1

Consider some Hamiltonian system $$ \begin{cases}\dot{y}=\frac{\partial H(y,p)}{\partial p}\\\dot{p}=-\frac{\partial H(y,p)}{\partial p}\end{cases}, $$ where $H(y,p)$ is a given smooth function at least $C^2$. Then it is well known that $H(y,p)$ is preserved by the system, i.e. $$ \frac{d}{dt}H(y,p)=\frac{\partial H}{\partial y}\dot{y}+\frac{\partial H}{\partial p}\dot{p}=0.~~~(*) $$

Today, in the lecture we had that $(*)$ only holds if the dimension $n$ of vector $y$ and the dimension $m$ of vector $p$ are equal, i.e. $m=n$.

As an explanation of that, it was written on the board that $$ H(y,p)=H(y_0,p_0)+\frac{\partial H}{\partial y_1}(y_1-y_{01})+\frac{\partial H}{\partial y_2}(y_2-y_{02})+\cdots+\frac{\partial H}{\partial y_n}(y_n-y_{0n})+\frac{\partial H}{\partial p_1}(p_1-p_{01})+\cdots+\frac{\partial H}{\partial p_m}(p_m-p_{0m})+\mathcal{O}_2.~~~~(**) $$

I think, this is the Taylor series at $(y_0,p_0)$, where $$ y_0=(y_{01},\ldots,y_{0n}),~~~~~p_0=(p_{01},...,p_{0m}). $$

But why does $(**)$ tell me that we need to have $m=n$ in order to have $(*)$?

M. Meyer
  • 639
  • 1
    I am not sure... But shouldn't the explanation come from the fact that $\dot{y}$ has the same dimension as $y$, $\frac{\partial H(y, p)}{\partial p}$ has the same dimension as $p$, therefore dimensions of $y$ and $p$ coincide? – Evgeny Nov 20 '15 at 16:01
  • I am not sure if I do understand what you are aiming at. I think you just mean that this follows directly by the ODE system. – M. Meyer Nov 20 '15 at 16:31
  • My point is that there is only one reasonable way to write the system in the beginning of your post, namely when $m=n$. So we can conclude that $m=n$ just from the form of system, it's some sort of prerequisite. – Evgeny Nov 20 '15 at 16:35
  • 1
    Basically I'm saying that the statement as you have written it is quite tautological. You can write system in this form only if $m=n$ (so it's Hamiltonian system), and the rest (that $H(y, p)$ is conserved quantity) could be checked by straight differentiation (even without using Taylor approximation). So, the condition "only when $m=n$" is looking strange here for me. – Evgeny Nov 20 '15 at 16:41
  • For me, it is strange too. In particular, because Hamiltonian systems are even defined with $m=n$, for example in the wikipedia, https://en.wikipedia.org/wiki/Hamiltonian_system. – M. Meyer Nov 20 '15 at 17:02
  • 1
    We share the same opinion about that piece of proof :) If you are comfortable with the easiest proof of Hamiltonian's conservation, then the best thing is to ask lecturer what he wanted to do with this proof (maybe it was the part of something else). – Evgeny Nov 20 '15 at 17:09

0 Answers0