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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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    I've just checked and $\frac{\mathrm d t}{\mathrm d\tau}$ is constant, but I still can't see how it works. Any chance of running through it step by step? Thank you.2011-09-13
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    @Peter4075: I thought that was step by step; could you say more specifically which part you're having trouble with? If $\frac{\mathrm d t}{\mathrm d\tau}$ is constant, then $\frac{\mathrm d^2 t}{\mathrm d\tau^2}=0$, and the second term vanishes, leaving the first term as expected.2011-09-13
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    joriki - don't worry about making it too simple! I can see now about the second term disappearing if $\frac{dt}{d\tau}$ is constant. I can see how you get from the first line to the second line, but I'm afraid I still can't see how you get from the second line (a product) to the third line (a sum).2011-09-13
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    @Peter4075: OK, thanks for the clarification -- it's much easier to answer questions at this level of specificity :-). By the way, I only get notified if you put the '@' in front of the user name -- in this case it's not necessary since I also get notified of all comments under my answers, but if you ever want someone to read something that's not under their own answer, you need to use the '@'.2011-09-13
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    @Peter4075: I've tried to make it a bit more explicit -- hope that helps? The product turning into a sum is due to the product rule, $(fg)'=fg'+f'g$.2011-09-13
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    Good grief, that is a long and winding road to the final answer. But now I can see it at last. I thought it might be the product rule but couldn't figure out how to fit it in. Thanks very much. The example is actually from Equations 4.18 to 4.19, page 106, in Carroll's pdf "Lecture notes on general relativity" where he's discussing the Newtonian weak field limit.2011-09-13
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    @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