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Can two particles be in equilibrium under the influence of their mutual gravitational forces alone?

Obviously, if the two particles are kept at rest at a distance apart, one will exert an attractive force on the other. As a result, both of them will accelerate towards each other. None of them can be in equilibrium under this condition.

Now consider the scenario in which the particles are moving in a circle, under the action of their mutual gravitational forces. Here again, both the particles will be accelerated towards the centre. Hence, no equilibrium.

In the second case, what happens if one particle is observed from the reference frame of the other particle? Will it be unaccelerated and hence in equilibrium?

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    $\begingroup$ What do you mean by equilibrium? In mechanics, equilibrium means not-moving. Dynamical systems can come to a steady state; stable orbits in Kepler's problem tends to be considered good enough, and this is proved mathematically to be stable. In GR, the corrections cause precession; precession might be considered stable too, but the giving off of gravitational radiation might turn all the stable orbits into unstable ones, if you care about infinite times. For all practical purposes, it is stable in the precessing sense; those times are loooongg. $\endgroup$
    – naturallyInconsistent
    Commented Jul 23 at 16:00
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    $\begingroup$ Two bodies orbiting each other have zero proper acceleration. What you are observing is just a coordinate acceleration and coordinates are arbitrary labels. $\endgroup$
    – John Rennie
    Commented Jul 23 at 16:54
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    $\begingroup$ @Solidification Accelerometers depend on experiencing reaction force. When you jump with an accelerometer on your wrist (i.e. apple watch, fitbit, etc.), your arm has to push on the accelerometer in order to accelerate it. The accelerometer detects that and calculates the acceleration. With the same watch in space, the watch and your arm are pulled (roughly) equally hard, so your arm doesn't push on the watch at all and it detects zero acceleration. At a certain point, it will start detecting tidal forces, but that's probably not what you're talking about. $\endgroup$
    – controlgroup
    Commented Jul 23 at 17:50
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    $\begingroup$ (contd) This is the same phenomenon that causes weight on Earth's surface/weightlessness aboard the ISS. There isn't "microgravity" on the ISS - it's about 60% as strong up there as on Earth's surface. When you're on the ground, though, the ground is constantly pushing up on you to keep you from falling through the planet (since we know that doesn't happen very often). But in space, the station and passengers are just constantly falling. They observe weightlessness because nothing is stopping them from falling. $\endgroup$
    – controlgroup
    Commented Jul 23 at 17:52
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    $\begingroup$ @SolomonSlow Yep, you're right. That's my mistake, hit a 6 instead of a 9 somehow. Apologies for the confusion. $\endgroup$
    – controlgroup
    Commented Jul 23 at 18:53

3 Answers 3

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There is no static equilibrium, but you can have a dynamic equilibrium (in Newtonian gravitational physics).

To prove that there is not static equilibrium, observe that this would require a local minimum in the gravitational potential. However in regions where there is no matter present the gravitational potential satisfies a differential equation named after Laplace. This equation reduces to the following form near any stationary point (a point where the potential is either max or min or a sadle shape): $$ a_x + a_y + a_z = 0 \tag{1} $$ where $a_x$, $a_y$ and $a_z$ are coefficients describing the gravitational potential $\phi$ near a point $(x_0, y_0, z_0)$, like this: $$ \phi(x,y,z) = a_x(x-x_0)^2 + a_y (y-y_0)^2 + a_z (z-z_0)^2. $$ To have a local minimum would require $a_x >0, \;a_y>0,\; a_z>0$ but equation (1) rules this out.

Coming now to the dynamical case, we have that two bodies can orbit in circular or elliptical orbits about their common centre of mass, and this situation is stable. So there can be a dynamic equilibrium of this kind.

With three or more bodies it is not so clear. This was a problem that famously bothered Newton. He thought that the solar system, for example, must be unstable and small perturbations would therefore grow, and therefore there must be a further contribution to the dynamics in order for the solar system to last so long without coming apart. Laplace analysed the situation more deeply and argued that the solar system was stable. It turns out that Newton was not entirely wrong, however, since in fact the tumbling motion of asteroids introduces an element of chaos into the solar system, such that it is unstable on very very long timescales (longer than the lifetime of the Sun).

(This question of stability is related to a famous anecdote that became an urban myth, one that still lives among those with amateur-level knowledge of philosophy and the history of science. The myth is that Laplace made a clever reply to a provocative question from Napoleon, in which (so the myth goes) God should be treated as a hypothesis, and the hypothesis is not required, in the opinion of Laplace. This reply was beloved by atheists so it got preserved and turned into a kind of knowing joke. In fact Laplace was merely questioning the need to invoke a further intermittent action by the real ground of all things (i.e. God), over and above making the solar system be the kind of physical system that it is, in order to achieve stability.)

