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

D H

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, for example (this is what JPL uses), or see equation 10.12 in http://www.iers.org/nn_11216/SharedD...f/tn36_151.pdf.

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 [itex] \mathbf{r} \cdot \mathbf{\dot{r}} = 0 [/itex] and that it's general relativity so [itex] \gamma = 1 [/itex] and we're left with just
[itex] \Delta |\mathbf{a}| = \frac{GMv^2}{c^2 r^2} [/itex] which is actually twice as big as you'd expect. Is that about right?

D H

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 [itex]\vec r \cdot \vec v[/itex], frame dragging, and solar effects we're left with
[tex]\Delta |\vec a| \approx \frac{GM_E}{c^2r^2} \left( 2(\beta+\gamma)\frac {GM_E}{r} - \gamma \, v^2\right)[/tex]
For a circular orbit, [itex]GM/r \approx v^2[/itex]. 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
[tex]\Delta |\vec a| \approx \frac{GM_E}{c^2r^2} (2\beta + \gamma)\,v^2[/tex]
Since β=γ=1 in general relativity, this becomes
[tex]\Delta |\vec a| \approx \frac{3GM_Ev^2}{c^2r^2}[/tex]
This is pretty much what you'll see as a given in a simplistic derivation of the relativistic precession Mercury.

PAllen

Maybe I'm blind, but I don't see any O(c^-4) corrections. I definitely agree we can dispense with the J terms for simplicity - consider the hypothetical non-spinning earth. I also agree it makes sense to pretend the sun doesn't exist, for simplicity, so that leaves only the first line.

Then, using your circular orbit suggestion, I get something different:

4(GM)^2/(c^2 r^3) - (GM/r^2) (v^2/c^2)



-------
One thing I don't understand is that versions of these approximate GR equations of motion I've seen before (like the first of DH references), have acceleration in a term on the right . This version does not. What gives?

[Edit: I see DH posted first. Fortunately, our comments are similar.]

D H

What gives is that those other references such as my first one are not valid for a satellite in Earth orbit. That equation in the second article I referenced gives the perturbative effect Δa. It does not say how to compute the non-relativistic aspects of the acceleration; that's an exercise left to the reader.

You can't pretend that the only perturbations to spherical gravity are due to other bodies and general relativity. For an artificial satellite, atmospheric drag is huge for vehicles in low Earth orbit, and the non-spherical nature of Earth's gravity field dominates over relativistic effects out to order 20 or so even out to geosynchronous orbit. Even solar radiation pressure swamps relativistic effects.

The same goes for orbits about the Earth's moon. Low lunar orbits are very bizarre.

pervect

Unfortunately, approximations that require that the speed be small (for instance the PPN formalism, i.e. Paramterized Post Newtonian approximations) can't really address the issue of how speed affects gravity, because the speed has already been assumed to be small.

The biggest pitfall to watch out for in trying to treat gravity as a "force" is how it behaves in different reference frames. I'm not aware of anything that plays the role of "force" in full GR that transforms like a tensor (much less like a 4-vector, which is what one expects a force to transform like).

The closest candidate I'm aware of are the Christoffel symbols - which are not tensors, and have the wrong rank as well.

This argument may sound technical, and it is - but not transforming properly ultimately leads to a lot of confusion. It breaks the usual way in which physics summarizes all the different things that different observers might measure into a single, unified, observer-independent framework.

In fact, I think that trying to understand how gravity transforms between different frames , how it appears to different observers, may be the root of the OP's question.

PAllen

Actually, on review, I see something much more basic. The acceleration terms on the right of general EIH equations are accelerations of bodies other than the 'test' body relative to the barycenter. For satellite and earth, in earth centered coords, this is zero by definitions.

PAllen

The EIH equations do have velocity dependent terms, so they indicate, even for low velocities, that there is v^2/c^ base term (where v is relative velocity of test body and gravitating mass).

Of course, more generally, I agree there is absolutely nothing in GR that has properties similar to gravitational force. My preferred analog is to match the physical situation. When we think of gravitational force in a Newtonian framework, we mean the force exerted on an object to resist free fall. This concept generalizes to GR except for the major complication that for a non-stationary geometry (e.g. a large mass moving by), there is no unique definition of a stationary world line. Yet, a reasonable answer is to take a world line that maintains a fixed position in Fermi-normal coords of a distant star (assuming asymptotically flat universe). Computing this, I get first order corrections of order gamma^2. In another thread, Bill_k addressed this problem in a different way, yet also got corrections of order gamma^2.

Jonathan Scott

There's a much simpler way of looking at gravity as a force, which is to look at the effective force (rate of change of coordinate momentum), rather than the acceleration (rate of change of coordinate velocity).

To keep the notation as Newtonian as possible, let all terms including the coordinate speed of light c be expressed relative to an isotropic coordinate system, so the coordinate momentum is [itex]E\mathbf{v}/c^2[/itex]. The equation of motion may then be obtained from the Euler-Lagrange equations. In the weak case where the metric factor for space is approximately the reciprocal of that for time, the equation is as follows:
[tex]
\frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )
[/tex]
This applies to motion in any direction. The Newtonian field is defined as follows:
[tex]
\mathbf{g} = - \frac{c^2}{\Phi_t} \nabla \Phi_t
[/tex]
where [itex]\Phi_t[/itex] is the time dilation term from the metric, approximately equal to [itex](1 - Gm/rc^2)[/itex]. For the Einstein vacuum solution in isotropic coordinates, it is equal to the following:
[tex]
\Phi_t = \frac{1 - Gm/2rc^2}{1 + Gm/2rc^2}
[/tex]
This approximation is sufficiently accurate to give the correct prediction for the Mercury perihelion precession (using the usual substitution u=1/r and orbit equations).

