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I've heard that relativistic mass can influence gravity, but this seems to create a paradox, unless I am missing something.

It seems to me that if there were two celestial bodies that are observed to be moving along approximately parallel trajectories at a relativistic speed, wouldn't the gravitational force between them also be larger than if the bodies were at rest, and therefore draw them closer together than they otherwise would be?

How would this attractive force be accounted for if one were observing the second celestial body from the reference frame of the first, where one would not otherwise see any significant relative motion to the other body and not have the apparent relativistic velocity to attribute the increased gravitational pull to?

What am I missing here?

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    $\begingroup$ Related: physics.stackexchange.com/q/3436/123208 & physics.stackexchange.com/q/3584/123208 $\endgroup$
    – PM 2Ring
    Commented Apr 6, 2021 at 18:55
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    $\begingroup$ Maybe time dilation?, even though the moving bodies have greater mass, the observed acceleration of them towards each other would be reduced. Not done as an answer as I'm not sure enough! $\endgroup$
    – John Hunter
    Commented Apr 6, 2021 at 19:16
  • $\begingroup$ Actually also the movement of the energy (mass) density changes the spacetime curvature, thus it has gravitational effect. But this is very small, close to be unmeasurable in all practical scenarios. More can you read about that ḣere. I think, the question boils down to: How does a moving point-like mass change the spacetime curvature in linearized GR?. $\endgroup$
    – peterh
    Commented Apr 7, 2021 at 16:17
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    $\begingroup$ Related: physics.stackexchange.com/q/122319/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Apr 9, 2021 at 3:06

6 Answers 6

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The answer is that "relativistic mass" does not produce gravity in the Newtonian sense. Newtonian gravity breaks down in the realm where velocities are relativistic, and you have to use general relativity to determine the answer. In GR gravity (i.e. curvature of spacetime) is produced by the stress-energy tensor, which is independent of coordinates. So if there's a frame of reference in which the two bodies are at rest relative to one another, then there is no additional gravitational attraction between them. This would be the case if they are moving on parallel paths in the same direction. If they were moving in opposite directions then there is no frame in which the relative kinetic energies disappear, and so that would produce an effect. But to determine the effect you would have to solve the Einstein field equations, I don't think the Newtonian approximation would work.

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    $\begingroup$ The an illustrative example of why various forms of "non-mass energy" must exert gravitational force is that almost all of the mass of a nucleus is in the binding energy, not the rest mass, of the components of the nucleus (strong nuclear force binding mostly, IIRC). And as converting to/from KE "happens all the time" in a sense (QM makes this strange); if KE didn't generate gravity, the gravitational mass of the nucleus would fluctuate. Which would be very strange. $\endgroup$
    – Yakk
    Commented Apr 7, 2021 at 16:04
  • $\begingroup$ I'm wondering about this answer. We know that if a relativistic rocket flies right past me, then for the moment that it passes my wristwatch will tick slower. So it is not merely the object itself that faces time dilation, but the surrounding space as well (in a sort of a bow wave situation). So if these two parallel masses are both causing time dilation in the surrounding space, and if "the reason your bum is stuck to your seat is because time runs faster at your head than your feet", then wouldn't it make sense that there is gravitational attraction between them? They are bending S-T. $\endgroup$
    – foolishmuse
    Commented Apr 9, 2021 at 20:54
  • $\begingroup$ Why do you think your wristwatch will tick slower because a relativistic rocket flies past you? $\endgroup$
    – Eric Smith
    Commented Apr 9, 2021 at 22:39
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    $\begingroup$ To supplement my answer a bit: if two masses are at rest relative to one another (or nearly so) then Newtonian gravitation will describe the attraction between them pretty accurately. If they are moving relative to one another at relativistic speeds, then Newtonian gravitation will break down -- for example, the finite speed of gravity becomes significant in these cases, as do effects due to momentum. $\endgroup$
    – Eric Smith
    Commented Apr 9, 2021 at 23:15
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The gravitational effect arising from a particular volume of space is proportional to (among other things) the energy density in that volume of space.

In the case of a spinning flywheel the rotational kinetic energy and the stress of having to provide centripetal force is confined to a finite volume of space, hence there is an energy density.

In the case of linear motion the motion is not confined and so there is nothing to give rise to a density.

More generally, kinetic energy is inherently relative. There is no such thing as attributing kinetic energy to a single object. The minimum is two objects. For any pair of objects the relative velocity between those two objects determines the amount of energy that is available for transformation (to other forms of energy) when those two objects collide.

