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I'm studying an introductory course in theoretical physics, I stumbled upon something I really can't understand.

So, in my book there is written the following statment:

Consider a Hamiltonian system $S$, with $N$ degrees of freedom. The generalised coordinates are $ q^{1}, q^{2},..,q^{N} $ and the conjugate momenta are: $p_1, p_2,..,p_N$. If the system $S$ has the following Hamiltonian: $$ \mathcal{H} = \mathcal{H}\left( f_1(q^{1}, p_1), f_2(q^{2}, p_2),.., f_N(q^{N}, p_N) \right) $$ then all the quantities $ f_i(q^{i}, p_i) $ are conserved.

The result is then used to proof another theorem. I don't understand how all these quantities are conserved? I tried to prove it, but I can't seem to find it.

We have to prove that:

$$ \frac{d}{dt}f_i(q^{i}, p_i) = 0 $$

I tried this:

$$ \frac{d}{dt}f_i(q^{i}, p_i) = \frac{\partial f_i}{\partial q^{i}}\dot{q}^{i} + \frac{\partial f_i}{\partial p_i}\dot{p}_i $$

But I can't get further than this, does anyone know why this derivative had to equal $0$?

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2 Answers 2

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Note that so far you are only taking the derivative of $f_i$ through basic chain rules, without using the fact that you are given a Hamiltonian, which governs the time evolution of variables in this problem.

Thus the next step would be to rewrite $\dot{q}_i$ and $\dot{p}_i$ with Hamilton's equation. Since $$ \dot{q}_i = \frac{\partial H}{\partial p_i}\ , $$ and by chain rules again, $$ \frac{\partial H}{\partial p_i} = \frac{\partial H}{\partial f_i}\frac{\partial f_i}{\partial p_i}\ . $$ Similarly $$ \dot{p}_i = - \frac{\partial H}{\partial q_i} = - \frac{\partial H}{\partial f_i} \frac{\partial f_i}{\partial q_i}\ . $$ Plugging these back in, you'll find that the two terms in the derivative neatly cancel.

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Hint: If \begin{align} \dot q_i &= \frac{\partial H}{\partial p_i} \\ \dot p_i &= -\frac{\partial H}{\partial q_i} \end{align}

then, for a compltely general $f$ and $H$, \begin{align} \frac{d}{dt} f &= \sum_{i = 1}^N \left( \dot q_i \frac{\partial f}{\partial q_i} + \dot p_i \frac{\partial f}{\partial p_i} \right) \\ &= \sum_{i = 1}^N \left( \frac{\partial H}{\partial p_i} \frac{\partial f}{\partial q_i} - \frac{\partial H}{\partial q_i} \frac{\partial f}{\partial p_i} \right) \\ &= \{f, H\}. \end{align} See what happens when you compute $\{f, H\}$ for $f$ and $H$ of the form you have given.

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