I originally asked this in the Physics SE, but the question was closed because it was not about the underlying physics of it, so I'll try asking here. Let $Q(q,p),P(q,p)$ be a canonical transformation with valence $\lambda$. The following is intended to be a proof of the following relation: $$\lambda = \frac{\partial(Q,P)}{\partial(q,p)}.$$ Let $F(q,Q,t)$ be the a generating function of the canonical transformation, which satisfies the following relations: $$p = \frac{1}{\lambda}\frac{\partial F}{\partial q},\hspace{1em} P = -\frac{\partial F}{\partial Q},\hspace{1em} K = \lambda H + \frac{\partial F}{\partial t},$$ with $K$ the Hamiltonian for the transformed coordinates. From those relations, it is easy to check that its differential is $$\mathrm{d}F = \lambda \hspace{0.25em}p\hspace{0.25em}\mathrm{d}q - P\hspace{0.25em}\mathrm{d}Q + (K-\lambda H)\hspace{0.25em}\mathrm{d}t$$ and the exterior differential of this is $$0 = \mathrm{d}^2F = \lambda \hspace{0.25em}\mathrm{d}p\wedge\mathrm{d}q - \mathrm{d}P\wedge\mathrm{d}Q + \mathrm{d}(K-\lambda H)\wedge\mathrm{d}t.$$ I think there is some argument for saying $\mathrm{d}(K-\lambda H)\propto\mathrm{d}t$, and $\mathrm{d}t\wedge\mathrm{d}t = 0$. Rearranging, we would get $$\mathrm{d}Q\wedge\mathrm{d}P = \lambda\hspace{0.25em}\mathrm{d}q\wedge\mathrm{d}p,$$ and that multiplicative factor must be the Jacobian determinant. QED.
My questions are:
- Is this proof rigorous? It involves manipulating differentials, and we physicists are known for doing that carelessly.
- Is there a simpler or a better-known proof of this? I have tried searched for one, although with no success.