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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.

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 $Q_0=Q_{\theta }+\theta T/(2\pi )$.