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

[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:

$$F' = \frac{GMm}{R^2} * \frac{\sqrt{1-v^2}}{(1-W^2)}$$

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:

$$F' = \frac{GMm}{R^2} * \frac{\sqrt{1-v^2}}{(1-W^2)}$$

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.

 

[Jonathan Scott]

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 non-circular so the speed will be varying so that should also be made clear.

As described in my previous post, that was the expression for the time dilation term in the metric, which is approximately 1 minus the Newtonian potential, and has nothing to do with the speed. If you call that $\Phi_t$ then the equivalent of the Newtonian gravitational field is given by the following:
$$\mathbf{g} = - \frac{c^2}{\Phi_t} \nabla \Phi_t$$
This doesn't depend on the speed either.

However, when something is moving in this field, the effect is roughly speaking as if the field were actually
$$\mathbf{g} (1 + v^2/c^2)$$
That's where the speed comes in. In this expression, v is the current coordinate speed in isotropic coordinates (and c is the speed of light in isotropic coordinates, not the standard value). In this simple case, because this is a ratio of speeds, it also works if v and c are the local values.

 

[PAllen]

First, I definitely agree with Pervect that the core statement to make is that in GR the question of whether relative motion between two bodies increases the 'force' between them (at closest approach) compared to not moving is just ill defined.

More below the following quotes:

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.

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:
$$\frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right )$$
This applies to motion in any direction. The Newtonian field is defined as follows:
$$\mathbf{g} = - \frac{c^2}{\Phi_t} \nabla \Phi_t$$
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:
$$\Phi_t = \frac{1 - Gm/2rc^2}{1 + Gm/2rc^2}$$
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:
$$\frac{d\mathbf{p}}{dt} = \frac{E}{c^2} \left ( \mathbf{g}_t + \frac{v^2}{c^2} \mathbf{g}_x \right )$$
$$\mathbf{g}_t = - \frac{c^2}{\Phi_t} \nabla \Phi_t$$
$$\Phi_t = \frac{1 - Gm/2rc^2}{1 + Gm/2rc^2}$$
$$\mathbf{g}_x = + \frac{c^2}{\Phi_x} \nabla \Phi_x$$
$$\Phi_x = \frac{1}{(1 - Gm/2rc^2)^2}$$

[Edit: can't read correction below]
This approach seems in general agreement with the second approach below, in that (1+v^2/c^2) := gamma^2

http://ajp.aapt.org/resource/1/ajpias/v53/i7/p661_s1?isAuthorized=no

https://www.physicsforums.com/showthread.php?p=3375529#post3375529

Note that while these agree in direction, they disagree in magnitude, which gets back to my agreement with Pervect.

(My approach seems to lead to a result closer to Bill_k's).

 

[yuiop]

I have found another reference http://www.rowan.edu/open/depts/math/NGUYEN/Newton's%20Greatest%20Blunder.pdf (Eq 5.9) that gives the General Relativistic force law (per unit mass) as:

$$F \approx \frac{GM}{r^2} + \frac{3GM h^2}{c^2r^4}$$

where h is the angular momentum per unit mass.

This at least agrees with another reference http://farside.ph.utexas.edu/teaching/336k/Newton/node116.html that I gave earlier.

 

[PAllen]

I have found another reference http://www.rowan.edu/open/depts/math/NGUYEN/Newton's%20Greatest%20Blunder.pdf (Eq 5.9) that gives the General Relativistic force law (per unit mass) as:

$$F \approx \frac{GM}{r^2} + \frac{3GM h^2}{c^2r^4}$$

where h is the angular momentum per unit mass.

This at least agrees with another reference http://farside.ph.utexas.edu/teaching/336k/Newton/node116.html that I gave earlier.

The issue here is how that is affected if you have a star moving by a test body at high speed? That's where I think, if you are comparing to the static case, the 'force at closest approach' can be considered to increase by a factor of gamma^2, along with smaller corrections.

