I just can't seem to find the answer to this seemingly simple question. Suppose we have a function $x$ of $t$, and we know that the following quantity is constant, i.e., independent of time: \begin{align} \frac{1}{2}\left(\frac{\text d x}{\text d t}(t)\right)^2 + V(x(t)) = E = \text{const.} \end{align}

This of course reminds one of the energy of a particle (of unit mass) in a potential $V$. Now I'm sure it should be the case that \begin{align} \frac{\text d^2 x}{\text d t^2}(t) = -\frac{\text d V}{\text d x}(x(t)), \tag{1} \end{align} but I can only prove this for those values of $t$ for which d$x/$d$t\neq 0$, in which case we simply have \begin{align} 0 = \frac{\text d E}{\text d t} = \frac{\text d x}{\text d t}\frac{\text d^2 x}{\text d t^2} + \frac{\text d V}{\text d x}\frac{\text d x}{\text d t} = \frac{\text d x}{\text d t}\left(\frac{\text d^2 x}{\text d t^2} +\frac{\text d V}{\text d x}\right) \end{align} from which the result follows.

So my question is, how does one show that $(1)$ holds, even if $(\text d x/\text d t)(t_0)=0$ for some $t_0$?

  • $\begingroup$ @CDCM I don't think that this gives us the answer, because we don't know d$x/$d$t$ as a function of $x$, precisely because the relation $x(t)$ is not invertible in general when d$x/$d$t=0$. $\endgroup$ – Sjorszini Dec 13 '17 at 17:26
  • $\begingroup$ In particular we cannot use d$/$dx = (d$t/$d$x)$d$/$d$t$ $\endgroup$ – Sjorszini Dec 13 '17 at 17:27
  • 2
    $\begingroup$ Take another time derivative. $\endgroup$ – Philo Dec 13 '17 at 17:29
  • $\begingroup$ @CDCM Nope, d$t/$d$x$ is simply not defined because $t$ does not exist as function of $x$. $\endgroup$ – Sjorszini Dec 13 '17 at 17:40
  • $\begingroup$ @Philo. That does the job. If you make your comment into an answer I'll accept it. $\endgroup$ – Sjorszini Dec 13 '17 at 17:41

Taking another derivative we get (dot being a time derivative and prime a spatial derivative)

\begin{equation} \ddot E = 0 = \ddot x(\ddot x + V') + \dot x (\dddot x +V''\dot x) \end{equation} in which the second term gives $0$ for $\dot x=0$. For $\ddot x\neq 0$ the desired equation follows.


For completeness, here is the proof for those values of $t_0$ for which $\dot x(t_0)= \ddot x(t_0) = 0$, as this is the only case not mentioned by Philo in his answer.

Since $\dot x(t_0) =0$, the constancy of $E=\frac{1}{2}\dot x^2 + V(x)$ shows that $V$ (as function of $t$) has a global maximum at $t_0$. But in fact this is also a global maximum at $x(t_0)$ of $V$ seen as function of $x$, since $V$ simply cannot get larger then $E$ because of the positivity of the term $\frac{1}{2}\dot x^2$. Since we also assume that $V$ is differentiable w.r.t. $x$ at $x(t_0)$, so that we implicitly are saying that $x(t_0)$ is an interior point of the domain of $V$, this global maximum is also a local maximum, which then implies that we have $($d$V/$d$x)(x(t_0))=0$. Since we also have $\ddot x(t_0)=0$, this yields the desired result at $t_0$.


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