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    $\begingroup$ This is exactly correct; I will add that there are no equilibria in GR gravity because of gravitational radiation constantly draining the system of energy. +1 $\endgroup$
    – controlgroup
    Commented Jul 23 at 17:10
  • $\begingroup$ Thanks for the answer! My vocabulary is this: Static equilibrium of a particle is a situation where the net force on it is zero and the particle is at rest. Dynamic equilibrium refers to a situation in which net force on the particle is zero but it is moving uniformly. Is this also how you're defining dynamic equilibrium (for any one particle of the two-particle system)? If so, none of the particles are moving uniformly. They're both accelerated. So they are not in dynamic equilibrium according to my understanding. Can you clarify this a bit? $\endgroup$
    – Solidification
    Commented Jul 23 at 17:25
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    $\begingroup$ Yes I used the term "dynamic equilibrium" for a case where the forces are non-zero and therefore the particles are accelerating, but the pattern of their motion is stable (e.g. going around a circle forever). Gravity can produce a point in space where there is no net force, but some of the nearby points will cause acceleration away from that point so that kind of equilibrium is unstable. $\endgroup$
    – Andrew Steane
    Commented Jul 23 at 19:00
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If you define equilibrium as having a net zero force on the system, then any gravitational system (or in fact any system as we know it, barring nonlinear systems like black holes or the strong force or EM above a certain energy) is in equilibrium, by Newton's third law (since you aren't accounting for quantum effects with either particle, I'll assume Newtonian mechanics). The force exerted by a satellite orbiting the Earth on the Earth and the force that the Earth exerts on the satellite are exactly the same (you can verify this by $F=Gm_1m_2r^{-2}$); hence, for the Earth-satellite system, the net force is zero and the system is in equilibrium. This is true of any two (non-relativistic non-extreme) masses, and is also true for any two electric charges (whose field strengths are below the Schwinger limit).

To answer your second question, it depends on what frame of reference you're using, but if you're talking about a comoving and zero-angular-momentum reference frame, then no, observers on one particle will see the other particle orbiting it. Think about the Earth-Moon system, which is actually identical to what you describe, two gravitating masses orbiting their mutual center of mass: we certainly do not see the Moon as static, and neither does an astronaut standing on the Moon. (I don't think the Apollo astronauts on the surface long enough to observe the motion of Earth, but I'm certain that with enough time they would have observed it.)

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    $\begingroup$ Here I am asking whether equilibrium is possible for individual particles of the system. Not of the system as a whole. For example, if you have a system of three particles, each of mass m, and placed on a line with equal distance, the particle on the middle will be in equilibrium. Not stable though. $\endgroup$
    – Solidification
    Commented Jul 23 at 16:59
  • $\begingroup$ @Solidification Single particles are always "in equilibrium". A particle (under the influence of only gravitation) always sees itself as experiencing zero net force and tracing out a perfectly-straight line in spacetime, so saying that a particle is then "in equilibrium" is meaningless. $\endgroup$
    – controlgroup
    Commented Jul 23 at 17:02
  • $\begingroup$ Please do not bring general relativity or spacetime. This question is about Newtonian gravity only where gravity is a force $\endgroup$
    – Solidification
    Commented Jul 23 at 17:04
  • $\begingroup$ @Solidification Even in Newtonian mechanics, a particle cannot "tell" that it is experiencing any force. You can trivially fix a reference frame either on an arbitrary point, where the particle is obviously accelerating, or on the particle itself, where the laws of physics still hold true. And particles don't have defined "equilibria" anyway. An equilibrium is a state of a system, not of an individual particle. Particles can be in stable or metastable or unstable states, but they can only be part of an equilibrium. $\endgroup$
    – controlgroup
    Commented Jul 23 at 17:06
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Yes they can. If one particle is a hollow structure of suitable shape such that it has a constant internal potential with the second particle small enough to fit in inside, there can be equilibrium. Also a torus could hold a small particle in its centre. This argument could be generalised to torus like shapes. This equilibrium would not be stable.

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    $\begingroup$ This is wrong owing to the fact that there cannot be a static local minimum (or maximum) in empty space since the potential satisfies Laplace's equation. In the torus example, the situation is unstable in the radial direction. $\endgroup$
    – Andrew Steane
    Commented Jul 23 at 16:43
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    $\begingroup$ @AndrewSteane I think, we must first address whether equilibrium is possible. After that, we may ask, if equilibrium posible, whether that is a stable or an unstable equilibrium. $\endgroup$
    – Solidification
    Commented Jul 23 at 16:52
  • $\begingroup$ @my2cts I have point particles in mind. $\endgroup$
    – Solidification
    Commented Jul 23 at 16:54
  • $\begingroup$ I don't think tori qualify as particles perse, except in the quantum sense if you could somehow get its wavefunction to that configuration. Besides there's a perfectly good argument that answers this question for all gravitating masses, point particles and larger shapes included. $\endgroup$
    – controlgroup
    Commented Jul 23 at 17:03
  • $\begingroup$ @controlgroup You should expand that into an answer. $\endgroup$
    – my2cts
    Commented Jul 23 at 18:25

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