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ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2015/paper-1/2/c/solution

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Past exam of the theoretical physics I course of the University of Cambridge / 2015 / Paper 1 / 2 / c / Solution by

Codex  0 2026-10-20

Time-translation symmetry conserves the energy

H

, and axial rotational symmetry makes

ϕ

a cyclic coordinate, conserving

J = p_{ϕ} = m r^{2} \dot{ϕ} + \frac{qB r ^{2}}{2} .

The radial Euler-Lagrange equation before elimination is

m \overset{r}{¨} = m r \dot{ϕ}^{2} + qB r \dot{ϕ} - 4 π^{2} k (r - r_{0}) .

Insert

\dot{ϕ} = J / (m r^{2}) - qB / (2 m)

. The cross terms linear in

J qB

cancel, leaving

m \overset{r}{¨} = \frac{J ^{2}}{m r ^{3}} - \frac{q ^{2} B ^{2}}{4 m} r - 4 π^{2} k (r - r_{0}) .

Equivalently use the reduced radial potential

U_{J} (r) = \frac{J ^{2}}{2 m r ^{2}} + \frac{q ^{2} B ^{2} r ^{2}}{8 m} + 2 π^{2} k (r - r_{0})^{2} - \frac{J qB}{2 m} .

The last term is constant at fixed

J

and has no radial derivative, but is part of the full conserved energy. Conservation of

J

is stated in the rotationally symmetric gauge; the conserved mechanical combination can be written in terms of magnetic flux in a gauge-independent way.

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