1
$\begingroup$

If $\frac{d^{2}x}{d\tau^{2}}=k\left(\frac{dt}{d\tau}\right)^{2}$ and I then multiply both sides by $\left(\frac{d\tau}{dt}\right)^{2}$ I get $\frac{d^{2}x}{dt^{2}}=k$ . Why does the $d\tau^{2}$ on the left hand side change to $dt^{2}$? $\frac{d^{2}x}{d\tau^{2}}$ is a second derivative, so surely I can't just cancel the d\tau's as in a normal fraction?

1 Answers 1

4

This only works if $\frac{\mathrm d t}{\mathrm d\tau}$ is constant. Otherwise

$\begin{eqnarray} \frac{\mathrm d^2x}{\mathrm d\tau^2} &=&\frac{\mathrm d}{\mathrm d\tau}\left(\frac{\mathrm dx}{\mathrm d\tau}\right) \\ &=& \frac{\mathrm d}{\mathrm d\tau} \left(\frac{\mathrm d t}{\mathrm d\tau}\frac{\mathrm dx}{\mathrm dt}\right) \\ &=& \frac{\mathrm d t}{\mathrm d\tau}\left(\frac{\mathrm d}{\mathrm d\tau}\frac{\mathrm dx}{\mathrm dt}\right) + \left(\frac{\mathrm d}{\mathrm d\tau} \frac{\mathrm d t}{\mathrm d\tau}\right)\frac{\mathrm dx}{\mathrm dt} \\ &=& \frac{\mathrm d t}{\mathrm d\tau}\left(\frac{\mathrm d t}{\mathrm d\tau}\frac{\mathrm d}{\mathrm dt}\frac{\mathrm dx}{\mathrm dt}\right) + \left(\frac{\mathrm d}{\mathrm d\tau} \frac{\mathrm d t}{\mathrm d\tau}\right)\frac{\mathrm dx}{\mathrm dt} \\ &=& \left(\frac{\mathrm d t}{\mathrm d\tau}\right)^2\frac{\mathrm d^2x}{\mathrm d t^2}+ \frac{\mathrm d^2 t}{\mathrm d\tau^2} \frac{\mathrm dx}{\mathrm d t}\;, \end{eqnarray} $

which differs by the second term if $\frac{\mathrm d t}{\mathrm d\tau}$ is not constant.

  • 0
    @Peter4075: This derivation is really only necessary if you're interested in the case where $\frac{\mathrm d t}{\mathrm d\tau}$ isn't constant. If it is, this is all just a simple scaling, and it becomes a lot easier to see the forest for the trees if you introduce a variable name for the scale $\frac{\mathrm d t}{\mathrm d\tau}$ instead of carrying it along as a derivative.2011-09-13