3
$\begingroup$

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 $t_0$?

$\endgroup$
  • $\begingroup$ @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
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $t_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.