# Changing the time parameter and finding the corresponding hamiltonian

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

$$\mathring{f} = \frac{\mathrm{d} f}{\mathrm{d} \tau} = \{ f,\tilde{H} \}~?$$

• Hi Álvaro. Is this from a reference? Which page? Which context? Commented Oct 18, 2023 at 6:55

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.