= Solution
Write the <rapidity> parameters as $\kappa_i=\cosh\theta_i$, $\beta_i=\sinh\theta_i$, and define $E_i=e^{X_i}$. The signed coefficient in the <Sine-Gordon multisoliton tau representation> is
$$
a_{12}=e^{B_{12}}=\frac{1-\cosh(\theta_1-\theta_2)}{1+\cosh(\theta_1-\theta_2)}=-\alpha_{12},\qquad \alpha_{12}=\tanh^2\frac{\theta_1-\theta_2}{2}.
$$
For distinct <rapidities>, $0<\alpha_{12}<1$. In particular, $B_{12}$ cannot be taken as a real logarithm of a positive coefficient. The finite sums defining the <Hirota tau functions> can instead be evaluated directly with the real, negative $a_{12}$. They give
$$
f=1-\alpha_{12}E_1E_2,\qquad g=E_1+E_2.
$$
The physical field is a continuous <branch of a multivalued function>, equivalently $\phi=4\arg(f+ig)$ with the argument followed continuously. The principal <inverse tangent> alone jumps when $f$ changes sign.
Follow the first <kink> with $x=v_1t+O(1)$. Then $X_1=O(1)$ and $X_2=\kappa_2(v_1-v_2)t+O(1)$. The two possible local limits are
$$
\begin{aligned}
E_2\longrightarrow0 &: \quad \phi\longrightarrow4\arctan E_1,\\
E_2\longrightarrow\infty &: \quad \frac gf\longrightarrow-\frac1{\alpha_{12}E_1},\quad \phi\longrightarrow2\pi+4\arctan(\alpha_{12}E_1),
\end{aligned}
$$
where the second field is written on the continuous <kink> branch. Thus both limits are single <Sine-Gordon kinks> of the same width and velocity, but their centers obey $X_1=0$ or $X_1+\log\alpha_{12}=0$. Following the second <kink> gives the same conclusion with labels exchanged. The incoming and outgoing velocities are therefore
$$
\boxed{v_i=\frac{\beta_i}{\kappa_i}=\tanh\theta_i\quad(i=1,2).}
$$
There is no change in the asymptotic <rapidities> or <kink> profiles.
Define the spatial shift as the outgoing center intercept minus the incoming center intercept. Since the large-$E_2$ limit occurs afterwards when $v_1>v_2$, and beforehand when $v_1<v_2$, the <soliton time delay> is
$$
\boxed{\Delta x_{1|2}=-\frac{\operatorname{sgn}(v_1-v_2)}{\kappa_1}\log\alpha_{12},\qquad \Delta^{(2)}t[v_1;v_2]=\frac{\operatorname{sgn}(v_1-v_2)}{\beta_1}\log\alpha_{12}.}
$$
The time formula uses $\Delta t=-\Delta x/v_1$ and requires $v_1\ne0$. Its dependence on the velocities is explicit on substituting
$$
\alpha_{12}=\frac{1-v_1v_2-\sqrt{(1-v_1^2)(1-v_2^2)}}{1-v_1v_2+\sqrt{(1-v_1^2)(1-v_2^2)}},\qquad \frac1{\beta_1}=\frac{\sqrt{1-v_1^2}}{v_1}.
$$
For a faster right-moving <kink>, $\Delta x_{1|2}>0$ and $\Delta t<0$: it arrives earlier than its freely continued incoming trajectory. If $v_1=0$, report the finite spatial shift; a fixed-position arrival-time delay for a stationary <kink> is undefined. Coincident velocities are excluded from a separated collision asymptotic.
For completeness, allowing <antikinks> means $\kappa_i=\sigma_i\cosh\theta_i$, $\beta_i=\sigma_i\sinh\theta_i$, with $\sigma_i=\pm1$. The velocities remain $\tanh\theta_i$. For opposite orientations, $|a_{12}|=\coth^2[(\theta_1-\theta_2)/2]$, and the general spatial shift is
$$
\Delta x_{1|2}=-\frac{\operatorname{sgn}[\kappa_2(v_1-v_2)]}{\kappa_1}\log|a_{12}|.
$$
This follows from the same two local limits; it makes explicit the orientation hypothesis behind the velocity-only all-<kink> answer.
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