# Is force of gravity dependent on speed of one of the body?

1. Mar 29, 2012

Is there any such relation as per general theory of relativity?

2. Mar 29, 2012

### PAllen

Yes, definitely, but there is no unique definition. For an isolated gravitating mass, and a test body stationary relative to it, there is a unique definition of 'stationary'. Thus you can ask what the proper acceleration is for a stationary world line near a massive body. This is analogous to Newtonian force of gravity.

However, for a moving body, there is no unique definition of a 'stationary' world line. Similarly, if you consider the gravitating mass to be stationary, and the test body moving, there is no unique definition of 'straight as if the body wasn't there'.

However, no unique definition doesn't mean no possible definition. For most any plausible definition, the moving gravitating mass exerts more force on the stationary test body ([edit: produces higher peak proper acceleration of the 'stationary' world line]); and the 'straight' moving test body experiences larger proper acceleration (at closest approach) than a stationary test body.

Unfortunately, the lack of uniqueness means you can't come up with universally agreed on numbers. Different definitions will lead to different amounts of increase in force [edit: peak proper acceleration of resisting world line] due to relative motion.

Last edited: Mar 30, 2012
3. Mar 30, 2012

### tom.stoer

Perhaps one could call this question is force of gravity dependent on speed of one of the body? a Category mistake.

In GR gravity is not a "force".

Let's assume we have rather heavy bodies (like stars) and a test body (like a satellite). The energy-momentum density T of the stars plus the metric g of spacetime manifold are defining a solution (g,T) of the Einstein field equations.

The curvature of the manifold defines geodesics C along which test bodies will move.

A different energy-momentum density of the stars T' with different positions, different motion comes with a different solution g' for the metric, i.e. a different spacetime with (g',T').

The geodesics C' of this new spacetime are 'different' in a certain sense. But be careful: in principle you are not allowed to compare C and C' b/c they are defined w.r.t. different (g,T) and (g',T').

It's like saying that a Depp is less stupid than a Twompe; that's dangerous b/c both terms are defined w.r.t. to two different languages; Depp is German for Twit, Twompe is (Haitian) Creole.

4. Mar 30, 2012

### julian

One answer is that if you are moving your mass is greater so the grav force increases, but...(thought experiment) say you have a speed boat at rest, the wieght (due to GRAVITY) equals the force due to water pressure. But what if the boat is moving at speed v with respect to the ocean? The area is reduced by $\gamma$ because of length conrtraction, plus the mass increases and hence the weight of the boat increases...the conclusion is that the boat will sink....

Last edited: Mar 30, 2012
5. Mar 30, 2012

### julian

The answer is that ALL transvesre forces change under Lorentz tranformations, not just with gravity. You are in motion with respect to a grav body, your mass and hence weight increases, but your transverse force will 'reduce' at the same time your mass and wieght increase.

Last edited: Mar 30, 2012
6. Mar 30, 2012

### tom.stoer

as I said, gravity is not a force!

7. Mar 30, 2012

### julian

Yeah but when the strengh of grav field, speed and masses are so small you got to conceed that such considerations are valid.

8. Mar 30, 2012

### tom.stoer

it's simple: when the velocity of the test particle changes, its geodesic changes as well

9. Mar 30, 2012

### julian

No offesnse, comming from someone who has asked questions that are difficult to understand, but what are you saying exactly?

10. Mar 30, 2012

### clamtrox

Can we please just agree to call mass and energy mass and energy, not mass and mass.

Also, you need to be more careful here because gravity does not couple to kinetic energy in the same way as it does to mass. It is not clear to me that your analysis is correct at all. Why would the boat driver see his boat sinking?

11. Mar 30, 2012

### julian

Length contraction (boat length) reduces and the force due to water pressure decreases and the weight of boat increases becauses the mass increases... factor of gamma^2 involved....this can only explained by transverse force changing...

12. Mar 30, 2012

### Mentz114

Are you sure about this ? After a proper acceleration the geodesic certainly does change but a body on a geodesic may have changing velocity.

