I'm currently working through Introduction to Mechanics and Symmetry, specifically the section on Generating functions of canonical transformations. I am having an issue with the following problem:
Show that $$S(q^i,s^j,t)=\frac{1}{2t}||\mathbf{q}-\mathbf{s}||^2$$ generates a canonical transformation that is the identity at $t=0$.
First off, I am somewhat confused as to the assumptions this problem is making. I think it would be right for me to assume that $$S:Q_1\times Q_2\times \mathbb{R}\cong \mathbb{R}^{n}\times \mathbb{R}^n\times\mathbb{R}\to \mathbb{R}$$ and if so, don't we have $||q-s||^2=||q||^2+||s||^2$ since $q$ and $s$ are orthogonal? Or am I supposed to take $||\mathbf{q}-\mathbf{s}||^2=\sum (q^i-s^i)^2$, like they share components.
I am also unsure how to deal with this time-dependent function, as the book only treated time-independent generating functions. I know that the canonical transformation generated by $S$ must satisfy $$i^*_\phi \Theta=dS$$ so if I go with $||\mathbf{q}-\mathbf{s}||^2=\sum (q^i-s^i)^2$ we get that $p_i=\frac{1}{t} q^i(q^i-s^i)$ and $r_i=\frac{1}{t}s^i(q^i-s^i)$, with $(q,p)$ and $(s,r)$ the coordinates on $T^* Q_1$ and $T^*Q_2$ respectively. I'm not sure what else I can do. Any suggestions would be greatly appreciated.
You have to treat $S=\frac{1}{2t}\sum(q^i-s^i)^2$ as a function of fixed $t$ and then use it to generate the proper diffeomorphism as follows:
We have $p_i=\frac{1}{t}q^i(q^i-s^i)$, which gives $$\frac{t p_i}{q^i}=q^i-s^i$$ Then we come to $s^i=q^i-\frac{tp_i}{q^i}$. If we substitute this into $r_i=\frac{1}{t} s^i(q^i-s^i)$ we get \begin{align}r_i&=\frac{1}{t}(q^i-\frac{tp_i}{q^i})(q_i-q^i+\frac{tp_i}{q_i})\\ &=\frac{1}{t}(q^i-\frac{tp_i}{q^i})\frac{tp_i}{q^i}\\ &=p_i-\frac{tp_i^2}{(q^i)^2}\end{align} So this generates a symplectomorphism $(q^i,p_i)\mapsto (q^i-\frac{tp_i}{q^i}, p_i-\frac{tp_i^2}{(q^i)^2})$ that is the identity at $t=0$, despite having a singularity there.