# Relative Mass Gravity

1. Jun 25, 2011

### TWest

Fg= Gm1m2/R2 and E=MC2(1/(1-(V2/C2)1/2))-MC2

So my question is, does the relative mass created by gamma,which I subbed for the relativity Equation in the above equation increase the gravitational pull that is exerted by a object near a body of mass? lets say we accelerated the moon to .99c or something, would this increase the pull on the earth by the moon or vice versa.

P.S. Sorry for terrible grammar

Last edited: Jun 25, 2011
2. Jun 25, 2011

### my_wan

In special relativity no. Because at such speeds it merely looks like the distance between the bodies have changed so that the difference in gravitational acceleration looks normal. In GR time dilation, where all observers agree on the acceleration to differing extents, it does. This is how Einstein explained the perihelion of Mercury. However, we do not say that gravity increased or decreased, we say that the apparent mass of the planets has changed. Same difference though.

3. Jun 25, 2011

### TWest

So then, it is merely a matter of how the observer sees it but, how can it be both? Logic states that One or the other has to happen right either the length truly gets smaller or the gravitational pull has truly become greater or is that flawed reasoning? So I guess my secondary question is the GR definition or SR definition the "Real" one.

Last edited: Jun 25, 2011
4. Jun 25, 2011

### Staff: Mentor

First, there is no gravity in SR.

Second, in GR, gravity is not given by Newton's equation but by the Einstein Field Equations. In the EFE the source of gravity is not mass (neither relativistic nor rest mass) but the stress energy tensor.

5. Jun 25, 2011

### pervect

Staff Emeritus
There are various issues that arise to confound what the "gravitational field" of a moving particle might be. One issue is related to the fact that you can't actually define gravity as a force on a test charge as you can in E&M, because everything is affected by gravity,there is no "gravity-neutral" reference object.

One of the easiest away around these issues is to avoid them by considering the total gravitational field of a box of particles. The boxes all start out identical, in a state where the particles are stationary and unmoving. One then adds energy to one of the boxes, by making the particles it contains move.

If you look at the whole system, box + particles, and measure its external gravitational field, the total field acts as if you'd added a Newtonian-type mass E/c^2 to the box.

So far, so good. But there's one other interesting effect. If your box was a sphere, the gravitational field due to the walls inside the box is zero, as one might expect from the simple Newtonian analysis. So one might expect that the probe readings inside the box, just below the surface, ignoring the walls would also increase to show that the gravitational field acted as if one had added a mass E/^c2 to it. However, this doesn't happen. The probe measures a larger internal gravitational field. It's about equal to what one would expect if the moving particles had a total newtonian mass of 2E/c^2.

Resolving this takes some effort, and it shows that the caution about talking about "the" mass of a moving particle was well deserved. One of the points in a resolution to this issue is that pressure, and tension, as well as mass and energy, affects the gravitational field of an object. In this case, the tension in the walls of the box is important to the problem, and can't be neglected in determining its gravitational field. It's an illustration of the need for using the "stress-energy-tensor" that the other posters have mentioned, in a complete treatment of the problem.

6. Jun 25, 2011

### Bill_K

TWest, Sorry I didn't see the question sooner. It's a good question, and has a simple easily-derived answer! Working out the details now and will post in the morning.

7. Jun 26, 2011

### Bill_K

The appropriate framework to answer this question is general relativity, which has the correct relativistic description of gravitational fields. But linearized general relativity since we are in the weak field region. The field of the source ("the Moon") is the Schwarzschild solution, and the most convenient coordinate system to write it in is Eddington coordinates. (The choice of coordinates is arbitrary, and a different choice would correspond to a gauge transformation when we go to the linearized theory.)

ds2 = - (1 - 2GM/r) dt2 + 4GM/r dt dr + (1 + 2GM/r) dr2 + r2 d2Ω
= (- dt2 + dr2 + r2 d2Ω) + 2GM/r (dt + dr)2

This shows that the Schwarzschild solution in Eddington coordinates is a Kerr-Schild metric: gμν = ημν + 2GM/r σμσν where σμ is a retarded null vector pointing in the direction from the source point S on the moon to the observation point P on Earth. The fact that it's a retarded null vector reminds us that the influence of gravitation travels at the speed of light: the gravitational field that we experience now results from the moon's position a time t = r/c ago.