 

[yuiop]

The issue here is how that is affected if you have a star moving by a test body at high speed? That's where I think, if you are comparing to the static case, the 'force at closest approach' can be considered to increase by a factor of gamma^2, along with smaller corrections.

I agree. This is essentially what the equation I gave in #37 https://www.physicsforums.com/showpost.php?p=3844584&postcount=37 says. The gravitational force on the test particle in the rest frame of the star is increased by a factor of gamma, while the force on the test particle as measured in the rest frame of the test particle is increased by a factor of gamma squared. My equation is the only one in this thread that allows for the case when the gravitational body to be moving relative to the observer. On the other hand my equation does not allow for circular motion.

 

[Jonathan Scott]

The issue here is how that is affected if you have a star moving by a test body at high speed? That's where I think, if you are comparing to the static case, the 'force at closest approach' can be considered to increase by a factor of gamma^2, along with smaller corrections.

As you're presumably aware (but others may not be), what actually happens in that case can easily be determined by applying the appropriate Lorentz transformations to the various quantities in the rest frame of the gravitational source, at least in the weak approximation case.

 

[Mentz114]

I can't make out if the OP is referring to geodesic motion or the other. I haven't seen any mention of the GR predictions for time-like orbits in the Schwarzschild spacetime. For orbits in the $\phi$ direction in the plane $\theta=\pi/2$ the radius must be greater than 3M ( r > 3M) and the angular velocity is $\sqrt{M}/r^{3/2}$.

Clamtrox - earlier you mentioned a perturbative calculation. Did you try accelerating a Hagihara frame to get your extra radial acceleration ?

[later]
I Lorentz boosted a Hagiara frame in the $\phi$ direction and found it generates a radial acceleration of $2\beta\sqrt{M}/r^{3/2}$, assuming $\mid \beta \mid <<1$ to first order in $\beta$. This makes sense because increasing speed will lead to a larger r and diminshing speed will lead to a smaller r for the orbit.

However, I don't know if this is relevant to the original question.

 

[pervect]

The scenario of a star passing by a test body at high speed has been discussed in the literature and on the forums. I'll post a few references when I'm through making a few other points.

Part of the reason the discussion tends to go around in circles is that there isn't any way of actually measuring "gravitational force" with a local measurement. Since there isn't any agreement on how one would go about measuring the "force" (and currently people are trying to divine it's value by various non-local measurements, such as orbital precession), it's not surprising that there isn't any agreement on what it's value is.

Physics is ultimately supposed to be about things you can measure. If you start to talk about things you can't measure, the discussions generally aren't very productive.

Let's look at the question of why you can't measure the gravitational force directly with local measurements, when you can measure electrical forces.

To measure an electric field, you can compare the motion of a neutral, uncharged particle to a charged one. Their relative acceleration (measured when they're both close to each other) tells you the value of the electric field at that point. However, you can't do the same with gravity - because gravity affects everything - there isn't such a thing as a gravitationally neutral particle.

What you can actually measure, unambiguously, is the gradient of the gravitational field, otherwise known as the tidal forces. This is basically one of the components of the Riemann curvature tensor. It's occasionally called the "Electric part of the Riemann" or the "Electrogravitic Tensor" if you study the Bel decomposition. http://en.wikipedia.org/w/index.php?title=Bel_decomposition&oldid=325505109

You can unambiguously both measure and compute this tensor quantity - and it's got a perfectly fine Newtonian analogue, the "tidal tensor". http://en.wikipedia.org/w/index.php?title=Tidal_tensor&oldid=332450104

It's a matrix (a rank 2 tensor) rather than a force, however. You mulitply the tidal tensor by the displacement vector - the result is another vector, the relative acceleration.

Unsurprisingly, it's also related to what you'd experience. If you were on the Earth, and a relativistic black hole flew by, you would not actually directly feel the black hole's gravity - anymore than you directly feel the sun's gravity. What you would feel would be the tidal effects. Those would be what you could actually measure (and what you feel from the sun, and the moon).

