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?
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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. |
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