I'm dealing with a problem where I have a (classical) Hamiltonian $H(q,p)$ such that, for any scalar function $f(p,q)$, $$ \dot{f} = \frac{\mathrm{d} f}{\mathrm{d}t} =\{ f,H \} $$
If I change the time parameter such that $\mathrm{d}t = N(p,q) \mathrm{d}\tau$, where $\tau$ is the new time coordinate, is there any (easy) way of finding the corresponding Hamiltonian $\tilde{H}$ such that
$$ \mathring{f} = \frac{\mathrm{d} f}{\mathrm{d} \tau} = \{ f,\tilde{H} \}~? $$
Just an ancillary remark: the ease of solving $$ \{f,H\}= N(p,q)~ \{f,\tilde H \} $$ may well depend on the particular form of N(p,q), which must, predictably, satisfy the integrability condition(s) $$ \{N,H\}= 0=\{N,\tilde H\}. \tag{I} $$
The reason is that, taking f to be q and p, respectively, you need the (sufficient) system of PDEs, $$ \frac{1}{N} \partial_p H = \partial_p \tilde H ,\\ \frac{1}{N} \partial_q H = \partial_q \tilde H , $$ hence, $$ \partial_p H = N \partial_p \tilde H ,\\ \partial_q H = N \partial_q \tilde H . $$
The integrability of this system of equations follows differentiation of the upper equations by $\partial_q$ and the lower ones by $\partial_p$, and equating the two, thus obtaining the consistency condition (I).
Choosing suitable integral limits for the boundary conditions, you might get $$ \tilde H = g(q)+ \int \!\! dp {1\over N} \partial_p H = h(p)+ \int \!\! dq {1\over N} \partial_q H . $$
For instance, for $N=H$, you get $\tilde H= \ln H$; if $N=\exp (-H)$, you get $\tilde H= \exp H$, and so on.
It sure looks like a homework problem.
