Suppose I have a smooth potential $U:\mathbb R\to\mathbb R$ with $U(x)=U(-x)$, $U(0)=0$, and $U'(x)>0$ for $x>0$. A particle of mass $1$ at rest at position $x=x_0$ has total energy $U(x_0)$, and if allowed to move freely in the potential well will have periodic motion with period $$T(x_0)=4\int_0^{x_0}\frac{dx}{\sqrt{2(U(x_0)-U(x))}}.$$ So if $U(x)$ is known, we can calculate $T(x)$. I want to invert this process to find $U(x)$ if $T(x)$ is known for all $x$. I would appreciate a hint on where to start this derivation.

One idea is to take the derivative of $T(x_0)$. Naively, we would expect the derivative to take the form $$\frac{dT}{dx_0}=4\left(\int_0^{x_0}\left(\frac{d}{dx_0}\frac{1}{\sqrt{2(U(x_0)-U'(x))}}\right)\,dx+\frac{1}{\sqrt{2(U(x_0)-U(x))}}\Bigg|_{x=x_0}\right).$$ But we notice that the integral doesn't converge, and the other term is undefined. Hopefully, the two singularities cancel. I think the right way to deal with this is to write $$T(x_0)=\lim_{a\to x_0^-}4\int_0^a\frac{dx}{\sqrt{2(U(x_0)-U(x))}},$$ and then take the derivative. However, I am unable to evaluate the derivative, and even if I did, I don't know if this method will ultimately work.

  • $\begingroup$ How did you apply the chain rule here? $\endgroup$ Mar 1, 2018 at 2:14
  • $\begingroup$ Let $f(x_0,x)=1/\sqrt{2(U(x_0)-U(x))}.$ Let $F(x_0,x)$ be the antiderivative of $f(x_0,x)$ with respect to $x$. Then we are looking for $$\frac{d}{dx_0}\int_0^{x_0} f(x_0,x)=\frac{d}{dx_0}(F(x_0,x_0)-F(x_0,0)).$$ Now applying the derivative in each coordinate, $$F^{(1,0)}(x_0,x_0)-F^{(1,0)}(x_0,0)+F^{(0,1)}(x_0,x_0)=\int_0^{x_0}\frac{d}{dx_0} f(x_0,x)\,dx+f(x_0,x_0).$$ $\endgroup$
    – Plutoro
    Mar 1, 2018 at 2:19

1 Answer 1


This is a classical version of the inverse scattering problem Can you hear the shape of the drum? which is answered e.g. in Refs. 1 & 2.

Here is the main result: If $\ell(V)$ denotes the accessible length at potential energy-level $V$, then the bijective relation with the period $T(E)$ (as a function of energy-level $E$) are given by an Abel transform$^1$ $$T(E) ~=~A^{\prime}(E)~=~\sqrt{2m}\frac{d}{dE}\int_{V_0}^E \frac{\ell(V)~dV}{\sqrt{E-V}},\tag{A} $$ $$ \ell(V) ~= ~\frac{1}{\pi\sqrt{2m}}\int_{V_{0}}^V \frac{T(E)~dE}{\sqrt{V-E}}.\tag{B} $$

Here we have defined the action/area variable

$$ A(E)~=~\oint_{H(x,p)=E} \! p~dx ~=~\iint_{H(x,p)\leq E}\! dx~dp . $$

Proof of eq. (A): Use that the speed is $v~:=~\frac{p}{m}~=~\sqrt{\frac{2(E-V)}{m}}$ and that the period is $$\begin{align} T(E)~=~& 2\int_{x_-}^{x_+}\! \frac{dx}{v}\cr ~=~&\sqrt{2m} \left(\int_{V_0}^E \frac{\ell^{\prime}(V)~dV}{\sqrt{E-V}}+\frac{\ell(V_0)}{\sqrt{E-V}}\right)\cr ~\stackrel{(9)}{=}~&A^{\prime}(E)\cr ~\stackrel{(2)}{=}~&\sqrt{2m}\frac{d}{dE}\int_{V_0}^E \frac{\ell(V)~dV}{\sqrt{E-V}}, \end{align}$$ where the eq. numbers refer to Ref. 2. $\Box$

Proof of eq. (B): Use eq. (A) and eq. (7). $\Box$

If we furthermore assume that the potential $\Phi(x)=\Phi(-x)$ is even, then the sought-for potential $\Phi$ is the inverse function of $V \mapsto \ell(V)/2$ of half the accessible length.


  1. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; $\S$12 + $\S$49.

  2. My Phys.SE answer here.


$^1$ Here we assume that the potential $\Phi$ does not have flat plateaus except for possibly at the bottom.


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