Last edited: Mar 30, 2012
13. Mar 30, 2012

### tom.stoer

sorry for the confusion, the formulation is missleading; what I want to say is that test masses at one spacetime point P starting in one direction but with different initial velocities v, v', v'', ... will follow different geodesics C, C', C'', ...

14. Mar 30, 2012

### Mentz114

Sure thing. Sorry to nit-pick ...

15. Mar 30, 2012

### D H

Staff Emeritus
In a sense, yes.

One way to answer this question is to look at things from the perspective of a parameterized post-Newtonian (PPN) formalism. Here gravity is still a "force" but it isn't quite Newtonian gravity. In addition to good old F=GMm/r2, there are some first order perturbation terms that result from general relativity in such a formalism. These perturbative effects on Newtonian gravity involve velocity.

This approach leads to a very accurate description of the relativistic precession of Mercury. This is also how JPL, the Russian Academy of Sciences, and the Paris Observatory (the organizations that produce the three leading ephemerides) now model the solar system.

16. Mar 30, 2012

### clamtrox

Do you know if the corrections are in first order what you would expect by naive thinking, ie. F->F(1+v2/c2/2), or something different?

17. Mar 30, 2012

### tom.stoer

I am not sure whether this is what you are looking for, but here's a paper discussing Einstein’s paper “Explanation of the Perihelion Motion of Mercury from General Relativity Theory”: http://www.wbabin.net/eeuro/vankov.pdf

You will recognize the simplest post-Newtonian terms (for this special case).

I am not sure if this is what you are looking for; I guess instead of "slow motion + strong gravitational field" you are interested in "fast motion + weak gravitational field"; anyway - the starting point is always the geodesic e.o.m.

18. Mar 30, 2012

### D H

Staff Emeritus
They're a bit more complex (well, more than a bit more complex) than that. See equation 8.1 in http://iau-comm4.jpl.nasa.gov/XSChap8.pdf [Broken], for example (this is what JPL uses), or see equation 10.12 in http://www.iers.org/nn_11216/SharedDocs/Publikationen/EN/IERS/Publications/tn/TechnNote36/tn36__151,templateId=raw,property=publicationFile.pdf/tn36_151.pdf [Broken].

Last edited by a moderator: May 5, 2017
19. Mar 30, 2012

### clamtrox

So basically yes, except it's hard to decide what the coefficient in place of 1/2 is because they are using vectors :) The equation in IERS document is pretty simple: first you have the correction from Schwarzschild metric, then two terms which look like v^2/c^2, and then corrections O(c^-4) which contain angular momentum and then it gets properly confusing, but we can drop those for now. The last term looks like some kind of angular momentum correction from the Sun, so lets drop that one too because we don't care about that... Finally, lets just be lazy and say we're on a circular orbit so $\mathbf{r} \cdot \mathbf{\dot{r}} = 0$ and that it's general relativity so $\gamma = 1$ and we're left with just
$\Delta |\mathbf{a}| = \frac{GMv^2}{c^2 r^2}$ which is actually twice as big as you'd expect. Is that about right?

Last edited by a moderator: May 5, 2017
20. Mar 30, 2012

### D H

Staff Emeritus
That's off by a factor of three.

Those "corrections which contain angular momentum" are frame dragging (Lens-Thirring effect), and they are small -- as is the solar influence (and where's the Moon?)

After dropping the $\vec r \cdot \vec v$, frame dragging, and solar effects we're left with
$$\Delta |\vec a| \approx \frac{GM_E}{c^2r^2} \left( 2(\beta+\gamma)\frac {GM_E}{r} - \gamma \, v^2\right)$$
For a circular orbit, $GM/r \approx v^2$. Note: This is exact in Newtonian mechanics; since we're dealing with a perturbation on top of a perturbation, we can consider this to be essentially correct in a PPN formulation as well. Thus the relativistic correction reduces to
$$\Delta |\vec a| \approx \frac{GM_E}{c^2r^2} (2\beta + \gamma)\,v^2$$
Since β=γ=1 in general relativity, this becomes
$$\Delta |\vec a| \approx \frac{3GM_Ev^2}{c^2r^2}$$
This is pretty much what you'll see as a given in a simplistic derivation of the relativistic precession Mercury.