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

[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:
<br /> \mathbf{g} = - \frac{c^2}{\Phi_t} \nabla \Phi_t<br />
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
<br /> \mathbf{g} (1 + v^2/c^2)<br />
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:
<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 />

[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 &lt;&lt;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...