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

[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...