# Showing time it takes a particle to reach a point is finite

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 $$V(x) for all $$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 $$x_1$$.

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

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, $$\dfrac{V'(x_1)}{2}=V'(x_1)-\epsilon<\frac{V(x_1)-V(x)}{x_1-x} Hence, \begin{align}\int_{x_1-\delta}^{x_1}\,\sqrt{\frac{m}{2(x_1-x)}}\,\left(\frac{V(x_1)-V(x)}{x_1-x}\right)^{-\frac12}\,\text{d}x &\leq \int_{x_1-\delta}^{x_1}\,\sqrt{\frac{m}{2(x_1-x)}}\,\left(\frac{V'(x_1)}{2}\right)^{-\frac12}\,\text{d}x \\&=2\,\sqrt{\frac{m}{V'(x_1)}}\,\sqrt{\delta}<\infty\,.\end{align} Also, clearly, $$\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 $$V'(x_1)=0$$.

• 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. Jul 15, 2020 at 1:23
• @zbrads2 I made a small mistake. Thanks for spotting that. Jul 15, 2020 at 1:30
• @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. Jul 15, 2020 at 1:32
• @zbrads2 I tried to find the old thread, but couldn't find it. So, I gave you a counterexample. See my updated answer. Jul 15, 2020 at 2:48
• I completely agree with your counterexample. Thanks again for your answer. Jul 15, 2020 at 3:12