This answer proved that $$\lim_{E\to E_0}2\int_{x_1}^{x_2}\frac{\mathrm dx}{\sqrt{2\left(E-U\!\left(x\right)\right)}}=\frac{2\pi}{\sqrt{U''\!\left(x_0\right)}},$$ where $E_0:=U\!\left(x_0\right)$ is a minimal value of the second-differentiable potential $U$.

However, I don't think this proof is rigorous enough in that it just throws away terms with order higher than $\left(x-x_0\right)^2$ in $U$. There are some skeptical points. We cannot just expand the integrand in Taylor series. We cannot guarantee that the integrand is analytic at $x_0$ (we cannot even assume that $U$ has third or higher derivatives). Also, we haven't argued that those terms do not have significant effects on the integral boundaries (i.e. the value of $x_1$ and $x_2$).

I hope there will be an $\varepsilon$-$\delta$ version of the proof.

  • 3
    $\begingroup$ In physics, we usually assume the functions are "nice enough" that the heuristic derivations we use can easily be made mathematically rigorous, and then we stop, because it doesn't really matter. Sometimes mathematical rigor matters in physics, but most of the time it doesn't, so we don't waste time on it. $\endgroup$
    – march
    Nov 14, 2022 at 21:58
  • $\begingroup$ If you want, this might be a better question for Mathematics, provided you turn it into something like, "What are the conditions on $U(x)$ that make this identity true?" $\endgroup$
    – march
    Nov 14, 2022 at 21:59
  • $\begingroup$ Throwing away higher order terms than $(x-x_0)^2$ is the word-to-math translation of 'small oscillations', so in that sense the proof is rigorous. I think the regularity requirement might be a more serious issue. $\endgroup$ Nov 14, 2022 at 22:05

2 Answers 2


If $U$ is twice-differentiable at the point $x_0$ and $U''(x_0)\neq 0$, then there exists a function $R$ such that $$U(x) = U(x_0) + U'(x_0) (x-x_0) +\frac{1}{2} U''(x_0)(x-x_0)^2 + R(x)(x-x_0)^2$$ and $\lim_{x\rightarrow x_0} R(x) = 0$. Given that $x_0$ is taken to be a stable equilibrium point of $U$, we have that $U'(x_0)= 0$ and $U''(x_0)>0$. We therefore have that

$$\sqrt{E-U(x)} = \sqrt{E-E_0 - \frac{1}{2} U''(x_0)(x-x_0)^2 - R(x)(x-x_0)^2}$$ where in accordance with OP's notation we define $U(x_0) \equiv E_0$. For $E>E_0$, we may define $\epsilon \equiv E-E_0$ and write $$\sqrt{E-U(x)} = \sqrt{\epsilon} \sqrt{1- \frac{1}{2\epsilon}\big[U''(x_0)+2R(x)\big] (x-x_0)^2}$$ From this it follows that we may write $$\frac{\mathrm dx}{\sqrt{E-U(x)}}= \frac{dx/\sqrt{\epsilon}}{\sqrt{1-\frac{1}{2}\big[U''(x_0)+2R(x)\big]\left(\frac{x-x_0}{\sqrt{\epsilon}}\right)^2}}$$ $$= \frac{d\xi}{\sqrt{1-\frac{1}{2}\big[U''(x_0)+2R\left(\xi\sqrt{\epsilon}+x_0\right)\big] \xi^2}}$$ where we've defined $\xi \equiv (x-x_0)/\sqrt{\epsilon}$. The turning points between which we are integrating occur when $U(x)=E$, which can be rewritten implicitly as $$\xi= \pm\frac{1}{\sqrt{\frac{1}{2}\big[U''(x_0)+2R\big(\xi\sqrt{\epsilon}+x_0\big)\big] }}$$ Let the solutions to this equation be $\xi_1$ and $\xi_2$, respectively. Our integral then becomes

$$T_{\epsilon} = \frac{2}{\sqrt{2}}\int_{\xi_1}^{\xi_2} \frac{\mathrm d\xi}{\sqrt{1-\frac{1}{2}\big[U''(x_0)+2R\left(\xi\sqrt{\epsilon}+x_0\right)\big] \xi^2}}$$

$\epsilon$ appears in the argument of $R$ as well as in the integral bounds, so taking the $\epsilon\rightarrow 0$ limit is a bit subtle - however, I claim that we obtain the correct answer by naively taking all three limits simultaneously (see below for a more rigorous justification). If we do so, we may note that $\lim_{\epsilon\rightarrow 0} R\big(\xi\sqrt{\epsilon}+x_0\big) = 0$ and $\lim_{\epsilon\rightarrow 0} \xi_{1/2} = \pm \xi_0$ where $\xi_0 \equiv \sqrt{2/U''(x_0)}$. Plugging all of this in, we obtain

$$\lim_{\epsilon\rightarrow 0} T_\epsilon =\frac{2}{\sqrt{2}}\int_{-\xi_0}^{\xi_0} \frac{1}{\sqrt{1-\frac{1}{2}U''(x_0) \xi^2}}$$ We conclude in the obvious way - define $\sigma \equiv \xi \sqrt{U''(x_0)/2}$ to obtain $$\lim_{\epsilon\rightarrow 0}T_\epsilon = \frac{2}{\sqrt{U''(x_0)}}\int_{-1}^1 \frac{d\sigma}{\sqrt{1-\sigma^2}} = \frac{2\pi}{\sqrt{U''(x_0)}}$$

For those interested, I'll justify my naive limit-taking more rigorously. Let $\{\epsilon_n\}$ and $\{\alpha_n\}$ be two positive sequences tending to zero such that $0<\epsilon<\epsilon_n \implies |R\big(\xi\sqrt{\epsilon} + x_0\big)|<\alpha_n$ for all each $\xi\in[\xi_1,\xi_2]$.