There's one other thing you could measure - the residual motion of the Earth (or in general test bodies) after the black hole had passed through. This is discussed in http://ajp.aapt.org/resource/1/ajpias/v53/i7/p661_s1?isAuthorized=no , "Measuring the active gravitational mass of a moving object"

(Unfortunately you need a subscription to see it).

Going back to tidal forces:
In the ultra-reltavistic limit, published results include http://arxiv.org/abs/gr-qc/0110032, the key words to look up for a literature search are Aichelburg-Sexl boost.

For the case of a boosted Schwarzschild metric (i.e. a black hole at an arbitrary velocity), I've done some calculations in the past for the tidal forces (though as far as I know nobody has verified them).

https://www.physicsforums.com/showthread.php?t=83157&page=1

Basically, the results for the tidal force in the r direction are:
$$\frac{2GM}{r^3} \left( \frac{1 - \frac{\beta^2}{2}}{1-\beta^2} \right)$$

This is the radial stretching tidal force, Newtons / meter, in the "r" direction.

This should be compared to the Newtonian value -2GM/r^3 - which is also the value seen by a stationary Schwarzschild observer.

Note: it's best to actually boost the 4x4 Riemann, though it's a pain to calculate. Early attempts took a shortcut via the geodesic equation that didn't work as expected, because the basis vectors were not in the end the basis vectors of the moving observer.

You can integrate the above to get a "force", but beware, there's no guarantee that it will be path indepedent).

If a candidate for gravitational "force" does not predict the correct tidal effects, given that the tidal effects are what we can actually measure, I would remark that it's not a good idea to rely on said expression for "force" overmuch - even if it does give you the correct value for, say, orbital precession.

 

[PAllen]

If you can define what 'staying in place' means, then the force required to do that as a star goes by is locally measurable and is an invariant (proportional to norm of 4-acceleration). The problem, as I see it, is ambiguity of definition 'staying in place'. For any given definition you come up with, you do have locally measurable, invariant quantity.

Given a definition of staying in place, you also (with coordinate transformation) have a (non-unique) definition of 'moving straight by a body as if it wasn't there'. Obviously, the coordinate transform won't change the 4-acceleration, so the result is symmetric.

So I would say the problem is not at all that you can't make local measurements. The problem is you can't determine which possible local measurement is correct - a slightly different definition of the above world line can yield a substantially different proper acceleration.

 

[pervect]

If you can define what 'staying in place' means, then the force required to do that as a star goes by is locally measurable and is an invariant (proportional to norm of 4-acceleration). The problem, as I see it, is ambiguity of definition 'staying in place'. For any given definition you come up with, you do have locally measurable, invariant quantity.

I'd tend to agree - in fact if you had an object that was unaffected by gravitational forces, and you shielded it from all other forces, it would define "staying in place". But there isn't any such object, alas.

Other attempts I"ve seen also - for instance, if you try to define "staying in place" by observing fixed, distance, stars, the light signals you're using to observe the distant stars will be distorted by the gravitational fields of any moving objects (such as the heavy, flyby object).

This does give a means to determine the amount of velocity that you pick up as a result of a gravitational flyby - you can make your observations to determine your velocity relative to the distant stars before the disturbance (well, as long as your objects aren't so distance that you run into cosmological redshift issues), and after the disturbance, and compare them - but attempting to maintain your observations during the flyby won't work.

Given a definition of staying in place, you also (with coordinate transformation) have a (non-unique) definition of 'moving straight by a body as if it wasn't there'. Obviously, the coordinate transform won't change the 4-acceleration, so the result is symmetric.

So I would say the problem is not at all that you can't make local measurements. The problem is you can't determine which possible local measurement is correct - a slightly different definition of the above world line can yield a substantially different proper acceleration.

I'm not sure I quite see the difference, at least in operational terms, since we've agreed there isn't any way to define "staying in place"...

 

[PAllen]

I'd tend to agree - in fact if you had an object that was unaffected by gravitational forces, and you shielded it from all other forces, it would define "staying in place". But there isn't any such object, alas.

