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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.

References:

  1. David Albert, HOW TO TEACH QUANTUM MECHANICS, notes; p. 3 eq. (1).
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    $\begingroup$ Why does your Hamiltonian has units of force? $\endgroup$ Dec 27 '19 at 19:07
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    $\begingroup$ I think your Hamiltonian is unphysical. $\endgroup$
    – garyp
    Dec 27 '19 at 19:09
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    $\begingroup$ 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
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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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