Define $\gamma_n = \sqrt{\frac{2}{U''(x_0)+2\alpha_n}}$ and observe that $\xi_1 < -\gamma_n$ and $\gamma_n < \xi_2$. We may now define the sequence $$t_n = \frac{2}{\sqrt{2}}\int_{-\gamma_n}^{\gamma_n} \frac{\mathrm d\xi}{\sqrt{1-\frac{1}{2}\big[U''(x_0)+2R\big(\xi\sqrt{\epsilon_n}+x_0\big)\big]\xi^2}}$$ and note that $\lim_{n\rightarrow \infty} t_n = \lim_{\epsilon\rightarrow 0}T_\epsilon$.

Note that the integrand is bounded below by $\left(1-\frac{1}{2}\big[U''(x_0)-2\alpha_n\big]\right)^{-1/2}$. The integral of this lower bound is $$L_n = \frac{2}{\sqrt{2}}\int_{-\gamma_n}^{\gamma_n} \frac{\mathrm d\xi}{\sqrt{1-\frac{1}{2}\big[U''(x_0)-2\alpha_n\big]\xi^2}}$$ $$= \frac{4}{\sqrt{U''(x_0)-2\alpha_n}}\sin^{-1}\left(\sqrt{\frac{U''(x_0)-2\alpha_n}{U''(x_0)+2\alpha_n}}\right)$$

Similarly, the integrand is bounded above by $\left(1-\frac{1}{2}\big[U''(x_0)+2\alpha_n\big]\right)^{-1/2}$, and the integral by $$U_n = \frac{2}{\sqrt{2}}\int_{-\gamma_n}^{\gamma_n} \frac{\mathrm d\xi}{\sqrt{1-\frac{1}{2}\big[U''(x_0)+2\alpha_n\big]\xi^2}} = \frac{2\pi}{\sqrt{U''(x_0)+2\alpha_n}}$$

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Because $L_n < t_n < U_n$ and $\lim_{n\rightarrow \infty} L_n = \lim_{n\rightarrow \infty} U_n = \frac{2\pi}{\sqrt{U''(x_0)}}$, we have by the squeeze theorem that $\lim_{n\rightarrow \infty} t_n = \lim_{\epsilon\rightarrow 0} T_\epsilon = \frac{2\pi}{\sqrt{U''(x_0)}}$.

  • 1
    $\begingroup$ Minor nitpicks: $x_0$ being a stable equilibrium point only implies $U''(x_0)\geq 0$ (e.g for a quartic potential $U(x)=x^4$), so it should be part of the hypothesis that $U''(x_0)>0$ (though one could follow essentially the same proof to deal with the case of say a $2k$-times differentiable $U$ with $U^{(2k)}(x_0)>0$). Also, "since a definite integral is a continuous function of its limits and of its integrand" is of course a little quick. Continuity in the limits is elementary, but continuity in the integrand requires some more justification (e.g DCT but I'm just being picky here) $\endgroup$
    – peek-a-boo
    Nov 15, 2022 at 3:56
  • $\begingroup$ @peek-a-boo Thank you for your nitpicking! I've added the assumption that $U''(x_0)\neq 0$. I've also added an explicit justification of the limit-taking - it may be a bit clunky, but I wanted the justification to be self-contained rather than simply referencing another theorem. Please do let me know if a more elegant but self-contained proof comes to mind, though. $\endgroup$
    – J. Murray
    Nov 16, 2022 at 2:00

While I won't try to make a rigorous proof in this answer, I'll point out some important remarks that I believe are sufficient to address this question.

  1. As mentioned in the comments, "small oscillations" means expanding the potential to second-order. We assume the oscillations are small enough to neglect higher order terms.

  2. This might be the most important point. The function doesn't need to be analytic to be expanded to second order. You don't need a Taylor series. A Taylor polynomial with a convenient form of the remainder (see Wikipedia on Taylor's Theorem) is enough. This requires way weaker assumptions.

  3. Due to the small oscillations hypotheses, higher order terms can't make a difference on the boundary. "Small oscillations" means $x_1$ and $x_2$ are sufficiently close to $x_0$ such that those terms are negligible.

  4. I don't believe this expression holds exactly, but using Taylor's theorem you'll probably be able to derive an estimate on the error. Again, this doesn't require analyticity of the involved quantities, only being differentiable/continuously differentiable a couple of times.

  • 3
    $\begingroup$ Notice that the integral form of the remainder requires only that $U^{(2)}$ is absolutely continuous (which implies the existence of $U^{(3)}$). This is definitely way more general than analyticity. Furthermore, most physical applications will have potentials that are considerably better behaved. $\endgroup$ Nov 14, 2022 at 22:47

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