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

For $T=1$, choose the positive Bogomolny equation and constant angular orientation $\alpha ={\alpha }_0$. Integrating $d\varphi /\sin \varphi =mdx$ givesThis interpolates from $\varphi (-\infty )=0$ to $\varphi (+\infty )=\pi$, so $T=-(-1-1)/2=1$. Its useful profiles are $\sin \varphi =\mathrm{sech}[m(x-x_0)]$ and $\cos \varphi =-\tanh [m(x-x_0)]$. It obeys the second-order field equation because differentiating ${\varphi }_x=m\sin \varphi$ gives ${\varphi }_{xx}=m^2\sin \varphi \cos \varphi$; the angular field equation holds since $\alpha$ is constant. ThusBoth $x_0$ and ${\alpha }_0$ are collective coordinates: translation and global angular rotation do not change the energy. This static solution is the zero-charge member of the rotating charged kink in a spherical sigma model.