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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: \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} 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.

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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 $$ \Phi_{+-}(x,t)\approx\Phi_+\left[\gamma\left(x+u\left(t-\frac{\Delta}{2}\right)\right)\right]+\Phi_-\left[\gamma\left(x-u\left(t-\frac{\Delta}{2}\right)\right)\right] \ . $$

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