 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2013/paper-1/2/a/solution

Use the magnetic-coupling convention of the supplied Hamiltonian; electromagnetic unit factors are absorbed into the stated

e

and

B

. Let

α = m_{1} / M

and

β = m_{2} / M

, so

r_{1} = R - β r, r_{2} = R + α r .

The kinetic energy separates into

M \dot{R}^{2} /2 + μ \dot{r}^{2} /2

\sum_{i} e_{i} \dot{r}_{i} \cdot (B \times r_{i}) /2

becomes

- \frac{e}{2} [\dot{R} \cdot (B \times r) + \dot{r} \cdot (B \times R)] - e^{*} \dot{r} \cdot (B \times r), e^{*} = \frac{e ( m _{1} - m _{2} )}{2 M} .

\dot{r} \cdot (B \times R) = \frac{d}{d t} [r \cdot (B \times R)] + \dot{R} \cdot (B \times r) .

Discarding

- e /2

times that total derivative, which changes endpoint terms and canonical momenta but not the trajectories, leaves

L = \frac{M}{2} \dot{R}^{2} + \frac{μ}{2} \dot{r}^{2} + \frac{e ^{2}}{r} - e^{*} \dot{r} \cdot (B \times r) - e \dot{R} \cdot (B \times r) .

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