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

[aditya23456]

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

 

[PAllen]

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

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.

 

[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.

 

[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...

 

[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.

 

[tom.stoer]

as I said, gravity is not a force!

 

[julian]

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

 

[tom.stoer]

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

 

[julian]

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

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

 

[clamtrox]

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...

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?

 

[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...

 

[Mentz114]

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

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

 

[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'', ...

 

[Mentz114]

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'', ...

Sure thing. Sorry to nit-pick ...

 

[D H]

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

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.

 

[clamtrox]

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.

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?

 

[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.

 

[D H]

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?

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/SharedDocs/Publikationen/EN/IERS/Publications/tn/TechnNote36/tn36__151,templateId=raw,property=publicationFile.pdf/tn36_151.pdf .

 

[clamtrox]

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/SharedDocs/Publikationen/EN/IERS/Publications/tn/TechnNote36/tn36__151,templateId=raw,property=publicationFile.pdf/tn36_151.pdf .

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 let's drop that one too because we don't care about that... Finally, let's 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?

 

[D H]

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 let's drop that one too because we don't care about that... Finally, let's 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?

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.

 

[PAllen]

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 let's drop that one too because we don't care about that... Finally, let's 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?

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]

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?

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]

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

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]

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.

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]

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.

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 E\mathbf{v}/c^2. 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:
<br /> \frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )<br />
This applies to motion in any direction. The Newtonian field is defined as follows:
<br /> \mathbf{g} = - \frac{c^2}{\Phi_t} \nabla \Phi_t<br />
where \Phi_t is the time dilation term from the metric, approximately equal to (1 - Gm/rc^2). For the Einstein vacuum solution in isotropic coordinates, it is equal to the following:
<br /> \Phi_t = \frac{1 - Gm/2rc^2}{1 + Gm/2rc^2}<br />
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:
<br /> \frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \left ( \mathbf{g}_t + \frac{v^2}{c^2} \mathbf{g}_x \right )<br />
<br /> \mathbf{g}_t = - \frac{c^2}{\Phi_t} \nabla \Phi_t<br />
<br /> \Phi_t = \frac{1 - Gm/2rc^2}{1 + Gm/2rc^2}<br />
<br /> \mathbf{g}_x = + \frac{c^2}{\Phi_x} \nabla \Phi_x<br />
<br /> \Phi_x = \frac{1}{(1 - Gm/2rc^2)^2}<br />

 

[clamtrox]

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.]

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 can't ask my question completely for having a genuine research ;)

 

[Naty1]

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

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 traveling along side by side in the same direction compared with traveling 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]

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?

See my answer here:http://physics.stackexchange.com/qu...al-gravitation-is-really-a-square/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 GM/r \ll v^2, (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
\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}})

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 -\frac{GMv^2}{c^2 r^2}. There is also an additional effect that accelerates you in the direction of your motion.

 

[clamtrox]

See my answer here:


http://physics.stackexchange.com/qu...al-gravitation-is-really-a-square/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.

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/SharedDocs/Publikationen/EN/IERS/Publications/tn/TechnNote36/tn36__151,templateId=raw,property=publicationFile.pdf/tn36_151.pdf

 

[clamtrox]

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 traveling along side by side in the same direction compared with traveling 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.

If you know the answer, feel free to give it. As no one seemed to come out with one, I thought it would be fun to try and figure it out, because it's a good question and it certainly deserves one.

 

[Naty1]

If you know the answer, feel free to give it.

I think this thread has at times veered far from the OP question...I am not sure I understand what the OP was asking...I want to know if it was the kind of questions I posted...I don't like distracting from OP questions...

The key point so far for me is right up front from PAllen:
"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'..."

Another foundational concept, not posted that I saw, is that the gravitational spacetime curvature is frame independent.

 

[yuiop]

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 GM/r \ll v^2, (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
\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}})

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 -\frac{GMv^2}{c^2 r^2}. There is also an additional effect that accelerates you in the direction of your motion.

I would be interested in how your equation compares to the one I formulated many years ago here --> https://www.physicsforums.com/showthread.php?t=363426&highlight=submarine

The equation I posted there can also be expressed as:

F&#039; = \frac{GMm}{R^2} * \frac{\sqrt{1-v^2}}{(1-W^2)}

where v is the velocity of the smaller test particle with mass m and W is the relativistic velocity difference of v and V where V is the velocity of the gravitational body with mass M.

W is obtained by the equation:

W = \frac { V-v}{1-Vv/c^2}Also, has anyone any response to the though experiment I posted in #32 ?