Other attempts I"ve seen also - for instance, if you try to define "staying in place" by observing fixed, distance, stars, the light signals you're using to observe the distant stars will be distorted by the gravitational fields of any moving objects (such as the heavy, flyby object).

That's just a 'complication' Pick an algorithm for factoring out gravitational effects to define hovering stationary near a body not moving relative to stars (I think we all agree hovering in this case is well defined). Blindly apply this to a flyby object assuming that at any moment you can treat it as stationary as far as factoring out gravity effects. Valid? Not really. A definition that can be arbitrarily agreed to and carried out? Yes.

Mathematically there are many choices easy to specify (but hard to compute). Pick a distant star. Establish a Fermi-Normal coordinate system based on its world line. Extend it as globally as you can (assume asymptotic flatness). Define that a line of constant position in these coordinates is hovering (which guarantees that proper distance to the chosen star in the star's natural spacetime folation, remains constant). Adopting this (or some other mathematical definition) you can compute what observations a pilot would need to make to stay on this path. Then, the thrust needed to do so is a local observable.

This does give a means to determine the amount of velocity that you pick up as a result of a gravitational flyby - you can make your observations to determine your velocity relative to the distant stars before the disturbance (well, as long as your objects aren't so distance that you run into cosmological redshift issues), and after the disturbance, and compare them - but attempting to maintain your observations during the flyby won't work.
I'm not sure I quite see the difference, at least in operational terms, since we've agreed there isn't any way to define "staying in place"...

I do see a difference between can't make a local measurement and can't justify why one of many local measurements is any better than another. To me, the problem of extending Newtonian force of gravity to GR is a fundamental problem of the latter category rather than the former.

 

[pervect]

I do see a difference between can't make a local measurement and can't justify why one of many local measurements is any better than another. To me, the problem of extending Newtonian force of gravity to GR is a fundamental problem of the latter category rather than the former.

Consider the situation where you have a flat space-time, far away from any mass. And what we want to do is to define what it means to be "not moving" as we bring a large mass close.

But the large mass distorts space-time, and in particular it also distorts space itself (using the usual time-slice).

If we use one of the usual analogies of space as being a flat rubber sheet, we can think of the problem as putting a mark on the rubber sheet, a mark that we want to use to define "stationary". But we know that the sheet is being deformed and stretched by the approach of our large mass. Initially the rubber sheet was flat - afterwards the sheet will not be flat, it will have to cover a sphere, or a saddle surface. (I'm not sure of the sign of the spatial curvature offhand).

It's not surprising that this idea has conceptual problems, I think. It turns out you can come up with various schemes to put "marks" on the rubber sheet, but what happens is that while some observers will say the mark is stationary, other observers will say the mark is moving - because the sheet itself is stretching as it changes shape, from something that was initially flat to something that is no longer flat, there's no way to put a mark on it that everyone will think is stationary.

 

[PAllen]

It's not surprising that this idea has conceptual problems, I think. It turns out you can come up with various schemes to put "marks" on the rubber sheet, but what happens is that while some observers will say the mark is stationary, other observers will say the mark is moving - because the sheet itself is stretching as it changes shape, from something that was initially flat to something that is no longer flat, there's no way to put a mark on it that everyone will think is stationary.

I completely agree with this (esp my bold, and it is implicit in my prior answers). However, my response is still the same: this prevents any preferred or unique definition; it does not prevent making some definition. Coordinates are arbitrary, but we can still choose and compute in them. A choice of convention for 'stationary' is completely analogous to a coordinate choice (as one of my example conventions clearly shows).

 

[clamtrox]

Put the two equations together and we get:

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

Don't do that! The term you added already contains the second term. There are many ways of writing the same thing, and I think everyone benefits if we don't try to confuse things further by using angular momentum density in one term and velocity in the next.

So I will stick to my earlier assumptions that particle velocity is relativistic, but the mass of the gravitating object is not: this means we don't get a circular orbit, and you can't use the "usual" formalism that one uses with for example Mercury precession.

