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The diagram in Why are L4 and L5 lagrangian points stable? shows that the gravitational potential decreases outside the ring of Lagrange points — this image shows it even more clearly: Potential surface

If I understand correctly, using the rubber-sheet model analogy, an object placed in the field moves as if a marble rolls on the sheet with downward gravity. That's fine for objects inside the Lagrange ring: They move towards either mass.

But outside it, it implies that they move away from both masses. Is that really what happens? If so, why? If not, why does the surface slope downwards?

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Where did you get the image from? I would guess that (as Chay Paterson said) the image somehow includes the Coriolis force as well as the gravitational potential, but it's difficult to give a clearer answer without a little more context about what precisely the image represents. – Nathaniel Apr 21 '13 at 15:23
It was from seemingly from a University of Tennessee lecture. The accompanying text states specifically, "The gravitational potential energy for a binary system is plotted in the adjacent diagram." – Gnubie Apr 30 '13 at 18:55
up vote 6 down vote accepted

Yes the free body moves outward, but there are two critical things you have to know to interpret this statement correctly.

First, this is the effective potential, taking into account gravity and centrifugal force. It has this form because we went into the non-inertial frame co-rotating with the two masses. Mathematically, the potential is $$ \Phi_\mathrm{eff}(\vec{r}) = -G \left(\underbrace{\frac{M_1}{\lvert \vec{r}-\vec{r}_1 \rvert}}_\text{potential from mass 1} + \underbrace{\frac{M_2}{\lvert \vec{r}-\vec{r}_2 \rvert}}_\text{potential from mass 2} + \underbrace{\frac{M_1+M_2}{2\lvert \vec{r}_1-\vec{r}_2 \rvert^3} \lvert \vec{r} \rvert^2}_\text{centrifugal component}\right), $$ and it only decreases far away because of that last term.

Physically, this is because placing an object "at rest" in this frame corresponds to having it move with the same angular frequency as $M_1$ and $M_2$ about the center of mass. If you initialize an object $5\ \mathrm{AU}$ on a tangential path having the same angular velocity as the Earth, it will be moving too fast for a circular orbit at that distance, and so it will move away from the Sun.

This does not mean the object will go away forever, and that brings us to the second point, explained in Chay's response: Not all effective forces have been accounted for; in particular, the Coriolis force does not arise from $\Phi_\mathrm{eff}$. The Coriolis force depends on velocity, so it has no scalar potential depending solely on position, and so it is not included in the analysis so far. Once your test object starts moving in your rotating frame, it will experience a perpendicular deflection that will eventually force it to turn around.

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Then what is the graph shown in the OP illustrating? What about the Wikipedia article? It clearly shows a potential field. Your answer seems to establish that those diagrams with a L2 local min/max are definitely not showing $\Phi_\mathrm{eff}(\vec{r})$, so what are they showing? I think your math shows this potential:… – Alan Rominger Apr 21 '13 at 15:50
All these diagrams are the same, and they plot the same $\Phi_\mathrm{eff}$ I wrote down. L4 and L5 are local maxima, the masses themselves are singular "minima," and L1-3 are all saddle points. These diagrams only establish the potential for a corotating object at fixed radius. Any motion with respect to the corotating frame invalidates the use of this (or any scalar) potential for showing movement. – Chris White Apr 21 '13 at 16:15
But in the math you gave L4 and L5 aren't local maxima, only L3 is a maximum. This is my plot of what you wrote, it's different from Wikipedia that has L4 and L5 as local maxima:^2%2By^2%29+-+1%2Fsqrt‌​%28%28x-1%29^2%2By^2%29+-+%28x^2%2By^2%29%29+for+x+from+-2+to+2%2C+y+from+-2+to+2 – Alan Rominger Apr 21 '13 at 16:41
@AlanSE That link isn't working, but I think I see the problem. This formula only holds if the origin is the center of mass, so if you set the two masses equal, they have to be located equidistant from the $y$-axis. Try a 9:1 mass ratio with this code: plot (-9/sqrt((x+1/10)^2+y^2) - 1/sqrt((x-9/10)^2+y^2) -5(x^2+y^2)) for x from -2 to 2, y from -2 to 2 – Chris White Apr 21 '13 at 17:08
Got it. That math clears up the mathematical issue. – Alan Rominger Apr 21 '13 at 23:18

The stability of L4 and L5 isn't at all obvious from looking at the scalar gravitational potential, because the stability of orbits near those points is very much dynamical: in a rotating frame like the one in the image, you have an additional velocity-dependent Coriolis potential as well as the pictured position-dependent scalar gravitational and centripetal one. So there's a second potential here, a vector field that goes as $\Omega\hat{z}$ (up to a sign).

When an object is perturbed radially from L4, it accelerates outwards, but this isn't the same as moving outwards monotonically. Because we're in a rotating frame, moving radially results in a Coriolis acceleration acting normally to it's velocity, which eventually pulls it back around to where it started.

I don't know of any proof that this definitely stabilises the problem in the case of large test masses, but for small test masses it can be linearised and shown to be dynamically stable.

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Thanks for the explanation. I'm not much concerned about the stability at L4 or L5, but rather the implications on movement far away. The surface slopes downwards (in fact at an increasing rate) far from the ring. Does that mean that a free body there will move away from the masses? – Gnubie Apr 8 '13 at 18:59
I have trouble with this concept of "Coriolis potential". In order to get the concept of potential in the first place you need a conservative vector field. It seems that the gravitational fields in this case clearly wouldn't qualify. But the bigger question is how you get a potential from a field that depends on velocity? – Alan Rominger Apr 21 '13 at 15:43
@AlanSE: just like the vector potential in electrodynamics, it turns out. See my answer to the question referenced in this question. – Art Brown Apr 21 '13 at 16:11

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