 

[clamtrox]

I would be interested in how your equation compares to the one I formulated many years ago here --> https://www.physicsforums.com/showthread.php?t=363426&highlight=submarine

The equation I posted there can also be expressed as:

F&#039; = \frac{GMm}{R^2} * \frac{\sqrt{1-v^2}}{(1-W^2)}

where v is the velocity of the smaller test particle with mass m and W is the relativistic velocity difference of v and V where V is the velocity of the gravitational body with mass M.

W is obtained by the equation:

W = \frac { V-v}{1-Vv/c^2}





Also, has anyone any response to the though experiment I posted in #32 ?

Yeah this was also my first "naive guess", but it seems like doing the actual perturbative GR calculation, you get something different. It seems that particle moving horizontally in a weak gravitational field feels twice as large correction as predicted by your formula -- you can compare them easily by setting r \cdot \dot{r} = 0 and expanding in your expression, setting V=0 and expanding in power series with respect to v/c. I'm not completely sure where the extra bit is coming from tho.

 

[yuiop]

Yeah this was also my first "naive guess", but it seems like doing the actual perturbative GR calculation, you get something different. It seems that particle moving horizontally in a weak gravitational field feels twice as large correction as predicted by your formula -- you can compare them easily by setting r \cdot \dot{r} = 0 and expanding in your expression, setting V=0 and expanding in power series with respect to v/c. I'm not completely sure where the extra bit is coming from tho.

It might be that the factor of 2 is due to the curvature around a spherical body similar to how the deflection of light by a massive body is greater by a factor of 2 than Einstein's initial prediction. My naive guess is based on a simplification of a long flat gravitational body that tries to eliminate the curvature of a curved body such a sphere. Also a factor of 2 puts it the same qualitative ball park and my formula is just intended to give a rough idea of whether the force is increasing or decreasing with relative motion (it can do both). Have you tried different combinations of V and v such as V=v, V=0 & v>0 or v=0 & V>0 to see how they compare?

 

[yuiop]

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 GM/r \ll v^2, (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
\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}})

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 -\frac{GMv^2}{c^2 r^2}. There is also an additional effect that accelerates you in the direction of your motion.

When v=0 the equation should reduce to the familiar Newtonian result of a = GM/r^2 but your equation reduces to a 4GM/r^2. This additional factor of 4 would be noticeable even in a casual measurement of acceleration in the Earth's gravitational field, so maybe there is a mistake there somewhere. Also, I don't think the c^2 factor should be there, as your equation does not have the required units of m/s^2. Also, since the "extra bending" adds to the acceleration, the sign in the expression in brackets probably should be a + sign.

 

[pervect]

OK, here is a very simple thought experiment that you can use to

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

What you've basically demonstrated with your thought experiment is that acceleration is not a tensor. It's got nothing to do with "velocity" that I can see, and everything to do with how acceleration transforms.

Basically, if you don't use tensors (and my impression, for what it's worth, is that you're not really all that interested in learning about them), you need to specify the complete coordinate system you are using in explicit detail for any of the values you communicate to have any meaning to anyone else. Though if your system is standard enough, you can do something fairly simple, like say "Schwarzschild coordinates", or "isotropic coordinates".

I do frequently get the impression you THINK these numbers have some sort of significance, though I can't quite imagine what significance you think that coordinate dependent quantities could have if you don't tell people what coordinates you're using to measure them.

 

[pervect]

See my answer here:


http://physics.stackexchange.com/qu...al-gravitation-is-really-a-square/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.

I took a quick look at this. It was interesting, but it was lacking in a lot of important detail.

My general impression of what was going on is that you found a Newtonian force law that would have the same perihelion precession that you get from Einstein's equations.

Perhaps I misunderstood what you were doing, is this basically accurate?

Also, if I'm interpreting the equations correctly (and I'm not sure I am, I was assuming h=rv and it wasn't clear what h really was), at the order at which the answer is given, it's predicting that light deflections of 4 times the Newtonian value, rather than the correct answer of twice the Newtonian value.

 

[Creator]

Is force of gravity dependent on speed of one of the body? Is there any such relation as per general theory of relativity?

So I guess the general consensus to answer the original poster's question is yes, there is a post Newtonian extra term added to the Newtonian acceleration that goes as v^2 / c^2.
But this is for relatively low velocity limit.

However, for relativistic velocities (which is what came to my mind when I read the OP), it is not generally appreciated that there is quite a different and unusual Gen Relativistic result... ...in which the acceleration of a test mass (coming in from 'infinity' toward a slowly rotating gravitational mass) can accelerate or decelerate depending on whether it is below or above a particular 'critical' relativistic velocity.

The general relation is given here in eqns. #17,#18 and #20 in the PPN approx.
http://arxiv.org/pdf/astro-ph/0510002v1.pdf

Maybe this has been discussed before...just thought I'd bring it up.
Creator

 

[pervect]

So I guess the general consensus to answer the original poster's question is yes, there is a post Newtonian extra term added to the Newtonian acceleration that goes as v^2 / c^2.
But this is for relatively low velocity limit.

