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Imagine two massive objects, with the same mass (M) circling around their center of mass (C.M.). Let's assume that the distance between them is 1 light hour. Don´t the two bodies get accelerated and move away from each other because they feel the gravity of each other as it was 1 hour ago, because of which a force tangent to the direction of the speed develops?

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    $\begingroup$ I think you mean that the masses circle around the centre of mass, not each other. $\endgroup$ Jun 17, 2016 at 21:18
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    $\begingroup$ That's a really interesting question! I have no clue how to do this in GR, but it may be solvable for charges in E&M. I would expect this is closely related to the rate of EM/gravitational wave radiation. $\endgroup$
    – knzhou
    Jun 17, 2016 at 21:38

3 Answers 3


They actually don't separate in distance, but Laplace himself asked the same question and made the exact same mistaken assumption that you did, so don't feel bad - you're in good company.

The best way to approach the problem is through an expansion in powers of $v/c$, where $v$ is the velocity at which the particles move and $c$ is the speed of light (or, more relevantly for this problem, the speed of gravity). At order $(v/c)^2$, we need to factor in the gravitational radiation emitted because the masses are accelerating. The $o((v/c)^2)$ contribution to the gravitational interaction actually causes the the particles to spiral inward as they lose kinetic energy due to the radiated gravitational energy. However, seeing this analytically requires the full machinery of general relativity, and it's a quite complicated calculation - in fact, it can't be analytically solved exactly.

However, the $o(v/c)$ effect is more tractable, and analyzing the problem at this order is enough to clear up your misconception. The particles' acceleration is $o(v^2/c^2)$, so to this order we can neglect it and only consider the effect of its velocity. In effect, we can consider the effect on particle B, assuming that particle A had always had its current instantaneous velocity ${\bf v}$ (i.e. assuming it had always been traveling in a straight line tangent to the orbit circle.).

It turns out that at this order the problem becomes mathematically equivalent to a familiar problem in special relativity: that of the electric field generated by a charge moving at constant relativistic velocity. It's a standard result from classical E&M (see, for example, Purcell) that in this situation, the electric field actually points exactly to the particle's present position, not to its retarded position as one might expect.

(This is because in a relativistic context, the fields generated by a particle depend on its velocity and acceleration as well as its position (as can be seen from Jefimenko's equations). It turns out that in the case where acceleration can be neglected, the fields simply point to the particle's "future projected position" based on its position and velocity at the retarded time, which is also its instantaneous present position.)

So it turns out that the $o(v/c)$ correction from general relativity is exactly zero: the acceleration on particle B is $GM/(2R)^2 + o((v/c)^2)$, and its direction is precisely radially inward, despite the apparent violation of causality.

See here for more information.

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    $\begingroup$ Can I double check the punch line? To first order - o(v/c) - two bodies of equal mass in circular orbits around their center of mass will rotate at the same speed weather gravity were instantaneous or travels at the speed of light? $\endgroup$
    – uhoh
    Jun 18, 2016 at 3:14
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    $\begingroup$ @uhoh Yes, that's correct - the corrections from general relativity first show up at $o((v/c)^2)$. $\endgroup$
    – tparker
    Jun 18, 2016 at 3:22
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    $\begingroup$ @tparker Very clarifying! $\endgroup$ Jun 19, 2016 at 19:54
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    $\begingroup$ @descheleschilder If I recall correctly, the history behind Lapace's calculation is interesting. He considered the possibility of a finite speed of gravity just for fun (remember, this was before special relativity), and (the same way you did) he just added in the simplest possible retardation effect into Newton's law of gravitation, which did indeed yield an (apparent) $o(v/c_g)$ effect. He then used observational astronomical data and his theory to (incorrectly) lower-bound the speed of gravity to a speed that was orders of magnitude faster than the speed of light. $\endgroup$
    – tparker
    Aug 31, 2020 at 3:05
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    $\begingroup$ I'm not sure how widely known this calculation was, but it may have led to all kinds of confusion later on, when people tried to reconcile gravity and special relativity between 1905 and 1915. $\endgroup$
    – tparker
    Aug 31, 2020 at 3:06

This isn't a complete answer, but you might find it helpful anyway. Any corrections are appreciated.

Whether this happens in theory I can't say, but I'm pretty sure it doesn't matter in practice. First notice that the effect is going to be negligible unless the bodies are moving close to the speed of light; in other words, the delay in gravity has to be comparable to their orbital period.

Solving the two-body problem in general relativity is pretty hard. According to Wikipedia if the gravitational field is weak one can do a post-Newtonian expansion, which leads to the result that we can pretend that there is only one orbit around a stationary object with mass $M_1+M_2$.

Making this approximation lets me (and Wikipedia) use the Schwarzschild solution, and we can solve for the orbits. It turns out that Kepler's third law is still valid, and we can find the orbital speed as a function of radius:

$$v^2 = \frac{GM}{r}$$

So if we want $v$ to be close to the speed of light, $r$ has to be close to the Schwarzschild radius $2GM/c^2$. This can only happen for extremely compact objects (black holes or neutron stars) orbiting very close to each other. But in this case, we know that our objects lose so much energy to gravitational waves that they spiral in and collide. So it doesn't seem that the effect would be relevant here either.

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    $\begingroup$ It's always a question of how small is the effect, quantitatively. As in tparker's answer, the fact is the sought-for effect is exactly 0. Suppose though things didn't exactly cancel and there was some effect proportional to $v/c$. This might seem small, and indeed Earth orbits the Sun at $v/c = 10^{-4}$, but we know the system has been pretty much the same for $5\times10^9$ orbits. If there was a $10^{-4}$ effect, we would only have about $10{,}000$ years before we were flung away from the Sun. $\endgroup$
    – user10851
    Jun 18, 2016 at 8:34
  • $\begingroup$ If $r$ equals the Schwarzschild radius then according to the last two equations (the only two equations) in your answer (which I wanted to upvote again) $v=\sqrt{2} c=0.7c$. How come? $\endgroup$ Aug 31, 2020 at 4:01
  • $\begingroup$ @descheleschilder I said close to the speed of light, not exactly equal. $\endgroup$
    – Javier
    Aug 31, 2020 at 13:32
  • $\begingroup$ I meant $v=\sqrt{\frac{1}{2}}c$ when you take $r$ to be the Schwarzschild radius. But isn't $v=c$ on this radius? $\endgroup$ Apr 27, 2021 at 12:15

This is a covariant motion you describe therefore both objects are inside the same inertial frame of reference and at rest.

It is no different from the two objects being absolutely still and not rotating. Thus, a static gravitational field between them.

Static fields by definition are instantaneous if not disrupted to a point and a static scalar amplitude field is attached to the source and moves instantaneously with the source.

So, the answer to your question why the two co-rotating bodies do not drift with time is because they are subjects to a static field that is invariant to time.

"Invariant with time" as an effect is equivalent to "instantaneous".

Gravitational waves caused by disruptions in a static gravity field propagate at c but a static field by definition is instantaneous otherwise it cannot exist.

LIGO did not and could not measure any static field but gravitational waves on a static field of the two co-rotating and approaching gravitational masses. You can have a fluctuation (wave) sitting around a scalar value and you can measure only the fluctuation without the scalar value.


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