Asymptotic form of soliton-antisoliton solution for S.G eqn.

[Othin]

I'm not sure this is the best place for this question, and apologize if it isn't. I'm studying the classical field solutions on the first few chapter of Rajaraman's Solitons and instantons : an introduction to solitons and instantons in quantum field theory. Well, my question is about one of those intermediate steps that authors love to skip on the grounds that the reader should be able to do it without much difficulty. It may be silly, but I just can't see it. Anyway, we know that soliton-antisoliton scattering is allowed for the sine-gordon equation, and this solution has the form:
\begin{equation}
\phi(x,t)=4arctan\left( \frac{sinh(ut/\sqrt{1-v^2})}{u \ cosh(x/\sqrt{1-v^2})}\right)
\end{equation}
He argues that in the limit that t goes to minus infinity, for example, this becomes
\begin{equation}
\phi\rightarrow 4arctan\left[exp\left(\frac{x+v(t+\Delta/2)}{\sqrt{1-v^2}}\right) \right] -
4arctan\left[exp\left(\frac{x-v(t+\Delta/2)}{\sqrt{1-v^2}}\right) \right]
\end{equation}
where
\begin{equation}
\Delta\equiv \frac{1-v^2}{v}lnv
\end{equation}
and a similar solution for the positive infinity case(Page 38 of Rajaraman's forementioned book). I tried to put the solution in a form in which I can use the arctangent addiction formula, but thus far no success. Closer I got was
\begin{equation}
4arctan\left \{\left[\frac{exp\left(x + \gamma v(t+(lnv)/(v\gamma))\right)}{1+e^{2\gamma x}} \right] - \left[\frac{exp\left(x - \gamma v(t-(lnv)/(v\gamma))\right)}{1+e^{2\gamma x}} \right] \right \}
\end{equation}
Thanks in advance!

 

[LightHero]

A:You're on the right track, but you have a sign error and your definition of $\Delta$ is incorrect. The correct definition is\begin{equation}\Delta \equiv \frac{1-v^2}{v} \ln \left( \frac{1+v}{1-v} \right) = 2 \gamma \ln \left( \frac{1+v}{1-v} \right),\end{equation}where $\gamma$ is as defined in your post. Then the limit is\begin{equation}\lim_{t \to -\infty} 4 \tan^{-1}\left[ \frac{\exp \left( x + \gamma v \left( t + \Delta/2 \right) \right)}{1 + \exp \left( 2 \gamma x \right)} \right] - 4 \tan^{-1}\left[ \frac{\exp \left( x - \gamma v \left( t + \Delta/2 \right) \right)}{1 + \exp \left( 2 \gamma x \right)} \right].\end{equation}The key trick is to use the addition formula for arctangent:\begin{equation}\tan^{-1}\left( x \right) + \tan^{-1}\left( y \right) = \tan^{-1}\left( \frac{x + y}{1 - xy} \right).\end{equation}Applying this to above yields\begin{equation}4 \tan^{-1}\left[ \frac{\exp \left( x + \gamma v \left( t + \Delta/2 \right) + \exp \left( x - \gamma v \left( t + \Delta/2 \right) \right)}{1 - \exp \left( 2 \gamma x \right)} \right] = 4 \tan^{-1} \left[ \exp \left( x + \gamma v \left( t + \Delta/2 \right) \right).\end{equation}