# Gravitational Force between Moving Bodies

1. Sep 9, 2009

### nickyrtr

The Einstein field equation for gravity uses the stress-energy tensor, rather than just mass, to determine the gravitational force acting between two bodies. When spacetime curvature is small and relative velocities are much less than the speed of light, this force is well-approximated by Newton's law of gravity (F = G*M1*M2/R^2)

My question is, how is the force modified when the bodies' relative velocity approaches the speed of light? I presume the other terms in the stress-energy tensor (T11, T12, etc.) would come into play, but I am not sure exactly how. Would the bodies' relative motion amplify (or reduce?) the gravitational force that each one feels from the other?

To clarify, I'm not thinking of curved spacetimes such as near a black hole. This would be a case where spacetime is still nearly flat, only with high relative velocity, such as a spacecraft approaching a planet at nearly the speed of light. Does the planet's gravity well appear any stronger or weaker due to the craft's velocity?

2. Sep 9, 2009

### javierR

There won't be a difference in strength. This is related to a common question asked by those new to special relativity (and who have poor source material): If the mass of an object increases with its speed, could a planet, e.g., ever become a black-hole by traveling closer and closer to the speed of light? Well, as we know, this is mistaken reasoning that goes to the heart of relativity. The rest-mass of the object is the physically relevant quantity here (which is the whole of the energy-momentum of the object in its rest frame). The velocity of an object is only defined *relative* to an observer, and different observers in more and more Lorentz-boosted reference frame, so how could some say there is a black hole and others say it's just a planet, including the observers on the planet itself.

But we can get a general relativistic description.
Let's first restrict our attention to test particle observers (that is, ignoring their contribution to stress-energy). Then the stress-energy tensor of the source object, like a passing planet, is a frame-independent thing. Each observer will have their own idea of what the components of the tensor are (different mixes of energy and momentum flows), but the tensor itself is an object that is independent of these differing viewpoints. Therefore the spacetime curvature and geodesics on this spacetime are geometric objects (frame-independent) even though different observers will naturally *describe* the components of these quantities *in different ways*.

Now instead of being a test particle, suppose you're an observer on one planet passing another. Now your planet is contributing to the stress-energy distribution in spacetime, but with respect to the reference frames issue, nothing has changed. If you are on the planet, that is a particular observer frame and you measure the contribution due to your planet's mass-energy. That gives the components of your planet's stress-energy tensor in your frame. You then turn to the other planet, which is moving by at that moment, and you measure its apparent energy content and momentum through your frame. That gives the components of the stress-energy tensor of the second planet in your frame.

If that same planet came by at a higher relative speed, it's like being a test particle in a new Lorentz frame (a boosted frame). The physical results encoded in Einstein's equations should therefore be the same. Your stress-energy components haven't changed, while those of the passing planet *have changed* in the usual way for a boosted reference frame. The geometrical objects that are the stress-energy tensors of the two planets are the same objects after boosting, though that of the passing planet will look "distorted" from your point of view (that is, its components change). This is the same idea behind the 4-velocity. No matter what frame you go to the 4-velocity vectors of the two planets are what they are...it's just that they look relatively rotated from your particular reference frame of choice. Anyway, since the fundamental stress-energy tensors of the planets have not changed in this sense, the spacetime curvature and therefore gravitational strength will be the same, though you describe the two situations with different component values in your coordinate frame. In contrast, if you add more rest-mass to the other planet, the stress-energy tensor itself is changing (in any fixed Lorentz frame), and so the spacetime curvature would change.

Despite these arguments surrounding spacetime objects (4-velocity, stress-energy tensor, curvature tensor,...) being frame-independent, note that relatively boosted test particle reference frames obviously have different trajectories in space (just consider the various trajectories of objects near Earth's surface).

3. Sep 9, 2009

### meopemuk

I would like to disagree with javierR.

The gravitational interaction between two massive bodies does depend on their velocity. It is well-know that for low velocities (in the $(v/c)^2$ approximation) the gravitational dynamics is accurately described by the so-called Einstein-Infeld-Hoffmann Hamiltonian. In addition to the usual Newtonian term -GMm/r this Hamiltonian contains velocity-dependent interactions as well.

4. Sep 10, 2009

### nickyrtr

Thank you very much for indicating the Einstein-Infeld-Hoffmann (EIH) theory. After some googling I found this equation on Wikipedia:

http://en.wikipedia.org/wiki/Einstein–Infeld–Hoffmann_equation

The equation is nicely stated there, but one bit of notation I don't understand. What is meant by (vAi)? I cannot figure out what the "i" superscript is supposed to denote.

P.S. Also, Javier, thank you for your reply but perhaps part of my question was misleading (the bit about the planet's gravity well). You are right that the total strength of the planet's gravity well is in some sense Lorentz invariant; however, I was wondering more about the acceleration experienced by the spacecraft, which does appear to be velocity-dependent as per the EIH equation.

Last edited: Sep 10, 2009
5. Sep 10, 2009

### nickyrtr

After reading the equation more closely, I believe (vAi) is just another way of writing the vector (vA), with (i) indexing the three spatial components of the vector.

In the simple case of two bodies moving only along the line passing through them both (no angular momentum), I believe the EIH equation reduces to this (in units where c=1):

a = -(GM/r2) (1 + v2/2 +/- 3v2)

Where a is the acceleration due to gravity on some test particle due to a body of mass M, at distance r, moving radially at velocity v = dr/dt.

The velocity-independent term is the same as Newton's law of gravitation. The v2/2 term seems to show that, in a sense, kinetic energy is added to the body's gravitational mass, regardless of the direction of motion.

The 3v2 term is interesting; it looks to me like the body's gravitational pull is enhanced if it is moving toward the test particle, and reduced if it's moving away from the test particle. When v exceeds approximately 0.632*c, the net force would be repulsive! However, that speed is clearly beyond the "small velocity" approximation on which the EIH equation is based.

Here is a paper I found where higher order terms of the approximation are worked out:

http://arxiv.org/PS_cache/gr-qc/pdf/0108/0108086v1.pdf