0
$\begingroup$

A dynamical system with generalised particle position $q$ and generalised momentum $p$, described by:

$$\dot{q}=F_1(q,p)\quad\text{and}\quad\dot{p}=F_2(q,p)\tag{1}$$

is a Hamiltonian system if:

$$\frac{\partial F_1}{\partial q} + \frac{\partial F_2}{\partial p} = 0\tag{2}.$$

But from Hamilton's equations: $$\frac{\partial H}{\partial q} = -\dot{p} = -F_2\tag{3}$$ and $$\frac{\partial H}{\partial p} = \dot{q} = F_1.\tag{4}$$

This gives the condition for a Hamiltonian dynamical system as: $$\frac{\partial^2 H}{\partial q \partial p} = \frac{\partial^2 H}{\partial p \partial q}.\tag{5}$$

Why is the symmetry of the second partial derivatives of the Hamiltonian function a necessary condition? What are the properties of a function $F$ for which $\frac{\partial^2 F}{\partial q \partial p} \neq \frac{\partial^2 F}{\partial p \partial q}$ in the context of Hamiltonian mechanics? And what physical property of the Hamiltonian does the symmetry condition highlight (in terms of the physical intuition of the total energy of a system, for example)?

$\endgroup$

2 Answers 2

0
$\begingroup$

The order of the derivatives in mixed second order partial derivatives does not matter if $F$ is has continuous second order partial derivatives. You can refer to a post on the Maths Stack Exchange.

It is usually an unstated assumption in classical mechanics that the functions are as smooth as needed. That is, as many derivatives as we need as assumed to exist and are also assumed to be continuous.

In the case of the Hamiltonian function it means that the velocity is a continuously differentiable function of $q$ and force is a ontinuously differentiable function of $p$. Both these conditions are usually true.

$\endgroup$
0
$\begingroup$
  1. If we consider the 1-form $$F~:=~F_1\mathrm{d}p-F_2\mathrm{d}q~=~\dot{q}\mathrm{d}p-\dot{p}\mathrm{d}q~=~\mathrm{d}H,\tag{A}$$ we see that it exact, and hence closed. OP's eq. (2) is this closedness condition. See also this related Phys.SE post.

  2. In physics it is often implicitly assumed that functions (like the Hamiltonian $H$) of phase space are sufficiently differentiable, say of class $C^2(\mathbb{R}^2)$. Then OP's eq. (5) becomes a triviality.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.