Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 47 / 2 / iv / Solution

At $\theta =0$, the field equations areWith $\alpha =\omega t+{\alpha }_0$ and static $\varphi$, the second equation holds automatically and the first reduces toFor $|\omega |<m$, put $\kappa =\sqrt{m^2-{\omega }^2}$. The finite-energy $T=1$ solution isThis is a rotating charged kink in a spherical sigma model. Its orientation moves around a circle while its energy profile stays at rest. The limiting value $|\omega |=m$ gives no localized finite-width kink; larger $|\omega |$ does not give this finite-energy interpolation.

Using ${\varphi }_x=\kappa \sin \varphi$ and $\int {\sin }^2\varphi dx=2/\kappa$, its rest energy and mechanical charge areEliminating $\omega$ givesFor this part $Q=Q_0$ since $\theta =0$. The static mass is recovered at $Q_0=0$, and the mass grows with the magnitude of the global charge. If theta is restored, the energy remains the same function of $Q_0$, but $Q_0=Q_{\theta }+\theta T/(2\pi )$.