I'm looking at a book by VI Arnold on mathematical physics and I've hit a roadblock pretty early on. I'll quote the question:

"Let $S(E)$ be the area enclosed by the closed phase curve corresponding to the energy level E. Show that the period of motion along this curve is equal to $T=\frac{dS}{dE}$."

Here the phase curve is a plotting the solutions to a system with one degree of freedom. An example of this would be the equation for simple harmonic motion (mass and spring constant k are set to one): $$\ddot{x}=-x$$ It's solution is defined by concentric circles about the origin, where each circle defines a particular energy level E. Anything that could possibly point me in the right direction would be very helpful.


Notice that the "speed" in the phase space is given by \begin{equation} v_{\mathrm{ps}} = \sqrt{\dot{q}^2 + \dot{p}^2} = \sqrt{H_{p}^2 + H_q^2}. \end{equation} Here, $H=H(p,q)$ is the Hamiltonian, and $H_p$ and $H_q$ are shorthand for $\partial H/\partial p$ and $\partial H/\partial q$. Provided that there exists a closed constant-energy curve in the phase space corresponding to $H(p,q) = E$, the period is given by \begin{equation} T(E) = \oint_{\partial D(E)}\frac{dl}{v_{\mathrm{ps}}} = \oint_{\partial D(E)}\frac{dl}{\sqrt{H_{p}^2 + H_q^2}} , \end{equation} where $dl$ is an infinitesimal length element along the curve. The symbols $D(E)$ and $\partial D(E)$ respectively denote the region bounded by the constant-energy curve and the curve itself.

Next, the area of the region $D(E)$ is given by \begin{equation} S(E) = \int_{D(E)} dpdq. \end{equation} One can make a change of variables from ($p,q$) to ($E,l$), where $l$ for a given $E$ parametrizes the constant-energy curve such that $dl$ is its infinitesimal length element. One should in principle calculate the associated Jacobian determinant. But an easy way to work around this is to first note that the area element of the phase space can be written as $dl_E\, dl$, where $dl_{E}$ is the length element in the direction perpendicular to a constant-energy curve. It is related to $dE$ as follows: \begin{equation} dl_{E} = \frac{dE}{|\nabla H|}=\frac{dE}{\sqrt{H_p^2 + H_q^2}}. \end{equation} Therefore, \begin{equation} S(E) = \int_{E_{\min}}^{E}dE^{\prime} \oint_{\partial D(E^{\prime})} \frac{dl}{\sqrt{H_p^2 + H_q^2}} . \end{equation} Comparing the expressions for $T(E)$ and $S(E)$ leads to the desired result.


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