I think the general consensus is (or should be) that question is the question is ambiguous.

One can certainly draw analogies between the differential equations of motion that result from the geodesic equations, and interpret these equations of motion as if they were the equations of motion of a Newtonian system.

However, the results of this interpretation are not going to necessarily transform in the way that Newtonian forces would. Thus the "forces" you come up with by utilizing this method may (and probably will) not have the same values if you use PPN coordinates (based on isotropc coordinates), or Schwarzschild coordinates. It's a mistake to think that the "forces" one comes up with this manner are necessarily coordinate independent - though it certainly may aid one's intuition to interpret the differential equations of motion in some specific coordinate system as if the system were Newtonian, the details of the analogy will not necessarily be independent of one's initial coordinate choice.

And the opportuity for making mistakes as a consequence is very large. I can easily see someone taking PPN results and being puzzled why they don't work for the Schwarzschild metric. (That's assuming they don't go off the deep end and starrt mumbling about how GR is inconsistent.).

 

[clamtrox]

When v=0 the equation should reduce to the familiar Newtonian result of a = GM/r^2 but your equation reduces to a 4GM/r^2. This additional factor of 4 would be noticeable even in a casual measurement of acceleration in the Earth's gravitational field, so maybe there is a mistake there somewhere. Also, I don't think the c^2 factor should be there, as your equation does not have the required units of m/s^2. Also, since the "extra bending" adds to the acceleration, the sign in the expression in brackets probably should be a + sign.

Sorry, my notation is very bad. \mathbf{v} = \dot{\mathbf{r}}. This is the correction to Newtonian result, so when v=0, the correction is 0 as well. The extra acceleration is to the direction of \mathbf{r}, so a negative sign means the orbit is bent towards the gravitating mass.

 

[Jonathan Scott]

As I previously mentioned, the motion of a test body within the field of a static central mass is most easily described in terms of the rate of change of coordinate momentum, which is very similar to the Newtonian expression g(E/c²) except for an extra factor of (1+v²/c²), where the velocity-dependent term is effectively due to the shape of space.

As momentum is Ev/c² and E is constant, the resulting motion depends on whether the coordinate value of the speed of light c changes.

(I find it more readable to use the standard Newtonian variable names for the coordinate values, including c, but if you have any objections, feel free to replace all copies of c in the above with c' or similar. If you really want to be picky, the use of c in Gm/rc² requires the other terms including G to be coordinate values, but this doesn't actually make any difference at least in the weak approximation where frequency and ruler sizes both vary in the same way with potential).

If the test body is moving horizontally, c is constant so the acceleration is simply (1+v²/c²) times the Newtonian acceleration.

If the test body is moving vertically downwards, c decreases, so that means the momentum increases even when the speed is nearly the speed of light. Similarly, if it is moving upwards, c increases, so the momentum decreases. This applies even if the speed is v = c, for a photon.

The definition of g in this context is as given in my previous post, relating to the gradient of the time-dilation factor. This factor is approximately (1-Gm/rc²) but to get the correct result for Mercury's perihelion precession it is also necessary to include an additional term, which for GR in isotropic coordinates to Post-Newtonian accuracy is as follows:

(1-Gm/rc²+(1/2)(Gm/rc²)²).

As previously mentioned, in strong field cases where the gradient of the space term may be different from the gradient of the time term, it is still possible to get a completely accurate result by splitting the field into the time and space parts.

 

[yuiop]

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/SharedDocs/Publikationen/EN/IERS/Publications/tn/TechnNote36/tn36__151,templateId=raw,property=publicationFile.pdf/tn36_151.pdf .

Pervect is right that the question is ambiguous. For example for an orbiting object, there are no forces and the object is simply following a geodesic. The old Newtonian concept of centripetal force balanced by centrifugal force is not valid although it a useful mathematical shortcut. It could be said that the acceleration of an orbiting object is zero because for a circular orbit the radial height remains constant and certainly no forces can actually be measured onboard the satellite. However, there are some cases where it reasonable to ask what the gravitational forces are. For example for an extended object such as a moon, the tidal forces due to the different forces on the extended object are real and can tear the moon apart at the Roche limit. Also, when asking about the buoyancy forces acting on a submarine in the Supplee-Matsas paradox require that we think in terms of forces rather than objects simply moving along geodesics. It would be helpful if the OP let us know what he had in mind. Then we might be able to produce a reasonable approximation of equation in quasi Newtonian terms that is of some practical use in a weak field situation. Come to that, the OP has not even specified if meant a weak or strong field situation.

In this thread there are a variety of solutions which can be put down to the ambiguity of the question and due to the difficulties of converting GR equations to extended Newtonian form because GR does not really deal with forces in the traditional way.

