 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2015/paper-1/2/b/solution

p_{r} = m \overset{r}{˙}, p_{ϕ} = m r^{2} \dot{ϕ} + \frac{qB r ^{2}}{2} .

The Legendre transform cancels the velocity-linear magnetic term, giving the energy expressed in velocities,

H = \frac{m}{2} \overset{r}{˙}^{2} + \frac{m r ^{2}}{2} \dot{ϕ}^{2} + 2 π^{2} k (r - r_{0})^{2} .

H (r, ϕ, p_{r}, p_{ϕ}) = \frac{p _{r}^{2}}{2 m} + \frac{( p _{ϕ} - qB r ^{2} /2 ) ^{2}}{2 m r ^{2}} + 2 π^{2} k (r - r_{0})^{2} .

It explicitly contains

B

because canonical angular momentum differs from mechanical angular momentum. This reconciles magnetic forces with the absence of magnetic work: the magnetic field can exchange radial and tangential motion without changing the total mechanical energy.

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