# Does the speed of moving object curve spacetime?

1. May 2, 2012

### M8M

Assume we have an object (o) at rest with a mass = m. Hence, we can calculate by general relativity the gravitational force (g) on this mass. Now, assume we remain stationary, at the origin, and a velocity (v) is imparted to the mass along the x-axis. The velocity (v) is substantially near the speed of light (c).

This situation begs the question:

Is the gravitational force (g) on the fast moving mass (m) now proportionally higher by a magnitude γ? Is speed linked to gravitational effects? Please provide credible citations which address this issue.

Thanks.

2. May 2, 2012

### pervect

Staff Emeritus
What curves space-time is not mass, but the stress-energy tensor.

The gravitational force on a test-mass is a well defined concept in Newtonian mechanics, and in special circumstances in general relativity, but isn't well defined in general circumstances in general relativity. What is well defined in general in GR s the Riemann curvature tensor, which is more or less equivalent to the Newtonian idea of tidal force, i.e. force / unit length.

Because gravitational force isn't really well defined, I doubht you will find much in the way of references on the topic. There is a treatment of something that is well defined, the velocity induced by a flyby, however.

Olson, D.W.; Guarino, R. C. (1985). "Measuring the active gravitational mass of a moving object". American Journal of Physics 53 (7)

To quote from the abstract:

3. May 2, 2012

### Matterwave

I think the easiest way to see if the velocity has an effect would be to use the Schwarzschild solution. One takes 2 observers, one which is stationary at Schwarszchild radius r (i.e. has 4-velocity u=(1,0,0,0)), and one observer which is moving but instantaneously at that same radius r (i.e. has 4-velocity u=(u0, u1, u2, u3)) and see which observer, if any, has a greater 3-acceleration.

Someone correct me if there's something wrong (something subtle I missed) with this scheme.

I have not carried out these calculations, but they shouldn't be too terribly difficult to do.

4. May 2, 2012

### M8M

pervect:

Thanks for your reply. It is a bit of a concern that the Olson reference is listed as having a mere 2 citations. Clearly, it is more popular on the forums than in the scientific community. I wonder why? Further, if what you say is true this seems to be a glaring disconnect between Special Relativity and the General Theory.

5. May 2, 2012

### Staff: Mentor

Do you mean the fact that gravitational effects do not in fact depend on the so-called "relativistic mass" $\gamma m_0$?

I would say rather that this is a glaring disconnect between popular and superficial introductory treatments of relativity that perpetuate the notion of "relativistic mass", and the professional physics community which mostly abandoned that notion decades ago.

6. May 3, 2012

### clamtrox

A moving object does not curve space-time any differently from a stationary object. This is because the curvature is a geometric quantity, and therefore invariant under coordinate transformations. However, like said above, moving objects do generate different apparent forces onto test particles through the source term (stress-energy tensor) in the Einstein equation.

7. May 3, 2012

### A.T.

Is it a correct approach to take the static field (say Schwarzshild metric) and apply length contraction to it, to obtain the field in a frame where the gravity source moves?

8. May 3, 2012

### M8M

"The gravitational force on a test-mass is a well defined concept in Newtonian mechanics, and in special circumstances in general relativity, but isn't well defined in general circumstances in general relativity"

In studying relativity, I have noticed the following:

In Einstein's treatment of Special Theory we are always concerned only with objects with constant velocity. Then, supposedly acceleration is GR as being indistinguishable, in some respects, to a gravitational force.

But, in the General theory -- we can now talk about an object moving at any speed and any acceleration. If one conceptualizes space as behaving like a viscous fluid then it can be imagined that a stationary object would not produce waves in the medium; however, a rapidly moving object would generate a wave pattern in the fluid -- i'm guessing equations would be found in acoustics and fluid dynamics for this type of behavior.

If the following is true -- energy-momentum tells space-time how to curve and space-time tells energy-momentum how to move/distribute. Why? My understanding is that GR tries to kill the idea of a gravitational force originally, by equating with acceleration, then "injects" it somehow later in the theory. Is that true? Thanks.