The reason that Eddington coordinates are convenient is that it has let us write Schwarzschild in the form Minkowski space + covariant term. What we really need is the field of a moving Schwarzschild, and now we can get it by a simple Lorentz transformation. That is, a moving Schwarzschild field is described by exactly the same form except that σμ has been Lorentz transformed.

The acceleration of test particles due to gravity is given by the geodesic equation, d2xμ/dτ2 + Γμνσ dxν/dτ dxσ/dτ = 0. For linearized gravity this reduces to an acceleration ai = 1/2 (h00,i - 2 hi0,0). The h00,i term is the usual gradient of the Newtonian potential, while the hi0,0 term is an additional gravitomagnetic effect.

Now suppose the Moon moves in the x-direction with velocity v, with distance of closest approach in the z-direction given by b. Then σμ = (γ, γv/c, 0, 1). The quantity in the metric that was r in the source frame is √(b2 + γ2v2t2) in our frame, and we finally get a time-dependent acceleration in the z-direction of

a = GMγ2(b - 2γvt)/(b2 + γ2v2t2)3/2

The point of all this is the γ2 factor in front. The acceleration caused by a moving Schwarzschild is larger. Moreover, for the corresponding problem in electromagnetism the field of a moving charge is larger by a single factor γ, but here we get two factors of γ since gravity is a tensor rather than a vector.

8. Jun 26, 2011

### bcrowell

Staff Emeritus
Interesting post, Bill! I haven't worked through all the details of your calculation, but I'm trying to get the flavor of it. Since you use linearized gravity approximations, am I right in thinking that this only gives the leading-order dependence of a on v? Have you worked out the result when the motion is radial? I'm thinking that when you average over all directions for the velocity vector, you should probably recover dependence on gamma to the first power as expected from mass-energy equivalence. I.e., if we simply heated the moon up, its gravity should go up in proportion to the average gamma of the molecules.

9. Jun 26, 2011

### PAllen

Of course, you can turn this around and analyze the effect of a Schwarzschild field on a rapidly moving test body instead. You should get an equivalent result, that there is somehow 'more effect than expected' for the rapidly moving test body.

10. Jun 27, 2011

### TWest

Oh, I was just wondering because I have been attempting to find a way to have high kinetic energy without increasing the mass of the Object or changing the rate of time passage in the area that is described by these theories and I thought I may have found a loop hole but I guess not...... After Seeing that EQ. BTW thanks for the detailed explanation Gentlemen

11. Jun 27, 2011

### Bill_K

Yes equivalent, but I believe you will only get one γ if you do it that way. In the frame where the Schwarzschild mass is standing still and the test particle is moving, the acceleration you measure will be the three-dimensional quantity a = dv/dt, which is smaller than the comoving quantity dv/dτ by a factor of γ.

12. Jun 27, 2011

### PAllen

Shouldn't it be possible to do this in a coordinate independent way? Compute the geodesic, define an acceleration in terms of deviation relative to a path as if the connection were flat Minkowski, using tau along the geodesic. I don't know how to formalize this. Any ideas?

Basically, if something is measurable, there must be some way to express it such that it is invariant. Changing coordinates cannot change the result of an actual measurement. And if it can't be measured, who cares?

Last edited: Jun 27, 2011
13. Jun 27, 2011

### PAllen

I can partially see how to do this in coordinates in which the massive body is stationary. Consider a worldline (t,r,theta,phi) = (t, r0,0,0). This has a proper acceleration. That would be effective gravity for a stationary test body. What we need for a rapid test body is the definition of world line that is 'Newtonian (or Minkowski) straight' towards a distant point (e.g. star). It is clear that this path should become geodesic as distance from source of gravity increases, the spacetime being asymptotically flat. Given such a world line, the proper acceleration at closest r should be a good definition of effective gravity for a rapidly moving test body.

The only problem is that I don't see an unambiguous way to define such a 'Newtonian straight' path near the massive body. Any ideas?

14. Jun 27, 2011

### PAllen

Here is a possible answer that is useless complicated. Consider again the static case. It is characterized by static Doppler shift of light from the massive body, and *also* by static Doppler shift of very distant stars. So we can say a 'Newtonian straight path' for a rapidly moving test body is one that maintains static Doppler shift to all sufficiently distant stars. The geodesic, orbital, path would not do this. (This definition is still only good in weak gravity, else we must factor in light bending by the massive body; perhaps consider only stars in the hemisphere away from the massive body).

