 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s7/3/2/solution

p = m \dot{x} + \frac{q}{2} B \times x, \frac{\partial L}{\partial x} = \frac{q}{2} \dot{x} \times B .

Since

B

is uniform and time independent, its Euler-Lagrange equation becomes

m \ddot{x} + \frac{q}{2} B \times \dot{x} - \frac{q}{2} \dot{x} \times B = 0, m \ddot{x} = q \dot{x} \times B .

This is the magnetic Lorentz force, which is unchanged by constant spatial translations. The Lagrangian itself changes by

δ L = ϵ \frac{q}{2} \dot{x} \cdot (B \times \hat{n}) = ϵ \frac{d F _{n}}{d t}, F_{n} = \frac{q}{2} x \cdot (B \times \hat{n}) .

A total time derivative changes the action only by

ϵ [F_{n} (t_{f}) - F_{n} (t_{i})]

. For variations with fixed endpoints this boundary contribution has zero variation, so the equations of motion are unchanged. The translation is therefore a symmetry of the dynamics even though this particular gauge makes the Lagrangian only quasi-invariant.

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