This question comes from a physics text, but I believe my issue is a mathematical one.

Suppose a particle is moving along the real line starting at $x_0$ subject to the potential $V$ with $V(x)<E_0$ for all $x\in[x_0,x_1)$ and $V(x_1)=E_0$ for some constant $E_0$ (the energy of the particle). Further suppose that $V'(x_1)\ne0$. Show that the particle reaches the point $x_1$ in finite time.

Let $x\in[x_0,x_1)$. Then the time it takes to reach $x$ is given by $$t(x)=\int_{x_0}^x\sqrt{\frac{m}{2(E_0-V(y))}}dy.$$ So the time it takes to reach $x_1$ is $$t(x_1)=\lim_{x\to x_1}\int_{x_0}^x\sqrt{\frac{m}{2(E_0-V(y))}}dy.$$ I need to show that this is finite. How can this be done with the limited knowledge of the potential? One thing I believe should be noted is that $$\frac{dt}{dx}=\sqrt{\frac{m}{2(E_0-V(x))}}$$ in the interval $[x_0,x_1)$, which blows up near $x_1$.

  • $\begingroup$ How do you get $E_0$ in the expression for $t(x)$? Shouldn't it be $V(x_0)$? $\endgroup$
    – sudeep5221
    Jul 14, 2020 at 23:37

1 Answer 1


Clearly, if $V'(x_1)\neq 0$, then $V'(x_1)>0$. Rewrite the integral as $$\tau:=\int_{x_0}^{x_1}\,\sqrt{\frac{m}{2\big(V(x_1)-V(x)\big)}}\,\text{d}x=\int_{x_0}^{x_1}\,\sqrt{\frac{m}{2(x_1-x)}}\,\left(\frac{V(x_1)-V(x)}{x_1-x}\right)^{-\frac12}\,\text{d}x\,.$$ Since $V$ is differentiable at $x_1$, we see that, if $\epsilon:=\dfrac{V'(x_1)}{2}$, then there exists a positive real number $\delta<\dfrac{x_1-x_0}{2}$ such that $$\left|\frac{V(x_1)-V(x)}{x_1-x}-V'(x_1)\right|<\epsilon\,.$$ Therefore, $$\dfrac{V'(x_1)}{2}=V'(x_1)-\epsilon<\frac{V(x_1)-V(x)}{x_1-x}<V'(x_1)+\epsilon=\frac{3V'(x_1)}{2}\,.$$ Hence, $$\begin{align}\int_{x_1-\delta}^{x_1}\,\sqrt{\frac{m}{2(x_1-x)}}\,\left(\frac{V(x_1)-V(x)}{x_1-x}\right)^{-\frac12}\,\text{d}x &\leq \int_{x_1-\delta}^{x_1}\,\sqrt{\frac{m}{2(x_1-x)}}\,\left(\frac{V'(x_1)}{2}\right)^{-\frac12}\,\text{d}x \\&=2\,\sqrt{\frac{m}{V'(x_1)}}\,\sqrt{\delta}<\infty\,.\end{align}$$ Also, clearly, $$\int_{x_0}^{x_1-\delta}\,\sqrt{\frac{m}{2(x_1-x)}}\,\left(\frac{V(x_1)-V(x)}{x_1-x}\right)^{-\frac12}\,\text{d}x<\infty\,.$$ Thus, $\tau<\infty$.

P.S. The converse (namely, if $\tau<\infty$, then $V'(x_1)\neq 0$) is not true. For $\tau$ to be finite, it suffices to have $V(x_1)-V(x) =\omega\big(|x-x_1|^2\big)$ for $x$ near $x_1$ (or $V'(x)=\omega\big(|x-x_1|\big)$). (Here, $\omega$ is the small omega notation. See https://en.wikipedia.org/wiki/Big_O_notation.)

According to the OP, the book claims that $V'(x_1)=0$ implies $\tau=\infty$. We shall prove that this is false (unless $V$ is assumed to be analytic at $x_1$).

Counterexample. Pick a real number $\lambda$ such that $1<\lambda<2$. Suppose that $k:=V(x_1)$. Take $$V(x):=k\left(1-\left(\frac{x_1-x}{x_1}\right)^\lambda\right)\,.$$ Therefore, $$V(x_1)-V(x)=k\left(\frac{x_1-x}{x_1}\right)^\lambda\,.$$ That is, $$\tau=\int_{x_0}^{x_1}\,\sqrt{\frac{m}{k\left(\frac{x_1-x}{x_1}\right)^\lambda}}\,\text{d}x\,.$$ By setting $\xi:=\dfrac{x_1-x}{x_1}$ and $\xi_0:=\dfrac{x_1-x_0}{x_1}$, we get $$\tau=\sqrt{\frac{m}{k}}\,x_1\,\int_0^{\xi_0}\,\xi^{-\frac{\lambda}{2}}\,\text{d}\xi=\sqrt{\frac{m}{k}}\,x_1\,\frac{\xi_0^{1-\frac{\lambda}{2}}}{1-\frac{\lambda}{2}}<\infty\,.$$ However, $$V'(x)=\frac{\lambda}{x_1}\,k\,\left(\frac{x_1-x}{x}\right)^{\lambda-1}\,,$$ whence $V'(x_1)=0$.

  • $\begingroup$ Nice answer! For the inequality after "Hence" I am getting $2\frac{m}{V'(x_1)}\sqrt{\delta}$ on the right hand side, but this is finite as well. However, the next part of the problem is to show that $V'(x_1)=0$ implies $\tau=\infty$, which is the contrapositive of the statement you claim is false. I don't immediately see how having $V(x_1)-V(x)=\omega\big(|x-x_1|^2)$ for $x$ near $x_1$ implies $\tau$ is finite. My attempt to verify it is giving me an inequality with infinity on the right hand side, which tells me nothing. Perhaps you could clarify if you have the time. $\endgroup$
    – zbrads2
    Jul 15, 2020 at 1:23
  • $\begingroup$ @zbrads2 I made a small mistake. Thanks for spotting that. $\endgroup$ Jul 15, 2020 at 1:30
  • $\begingroup$ @zbrads2 I remember that there was another user posted the same question regarding the case $V'(x_1)=0$. It was a long time ago, and the thread was probably deleted. After some long discussion, we believe that it was a mistake by the book. Either the exercise is wrong, or it is assuming (without saying explicitly) that $V(x)$ is analytic at $x=x_1$. In that case, if $V'(x_1)=0$, then $$V(x_1)-V(x)=\mathcal{O}\big(|x-x_1|^2\big)\,.$$ Physicists have a tendency to think that every function is analytic. $\endgroup$ Jul 15, 2020 at 1:32
  • $\begingroup$ @zbrads2 I tried to find the old thread, but couldn't find it. So, I gave you a counterexample. See my updated answer. $\endgroup$ Jul 15, 2020 at 2:48
  • $\begingroup$ I completely agree with your counterexample. Thanks again for your answer. $\endgroup$
    – zbrads2
    Jul 15, 2020 at 3:12

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