# Where does gravity pull?

1. Jun 19, 2009

### Chatt

My question is quite simple

Imagine a massive object speed by you. We know that everything including gravity abides the law that nothing can travel faster than light. Therefore it will take a little bit of time for the objects gravity to reach you, but in the time spent for the gravity to travel to you, the object has moved a bit.

My question is: are you pulled toward a point behind the object, or are you pulled toward the objects center of gravity?

I hope you understand my question, and thanks in advance :)

2. Jun 19, 2009

### mathman

You are pulled to where the object was, which will be where the object is seen to be.

3. Jun 19, 2009

### Chatt

Thanks :)

4. Jun 19, 2009

### Phrak

I don't think the answer is as simple as this. In Newtonian gravity, gravitational force is instantaneous. In the weak field limit the solution in general relativity approaches the Newtonian solution, where the mass acting on a test particle is considered stationary.

Now place an observer in an inertial frame in a distant region with a relative velocity wrt the mass under consideration. This introduces a time delayed 'center of mass.' It seems this would introduce contradictory results to the sationary solution.

To reconcile the differences we could note that a moving mass in general relativity hasn't the same solution as a stationary mass. I would guess that the force on a test particle is directed toward the future position of the massive object.

Edit. Oops. Make that current, non-time delayed, position

Last edited: Jun 19, 2009
5. Jun 19, 2009

### DrGreg

If we were talking about electrical attraction between charges instead of gravitational attraction, the answer would be as follows.

If you are one of the charges, look at the delayed visual image of the other charge, work out where the charge ought to be now (in your frame) if it were to continue at the same constant velocity that you observe, and that location is the direction from which you feel the attractive force acting on you. So if the charge did not accelerate between what you see and where it is "now", you'll feel the force from where it is now.

But I don't know enough about general relativity to know if the same principle applies to gravity. I suspect it might.

6. Jun 20, 2009

### granpa

you are pulled toward where the field at your location is pointing. the gravity field surrounding the massive object is already in equilibrium with that object so every part of its field will point directly at that object. you will therefore be pulled directly toward the object.

7. Jun 20, 2009

### Phrak

1) "the gravity field surrounding the massive object is already in equilibrium with that object so" 2) "every part of its field will point directly at that object."

But, unfortunately, 2 does not follow from 1.

8. Jun 20, 2009

### Combustible

Would it not be possible to use a retarded Newtonian gravitational potential using ρ(t-r/c) instead of ρ(t) as you would in electromagnetism. Obviously it wouldn't be as accurate as GR but surely it would allow the gravitational force to have a finite propagation speed in Newtonian theory.

9. Jun 20, 2009

### A.T.

My thoughts: If the big mass moves inertially you can switch to its frame: Here the field is static and every object is pulled directly towards the current mass position, no matter if the object is moving or at rest. Why should that be different in the frame of the moving object?

The gravity (space-time curvature) doesn't have to get from the mass to the object, to affect it. It is already there in the traversed region of space. If the big mass would suddenly accelerate the effect on the object would be delayed: For a while the object would be pulled towards the supposed position of the mass assuming inertial motion.

Light is different: In the objects frame the light hits it from an old position of the mass. In the masses frame the moving object experiences light aberration. So in both cases it sees a wrong postion of the mass. But it is pulled towards the current position, assuming inertial relative motion.

Last edited: Jun 20, 2009
10. Jun 20, 2009

### darkwood

You would be pulled to a point when the object was level with you and the gravitational field the strongest but not in a straight line as your motion would be affected by the objects approach and strengthening field and its exit and weakening field. As the approach and exit field should be respectively equal the will cancel each other out after the mass has passed heading you to the center of the object as it passed closest but not in a straight line.
This assumes the object is perfectly spherical and not a spinning uneven mass, and the faster the object the less noticeable the curve of your motion.

11. Jun 20, 2009

### Staff: Mentor

I agree, with the caviat that it only applies for a non-accelerating gravitating object. In the rest frame you can determine the force, it is towards where the object is. Then just Lorentz transform to determine where the force is in a moving frame.

12. Jun 20, 2009

### Naty1

Post #2 is a nice, neat, understandable, concise and accurate reply. well done.

It appears the original poster was interested in understanding gravitational delay and so Newtonian considerations would not apply. The only thing I'd clarify for the original poster is that the observable location and the center of gravitational attraction are coincident because both gravitational and visible light waves travel at the speed of light...so they should arrive together at the observers location....likely frequency shifted a bit....

13. Jun 20, 2009

### ZikZak

Except that it is completely wrong.

