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The Two-Body problem consists of 6 2nd-order differential equations

\begin{equation} \ddot{\mathbf{r}}_1 = \frac{1}{m_1}\ \mathbf{F_g} \\ \ddot{\mathbf{r}}_2 = -\ \frac{1}{m_2}\ \mathbf{F_g} \end{equation}

where $\mathbf{F_g}$ is the gravitational force between the two bodies.

This system of equation can be re-written as 12 1st-order differential equations by introducing velocities

\begin{equation} \dot{\mathbf{r}}_1 = \mathbf{v_1} \\ \dot{\mathbf{r}}_2 = \mathbf{v_2} \\ \dot{\mathbf{v}}_1 = \frac{1}{m_1}\ \mathbf{F_g} \\ \dot{\mathbf{v}}_2 = -\ \frac{1}{m_2}\ \mathbf{F_g} \end{equation}

That being said, it is known that analytic solutions to the Two-Body problem exist, which would require 12 constants of motion.

What are these 12 constants?

Energy and angular momentum alone are not enough.

A common route to solve this issue starts with noting that the motion will happen in a static plane, so you can kill some degrees of freedom and therefore you don't need as many as 12 constants. Nevertheless, I'm more interested in not doing that and finding what these constants are in any arbitrary reference frame, as it seems that 12 constants need to exist, because we know that every single one of the 12 differential equations has an analytical solution.

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    $\begingroup$ Related: Conserved Quantities in Kepler Problem? $\endgroup$
    – Qmechanic
    Oct 18, 2023 at 17:35
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    $\begingroup$ Just because a system is determined doesn't mean you can write out the conserved quantities. How is that supposed to work in a chaotic system like the double pendulum? $\endgroup$ Oct 19, 2023 at 17:00
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    $\begingroup$ @leftaroundabout You're right. I meant to say that an explicit solution exists $\endgroup$ Oct 23, 2023 at 0:48

2 Answers 2

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There's not much difference in having two bodies rather than one. For two bodies, you can separate the center of mass position/velocity, and the relative position/velocity. The six center of mass degrees of freedom are just specified by the initial position and velocity of the center of mass.

Now that those are taken care of, we can focus on the relative position/velocity, which is equivalent to a single body moving in an inverse square field. In this case, the conserved quantities are the total energy, the angular momentum vector, and the Laplace-Runge-Lenz vector. Technically, that's 7 in total, but they're not all independent because the Laplace-Runge-Lenz's direction is always perpendicular to the angular momentum, and its magnitude is fixed by the angular momentum and energy.

So we have 6 degrees of freedom and 5 conserved quantities, leaving behind 1. And that's exactly what we expect, because even once we fix all the conserved quantities, there's still one quantity left, which is where the body is in its phase space orbit (or, corresponding to its initial phase at $t = 0$). There needs to be some freedom left so that time evolution can happen at all.

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    $\begingroup$ Note that the general two-body problem (with an arbitrary radial potential between the two bodies) has only 4 conserved quantities associated with the relative motion: 3 components of the angular momentum about the center of mass, and the energy of the relative motion. Only if the potential has an additional special symmetry (as do the gravitational and harmonic oscillator potentials) is there an additional fifth independent conservation law. $\endgroup$
    – Buzz
    Oct 19, 2023 at 19:55
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    $\begingroup$ Thanks! Let me do a follow-up: I thought that the only conserved quantity of the center of mass was the momentum. If we sum up all of the forces acting on the system then $m_1\ a_1\ +\ m_2\ a_2\ +\ ...\ +\ m_N\ a_N = 0$ because of Newton's third law (no external forces), so we get our new constant, the momentum of the center of mass, i.e $m_1\ v_1\ +\ m_2\ v_2\ +\ ...\ +\ m_N\ v_N = constant$, which you mention. But the constant associated with the POSITION of the center of mass seems like a direct consequence of this, so not an independent constant. Am I wrong? $\endgroup$ Oct 23, 2023 at 1:13
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    $\begingroup$ @MatíasCerioni There are two symmetries of the system: symmetry under translations (which gives conservation of the center of mass velocity) and symmetry under Galilean boosts (which, in the case of zero total momentum, gives conservation of the center of mass position). The two symmetries give two different conservation laws. $\endgroup$
    – knzhou
    Oct 23, 2023 at 1:36
  • $\begingroup$ Thank you, I didn't know about Galilean boosts symmetries. Still, let's see, in the two-body case we have $m_1\ \textbf{v_1} + m_2\ \textbf{v_2} = \textbf{C}$, where $\textbf{C}$ is a constant vector determined by the masses and initial velocities. In consequence we get $m_1\ \textbf{r_1} + m_2\ \textbf{r_2} = \textbf{C}\ t + \textbf{D}$, where $\textbf{D}$ is a constant vector determined by the initial positions ($\textbf{D} = \textbf{r_{CM,0}}$). Are $\textbf{C}$ and $\textbf{D} = m_1\ \textbf{r_1} + m_2\ \textbf{r_2} = \textbf{C}\ t$ the conserved quantities associated with the CM? $\endgroup$ Oct 27, 2023 at 17:35
  • $\begingroup$ Yup, that's right! $\endgroup$
    – knzhou
    Oct 27, 2023 at 17:41
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Before getting started, it's worth mentioning that you don't actually need twelve constants of motion, you only need six, one per degree of freedom. And you get that total energy is conserved for free, so you actually only need five.