9. May 4, 2012

### PAllen

1) I would say it is more meaningful to calculate proper acceleration, not 3-acceleration (which corresponds to no measurable quantity, especially in SC coordinates for this purpose). Proper acceleration is what an accelerometer would measure.

2) While the u=(1,0,0,0) path has a well defined meaning as a static path (given the symmetry of the solution and the fact that it is static), you cannot come up with a preferred choice for the non-static path that aims to be 'straight as if the mass wasn't there', to compute proper acceleration for the path (I have tried this several times; there are plausible definitions - emphasis on the plural; each produces a different answer).

There is nothing subtle about the issue - there is no way at all to make your proposal correspond in any unique way to a measurable quantity.

10. May 4, 2012

### Mentz114

In SR, it is possible to deal with objects experiencing acceleration ( those that have curved worldlines). Non-accelerating (inertial) frames are distiguished as being equivalent
in that Newton's laws apply in all of them.

Yes, that is the principle of equivalence in one form. But replace the final 'force' with 'acceleration'.

There are two kinds of acceleration in the presence of gravity. 'Coordinate acceleration' is the kind you can't feel when you are in free-fall but accelerating, the other is 'proper acceleration' caused by an applied force. This you can measure with an accelerometer and f=ma applies.

Treating spacetime as a medium is not very productive and probably unnecessary. But gravitational waves can be deduced from GR.

Gravitational forces (except tidal effects) don't exist in GR because forces cannot be 'transformed away' by going into free fall. This was Einstein's great insight.

The huge difference between GR and Newtonian dynamics is that GR is a kinematic theory. The spacetime metric already contains all possible paths, so solving the equations of motion is different entirely.

I suggest you look at an introductory GR text, because going into GR with a Newtonian ideas to the fore is likely to cause serious confusion.

11. May 4, 2012

### Naty1

Clamtrox:

This is one appropriate explanation. What seems confounding, however, is that two objects of equal rest mass energy but traveling at different velocities follow different worldlines. So how do we explain that apparantly different 'curvature':

From my notes of other discussions in these forums:
Lots more here:

DrGreg explained this way:

12. May 4, 2012

### tom.stoer

Let's make a simple example.

Consider a spherical, stationary mass-distribution, e.g. a massive star, a study its spacetime metric, curvature and geodesics of test bodies.

Now consider a spherically symmetric collapse of this star. e.g. to a neutron star or a black hole. During this (spherically symmetric) collapse the spacetime metric and curvature do not change; the test bodies feel nothing else but the stationary spacetime as before.

Therefore at least in this situation the change in the energy momentum tensor due to the collapse and the huge additional kinetic energy of the collapsing matter does not act gravitationally.

13. May 4, 2012

### Passionflower

It is true that it is a simple example but to suggest that this situation is typical I find frankly misleading as an explanation whether a moving object curves spacetime.

14. May 4, 2012

### tom.stoer

It's not a typical scenario, but you can separate the effects cleary. You have a non-stationary mass distribution with stationary spacetime. So my conclusion is that the idea of an 'object moving through spacetime and distorting spacetime' (due to its velocity, not only due to its presence!) is misleading in GR.

15. May 4, 2012

### Staff: Mentor

My contribution to this for what it's worth: I think it's helpful to carefully distinguish different possible meanings for the term "moving object", since different meanings lead to different answers to the question of what is affected by "motion":

(1) Consider a spacetime containing a single gravitating body, such as a star. Observer A is at rest relative to the body; observer B is moving relative to it. Both observers are "test bodies", which do not contribute to the curvature of the spacetime. In this case, spacetime curvature is the same for both observers, even though one perceives the gravitating body to be "at rest" but the other perceives it to be "moving". The spacetime itself, the geometric object, is the same for both observers. So in this sense, the "motion" of the gravitating body does not change anything.