The equation for acceleration to order v²/c² is
$\mathbf{a} = -\frac{GM\hat{\mathbf{r}}}{r^2} + \frac{4GM(\hat{\mathbf{r}} \cdot \dot{\mathbf{r}}) \dot{\mathbf{r}}}{c^2 r^2} - \frac{GMv^2 \hat{\mathbf{r}}}{c^2 r^2}$

If you have difficulties with vectors, it's very easy to write it without them: Let me define
$a_r = \mathbf{a} \cdot \hat{\mathbf{r}}$ and
$a_{\perp} = |\mathbf{a} - a_r \hat{\mathbf{r}}|$ and likewise for velocity v. Then
$a_r = -\frac{GM}{r^2} (1 + \frac{v^2-4 v_r^2}{c^2}) = -\frac{GM}{r^2} (1 + \frac{v_{\perp}^2-3 v_r^2}{c^2})$ and
$a_{\perp} = \frac{4 GM v_r v_{\perp}}{c^2 r^2}$
Did that make it clearer?

 

[clamtrox]

Clamtrox - earlier you mentioned a perturbative calculation. Did you try accelerating a Hagihara frame to get your extra radial acceleration ?

I think that will just confuse the subject even more: I think it's beneficial to focus on a situation where the gravitational field is weak compared to the velocity, ie. $\frac{2GM}{r} \ll v^2$ ie. the test particle is moving much faster than the escape velocity. If we do this, we deviate away from the standard lore of orbiting planets etc. but I don't think Hagihara frame makes any sense in this setup. You could of course assume the particle is on a circular orbit (that's what I did initially too), but then you need to remember that due to the relation between velocity and gravitational mass which applies in a circular orbit, the situation is simplified to the standard perihelion-of-Mercury-problem.

 

[PAllen]

Don't do that! The term you added already contains the second term. There are many ways of writing the same thing, and I think everyone benefits if we don't try to confuse things further by using angular momentum density in one term and velocity in the next.

So I will stick to my earlier assumptions that particle velocity is relativistic, but the mass of the gravitating object is not: this means we don't get a circular orbit, and you can't use the "usual" formalism that one uses with for example Mercury precession.

The equation for acceleration to order v²/c² is
$\mathbf{a} = -\frac{GM\hat{\mathbf{r}}}{r^2} + \frac{4GM(\hat{\mathbf{r}} \cdot \dot{\mathbf{r}}) \dot{\mathbf{r}}}{c^2 r^2} - \frac{GMv^2 \hat{\mathbf{r}}}{c^2 r^2}$

If you have difficulties with vectors, it's very easy to write it without them: Let me define
$a_r = \mathbf{a} \cdot \hat{\mathbf{r}}$ and
$a_{\perp} = |\mathbf{a} - a_r \hat{\mathbf{r}}|$ and likewise for velocity v. Then
$a_r = -\frac{GM}{r^2} (1 + \frac{v^2-4 v_r^2}{c^2}) = -\frac{GM}{r^2} (1 + \frac{v_{\perp}^2-3 v_r^2}{c^2})$ and
$a_{\perp} = \frac{4 GM v_r v_{\perp}}{c^2 r^2}$
Did that make it clearer?

And all this simplifies, for a high speed flyby, at closest approach (where radial v=0) and all acceleration is radial:

a = -Gm/r^2 (1+v^2/c^) =(appx) -gamma^2 Gm/r^2

 

[clamtrox]

And all this simplifies, for a high speed flyby, at closest approach (where radial v=0) and all acceleration is radial:

a = -Gm/r^2 (1+v^2/c^) =(appx) -gamma^2 Gm/r^2

1 + v²/c² = gamma + O(v⁴/c⁴) -- remember this is just a power series wrt velocity

 

[PAllen]

1 + v²/c² = gamma + O(v⁴/c⁴) -- remember this is just a power series wrt velocity

gamma^2, not gamma. Expansion of gamma would be (1+ (1/2)(v^2/c^2) + ...).

 

[clamtrox]

gamma^2, not gamma. Expansion of gamma would be (1+ (1/2)(v^2/c^2) + ...).

oops, I keep miscalculating that...