Where does gravity pull?

**Chatt** · Jun19-09, 04:52 PM

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 :)

**mathman** · Jun19-09, 05:27 PM

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

**Chatt** · Jun19-09, 05:47 PM

Thanks :)

**Phrak** · Jun19-09, 06:17 PM

Originally Posted by mathman

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

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

**DrGreg** · Jun19-09, 06:59 PM

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.

**granpa** · Jun20-09, 01:28 AM

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.

**Phrak** · Jun20-09, 01:50 AM

Originally Posted by 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.

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.

**Combustible** · Jun20-09, 03:48 AM

Originally Posted by 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.

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.

**A.T.** · Jun20-09, 05:19 AM

Originally Posted by 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.

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.

**darkwood** · Jun20-09, 06:27 AM

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.

**DaleSpam** · Jun20-09, 08:55 AM

Originally Posted by 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.

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.

**Naty1** · Jun20-09, 10:43 AM

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....

**ZikZak** · Jun20-09, 10:59 AM

Originally Posted by Naty1

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

Except that it is completely wrong.

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

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.

**Naty1** · Jun20-09, 11:16 AM

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

**D H** · Jun20-09, 11:33 AM

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).

**ZikZak** · Jun20-09, 11:47 AM

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

This fellow also has a decent explanation of it in English. See also the Wikipedia article and references therein on "speed of gravity."