Suppose you've been given this Hamiltonian: $H = m\ddot{x}$ and are asked to find the equations of motion. (This is a simplified version of a question given here on page 3.)

This isn't a homework problem; I'm trying to learn Hamiltonian mechanics on my own "for fun." I'm having a hard time finding examples online to help me understand how to solve such problems, but I'll show you what I've tried.

$$-\dot{p} = \frac{\partial{H}}{\partial q}$$ $$\dot{q} = \frac{\partial{H}}{\partial p}$$

So first let's try to find $\dot{q}$. We can rewrite $H = m\ddot{x} = \dot{p}$, and now we have to find the derivative of $\dot{p} = \frac{\partial p}{\partial t}$ with respect to $p$. I'm not sure how to think about that.

I'm told that the answer is $x(t) = c_0 + c_1t$. Working backward, this gives us $\dot{q}=c_1$. But if $\dot{q} = \frac{\partial{H}}{\partial p} = c_1$ and $H = \frac{\partial p}{\partial t}$, then apparently we should have found that $\frac{\partial}{\partial p} \frac{\partial p}{\partial t} = c_1$. I'm not sure why we would have assumed such a thing (or if it even makes sense).

Trying the other equation first didn't lead anywhere better.

Clearly I'm misunderstanding some things here, and any guidance would be appreciated.


  1. David Albert, HOW TO TEACH QUANTUM MECHANICS, notes; p. 3 eq. (1).
  • 2
    $\begingroup$ Why does your Hamiltonian has units of force? $\endgroup$ Dec 27 '19 at 19:07
  • 1
    $\begingroup$ I think your Hamiltonian is unphysical. $\endgroup$
    – garyp
    Dec 27 '19 at 19:09
  • 1
    $\begingroup$ The Hamiltonian in the linked document is unphysical!! He says it consists of kinetic energy terms. Does that look like kinetic energy? (answer: no, it doesn't) I think he made a typo. More accurately, a brain fart. Look at the text following eq (3). $\endgroup$
    – garyp
    Dec 27 '19 at 19:13
  • $\begingroup$ @garyp Wow, okay. I suppose maybe Prof. Albert meant to write that $H=\frac{1}{2}m\dot{x}^2$, which is a particle with KE but no PE, so has no forces acting on it? (Edit: thanks for pointing out eq 3; this does seem to be what he meant.) $\endgroup$
    – A_P
    Dec 27 '19 at 19:28

From the sentence below eq. (3) it becomes clear that there is a typo in eq. (1). It was intended to read $$ H~=~\frac{m}{2}(\dot{x}^2_1+ \dot{x}^2_2). \tag{1'}$$ Even this is slightly wrong because the Hamiltonian is by definition a function of momenta rather than velocity, so it should read $$ H~=~\frac{1}{2m}(p^2_1+ p^2_2). \tag{1"}$$ This brings us back to OP's question: A Hamiltonian does by definition not contain any dots, let alone double dots.


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