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    $\begingroup$ "There is no such thing as attributing kinetic energy to a single object" I do it all the time, for example when I am on the road. $\endgroup$
    – my2cts
    Commented Apr 6, 2021 at 19:47
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    $\begingroup$ @my2cts I've never been on a collision course with a tree, hope I never will be, but if it does happen I will be acutely aware, in my final seconds alive, of the relative velocity between me and the Earth. $\endgroup$
    – Cleonis
    Commented Apr 6, 2021 at 20:06
  • $\begingroup$ "In the case of linear motion the motion is not confined and so there is nothing to give rise to a density." But of course. Namely in the volume in which the object is observed to be at time t. $\endgroup$
    – David Tonhofer
    Commented Apr 7, 2021 at 7:17
  • $\begingroup$ @my2cts Cleonis is right of course in your quote. For a single object kinetic energy or rather linear or angular momentum lose all meaning. Mach's principle actually inspired General Relativity $\endgroup$
    – R. Rankin
    Commented Apr 11, 2021 at 7:24
  • $\begingroup$ @R.Rankin About what is referred to as Mach's principle. Historians of physics describe that Ernst Mach did not formulate Mach's principle, and moreover, asserting Mach's principle goes against Mach's austere philosophy of science. The first to propose it, and name it, was Einstein. I prefer to call it Einstein's Mach's principle. Historians of physics describe that in the early 1920's Einstein abandoned Einstein's Mach's principle, as evidenced by the fact that Einstein ceased mentioning it in his publications, whereas earlier he had asserted it was central to GR. $\endgroup$
    – Cleonis
    Commented Apr 11, 2021 at 7:51
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I believe what matters is the relative curvature of spacetime between them. If they are both moving at relativistic speed in the same direction relative to some third observer, according to that observer the additional curvature of spacetime that this causes is the same for both bodies. Since it doesn't change the relative curvature between them, it shouldn't alter the gravitational pull the experience relative to each other.

I'm sorry if that's a tad hand-wavy.

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  • $\begingroup$ So if they are moving in the same direction at some relativistic speed relative to some third point, then gravity from their relativistic mass will not further pull them together than they otherwise would be if they were actually at rest relative too that point? $\endgroup$
    – Mark
    Commented Apr 6, 2021 at 20:04
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    $\begingroup$ what is relative curvature? I do not understand your answer, but GR is local theory. I.e. what determines my movement is curvature at my place, not some relative curvature (whatever it is). Also, curvature is a tensor, it does not depend on the observer. $\endgroup$
    – Umaxo
    Commented Apr 10, 2021 at 9:19
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There are two effects you need to consider.

One is that gravity depends on the full energy-momentum-stress tensor. Usually for slow moving objects the biggest component of this is the energy, all the others are negligible, and so we commonly say gravity depends on the mass (=energy) alone. But energy and momentum are just individidual coordinates of an invariant 4-dimensional quantity, and if you change to a moving reference frame it's just like rotating your coordinate system (only it is a hyperbolic rotation, it preserves the length $x^2-y^2$ instead of $x^2+y^2$, meaning that as one coordinate increases so does the other). The invariant magnitude of the 4-dimensional quantity is the energy squared minus the momentum squared. If you change to a moving coordinate system, momentum increases along with energy in synchrony, leaving the magnitude the same.

This is the same answer as to the question: when you accelerate and see the whole universe start moving past you, where does the kinetic energy of the whole universe come from? The answer is that energy and momentum are complementary coordinates of a 4-dimensional quantity, which is what is conserved, and the increase in energy is cancelled by the increase in momentum. Or more accurately, you are just rotating your coordinate system, and some of the enormous rest energy of the whole universe is appearing in the spatial direction as momentum. A lot of apparent paradoxes can be resolved because of this sort of compensatory-coordinates effect.

The other effect you need to consider is time dilation. The convergence of your moving celestial bodies can be considered to be a clock - they get so many metres closer in such and such a time, they orbit around each other with such and such a period - and relativistic velocity slows this down. The acceleration you measure in the moving frame will be different.

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In the example of the OP the force is observed from two different reference frames and just differs by a Lorentz transform. To answer the question in the title, a different thought experiment is needed. Assume one particle to be at rest and the other moving at speed v. Does the gravitational force depend on v? The answer is yes. Gravitation acts on energy, including kinetic energy. For example, the mass of a hydrogen atom consists of proton and electron mass plus binding energy, hence kinetic and potential energy. It is this total energy that feels gravitation, even in Newton gravity.

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Well I do not understand the other answers, but as the correct answer is simple, the other answers must be wrong.

An observer at rest relative to a force observes that the force is F.

Then the observer changes her frame, so that the force moves at speed v.

Now the force is F/gamma, according to the observer.

This is the correct way to calculate the force in this case, if the masses move side by side.

If the masses move along the same line, then the motion has no effect on the force.

Now I don't want to hear: "Can't use special relativity, must use general relativity". General relativity has nothing to do with this.

Like quantum mechanics has nothing to do with the force between moving electrons.

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  • $\begingroup$ "General relativity has nothing to do with this." Theory of gravity has nothing to do with question about gravity, while theory that is notorious about not being able to explain gravity is sufficient? $\endgroup$
    – Umaxo
    Commented Apr 10, 2021 at 9:04
  • $\begingroup$ Yes. Principle of relativity. Men's testicles don't fall of at high speeds. Stars' planets don't fall off at high speeds. Scales on high speed planets don't break , so gravity and spring forces transform the same way. $\endgroup$
    – stuffu
    Commented Apr 10, 2021 at 9:23
  • $\begingroup$ there are no gravity forces to begin with and principle of relativity was revised by GR. $\endgroup$
    – Umaxo
    Commented Apr 10, 2021 at 9:30
  • $\begingroup$ @Umaxo Unlike apparently you, I don't think GR is unable to answer questions about gravity forces. ; ) The subset, or part, of GR called SR is particularly good when we transform forces - gravity forces or other kind of forces. $\endgroup$
    – stuffu
    Commented Apr 10, 2021 at 12:38

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