Question

Show that the two-body hamiltonian\[H_{\text{sys}}=\frac{\mathbf{p}_1^2}{2m_1}+\frac{\mathbf{p}_2^2}{2m_2}+V( \mathbf{r}_1,\mathbf{r}_2)\]can be separated into centre of mass and relative hamiltonians\[H_{\text{sys}}=\frac{\mathbf{p}^2}{2M}+\frac{\mathbf{p}_{\text{rel}}}{2\mu}+V(r)\]where \(\mu\) is the reduced mass

Answer 1

(I hope this is just a normal Hamiltonian and not a Hamiltonian operator)

We can start out by defining the momenta in terms of mass and velocity (using the rdot nomenclature do denote dr/dt, where r is a generic position vector pointing from the origin)
\[ \textbf p_1 = m_1 \dot {\textbf r}_1 \quad ; \quad \textbf p_2=m_2 \dot {\textbf r}_2\]
The Center of Mass position vector is given by
\[ \textbf R = \frac{m_1\textbf r_1+m_2\textbf r_2}{M}\]
And taking the first derivative wrt time gives the velocity of the CoM as
\[ \dot{\textbf R}=\frac{m_1\dot{\textbf r}_1+m_2\dot{\textbf r}_2}{M}\]
And similarly the momentum of the CoM as
\[\textbf{P}_{CoM}=\textbf P = M\dot{\textbf R}\]
Next, the relative position vector r is defined to be
\[ \textbf r = \textbf r_1-\textbf r_2\]
Next you can rearrange the CoM position vector and the relative position vectors with a bit of algebra (about 4 lines) to give
\[ \textbf r_1 = \textbf R + \frac{m_2}{M} \textbf r \quad \quad \text{and} \quad \quad \textbf r_2 = \textbf R - \frac{m_1}{M} \textbf r\]
and the first derivative wrt time is
\[ \dot{\textbf r}_1 = \dot{\textbf R} + \frac{m_2}{M}\dot{\textbf r} \quad \quad \text{and} \quad \quad \dot{\textbf r}_2 = \dot{\textbf R} - \frac{m_1}{M}\dot{\textbf r}\]

Looking again at our Hamiltonian, we can first say that it is a function of r1dot and r2dot and r1 and r2

\[H_{(\dot{ \textbf r}_1 , \dot{ \textbf r}_2, \textbf r_1 , \textbf r_2)}=T_{(\dot{ \textbf r}_1 , \dot{ \textbf r}_2)}+V_{( \textbf r_1,\textbf r_2)}\]
Looking at the kinetic energy portion
\[T_{(\dot{ \textbf r}_1 , \dot{ \textbf r}_2)}=\frac{1}{2}\Big( m_1 \dot{ \textbf r}_1^2 + m_2 \dot{ \textbf r}_2^2 \Big)\]
Now we can substitute in our values for r1dot and r2dot in terms of the first derivative of the relative position terms
\[T_{(\dot{\textbf R}, \dot{\textbf r})} = \frac{1}{2} \left( m_1 \left[ \dot{ \textbf R} + \frac{m_2}{M} \dot{ \textbf r} \right]^2+ m_2 \left[\dot{ \textbf R} - \frac{m_1}{M} \dot{ \textbf r} \right]^2 \right) \]
\[=\frac{1}{2}\left((m_1+m_2)\dot{ \textbf R}^2+(m_1+m_2)\frac{m_1 m_2}{M^2}\dot{ \textbf r}^2 \right)\]
etc. Reduce it a few more times and that gives the kinetic energy in terms of the CoM and the relative momenta. Remember that
\[ M = m_1+m_2\]
\[\mu = \frac{m_1m_2}{M}\]
\[ \textbf P = M\dot{ \textbf R}\]
\[ \textbf P_{rel} = \mu \dot{ \textbf r}\]
\[T_{(\dot{\textbf R},\dot{\textbf r})}=T_{(\dot{\textbf R})}+T_{(\dot{\textbf r})}\]


As for the potential V, since our position vectors r1 and r2 are pointing from the origin (a requirement of the CoM equation) then the distance between 2 objects we're looking at - be it masses or charges - is merely the magnitude of the vector pointing from one object to the other - aka, the magnitude of our relative position vector

\[\textbf r = \textbf r_1-\textbf r_2\]

so for example

\[ V_{(\textbf r_1 , \textbf r_2)} = -G\frac{m_1m_2}{|r_1-r_2|}\]
gives
\[V_{(\textbf r_1 , \textbf r_2)} = V_{(|\textbf r_1 - \textbf r_2|)}\]
aka
\[V=V_{(r)}\]

So now we can separate the Hamiltonian into the CoM and relative components

\[ H = H_{CoM} + H_{rel}\]
\[ = T_{\dot{(\textbf R})} + (T_{(\dot{\textbf r})}+V_{(r)})\]


Just finish reducing the kinetic energy, and you're good to go ^_^