The question Is the Jacobi constant stationary along a periodic orbit? lead me to Wikipedia's Jacobi integral, which begins:

In celestial mechanics, Jacobi's integral (also known as the Jacobi integral or Jacobi constant) is the only known conserved quantity for the circular restricted three-body problem. Unlike in the two-body problem, the energy and momentum of the system are not conserved separately and a general analytical solution is not possible. The integral has been used to derive numerous solutions in special cases.

It was named after German mathematician Carl Gustav Jacob Jacobi.

The statement in bold confuses me because Newtonian gravity is a conservative force, and so I can't see how energy would not be conserved.

I'm probably missing something simple, but what is it? Is it the word "separately"? Is is only when the system's motion is treated in a synodic (rotating) frame and this is then ignored?

  • 1
    $\begingroup$ I agree that sentence is poorly worded. But Don'tPanic(TM) :-) they're not proposing a rift in the space-time continuum $\endgroup$ May 12, 2020 at 13:12
  • $\begingroup$ Yet another poorly written Wikipedia article, and in this case an incorrect Wikipedia article. "Momentum" without a qualifier almost universally refers to linear momentum rather than angular momentum, and linear momentum is not a separately conserved quantity in the two body problem. Angular momentum is however conserved both in total and separately in the two body problem. $\endgroup$ May 13, 2020 at 9:23
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    $\begingroup$ Note that the answer by @nanoman interpreted the unqualified "momentum" to mean linear momentum. There's nothing wrong with that interpretation as that is the standard meaning of a momentum without a qualification. It's easy to show that (linear) momentum is not a separately conserved quantity. The linear momentum of some orbiting object is diametrically opposed to its linear momentum half an orbit later. $\endgroup$ May 13, 2020 at 9:35
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    $\begingroup$ The mathematician Carl Gustav Jacob Jacobi was born and died within the kingdom Prussia. 20 years after his death, the unified German Empire was founded. $\endgroup$
    – Uwe
    May 13, 2020 at 20:15

3 Answers 3


In a two body system the total energy, the sum of kinetic energy and potential energy is constant for each body. If the total energy is constant for each body, the total energy for the whole system is also constant. So energy conservation is valid for each body alone as well as the system.

In a three body system energy may be exchanged between the bodys. Therefore the sum of kinetic energy and potential energy could not be constant for each of the three bodies alone. But when energy is exchanged lossless, the sum remains constant for all three bodies together.

So energy could not be conserved for each of the three bodies separately but for the whole system.

There are special solutions of the three body problem, the Lagrangian points. There is no energy exchange between the three bodies of a Lagrangian system. Energy conservation is valid for each body alone as well as for all three bodies together, but only for this special case of a three body system.

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    $\begingroup$ Oh! I misread the sentence thinking that separately applied to the distinction between energy and momentum, not the distinction between the three bodies. Duh! thanks $\endgroup$
    – uhoh
    May 12, 2020 at 10:59
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    $\begingroup$ The gravity-assist space probes use is a simple example of the type of energy transfer that can occur. $\endgroup$
    – antlersoft
    May 12, 2020 at 14:30
  • $\begingroup$ @antlersoft that's a three-body system, though. By flying by Jupiter, you can't get energy from Jupiter's point of view, only from Sun's point of view. In fact, without Sun, there's no in front of or behind Jupiter to fly through. The Sun doesn't facilitate energy exchange, but it does provide a reference point. $\endgroup$ May 13, 2020 at 13:15
  • $\begingroup$ @JohnDvorak Yes, it's a three-body system, and it provides an example of the type of energy transfer that can occur in a three-body system. $\endgroup$ May 13, 2020 at 19:46

I agree that the Wikipedia quote is confusing, but I disagree with Uwe's explanation of it.

Certainly, with the ordinary definitions, the total energy of the (two- or three-body) system is conserved and the total momentum of the system is conserved. Here total energy and momentum are given by the usual Newtonian formulas in terms of masses, coordinates, and velocities in an inertial reference frame.

Contrary to what Uwe's answer seems to say, in neither a two- nor a three-body system is the ordinary energy or momentum of each body separately conserved. Even in the two-body problem, energy and momentum flow between two bodies via the gravitational force. So I disagree with that interpretation of the Wikipedia quote.

In what sense, then, can it be that energy and momentum are conserved for the two-body problem (2BP) but not for the restricted three-body problem (R3BP)?

First, I propose that the Wikipedia author meant to say angular momentum (which is also conserved in the ordinary sense for the 2BP or R3BP, as for any closed system).

Second, the energy and angular momentum being spoken of are quantities defined in the one-body reduction of the problem.

The 2BP and the R3BP share the special attribute that the entire dynamics can be expressed via the trajectory of one of the bodies (the lightest one for the R3BP). Given this trajectory, the motion of the other body(ies) is readily inferred -- from inertial motion of the center of mass (in the 2BP) or from autonomous two-body motion of the two very massive bodies (in the R3BP).

