 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2012/paper-1/3/b/solution

E = \sum_{i} \overset{q}{˙}_{i} \partial L / \partial \overset{q}{˙}_{i} - L

E = \frac{m}{2} (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2} + \overset{z}{˙}^{2}) + e λ z^{2} .

The velocity-linear magnetic term cancels. Since

L

has no explicit time dependence,

d E / d t = - \partial_{t} L = 0

. Equivalently, magnetic force does no work, and electric work is exactly minus the change of

e λ z^{2}

. Energy here is defined in the supplied time-independent gauge and equals physical kinetic plus electrostatic potential energy.

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