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I have this formula to calculate the period of a motion in the phase space (plan, in this case) along a phase curve. \begin{equation} T(E)=\int_{x_1}^{x_2}\frac{dx}{\sqrt{2(E-U(x))}} \end{equation} where $E$ is the total energy of the system, $U(x)$ is the potential energy and $x_1$ and $x_2$ are the points of my phase curve that intercept the $x$-axis of the phase plan (points of inversion). Now, I'm asked to calculate the period of the small oscillations in a neighborhood of the point $x_0$, which is the point of potential energy minimum. So, basically, I have to find: \begin{equation} T_0=\lim_{E\to E_0}{T(E)} \end{equation} where $E_0=U(x_0)$.

My attempt: I have used Taylor to evaluate U(x) in a neighborhood of $x_0$. So, considering that E tends to $U(x_0)$ I got: \begin{equation} T(E)=\int_{x_1}^{x_2}\frac{dx}{\sqrt{2(U(x_0)-U(x_0)-U'(x_0)(x-x_0)-\frac{U''(x_0)(x-x_0)^2}{2} }} \end{equation} and, since $U'(x_0)=0$ (because $x_0$ is a minimum for $U(x)$) \begin{equation} T(E)=\int_{x_{1}}^{x_{2}}\frac{dx}{\sqrt{-U''(x_{0})(x-x_{0})^2} } \end{equation} So at this point I'm puzzled because $U''(x_{0})$ is positive for sure, and that leaves me with a negative number under a square root. I'm sure that somewhere i have applied some theorem where I shouldn't have. I feel like it's a mathematical issue.

My book (Arnold) says that the solution is $2\pi$/$\sqrt{U''(x_{0})}$ but I got totally stuck here. Can you help me? Where's my mistake?

P.S. Sorry for my lack in formalism, like omitting little-o notation, etc. but I think you got my point anyways :)

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Your expression is for the half-period, so you double it for oscillating back. You are right the potential is supposed to be concave, so positive second derivative, but you did not consider that if the energy of the system is at the bottom, no kinetic energy, it won't move: so you really need a small amount of energy above $E_0$, and, as any pendulum demonstrates, it does not matter how much, provided it not be zero! So take $E=U(x_0)+\epsilon/2$, holding your breath that the size of $\epsilon$ is irrelevant--spoiler alert.

The rest of your evaluation is fine, $$T(E)=\int_{x_1}^{x_2}\frac{\text{d}x}{\sqrt{2(U(x_0)+\epsilon/2-U(x_0)-U'(x_0)(x-x_0)-\frac{U''(x_0)(x-x_0)^2}{2}}}$$ hence $$T(E)=\int_{x_1}^{x_2}\frac{\text{d}x}{\sqrt{ \epsilon -\frac{U''(x_0)(x-x_0)^2}{2} }} \\ = \int_{x_1}^{x_2}\frac{\text{d}x/\sqrt{ \epsilon} }{\sqrt{ 1 - \frac{U''(x_0)(x-x_0)^2}{2\epsilon}}}$$

Now you see that the integration variably and points of inversion may be shifted by $x_0$, and $\sqrt{\epsilon} $ be absorbed into the normalization of the new variable, and disappear, as long as it is not 0. Moreover, the redefined dummy variable may be yet redefined by similarly absorbing $U''(0)$ in it, which does not disappear, and then further redefined as $=\sin \theta$, with the evident translation of the inversion points of the half-oscillation: $$T(E)= \frac{1}{\sqrt{U''(0)}} \int_{-\pi/2}^{\pi/2}\frac{\cos\theta \text{d}\theta }{\sqrt{ 1 -\sin^2\theta}} \\ = \frac{\pi}{\sqrt{U''(0)}}$$ for the half-period and twice that for the full period.

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  • $\begingroup$ Thank you Cosmas! Now I can finally understand where my mistake was! You're right, I simply put E=U(x_0), not considering that in this case TENDING TO and EQUAL TO are two completely different things! Thank you very much, you made everything more clear! $\endgroup$
    – Luthien
    Mar 24, 2016 at 23:24
  • $\begingroup$ How are the extrema points of the integral of $T(E)$ constant? they clearly depend on $E$ $\endgroup$
    – Pierto
    Nov 3, 2020 at 23:03
  • $\begingroup$ @Pietro No they don't! Do the scaling explicitly. If you followed the disappearance of $\epsilon$, a proxy for E, in the scaling, you see the scaled ones don't. $\endgroup$ Nov 3, 2020 at 23:17
  • $\begingroup$ @Pierre One should think of the inversion points $x_1$ and $x_2$ as the phase points $(x_0 \pm \frac{\epsilon}{2},0)$, corresponding to when the perturbed oscillator is at its peak and nadir. Then we apply the change of variables $\sin{\theta} = \frac{(x - x_0)}{\sqrt{2\epsilon}}$. We then get the $\pm \frac{\pi}{2}$ as the limits of integration. $\endgroup$ Aug 12, 2021 at 3:19

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