= Solution
Choose the positive-exponential branches from part (i) and set their additive constants to zero. First take $0<b<1$, and define
$$
v=\frac{1-b^2}{1+b^2}\in(0,1),\qquad\gamma=\frac{b+b^{-1}}2=\frac1{\sqrt{1-v^2}},\qquad a=-b^{-1}.
$$
Then $\tan(\phi_b/4)=e^{\gamma(x-vt)}$ and $\tan(\phi_a/4)=e^{-\gamma(x+vt)}$. The tangent subtraction formula gives
$$
\tan\frac{\phi_b-\phi_a}{4}=\frac{\sinh(\gamma x)}{\cosh(\gamma vt)},\qquad\frac{b+a}{b-a}=-v.
$$
Consequently the allowed <Sine-Gordon superposition formula> produces the smooth field
$$
\boxed{\phi_{a,b}(x,t)=-4\arctan\left[\frac{v\sinh(\gamma x)}{\cosh(\gamma vt)}\right].}
$$
This is the negative of the <Sine-Gordon two-kink solution>, and hence a two-<antikink> configuration. The auxiliary seeds have opposite <topological charges>, but their charges cannot simply be added to infer the charge of the nonlinear two-step <Bäcklund transformation>. Indeed, the displayed final field tends to $2\pi$ at the left spatial end and $-2\pi$ at the right, so its total <topological charge> is $-2$.
Let $T=|t|$ become large. Near the right transition, $x=vT+O(1)$, the tangent argument has the asymptotic form
$$
\frac{v\sinh(\gamma x)}{\cosh(\gamma vt)}=v e^{\gamma(x-vT)}+o(1),
$$
so the local field is $-4\arctan e^{\gamma(x-vT)+\log v}$, a single <antikink>. Near the left transition, the local field is $4\arctan e^{-\gamma(x+vT)+\log v}+o(1)$, again a decreasing <antikink>. The resulting asymptotic center lines are
$$
\begin{array}{c|cc}
& t\to-\infty&t\to+\infty\\
x_L(t)&vt+\gamma^{-1}\log v&-vt+\gamma^{-1}\log v\\
x_R(t)&-vt-\gamma^{-1}\log v&vt-\gamma^{-1}\log v.
\end{array}
$$
Thus two incoming <antikinks> with <topological charges> $(-1,-1)$ and <velocities> $(+v,-v)$ separate again with exactly the same <topological charges> and <velocities>. There is no radiative tail in these asymptotic profiles. Labeling the outgoing objects by their preserved <rapidities> makes this elastic <soliton> scattering; labeling the left and right lumps instead describes reflection with exchanged <velocities>.
For the right-moving <soliton>, its incoming intercept is $\gamma^{-1}\log v$ and its outgoing intercept is $-\gamma^{-1}\log v$. The spatial shifts are therefore $\Delta x_+=-2\gamma^{-1}\log v$ and $\Delta x_-=2\gamma^{-1}\log v$. Define the <soliton time delay> as the change in arrival time at a fixed distant spatial point relative to continuation of the incoming straight line, so $\Delta t=-\Delta x/v_{\mathrm{particle}}$. Both objects have \b[the same signed <soliton time delay>, which is an advance:]
$$
\boxed{\Delta t=\frac{2\log v}{\gamma v}<0.}
$$
This is the <Sine-Gordon two-kink time advance>. In physical coordinates $T_{\rm phys}=t/m$, the time shift is $2\log v/(m\gamma v)$. The explicit intercepts fix the sign convention unambiguously.
The remaining real parameter choices are covered without changing the calculation. For any $b\ne0$ with $b^2\ne1$, put $v_b=(1-b^2)/(1+b^2)$, $u=|v_b|$, $\gamma=(|b|+|b|^{-1})/2$ and $\epsilon=\operatorname{sgn}(b v_b)$. The same choice of zero additive constants gives
$$
\phi_{a,b}=-4\epsilon\arctan\left[\frac{u\sinh(\gamma x)}{\cosh(\gamma ut)}\right].
$$
Each scattered object's <topological charge> is $-\epsilon$, the <velocities> are $\pm u$, and the signed <soliton time delay> is $2\log u/(\gamma u)$. If $b=\pm1$, the superposition coefficient vanishes and this representative is the vacuum; there is no pair of separated moving <solitons> and no scattering delay to assign. Thus the scattering conclusion requires the nondegenerate case $0<u<1$.
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