Another variant: Consider test body far from massive object with high speed relative to it. There is some extremely distant point (stationary relative to massive body) for which test body has maximum blue shift (a point it is heading towards). Define the test body path that maintains this exact blue shift towards this distant reference point. Declare that to be a Newtonian straight path, and determine its proper acceleration at closest approach to massive body.

15. Jun 27, 2011

### pervect

Staff Emeritus
I hope I have the time to look this over more carefully, but I'd like to ask you a few things.

1) Does your linearized result agree or disagree with an exact result I calculated for the _tidal force_ by boosting the Riemann tensor?

I concluded long ago that the tidal forces are what one can easily measure. (I hope you'd agree? If not, we can talk about it, it's been awhile since I discussed the issue with anyone, I've found trying to convince laypeople of this point not very fruitful, usually).

For some past history, https://www.physicsforums.com/showpost.php?p=740307&postcount=2 might be useful....

The standard textook results are: (MTW,pg 821 and pg 860), using their notation)

For an unmoving mass, and in Schwarzschild coordinates, the tidal forces in a schwarzschild basis frame are:

in the $\hat{r}$ direction 2m/r^3
in the $\hat{\theta}$ direction -m/r^3
in the $\hat{\phi}$ direction -m/r^3

The frame field used is defined by the orthonormal basis of one forms
$\omega^{\hat{t}} = \left( 1 - \frac{2M}{r} \right) ^ {\frac{1}{2} } {\bf d}t$
$\omega^{\hat{r}} = \frac{{\bf d}r}{\left( 1- 2M/r\right)^{\frac{1}{2}} }$
$\omega^{\hat{\theta}} = r{\bf d}\theta$
$\omega^{\hat{\phi}} = r\, sin \theta \, {\bf d}\phi$

Basically the frame field, denoted by the hats, are what the Riemann (and hence the tidal forces) would be for an observer using local coordinates. The metric in these local "hattted" coordinates is Minkowskian, a diagonal metric with components +/- 1.

It's a textbook result that if you move directly towards the source, the tidal forces are not changed. The textbooks don't cover what happens if you move at right angles, however, as in a flyby. This I had to compute.

For a mass moving at right angles to the source (i.e. in the $\hat{\theta}$ direction) at velocity $\beta$ (as defined in the frame-field) at the exact same point, you need to boost the Riemann by the appropriate Lorentz transform (which is the usual one, as mentioned the metric coefficients are diagonal).

$$R_{abcd} L^{a}{}_{e}L^{b}{}_{f}L^{c}{}_{g}L^{d}{}_{h}$$

It's tempting - but incorrect! - to use R_{abcd}u^{b}u^{d}, one actually need to the boost, or something mathematically equivalent. It would be tedious and error-prone to do this by hand, but feasible with GRTensor.

The result for the tidal forces is:

$$\hat{r} = \frac{2m}{r^3} \,\left( \frac{1+\beta^2/2}{1-\beta^2} \right)$$
$$\hat{\theta} = -\frac{m}{r^3}$$
$$\hat{\phi} = -\frac{m}{r^3} \left( \frac{1+2 \beta^2}{1-\beta^2} \right)$$

interestingly enough, the component in the direction of motion doesn't change.

This hasn't been checked by another human, but the sum of the tidal forces is zero as it should be

Last edited: Jun 27, 2011
16. Jun 27, 2011

### pervect

Staff Emeritus
Let me put in a plug for using tidal forces.

The tidal forces can be measured in a local coordinate system, and they're already defined by a tensor, so we know how they transform.

Now, you'd think you could integrate the tidal force, which is just force / unit length, over some path to infinity to come up with a "force", but it seems the answer depends on the choice of path you use, whenever I try to dig into it. So far I haven't seen any good way to define the path.

Note that the tidal force is actually defined by geodesic deviation, so it's only equal to what you measure if you measure it with a non-rotating observer. Unfortunately, from what I can tell, non-rotating observers do not maintain their orientation with respect to the fixed stars when a mass goes whizzing by, making the problem more complex than it appears - a "non-rotating obserer" finds that his orientation has changed with respect to the fixed stars after the mass has whizzed by. I'm not sure what optical effects happen during the actual event.

17. Jun 27, 2011

### pervect

Staff Emeritus
Oh, the other question I had for Bill K - what order of accuracy do you expect the linear analysis to have, in terms of coefficeints of $\beta$. The lowest order of effect I see on the tidal forces is of order $\beta^2$.