Anyway, we know this is the Hamiltonian for the two body problem: $$H = \frac{m_1}{2}(\dot x_1^2 +\dot y_1^2 +\dot z_1^2) +\frac{m_2}{2}(\dot x_2^2+ \dot y_2^2+ \dot z_2^2) + \frac{Gm_1m_2}{\sqrt{(x_1-x_2)^2+(y_1 - y_2)^2+(z_1-z_2)^2}}$$.

To start with, we'll cut six degrees of freedom down to three by switching to Jacobi coordinates:

$$\mathbf{r} = \mathbf{x_1}-\mathbf{x_2} \quad and\quad \mathbf{R} = \frac{m_1}{m_1+m_2}\mathbf{x_1}+\frac{m_2}{m_1+m_2}\mathbf{x_2}$$

Where $\mathbf{x_1}$ and $\mathbf{x_2}$ are the bodies' positions in Cartesian coordinates, $\mathbf{R}$ is the location of the center of mass, and $\mathbf{r}$ is the separation between the two bodies. With some quick algebra, the Hamiltonian turns into: $$H = \frac{1}{2}(\dot{\mathbf{R}}^2 + \frac{m_1m_2}{m_1+m_2}\dot{\mathbf{r}}^2) + \frac{Gm_1m_2}{|\mathbf{r}|} $$

Hamilton's equations for $\mathbf{R}$ are $$\frac{\partial H}{\partial \mathbf{p}_{\mathbf{R}}} = \dot{\mathbf{R}}\quad\rightarrow \quad\mathbf{p}_{\mathbf{R}} = \dot{\mathbf{R}} $$ and $$\frac{\partial H}{\partial \mathbf{R}} = \dot{\mathbf{p}}_{\mathbf{R}} = 0\quad\rightarrow \quad\ddot{\mathbf{R}} = 0\quad$$

i.e. the velocity of the center of mass is constant. This accounts for three conserved quantities. The dynamics are the same if H varies by a constant, so we can drop the $\frac{1}{2}\dot{\mathbf{R}}^2$ term and we just have

$$H = \frac{m_1m_2}{2(m_1+m_2)}\dot{\mathbf{r}}^2 + \frac{Gm_1m_2}{||\mathbf{r}||} $$

Which is basically the Hamiltonian for a single particle in a stationary central potential, so we've succeeded in eliminating three degrees of freedom. Notice that $||\mathbf{r}||$ and $\dot{\mathbf{r}}^2 = ||\dot{\mathbf{r}}||^2$ are both invariant under a rotation in three dimensions, so the Hamiltonian (and therefore the Lagrangian) is invariant under such a transformation as well. A rotation in three dimensions can be decomposed into three rotations - one around the $x$ axis, one around the $y$ axis, and one around the $z$ axis. By Noether's theorem, if the Lagrangian is invariant under a one parameter differentiable transformation $g^\phi$, then there is a corresponding conserved quantity:

$$I(\mathbf{q}, \mathbf{\dot q}) = \frac{\partial L}{\partial \mathbf{\dot q}}\frac{dg^{\phi}(\mathbf{q})}{d\phi}|_{\phi=0}$$