(2) However, observers A and B in the above scenario *can* feel different "forces" due to the gravitating body. For example, suppose that A is "hovering" motionless at a given radius, while B is passing by on a free-fall trajectory at a speed close to the speed of light (relative to the gravitating body and A) with an impact parameter equal to the radius at which A is hovering. The effective "acceleration" of B towards the gravitating body will be *larger* than the "acceleration due to gravity" that is measured for test objects dropped radially by A--in the limit as B's speed goes to the speed of light, the "acceleration" for B will be twice as large. ("Acceleration" is in quotes because it's coordinate acceleration, not proper acceleration--both B and the test objects dropped by A are in free fall.) So in this sense, the "motion" of a *test object* does change things.

(3) Now consider a different scenario, where we have a gravitating body that is undergoing proper acceleration--say a star which is ejecting mass in a definite direction, so that it behaves as if it had a rocket engine attached to it. (Suppose a very advanced alien civilization has caused the star to do this for purposes unknown to us.) The spacetime around this star will be *different* than the spacetime in #1 above, and this could be observed by noting the different "forces" observed by A' and B', observers analogous to A and B in #1 and #2 above in their motion relative to the star (A' is at rest relative to the star, B' is passing by at near the speed of light). So again in this sense, "motion" does change things--but now it is "motion" of the gravitating body in an invariant sense, i.e., "motion" due to actual proper acceleration, as opposed to just a change in reference frame.

16. May 4, 2012

### Passionflower

However when A and B are not test bodies then certainly the spacetime is no longer stationary. Then relatively moving bodies certainly influence spacetime.

As in a previous writing of yours you cannot use test bodies to prove they have no gravitational influence because test bodies by definition have no gravitational influence.

It is like saying I want to prove that jumping in a swimming pool causes no waves because when a test swimmer jumps in we see no waves.

17. May 4, 2012

### Staff: Mentor

In the sense that a spacetime with multiple gravitating bodies in it is a different spacetime, a different geometric object, than a spacetime with only one, yes, certainly this is the case. I didn't mean to imply that I had covered all the possible spacetimes that might bear on the question; I was only trying to cover the simplest ones.

18. May 5, 2012

### tom.stoer

I only wanted to introduce two simple spacetimes a) w/o and b) w/ a "kinetic energy" of a mass distribution.

I can't do this using test bodies (as discussed) and I can't do this using one single object (a star) in motion, b/c this is fake. A single object in an otherwise empty spacetime can't move w.r.t. this spacetime. It defines the entire spacetime (typically Schwarzsschild).

19. May 5, 2012

### Naty1

M8M queried about gravitational forces resulting from different translational velocities between two objects. Suppose we set the one object rotating instead of moving translationally.

Would there be a way [or is it even useful] in GR to describe such spacetime curvature effects differently between these cases since the 'invariant SET based spacetime curvature' changes [from the stationary] with rotational energy and momentum but not with translational?? I'm wondering here about our descriptions of curvature, not the magnitude of the effects.

20. May 5, 2012

### Staff: Mentor

In M8M's OP, it looks to me like the "object with mass m" is supposed to be a test body; i.e., its effect on the overall curvature of the spacetime is too small to be significant. If that's the case, I'm not sure how to add internal angular momentum (or "spin") to the test body and have it have any effect. When I've seen angular momentum is ascribed to test bodies in GR problems, it's always orbital angular momentum, due to the trajectory of the body around the central mass.

If the "object with mass m" in the OP is *not* supposed to be a test body, if it is supposed to have significant effects on its own on the overall curvature of the spacetime, then yes, obviously those effects are different if the object has angular momentum as well as mass. If gravity throughout the spacetime were weak enough, one could approximate the solution as one object (the one that gives rise to the "gravitational force" in the OP) with some mass M, described by a Schwarzschild solution centered on that object, and another object (the "object with mass m" in the OP) with mass m and angular momentum L, described by a Kerr solution centered on *that* object. This would not be an exact solution, but it could be a starting point for an approximation scheme.