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aditya23456

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

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- Thread starter aditya23456
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In summary, according to general theory of relativity, there is no unique definition of 'stationary'. However, for a moving body, there is no unique definition of a 'stationary' world line. Additionally, the 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.

- #1

aditya23456

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

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- #2

PAllen

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aditya23456 said: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.

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- #3

tom.stoer

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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

The curvature of the manifold defines geodesics

A different energy-momentum density of the stars

The geodesics

It's like saying that

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julian

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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 [itex]\gamma[/itex] because of length conrtraction, plus the mass increases and hence the weight of the boat increases...the conclusion is that the boat will sink...

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- #5

julian

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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.

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- #6

tom.stoer

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as I said, gravity is not a force!

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julian

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- #8

tom.stoer

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

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julian

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aditya23456 said: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?

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clamtrox

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julian said:

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?

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julian

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- #12

Mentz114

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

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tom.stoer

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Mentz114

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tom.stoer said:

Sure thing. Sorry to nit-pick ...

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In a sense, yes.aditya23456 said:Is there any such relation as per general theory of relativity?

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/r

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.

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clamtrox

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D H said: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/r^{2}, 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+v

- #17

tom.stoer

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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.

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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 said:Do you know if the corrections are in first order what you would expect by naive thinking, ie. F->F(1+v^{2}/c^{2}/2), or something different?

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clamtrox

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D H said: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 [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?

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That's off by a factor of three.clamtrox said: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 [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?

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.

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PAllen

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clamtrox said: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 [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?

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.]

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- #22

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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.PAllen said: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?

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.

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julian said:speedand 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.

- #24

PAllen

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D H said: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

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PAllen

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pervect said: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.

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- #26

Jonathan Scott

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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]

- #27

clamtrox

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PAllen said: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.

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Naty1

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aditya: was your original questions answered?

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aditya23456

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Thing is that I can't ask my question completely for having a genuine research ;)

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Naty1

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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.

- #31

jimgraber

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See my answer here:http://physics.stackexchange.com/qu...al-gravitation-is-really-a-square/22025#22025clamtrox said:Do you know if the corrections are in first order what you would expect by naive thinking, ie. F->F(1+v^{2}/c^{2}/2), or something different?

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.

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- #32

yuiop

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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

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

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- #33

clamtrox

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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.

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.

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- #34

clamtrox

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jimgraber said: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

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- #35

clamtrox

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Naty1 said: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.

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