The equation of motion can also be made accurate for stronger fields as well by removing the assumption that the time and space terms in the metric are exact reciprocals of one another and using separate field values for the gradients of the time and space terms:
[tex]
\frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \left ( \mathbf{g}_t + \frac{v^2}{c^2} \mathbf{g}_x \right )
[/tex]
[tex]
\mathbf{g}_t = - \frac{c^2}{\Phi_t} \nabla \Phi_t
[/tex]
[tex]
\Phi_t = \frac{1 - Gm/2rc^2}{1 + Gm/2rc^2}
[/tex]
[tex]
\mathbf{g}_x = + \frac{c^2}{\Phi_x} \nabla \Phi_x
[/tex]
[tex]
\Phi_x = \frac{1}{(1 - Gm/2rc^2)^2}
[/tex]

clamtrox

You are right; I just misread the parentheses there. This result seems a little strange: in a circular orbit, the net effect is pushing, not pulling. But this is because assuming circular orbit makes the result weird: the velocity needed to maintain it is very high, and therefore the Schwarzschild correction is high as well. I missed that completely when I first thought about this. I wonder if you can have orbits where the pulling term dominates, and you have a net negative correction.

Naty1

aditya: was your original questions answered?

aditya23456

Welll..!!!! I just got to know that my question was valid ie answer to my question is yes and at the same time explaining this effect needs mathematics which is beyond my undergraduate level but surely I ll try to understand what u all meant as per part of my research on gravity :) But I dint think explaining my thought experiments mathematically is QUITE DIFFICULT AT MY PRESENT LEVEL and hope someone of this forum helps me for this.
Thing is that I cant ask my question completely for having a genuine research ;)

Naty1

Do you have an example relating to your question?

For example, as the Earth revolves around the sun is the 'force' on each different? Are the 'speeds' different? Would the force on each be different than the first case if the earth were plunging towards the sun? How about a pair of photons travelling along side by side in the same direction compared with travelling in opposite directions?


A bunch of the 'latter' answers here are pretty far afield from your question.... as experts are discussing stuff with each other.

jimgraber

See my answer here:


http://physics.stackexchange.com/que...re/22025#22025

The leading order PPN correction involves a v^2/c^2, but also a 3 and a 1/r^2. So its not quite the same as you would naively expect.

yuiop

OK, here is a very simple thought experiment that you can use to convince yourself that the acceleration of gravity is speed dependent. (Note that I am using acceleration to try avoid the arguments about gravity not being a force).

Let us say that we have a rocket that is sufficiently massive that it has a detectable gravity field but not so massive that the gravitational time dilation is significant. Let us say it takes an astronaut 60 seconds to fall in the y direction to the centre of gravity of the rocket when the rocket is at rest in ref frame S. When the rocket is moving at a velocity relative to frame S in the x direction such that the gamma factor is 10 it will now take approximately 600 seconds (as measured in frame S) for the astronaut to fall the same distance in the y direction. That is in very broad terms but I hope you get the general idea and if it was not true, it would be possible to detect absolute motion.

It is also worth noting that the effects of gravity are not only speed dependent but also sensitive to the direction of relative motion.

For stuff about sinking relativistic boats see Supplee's submarine paradox http://en.wikipedia.org/wiki/Supplee's_paradox

Here is another example. Consider objects falling towards a black hole. In coordinate terms, objects accelerate but as they approach the event horizon they decelerate eventually coming to a stop. Low down near the vent horizon, fast moving objects moving towards the event horizon are decelerating while objects that are released from stationary at the same altitude are accelerating. See http://www.mathpages.com/rr/s6-07/6-07.htm

clamtrox

Okay, forget the spherical orbit; that was a bad idea. Only way for getting an object moving on a spherical orbit at relativistic speeds is to have the object orbit relatively close to the Schwarzschild radius, which obviously messes up the thought process here. So let's switch the assumption to [itex] GM/r \ll v^2 [/itex], (and still forget all corrections from angular momentum). This means that the leading order correction really comes from relativistic speeds, not a strong gravitational field.

This leaves us with
[itex] \Delta \mathbf{a} = \frac{GM}{c^2 r^2} (4(\hat{\mathbf{r}} \cdot \dot{\mathbf{r}} )\dot{\mathbf{r}} - v^2 \hat{\mathbf{r}}) [/itex]

So from this it's a little difficult to see exactly what will happen, but for example you can see that the "extra bending" (in the direction of r) is [itex] -\frac{GMv^2}{c^2 r^2} [/itex]. There is also an additional effect that accelerates you in the direction of your motion.

clamtrox

What assumptions did you make to get that formula? You seem to be missing some terms compared for example to http://www.iers.org/nn_11216/SharedD...f/tn36_151.pdf