You can find a proof of this theorem on pages 88-89 of Mathematical Methods of Classical Mechanics by V.I. Arnold. We have the good fortune of being in possession of three one parameter differentiable transformations under which $L$ is invariant, one for each axis of rotation:

$$g_x^{\phi} = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos (\phi ) & -\sin (\phi ) \\ 0 & \sin (\phi ) & \cos (\phi ) \\ \end{array} \right) $$

$$g_y^{\phi} = \left( \begin{array}{ccc} \cos (\phi ) & 0 & \sin (\phi ) \\ 0 & 1 & 0 \\ -\sin (\phi ) & 0 & \cos (\phi ) \\ \end{array} \right)$$

$$g_z^{\phi} = \left( \begin{array}{ccc} \cos (\phi ) & -\sin (\phi ) & 0 \\ \sin (\phi ) & \cos (\phi ) & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$

Which gives us three more conserved quantities: $$I_x(\mathbf{r}, \dot{\mathbf{r}}) = \dot z y - \dot y z $$ $$I_y(\mathbf{r}, \dot{\mathbf{r}}) = \dot x z - \dot z x$$ $$I_z(\mathbf{r}, \dot{\mathbf{r}}) = \dot y x - \dot x y$$

And these are the components of the angular momentum vector $\mathbf{L}$. Now we can rotate the axes so that $z$ points in the $\mathbf{L}$ direction, confining motion to the new $(x,y)$ plane. We actually lose two conserved quantities doing this, but we still have $\dot y x - \dot x y $ which I will just refer to as $L$ from here on out. Our now Hamiltonian now is just down to two degrees of freedom:

$$H = \frac{m_1m_2}{2(m_1+m_2)}(\dot x^2 +\dot y^2) + \frac{Gm_1m_2}{\sqrt{x^2+y^2}}\quad and \quad L = x\dot y- \dot x y$$

Or in polar coordinates,

$$H = \frac{m_1m_2}{2(m_1+m_2)}(\dot r^2 +r^2\dot \theta^2) + \frac{Gm_1m_2}{r}\quad and \quad L = r^2\dot\theta$$

And we can use $L$ to get rid of the second-to-last degree of freedom so we only have to deal with a first order ODE: $$H = \frac{m_1m_2}{2(m_1+m_2)}(\dot r^2 +\frac{L^2}{r^2}) + \frac{Gm_1m_2}{r}$$ $$ \rightarrow\quad\frac{dr}{dt} = \sqrt{2(m_1+m_2)\left(\frac{H}{m_1m_2}-\frac{G}{r}\right)-\frac{L^2}{r^2}}$$

which can be integrated to get a relationship between $t$ and our last degree of freedom.

A few remarks. First, you may have noticed that the three angular momentum constants only let us reduce the number of degrees of freedom by two - this is because they are not functionally independent, which is the same as saying they do not form independent coordinates on the configuration space, which is the same as saying $\frac{\partial(I_x,I_y,I_z)}{\partial(x,y,z)} = 0$. This ends up not being a problem because with the Hamiltonian, we still have enough constants to reduce the problem to a first order ODE in a single degree of freedom.

The second thing is that there are actually other conserved quantities, namely the components of the eccentricity vector a.k.a the Laplace-Runge-Lenz vector which other commenters have mentioned. Unlike the constants of motion I've used above, the tactic of switching coordinate systems which reflect/reveal the symmetries of the system won't help you discover the LRL vector, at least not using simple coordinate transformations. That said, you can use the Hamilton-Jacobi method to solve this problem, in which case you'll find that the Hamilton-Jacobi PDE is solvable by separation of variables in more than one coordinate system (parabolic and spherical) which is how you can end up with more constants of motion than degrees of freedom. Evidently there is also a way to catch the LRL conserved quantities using a more involved treatment of Noether's theorem, but I haven't looked into it enough to comment on it. In any case, the LRL conserved vector unfortunately does not exist with different potentials or with 2+ bodies, whereas the conserved quantities outlined above are also conserved quantities of the $N$ body problem. Hope this helps.

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