= Solution
For $T=1$, choose the positive <Bogomolny equation> and constant angular orientation $\alpha=\alpha_0$. Integrating $d\phi/\sin\phi=m\,dx$ gives
$$
\boxed{\phi(x)=2\arctan e^{m(x-x_0)},\qquad\alpha(x)=\alpha_0.}
$$
This interpolates from $\phi(-\infty)=0$ to $\phi(+\infty)=\pi$, so $T=-(-1-1)/2=1$. Its useful profiles are $\sin\phi=\operatorname{sech}[m(x-x_0)]$ and $\cos\phi=-\tanh[m(x-x_0)]$. It obeys the second-order field equation because differentiating $\phi_x=m\sin\phi$ gives $\phi_{xx}=m^2\sin\phi\cos\phi$; the angular field equation holds since $\alpha$ is constant. Thus
$$
\boxed{E=M=rm.}
$$
Both $x_0$ 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>.
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