 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/8d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

L^{'} = \frac{1}{2} m \overset{r}{˙}^{2} - q (ϕ - \partial_{t} f - \overset{r}{˙} \cdot (A + \nabla f)) = L + q (\partial_{t} f + \overset{r}{˙} \cdot \nabla f) = L + q \frac{df ( r ( t ) , t )}{d t} .

Thus

L^{'} - L = q df / d t .

The action changes only by

q [f (t_{2}) - f (t_{1})]

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