# Asymptotic behaviour of soliton-antisoliton solution for the Sine Gordon equation

The question isn't about any actual homework, it's rather a (probably simple) intermediate step I've encountered on Rajaraman's Solitons and instantons : an introduction to solitons and instantons in quantum field theory, in the context of the Sine-Gordon equation. The full solution of the equation is a rather complicated problem, so we limit ourselves to particular solutions, one of which is the soliton-antisoliton scattering solution, which has the form: $$\phi(x,t)=4arctan\left( \frac{sinh(ut/\sqrt{1-v^2})}{u \ cosh(x/\sqrt{1-v^2})}\right)$$ He argues that in the limit that t goes to minus infinity, for example, this becomes $$\phi\rightarrow 4arctan\left[exp\left(\frac{x+v(t+\Delta/2)}{\sqrt{1-v^2}}\right) \right] - 4arctan\left[exp\left(\frac{x-v(t+\Delta/2)}{\sqrt{1-v^2}}\right) \right]$$ where $$\Delta\equiv \frac{1-v^2}{v}lnv$$ 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 $$4arctan\left \{\left[\frac{exp\left(x + \gamma v(t+(lnv)/(v\gamma))\right)}{1+e^{2\gamma x}} \right] - \left[\frac{exp\left(x - \gamma v(t-(lnv)/(v\gamma))\right)}{1+e^{2\gamma x}} \right] \right \}$$ Since we're studying an asymptotic case, I don't really understand why the first term on this last equation should even be there instead of going to zero.

I'm probably missing something pretty obvious, but whatever the reason is, I just don't get it.Thanks in advance for any help.

There is a typo in Rajaraman's book. The correct expression for the $$\Delta$$ time delay should be $$\frac{\Delta}{2}=\frac{\sqrt{1-u^2}}{u}\ln(u) \ ,$$ as you can check it in Rajaraman's 1975 review article (Rajaraman: Phys. Rep. 21 5 227-313 (1975)).
With this, the soliton-antisoliton solution can be written as \begin{aligned} \displaystyle\Phi_{+-}(x,t) &=4\arctan\left(\frac{\sinh(\gamma ut)}{u\cosh(\gamma x)}\right) \\ &=4\arctan\left(\frac{\exp{\left[\gamma\left(x+u\left(t-\frac{\Delta}{2}\right)\right)\right]-\exp\left[\gamma\left(x-u\left(t+\frac{\Delta}{2}\right)\right)\right]}}{1+\exp{(2\gamma x)}} \right) \ , \end{aligned} where $$\gamma=\frac{1}{\sqrt{1-u^2}}$$ is just the usual Lorentz factor, and we used $$\frac{1}{u}=\exp\left(-\gamma u \frac{\Delta}{2}\right) \ .$$ Note that nondimensionalized units are used; let $$x\rightarrow mx$$, $$t\rightarrow mt$$ and $$\Delta\rightarrow m\Delta$$ to restore the usual natural units.
In the $$t\rightarrow-\infty$$ limit the second term in the bracket is large and the first term is exponentially small, so we can make the substitution $$\exp\left[\gamma u\left(t-\frac{\Delta}{2}\right)\right]\approx\exp\left[\gamma u\left(t+\frac{\Delta}{2}\right)\right] \ .$$ This is a good approximation for large but finite $$t<0$$, and becomes exact as $$t\rightarrow-\infty$$.
So, in the $$t\rightarrow-\infty$$ limit, the solution is \begin{aligned} \Phi_{+-}(x,t) &\approx4\arctan\left(\frac{\exp{\left[\gamma\left(x+u\left(t+\frac{\Delta}{2}\right)\right)\right]-\exp\left[\gamma\left(x-u\left(t+\frac{\Delta}{2}\right)\right)\right]}}{1+\exp{(2\gamma x)}} \right) \\ &=4\arctan\left(\exp\left[\gamma\left(x+u\left(t+\frac{\Delta}{2}\right)\right)\right]\right)-4\arctan\left(\exp\left[\gamma\left(x-u\left(t+\frac{\Delta}{2}\right)\right)\right]\right) \\ &=\Phi_+\left[\gamma\left(x+u\left(t+\frac{\Delta}{2}\right)\right)\right]+\Phi_-\left[\gamma\left(x-u\left(t+\frac{\Delta}{2}\right)\right)\right] \ , \end{aligned} where we used the identity $$\arctan(x)-\arctan(y)=\arctan\left(\frac{x-y}{1+xy}\right) \ ,$$ and where the single soliton/antisoliton solution is $$\Phi_{\pm}(X)=\pm4\arctan(\exp(X)) \ .$$ The situation is analogous for the $$t\rightarrow+\infty$$ limit: $$\exp\left[-\gamma u\left(t+\frac{\Delta}{2}\right)\right]\approx\exp\left[-\gamma u\left(t-\frac{\Delta}{2}\right)\right] \ ,$$ leading to $$\Phi_{+-}(x,t)\approx\Phi_+\left[\gamma\left(x+u\left(t-\frac{\Delta}{2}\right)\right)\right]+\Phi_-\left[\gamma\left(x-u\left(t-\frac{\Delta}{2}\right)\right)\right] \ .$$