This special attribute sets the context for the discussion. The trajectory of the distinguished body is described relative to the center of mass of the other body(ies). This separates out the overall motion (of the center of mass of the entire system), and trivializes conservation of total (linear) momentum. Now, the goal is to understand the trajectory of the distinguished body, by treating it as simply a body in an external potential and applying one-body mechanics.

For the 2BP, this transformation results in a body with the "reduced mass" moving in an external gravitational potential corresponding to attraction by the total mass. Because this potential is time- and rotation-invariant, energy and angular momentum written for this one-body problem are conserved. In fact, they are equal to the ordinary total energy and angular momentum of both bodies.

For the R3BP, due to motion of the massive bodies about their center of mass, the gravitational potential governing the distinguished body is neither time- nor rotation-invariant. This is what Wikipedia means by "the energy and [angular] momentum of the system are not conserved separately" (again, "separately" does not mean "separately conserved for each body"), inserting the missing "angular" and understanding this to refer to quantities in the one-body reduction.

However, for the circular R3BP, the gravitational potential rotates rigidly with the orbit of the massive bodies. Thus, while it is not invariant under time translation or spatial rotation individually, it is invariant under a specific combination of the two corresponding to this uniform angular velocity. The corresponding combined conserved quantity is the Jacobi integral.

Though the ordinary total energy and angular momentum of the three bodies (which really means the two massive ones) are of course conserved in the R3BP, these conserved quantities are not useful because they do not involve the lightest body whose motion we are trying to understand. What matters is finding a conserved quantity for the one-body reduction.

  • $\begingroup$ "Even in the two-body problem, energy and momentum flow between two bodies via the gravitational force." Is this really true? Can you show this mathematically? $\endgroup$
    – uhoh
    May 13, 2020 at 8:21
  • $\begingroup$ @uhoh For momentum it's clearest -- the total momentum is $\mathbf{p} = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2$, and the only sensible definition for momentum of each body is $\mathbf{p}_1 = m_1 \mathbf{v}_1$ and $\mathbf{p}_2 = m_2 \mathbf{v}_2$. Since each body accelerates under the gravitational force, $\dot{\mathbf{p}}_1 \ne \mathbf{0}$ and $\dot{\mathbf{p}}_2 \ne \mathbf{0}$ whereas $\dot{\mathbf{p}} = \mathbf{0}$. $\endgroup$
    – nanoman
    May 13, 2020 at 8:42
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    $\begingroup$ The Wikipedia article is wrong. It should have addressed angular momentum rather than momentum (unqualified). Angular momentum is conserved both in total and separately in the two body problem. Angular momentum is not conserved separately in the three body problem, not even in the restricted version of the three body problem. $\endgroup$ May 13, 2020 at 9:30
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    $\begingroup$ On the other hand, it's easy to show that angular momentum is conserved not just in total but also individually in the two body problem. The angular momentum of an orbiting body with respect to the center of mass of a two body system is $\mathbf r \times m \mathbf v$. The time derivative of this is $\dot{\mathbf r} \times m \mathbf v +\mathbf r \times m \dot{\mathbf v} = m \mathbf v \times \mathbf v + \mathbf r \times \mathbf F = \mathbf r \times \mathbf F$, which is identically zero because the force $\mathbf F$ is radial. $\endgroup$ May 13, 2020 at 9:47
  • $\begingroup$ @DavidHammen The way you wrote your comments, I'm not sure if you noticed that I too concluded in my answer that the article should have said angular momentum. $\endgroup$
    – nanoman
    May 13, 2020 at 17:56

They say "It's never too late to forget high school physics" and that's no doubt what had happened here. I've revised this post based on comments below, we should keep them there to maximize my embarrassment. ;-)

Since there's discussion below this answer which currently states

Contrary to what Uwe's answer seems to say, in neither a two- nor a three-body system is the ordinary energy or momentum of each body separately conserved. Even in the two-body problem, energy and momentum flow between two bodies via the gravitational force. So I disagree with that interpretation of the Wikipedia quote.

and I called the "momentum flow between two bodies" into question I thought I would look at this in a less sophisticated, less cerebral way, and instead turn to python.

Here's a two body orbit calculator. I used $m_1, m_2 = 0.2, 0.8$ and balanced the velocities to zero out center of mass motion.

With the parameter f set to 1.0 they are circular orbits and each body's angular momentum is constant. Since these are circles it also means that the magnitude of each of their linear momenta are constant.

With f set to 0.5 they are in elliptical orbits, and while each separate body's angular momentum rises and falls, we know that the sum $m_1 \mathbf{v_1} \times \mathbf{r_1} + m_2 \mathbf{v_2} \times \mathbf{r_2} = 0$ must hold.

