 Solution

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

In Cartesian components,

\partial L / \partial v_{i} = m v_{i} + e A_{i}

and

\partial L / \partial x_{i} = e v_{j} \partial_{i} A_{j} - e \partial_{i} ϕ

m \overset{v}{˙}_{i} = e [- \partial_{i} ϕ - \partial_{t} A_{i} + v_{j} (\partial_{i} A_{j} - \partial_{j} A_{i})] = e [E_{i} + (v \times B)_{i}] .

Thus the minimally coupled particle Lagrangian yields the Lorentz force.

For an arbitrary smooth gauge function

χ

, take

A \to A + \nabla χ

and

ϕ \to ϕ - \partial_{t} χ

. Both fields are unchanged, while

L \to L + e (\partial_{t} χ + v \cdot \nabla χ) = L + e \frac{d χ}{d t} .

The action changes only by an endpoint term, so fixed-endpoint variation gives the same motion. Canonical momentum shifts by

e \nabla χ

m v

and the Lorentz force are gauge invariant. This is the relevant gauge invariance of the classical particle dynamics.

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