 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2015/s7/2/4/solution

Let

E = m^{2} c^{4} + c^{2} π^{2}

\overset{r}{˙}_{i} = \frac{c ^{2} π _{i}}{E}, \overset{p}{˙}_{i} = q j \sum \overset{r}{˙}_{j} \partial_{i} A_{j} - q \partial_{i} ϕ .

Subtracting

q d A_{i} / d t

from the second equation gives the familiar Lorentz force,

\overset{π}{˙} = q (E + \overset{r}{˙} \times B)

, with

E = - \nabla ϕ - \partial_{t} A

and

B = \nabla \times A

. This also checks the signs of both potential terms.

 New to topics? Read the docs here!