Last edited: Jun 27, 2011
18. Jun 27, 2011

### PAllen

Except that I don't see that tidal gravity describes what is being sought. For a stationary observer we have the proper acceleration of a world line with (r,theta,phi) constant. We want to see how this changes if either we consider the massive body boosted, or the observer boosted (and hopefully find they are the same). Clearly, the static observer case involves no tidal gravity at all.

I thought of another proposal for the 'straight but not geodesic' world line of rapidly passing test body. Consider a geodesic path representing rapid motion very far from the massive object. With asymptotic flatness, we can have this path be arbitrarily close to a Minkowski geodesic. Then consider a path defined relative to this reference path: for each proper time along the reference path, extend a spacelike geodesic 4-orthgonal to the reference path toward the massive body. Map the locus of events a fixed proper distance from the reference path along these geodesics. Then the maximum proper acceleration along this path is the desired quantity.

If you build fermi-normal coordinate from this test body path, and transform to those (the massive body now highly boosted in these coordinates), and compute the maximum proper acceleration of this coordinate stationary (non-geodesic) path, you must obviously get the same quantity.

This, I claim, is an ideal way to discuss the equivalent effect of either a boosted black hole or boosted test body.

19. Jun 27, 2011

### pervect

Staff Emeritus
It's something that we can actually measure that describes the "strength" of gravity. We can't measure the gravitational force directly, because gravity acts on everything, but we can measure the rate of change of the force. Not only that, we know how it transforms, because it's a tensor.

At the very least, I think that the gradient of any correct description of gravity in terms of forces (if such exists, I'm skeptical), will yield an expression for the tidal force that should match the theoretical calculations.

Hmmm? A stationary observer has a radial tidal gravity of -2M/r^3, in both GR and in the Newtonian theory, which they can easily observe. I

Why do you think it involves no tidal gravity at all, much less "clearly" involves no tidal gravity?

n the Newtonian case, you can just integrate the tidal force over a radial path to infinity and get -M/r^2, integrate again and get the potential -M/r.

Similarly, you can take the Newtonian force, and say that the radial tidal force is -2M/r^3 just by differentiating.

In the GR case, to get an exact answer you'd need to take into account the metric coefficients to correct for the length, i.e. sqrt(g_rr) dr is the length, and the time dilation factors to get "force at infinity", assuming that's where you're measuring the force (it's unclear, really). But if you say "r is large", then dr is "close enough to a unit length" and "time dilation is small, we don[t care about it", you just integrate, and get F = M/r^2. Which is all we're after, I think, something that works for large enough r.

I'm not sure what you're proposing here - the maximum proper acceleration of a fermi-normal observer at infinity?

How would you demonstrate that said force you compute in this manner "transforms as a 4-vector", when you've defined it by recourse to a particular set of coordinates?

20. Jun 27, 2011

### Ben Niehoff

In order to write down the metric of a boosted black hole, first use the coordinate transformation

$$r = R \left( 1 + \frac{M}{2R} \right)^2$$

to put the Schwarzschild solution in isotropic coordinates:

$$ds^2 = - \left( \frac{1 - \frac{M}{2R}}{1 + \frac{M}{2R}} \right)^2 \; dt^2 + \left( 1 + \frac{M}{2R} \right)^4 \; (dx^2 + dy^2 + dz^2)$$

where R is defined as $R = \sqrt{x^2 + y^2 + z^2}$.

Now obviously x, y, and z become the ordinary Cartesian coordinates in the asymptotic region. So, simply Lorentz boost as usual:

\begin{align*} t &\mapsto \gamma \; t - \beta \gamma \; z \\ z &\mapsto - \beta \gamma \; t + \gamma \; z \end{align*}

So now you have

$$ds^2 = - \left( \frac{1 - \frac{M}{2R}}{1 + \frac{M}{2R}} \right)^2 \; (\gamma \; dt - \beta \gamma \; dz)^2 + \left( 1 + \frac{M}{2R} \right)^4 \; \big( dx^2 + dy^2 + (\gamma \; dz - \beta \gamma \; dt)^2 \big)$$

where R is now given by

$$R = \sqrt{x^2 + y^2 + (\gamma \; z - \beta \gamma \; t)^2}$$

So as you can see, the relationship between the event horizon radius and the gamma factor is a little complicated. However, with a little algebra, you should be able to work out what happens along the direction of motion at least.