No, they are not. The center of attraction in the weak field limit in the Sun's rest frame is the instantaneous position of the Sun, and NOT the apparent (retarded) position of the Sun. This is exceedingly easy to verify: if it were the retarded position, then planetary orbits would rapidly decay and all the planets would spiral into the sun! The existence of Keplerian orbits absolutely depends on the vector gravity being directed at the instantaneous position of the sun, and will not work if the planets are attracted to the retarded position. I am sorry, but your claim is a naive application of the physical concepts.

Yes, gravitational influences travel at c in General Relativity. No, that does not mean that planets are attracted to the Sun's retarded position.

14. Jun 20, 2009

### Naty1

Zikzak..can you cite a reasonably reliable reference? I find your posted explanation difficult to understand. unless you are referring to a Newtonian explanation??

Last edited: Jun 20, 2009
15. Jun 20, 2009

### D H

Staff Emeritus
Zikzak is correct. Various crackpots (including some very well-educated crackpots) have made a false argument against general relativity based on the finite transmission of gravitational influence in GR. If this were the only affect one took into account, GR would yield worse predictions of the future state of an orbiting body than does Newtonian mechanics. The reason this is a false claim is that this is not the only affect predicted by GR. The finite transmission speed coupled with frame dragging act to nearly cancel one another, making the Newtonian approximation of an apparently infinite transmission speed pretty good in many cases (e.g., our own solar system).

16. Jun 20, 2009

### ZikZak

It is basically the same effect that magnetism has on a charged particle moving at constant speed. The total Lorentz force due to a uniformly moving charge is oriented towards or away from the instantaneous position of the particle, even though electromagnetic effects travel at c... because there is a radial (but retarded) electrostatic force but also a skewed magnetic force which makes up exactly for the retardation. In weak field Relativity, there is also a gravitomagnetic force which almost exactly corrects (except for gravitational radiation) for the position difference.

Aha! I have found the perfect reference for you: http://arxiv.org/abs/gr-qc/9909087" [Broken]

http://www2b.abc.net.au/science/k2/stn/archives/archive22/newposts/128/topic128772.shtm" [Broken] also has a decent explanation of it in English. See also the Wikipedia article and references therein on "speed of gravity."

Last edited by a moderator: May 4, 2017
17. Jun 20, 2009

### Jonathan Scott

In electromagnetism, a moving charge is affected not only by the electrostatic field but also by any magnetic field, which can in turn be caused by a moving charge. This means that you can analyze a situation involving a source moving with constant velocity in two different ways: You can transform it to the frame in which the source is at rest, in which case it is clear that the field points to the source, or you can describe the effective force in terms of a combination of an electrostatic field and a magnetic field. In general, the force points to where the source would be at the current moment given its velocity at the time when it was last seen.

There are similar effects in gravity, although they are not exactly equivalent to the electric and magnetic fields alone. The usual gravitational field is like the electric field. A moving or rotating mass also induces a "gravitomagnetic" field which looks like a rotating frame of reference in the same way that the ordinary gravitational field looks like an accelerating frame of reference. This induces rotational effects and forces related to the velocity of a test mass in the field. There is also a further significant effect which is related to the curvature of space that effectively leads to forces related to the square of the relative velocity, which has no equivalent in electromagnetism, and affects objects at rest in the field of the moving source.

In the gravitational case, as for the electromagnetic case, the overall force caused by a fast-moving source points to the extrapolated current position of the source. I can't remember the exact details, but I think that in the gravitational case the extrapolation process even effectively takes the expected acceleration into account too, because conservation and continuity of energy and momentum make it very difficult to create a gravitational "surprise". It is only in a very rapidly changing system that the field does not point exactly to the current location of the source; such systems gradually lose energy in the form of gravitational radiation.

18. Jun 20, 2009

### Naty1

Great sources, thanks!!! I'm not sure I understand the fine points, but the following gives a summary from the sources...Thanks....have to read more later...

In the Wiki article under LAPLACE:

ARXIV says in part:

and on page 5

19. Jul 1, 2009

### bobbyk

Intresting question, which gave rise to much discussion!

My only comment is about your assumption that it's an object's "center of gravity" that pulls you. Even in the static case you are not in general pulled toward the "center of
gravity" of an object but to its "center of attraction", which can differ widely from its "center of gravity".

Last edited: Jul 1, 2009
20. Jul 1, 2009

### stone1

There is no way to tell what is actually moving, you or the massive object that is passing by you. If you think of the object as stationary and you moving then the answer is obvious -- the gravity that you experience now depends on where you are now, not where you were in your past.