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.

  • $\begingroup$ First guess would be taking the derivative of both sides with respect to x0. You would have to consider the fact that you have x0 in the integrand as well though. $\endgroup$ – Aaron Stevens Mar 1 '18 at 1:44
  • $\begingroup$ I have tried that route. You use the chain rule to account for $x_0$ in the integrand and limit. But you end up with an integral that doesn't converge and a singular boundary term. I expect that the singularities cancel, but I do not see how. $\endgroup$ – Alex S Mar 1 '18 at 1:48
  • $\begingroup$ How did you apply the chain rule here? $\endgroup$ – Aaron Stevens Mar 1 '18 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, $$F^{(1,0)}(x_0,x_0)-F^{(1,0)}(x_0,0)+F^{(0,1)}(x_0,x_0)=\int_0^{x_0}\frac{d}{dx_0} f(x_0,x)\,dx+f(x_0,x_0).$$ $\endgroup$ – Alex S Mar 1 '18 at 2:19
  • $\begingroup$ Maybe I miss something...how you derive your first equation for T(x0)? Probably looking at the derivation will suggest a way to invert the problem. $\endgroup$ – sintetico Mar 1 '18 at 2:20

This is a classical version of the inverse scattering problem Can you hear the shape of the drum? which is answered e.g. in Ref. 1 and my Phys.SE answer here.

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 potential energy-level $E$) are given by an Abel transform $$T(E) ~= ~\sqrt{2m} \int_{V_0}^E \frac{\ell^{\prime}(V)~dV}{\sqrt{E-V}},\tag{A} $$ $$ \ell(V) ~= ~\frac{1}{\pi\sqrt{2m}}\int_{V_{0}}^V \frac{T(E)~dE}{\sqrt{V-E}}.\tag{B} $$

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.

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