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 Incoming links: Lorentz-force equation from the worldline action

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 37D / ii / Solution Created 2026-09-24 Updated 2026-09-29

The momentum conjugate to

x^{μ}

is

\frac{\partial L}{\partial x ˙ ^{μ}} = m c \frac{x ˙ _{μ}}{- x ˙ ^{2}} + q A_{μ},

while

\frac{\partial L}{\partial x ^{μ}} = q (\partial_{μ} A_{ν}) \overset{x}{˙}^{ν} .

The Euler-Lagrange equation therefore gives

m c \frac{d}{d λ} (\frac{x ˙ _{μ}}{- x ˙ ^{2}}) = q (\partial_{μ} A_{ν} - \partial_{ν} A_{μ}) \overset{x}{˙}^{ν} = q F_{μν} \overset{x}{˙}^{ν} .

Since

d o t x^{μ} = (d τ / d λ) u^{μ}

and

- \overset{x}{˙}^{2} = c, d τ / d λ

d τ / d λ

gives the Lorentz-force equation from the worldline action

m \frac{d u _{μ}}{d τ} = q F_{μν} u^{ν} .

Raising the free index yields exactly

m \frac{d u ^{μ}}{d τ} = q F^{μν} u_{ν} .

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 37A / b / Solution Created 2026-09-24 Updated 2026-09-25

Write

L = - m c - \overset{x}{˙}^{2} + q A_{ν} (x) \overset{x}{˙}^{ν} .

Since

- \overset{x}{˙}^{2} = c \frac{d τ}{d λ}, \overset{x}{˙}^{μ} = \frac{d τ}{d λ} u^{μ},

its derivatives are

\frac{\partial L}{\partial x ˙ ^{μ}} = m u_{μ} + q A_{μ}, \frac{\partial L}{\partial x ^{μ}} = q (\partial_{μ} A_{ν}) \overset{x}{˙}^{ν} .

The Euler-Lagrange equation therefore gives

0 = \frac{d}{d λ} (m u_{μ} + q A_{μ}) - q (\partial_{μ} A_{ν}) \overset{x}{˙}^{ν} = m \frac{d u _{μ}}{d λ} + q (\partial_{ν} A_{μ} - \partial_{μ} A_{ν}) \overset{x}{˙}^{ν} .

Using the electromagnetic field tensor

F_{μν} = \partial_{μ} A_{ν} - \partial_{ν} A_{μ}

and dividing by

d τ / d λ

gives

m \frac{d u _{μ}}{d τ} = q F_{μν} u^{ν} .

Raising the first index yields the equivalent requested form

m \frac{d u ^{μ}}{d τ} = q F^{μν} u_{ν} .

This is the Lorentz-force equation from the worldline action.

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