 

[yuiop]

This factor is approximately (1-Gm/rc²) but to get the correct result for Mercury's perihelion precession it is also necessary to include an additional term, which for GR in isotropic coordinates to Post-Newtonian accuracy is as follows:

(1-Gm/rc²+(1/2)(Gm/rc²)²).

As previously mentioned, in strong field cases where the gradient of the space term may be different from the gradient of the time term, it is still possible to get a completely accurate result by splitting the field into the time and space parts.

In the last equation you quote, there is no variable for the speed of the object so it is not really adressing the question in the OP, namely "Is force of gravity dependent on speed". It may be that you are assuming an orbiting particle and for a given orbital radius we can assume a given average speed but you should make that clear. Also, for precession to occur, the orbit has to be non-circular, so the speed will be varying so that should also be made clear.

 

[yuiop]

This leaves us with
\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}})

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 -\frac{GMv^2}{c^2 r^2}. There is also an additional effect that accelerates you in the direction of your motion.

Sorry, my notation is very bad. \mathbf{v} = \dot{\mathbf{r}}. This is the correction to Newtonian result, so when v=0, the correction is 0 as well. The extra acceleration is to the direction of \mathbf{r}, so a negative sign means the orbit is bent towards the gravitating mass.

OK, that clears that up. Sorry, but I am not very good with vector notation. Could you state your equation in a non vector form with v for radial velocity and h for angular momentum per unit mass which takes care of horizontal velocity. Also, I do not think it is good to ignore the angular momentum terms when we are talking about orbiting objects.

My best guess including the Newtonian term, for radial velocity (v) only, is:

\Delta \mathbf{a} = -\frac{GM}{r^2} + \frac{3GMv^2}{c^2 r^2}

but that might not be right.

backtracking a little your equation with the Newtonian term added in is:

\Delta \mathbf{a} = -\frac{GM}{r^2} + \frac{4GM(\hat{\mathbf{r}} \cdot \dot{\mathbf{r}}) \dot{\mathbf{r}}}{c^2 r^2} - \frac{GMv^2 \hat{\mathbf{r}}}{r^2}

Could you remove the vector notation from that and make it clear when you are talking about vertical or horizontal velocities?

 

[yuiop]

... 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?

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.

For a circular orbit, GM/r \approx v^2 but since you are ignoring angular momentum terms and seem to be only considering radial motion then the escape velocity 2GM/r \approx v^2 might be a more appropriate substitution and in that case your final equation becomes:

\Delta |\vec a| \approx \frac{GMv^2}{c^2r^2}

which is the same as what Clamtrox got.

You say this is just a correction term (for vertical or radial velocity) so the main terms for gravitational acceleration where the horizontal velocity is negligible are:

\Delta |\vec a| \approx \frac{GM}{r^2} + \frac{GMv^2}{c^2r^2}

where v is the radial velocity.

This reference http://farside.ph.utexas.edu/teaching/336k/Newton/node116.html gives the main terms for gravitational acceleration with negligable vertical motion as:

\Delta |\vec a| \approx \frac{GM}{r^2} + \frac{3GM h^2}{c^2r^4}

where h is the angular momentum per unit mass.

Put the two equations together and we get:

\Delta |\vec a| \approx \frac{GM}{r^2} + \frac{GMv^2}{c^2r^2} + \frac{3GM h^2}{c^2r^4}

If we use the orbital GM/r \approx v^2 substitution you originally used, then the result would be:

\Delta |\vec a| \approx \frac{GM}{r^2} + \frac{3GMv^2}{c^2r^2} + \frac{3GM h^2}{c^2r^4}

I am not sure the signs are correct. Are all the corrections operating in the same direction? Since most of the equations give only the corrections and omit the Newtonian term it is hard to tell if corrections are in the same direction as the Newtonian term or in the opposite direction. Hopefully people here will post equations that include the Newtonian term so that there is no ambiguity over the sign conventions used. We all know that the Newtonian term works downwards ;)

This reference http://www.biblioteca.org.ar/libros/90154.pdf gives the first order post Newtonian gravitational force as:

F \approx -\frac{GMm}{r^2} + \frac{G^3M^3m(rc^2 + 1.5 GM)}{h^2r^2(rc^2 + 6GM)}

Note in this case this reference suggests that the angular momentum per unit mass gives a force contribution in the opposite direction to the Newtonian term in contrast to the other equations here. Note that this effective centrifugal force reduces with increasing angular momentum which is sort of the opposite of what you might ordinarily expect, but from my recollection of orbiting objects very close to a black hole, that might well be correct.

Hopefully we might eventually reach a consensus on how gravity depends on vertical and horizontal motion in the weak field approximation.