Let $(q,p) \mapsto (Q,P)$ be a diffeomorphism of phase space. Then this is a canonical transformation (CT) if

$$p\dot{q}-H(q,p,t)=P\dot{Q}-K(Q,P,t) + \frac{dM}{dt}\tag{1}$$

for some $M=M(q,p,Q,P,t)$.

  1. The case when $M$ only depends on the old and new positions (and time $t$) gives conditions on the transformations to be canonical. This is known as a CT of type 1.

  2. But $M$ depending on $q$, $P$, $t$ for example does not work, we need $$M=M_2(q,P,t)-QP.\tag{2}$$ Why?

  1. If the generator $M$ was a function $M(q,P,t)$ only, eq. (1) would e.g. imply $P=0$, which is bad if we are trying to find a bijective transformation $(q,p,t)\to (Q,P,t)$.

  2. On the other hand, a function $M=M_1(q,Q,t)$ works just fine because the $\dot{M}$ term then only produces dot-variables $\dot{q}$ and $\dot{Q}$, which are already present in eq. (1) to match.

  3. The transformation (2) can be viewed as a Legendre transformation $Q\leftrightarrow P$. The function $M_2(q,P,t)$ is known as a type-2 generating function of a canonical transformation (CT).

  • $\begingroup$ Yes, but Why is this true? Can it only be seen from the calculation? $\endgroup$ – user242318 May 11 '18 at 19:42
  • $\begingroup$ After a VERY short calculation, if you want to phrase it that way. In fact, it is more like an identification of terms. $\endgroup$ – Qmechanic May 11 '18 at 19:53
  • $\begingroup$ How could I see it without that calculation? What's the reason behind the structural difference of generating functions of first and second kind? $\endgroup$ – user242318 May 11 '18 at 19:55
  • $\begingroup$ The structural difference is that there is a $\dot{Q}$ but no $\dot{P}$ in eq. (1) besides of possibly inside the $\dot{M}$ term. $\endgroup$ – Qmechanic May 11 '18 at 20:20

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