Though I'm still not 100% comfortable with momentum flow between two bodies nor energy flow, it's certainly true that the linear momenta move in opposite manners in order to maintain conservation of momentum.

Likewise there is exchange between each body's kinetic energy and their shared potential energy, but I don't necessarily see energy "flowing" from one body to the other.

two body circular orbit

two body elliptical orbit

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

def deriv(X, t):
    x1, x2, v1, v2 = X.reshape((4, -1))
    a1 = -(x1-x2) * m2 * (((x1-x2)**2).sum())**-1.5
    a2 = -(x2-x1) * m1 * (((x2-x1)**2).sum())**-1.5
    return np.hstack((v1, v2, a1, a2))

m1, m2 = 0.2, 0.8
f = 0.5
X0 = np.array([0.8, 0, -0.2, 0, 0, f*0.8, 0, -f*0.2])

times = np.arange(0, 20, 0.01)
answer, info = ODEint(deriv, X0, times, full_output=True)

x1, x2, v1, v2 = answer.T.reshape(4, 2, -1)
p1, p2 = m1 * v1, m2 * v2
L1, L2 = m1 * np.cross(x1, v1, axisa=0, axisb=0), m2 * np.cross(x2, v2, axisa=0, axisb=0)
KE1, KE2 = 0.5 * m1 * (v1**2).sum(axis=0), 0.5 * m2 * (v2**2).sum(axis=0)
PE = - m1 * m2 / np.sqrt(((x2-x1)**2).sum(axis=0))
Etot = KE1 + KE2 + PE

if True:
    plt.subplot(5, 1, 1)
    plt.plot(x1[0], x1[1])
    plt.plot(x2[0], x2[1])
    plt.plot([0], [0], '.k')
    plt.plot(x1[0][0], x1[1][0], 'ok')
    plt.plot(x2[0][0], x2[1][0], 'ok')
    xmin, xmax = plt.xlim()
    plt.xlim(xmin-0.05, xmax+0.05)
    ymin, ymax = plt.ylim()
    plt.ylim(ymin-0.05, ymax+0.05)
    plt.subplot(5, 1, 2)
    for thing in (x1[0], x1[1], x2[0], x2[1]):
        plt.plot(times, thing)
    plt.title('x1, y1, x2, y2')
    plt.subplot(5, 1, 3)
    for thing in (p1[0], p1[1], p2[0], p2[1]):
        plt.plot(times, thing)
    plt.title('px1, py1, px2, py2')
    plt.subplot(5, 1, 4)
    plt.plot(times, L1)
    plt.plot(times, L2)
    plt.ylim(0, 0.14)
    plt.title('L1, L2')
    plt.subplot(5, 1, 5)
    plt.plot(times, KE1)
    plt.plot(times, KE2)
    plt.plot(times, PE)
    plt.plot(times, Etot)
    plt.title('E1, E2, PE, Etot')
  • $\begingroup$ 1. You refer to x and y components of momentum (presumably linear momentum), but they aren't plotted. Momentum is a vector and each component is conserved. Looking at say the y component of momentum, you'll see that first body 1's is positive while body 2's is negative, then it goes the other way, while the sum (y component of total momentum) remains constant (zero). This is what I mean by momentum flowing between the bodies. It does not makes sense to say "momentum in x is exchanged with momentum in y" because these are two different conserved quantities and one cannot convert to the other. $\endgroup$
    – nanoman
    May 14, 2020 at 7:50
  • $\begingroup$ 2. Your calculation of angular momentum is incorrect -- missing the dependence on the angle between the position vector and the momentum vector (it's a cross product -- angular momentum is also a vector but only the z component is nonzero here). When this is fixed, you'll find that the angular momentum is conserved for each body (as David Hammen noted). Contrary to what you say, the total angular momentum of both bodies is not zero, but is constant. $\endgroup$
    – nanoman
    May 14, 2020 at 7:51
  • $\begingroup$ 3. The energy plot (especially for an eccentric orbit) illustrates that it's not clear what is even meant by Uwe's claim that energy is conserved for each body separately. As you note, kinetic energy is naturally defined for each body but potential energy is a mutual quantity. The default way to allocate potential energy is half and half. This does not result in conservation of kinetic plus potential energy for each body (rather, one body loses energy while the other gains, then vice versa). The results do not support Uwe's claim of separate conservation of energy for each body in the 2BP. $\endgroup$
    – nanoman
    May 14, 2020 at 7:53
  • $\begingroup$ @nanoman yikes!!! This is what happens when I do stuff before my morning coffee! Yes of course once the orbits are not circular I can no longer multiply scalar speed by scalar distance to get absolute value of angular momentum. Okay I'll fix these up in a few minutes. Thanks! $\endgroup$
    – uhoh
    May 14, 2020 at 7:57
  • $\begingroup$ @nanoman revised, thanks! Please keep your comments here if you don't mind. $\endgroup$
    – uhoh
    May 14, 2020 at 8:37

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