Hamiltonian for the two-body problem

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.5; Exercise 2.5.4.

Here we derive the equations of motion of the two-body problem using the Hamiltonian formalism.

The Hamiltonian is given by

\displaystyle H\left(q,p\right)=\sum_{i}p_{i}\dot{q}_{i}-L\left(q,\dot{q}\right) \ \ \ \ \ (1)

where the velocities {\dot{q}_{i}} are expressed in terms of the positions {q_{i}} and momenta {p_{i}}. In this case, we start with the Lagrangian in terms of the centre of mass position {\mathbf{r}_{CM}} and the relative position {\mathbf{r}} of mass 2 to mass 1.

\displaystyle L \displaystyle = \displaystyle \frac{1}{2}\left(m_{1}+m_{2}\right)\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{1}{2}\frac{m_{1}m_{2}}{m_{1}+m_{2}}\left|\dot{\mathbf{r}}\right|^{2}-V\left(\mathbf{r}\right)\ \ \ \ \ (2)
\displaystyle \displaystyle = \displaystyle \frac{M}{2}\left|\dot{\mathbf{r}}_{CM}\right|^{2}+\frac{\mu}{2}\left|\dot{\mathbf{r}}\right|^{2}-V\left(\mathbf{r}\right) \ \ \ \ \ (3)

where {M=m_{1}+m_{2}} is the total mass and {\mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}} is the reduced mass.

There are potentially 6 velocity components and 6 coordinate components in the Lagrangian, but the 3 components of {\mathbf{r}_{CM}} do not appear, which simplifies things a bit. To convert to a Hamiltonian, we need the momenta

\displaystyle p_{i}=\frac{\partial L}{\partial\dot{q}_{i}} \ \ \ \ \ (4)

The {x} component of momentum of the centre of mass is

\displaystyle p_{CM,x}=\frac{\partial L}{\partial\dot{r}_{CM,x}}=M\dot{r}_{CM,x} \ \ \ \ \ (5)

The other two components of the centre of mass velocity, and of the relative velocity, have a similar form, and in general we can write

\displaystyle p_{CM,i} \displaystyle = \displaystyle M\dot{r}_{CM,i}\ \ \ \ \ (6)
\displaystyle p_{i} \displaystyle = \displaystyle \mu\dot{r}_{i} \ \ \ \ \ (7)

In vector notation, this becomes

\displaystyle \dot{\mathbf{r}}_{CM} \displaystyle = \displaystyle \frac{\mathbf{p}_{CM}}{M}\ \ \ \ \ (8)
\displaystyle \dot{\mathbf{r}} \displaystyle = \displaystyle \frac{\mathbf{p}}{\mu}\ \ \ \ \ (9)
\displaystyle \left|\dot{\mathbf{r}}_{CM}\right|^{2} \displaystyle = \displaystyle \frac{\left|\mathbf{p}_{CM}\right|^{2}}{M^{2}}\ \ \ \ \ (10)
\displaystyle \left|\dot{\mathbf{r}}\right|^{2} \displaystyle = \displaystyle \frac{\left|\mathbf{p}\right|^{2}}{\mu^{2}} \ \ \ \ \ (11)

The Lagrangian thus becomes

\displaystyle L=\frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}-V\left(\mathbf{r}\right) \ \ \ \ \ (12)

The Hamiltonian is

\displaystyle H \displaystyle = \displaystyle \mathbf{p}\cdot\dot{\mathbf{r}}+\mathbf{p}_{CM}\cdot\dot{\mathbf{r}}_{CM}-L\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle \frac{\left|\mathbf{p}\right|^{2}}{\mu}+\frac{\left|\mathbf{p}_{CM}\right|^{2}}{M}-\left[\frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}-V\left(\mathbf{r}\right)\right]\ \ \ \ \ (14)
\displaystyle \displaystyle = \displaystyle \frac{\left|\mathbf{p}_{CM}\right|^{2}}{2M}+\frac{\left|\mathbf{p}\right|^{2}}{2\mu}+V\left(\mathbf{r}\right) \ \ \ \ \ (15)

Once we’ve got the Hamiltonian, we can apply Hamilton’s canonical equations to get the equations of motion.

\displaystyle \frac{\partial H}{\partial p_{i}} \displaystyle = \displaystyle \dot{r}_{i}\ \ \ \ \ (16)
\displaystyle -\frac{\partial H}{\partial r_{i}} \displaystyle = \displaystyle \dot{p}_{i} \ \ \ \ \ (17)

Since {\mathbf{r}_{CM}} does not appear in the Hamiltonian, we have

\displaystyle \dot{\mathbf{p}}_{CM} \displaystyle = \displaystyle 0\ \ \ \ \ (18)
\displaystyle \mathbf{p}_{CM} \displaystyle = \displaystyle \mbox{constant} \ \ \ \ \ (19)

so the momentum of the centre of mass does not change, as expected.

For {\mathbf{r}}, we have

\displaystyle \frac{\partial H}{\partial p_{i}} \displaystyle = \displaystyle \frac{p_{i}}{\mu}=\dot{r}_{i}\ \ \ \ \ (20)
\displaystyle \frac{\partial H}{\partial r_{i}} \displaystyle = \displaystyle \frac{\partial V}{\partial r_{i}}=-\dot{p}_{i} \ \ \ \ \ (21)

The first equation tells us nothing new, while the second is just Newton’s law for a central force: {\mathbf{\dot{p}}=-\nabla V}.

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