to be a fact.

5. Dec 5, 2014

### Staff: Mentor

What you are describing is not tidal gravity. For tidal gravity, you need at least two test masses, or a single extended object (a test mass is assumed to be point-like). The way you detect tidal gravity is to start the two test masses off at rest relative to each other, and close together spatially; then you let them move inertially (i.e., in free fall), and see what happens. If they do not remain at rest relative to each other, then tidal gravity is present.

For an extended mass that is spherically symmetric, the tidal gravity observed external to the mass depends on how the pair of test masses is oriented. If they are oriented radially (i.e., both along a single radial line, one slightly higher than the other), the tidal gravity will cause them to move apart; if they are oriented tangentially (i.e., both at the same height but on slightly different radial lines), the tidal gravity will cause them to move together. The full description of tidal gravity, including all possible orientations of the test masses, requires a tensor, the Riemann curvature tensor, which has twenty independent components at any point in a 4-D spacetime.

As above, tidal gravity can't be described by a vector, and the vector you are describing is not tidal gravity; it's the "acceleration due to gravity", which is coordinate-dependent. However, even leaving that aside, your definition of what it means for a vector to "point" at the center of the EM is not correct, because you are not using the term "parallel transport" correctly. Any vector can be parallel transported. What I think you actually mean to say is more correctly stated like this:

A vector "points" towards the center of the EM at a given event if the spacelike geodesic that has that vector as its tangent vector at the given event, and lies entirely within a spacelike hypersurface of constant time containing the given event, passes through the center of the EM. The fact that you have to specify a hypersurface of "constant time" in order to uniquely specify the geodesic makes this definition coordinate-dependent.

6. Dec 5, 2014

### James_Harford

Thank you for these observations about the use and misuse of the notion of tidal gravity. First of all, yes, I am asking about phenomena whose measurements are specific to particular idealized reference frames. And yes, the test masses were meant to be extended objects I should have used, instead, small regions of dust instead, where initially all particles in the region are initially motionless relative to one another. Then by observing their instantaneous relative accelerations the radial direction you spoke of can be ascertained. This is encouraging, because such a unique direction evidently exists. Whether, and under what conditions, this direction is tangent to the start of a path as you so clearly restated, is the heart of my question.

7. Dec 5, 2014

### Staff: Mentor

It uniquely exists for the idealized, perfectly spherically symmetric case. With some complications, it uniquely exists for the slightly more general idealized case of a perfectly axisymmetric object (i.e., one which is rotating in a perfectly uniform way and whose shape is perfectly symmetrical about the axis of rotation).

In the spherically symmetric (i.e., non-rotating) case, I think the radial direction will "point" towards the center of the object even if we adopt coordinates in which the object is moving. (I have not done a detailed calculation to confirm this, though.)

In the axisymmetric (i.e., rotating) case, the radial direction (as defined by tidal gravity) does not point towards the center of the object everywhere, even in coordinates in which the object is at rest. It only does so in the "equatorial plane" of the object, assuming it has one--i.e., assuming the object is not only symmetrical around its axis of rotation, but also has an additional plane of symmetry perpendicular to that axis--this plane is the "equatorial plane". For example, an idealized rotating body composed of a perfect fluid will be an oblate spheroid, and will have an equatorial plane (the equatorial plane of the oblate spheroid).

If we restrict to rotating objects with equatorial planes, then I think (but again, I have not done a detailed calculation to confirm) that, in the equatorial plane, the radial direction continues to point towards the center of the object in coordinates in which the object is moving, as long as the motion is perpendicular to the axis of rotation (i.e., parallel to the equatorial plane).

Also, as bcrowell pointed out, for any of the above to be well-defined, the spacetime has to be asymptotically flat (which basically means the gravitating body in question has to be isolated, with no other significant masses present).

8. Dec 6, 2014

### James_Harford

Thank you PeterDonis for these very interesting comments -- caveats included.

It is interesting that the "convergent-pointing-to-center" effect occurs in so many cases, and in particular under what to me seem the somewhat generous conditions described in your second-to-last-paragraph. I would have thought that frame-dragging would have introduced a skew symmetry to the radial directions that would destroy the effect entirely. Or, alternatively, that it would be destroyed by the "off-center" displacement of the rotating EM's center of mass (a SR effect defined in an asymptotically flat spacetime). Thank you!

Question: In the case of rotating objects with equatorial planes, is there any restrictions on the nature of the rotation, such as "uniform rotation" -- although I am not sure what that means?

9. Dec 6, 2014

### Staff: Mentor

There is a subtle point here which I didn't mention before. The "radial direction" actually changes (because of frame dragging) if you are moving radially, compared to what it is if you are "hovering" at a constant radius. So, for example, if you release an object from rest at a given radius, it will start falling "straight down", but as it falls, frame dragging will start to curve its trajectory in the prograde direction (the same direction as the object's rotation). (I'm describing this heuristically, hopefully you can see what I mean.) Similarly, a small sphere of test particles that is falling radially will describe a different "radial direction" with its tidal effects than a sphere which starts at rest at a constant radius.

"Uniform rotation" means, basically, that the angular velocity of the object, as measured at infinity, is constant in both time and space (i.e., it doesn't change with time and it's the same for every part of the object). (Note that the angular velocity as measured locally, by an observer riding along with a small piece of the object, does change with position, because of time dilation; that's why I specified "at infinity" above.)

If you are asking what the conditions are for an object to have a well-defined "equatorial plane", AFAIK the ones I gave (uniform rotation, as above, plus a plane of symmetry perpendicular to the axis of rotation) are necessary and sufficient; but I don't have a proof handy.

10. Dec 7, 2014

### James_Harford

Points taken. If I understand correctly, the conditions you gave (with caveats) could be stated as follows: Given an EM that rotates uniformly, is axisymmetric, has a well-defined equatorial plane of symmetry perpendicular to its axis of rotation, there exists for every inertial frame at every point external to the EM in the EM's equatorial plane a "radial" direction tangent to the start of a space-like geodesic that passes through the well-defined center of the EM.

11. Dec 7, 2014

### James_Harford

Ah, I should add that this radial direction is ascertained from the relative accelerations of a small sphere of test particles.

12. Dec 7, 2014

### James_Harford

Ach! Instead of space-like geodesic, I mean geodesic within the sheet of 3-space of constant time, and instead of point, point-event.

13. Dec 7, 2014

### Staff: Mentor

James_Harford, yes, posts #10, #11, and #12 look OK to me. (Re post #12, "point" and "point-event" mean the same thing when referring to spacetime, so you didn't really need to make that last clarification.)

14. Dec 7, 2014

### James_Harford

Ok, thanks PeterDonis.

15. Dec 7, 2014

### James_Harford

Another thought...

If the point event outside of the EM is not in its equatorial plane, the radial path will not cross the center of the rotating EM -- but will it cross the axis at some point within the EM? If so, then an interesting special case is an infinitely thin disk, in which case the radial path should cross through its center?

16. Dec 7, 2014

### Staff: Mentor

I think so, yes.

This case is actually just a limiting case of an object with an equatorial plane--the equatorial plane just contains the entire object. Yes, the radial path will cross through its center.