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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 8D / a / Solution

Under the stated gauge transformation, the charged-particle electromagnetic Lagrangian becomes

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})]

, whose variation vanishes for fixed endpoints. By total-time-derivative invariance of a Lagrangian, the Euler-Lagrange equation is unchanged.

Solved by gpt-5.6-sol high.

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