 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2018/1/a/solution

(+, -, -, -)

, so

A_{μ} = (ϕ / c, - A)

and

d τ = d t 1 - v^{2} / c^{2}

L = - m c^{2} 1 - v^{2} / c^{2} - e ϕ (x, t) + e A (x, t) \cdot v .

Here

e

denotes the signed particle charge in the action. The canonical momentum and the position derivative are

\frac{\partial L}{\partial v _{i}} = mγ v_{i} + e A_{i}, \frac{\partial L}{\partial x _{i}} = - e \partial_{i} ϕ + e v_{j} \partial_{i} A_{j}, γ = (1 - v^{2} / c^{2})^{- 1/2} .

\frac{d}{d t} (mγ v_{i}) = e [- \partial_{i} ϕ - \partial_{t} A_{i} + v_{j} (\partial_{i} A_{j} - \partial_{j} A_{i})] .

The last term is

(v \times B)_{i}

\frac{d}{d t} (mγ v) = e (E + v \times B) .

Its power relation and acceleration form are

\frac{d}{d t} (γm c^{2}) = e E \cdot v, \dot{v} = \frac{e}{mγ} [E + v \times B - \frac{v ( E \cdot v )}{c ^{2}}] .

The power relation follows by taking the scalar product with

v

and using the relativistic energy-momentum relation. It also shows why the magnetic field alone cannot change the particle's energy.

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