= Solution
At $\theta=0$, the field equations are
$$
\Box\phi-\sin\phi\cos\phi[(\partial\alpha)^2-m^2]=0,
\qquad\partial_\mu(\sin^2\phi\,\partial^\mu\alpha)=0.
$$
With $\alpha=\omega t+\alpha_0$ and static $\phi$, the second equation holds automatically and the first reduces to
$$
\phi_{xx}=(m^2-\omega^2)\sin\phi\cos\phi.
$$
For $|\omega|<m$, put $\kappa=\sqrt{m^2-\omega^2}$. The finite-energy $T=1$ solution is
$$
\boxed{\phi(x)=2\arctan e^{\kappa(x-x_0)},\qquad
\alpha(x,t)=\omega t+\alpha_0.}
$$
This 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 $\phi_x=\kappa\sin\phi$ and $\int\sin^2\phi\,dx=2/\kappa$, its rest <energy> and mechanical charge are
$$
\begin{aligned}
M&=\frac r4\int\left[\phi_x^2+(m^2+\omega^2)\sin^2\phi\right]dx
=\boxed{\frac{rm^2}{\kappa}},\\
Q_0&=\frac{r\omega}{2}\int\sin^2\phi\,dx
=\boxed{\frac{r\omega}{\kappa}}.
\end{aligned}
$$
Eliminating $\omega$ gives
$$
\boxed{M=m\sqrt{r^2+Q_0^2},\qquad
\omega=\frac{mQ_0}{\sqrt{r^2+Q_0^2}}.}
$$
For 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 $Q_0=Q_\theta+\theta T/(2\pi)$.
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