# $\frac{dt}{d\tau}=\gamma$ in special relativity

I hope this is not too silly a question: We often see

$$\frac{dt}{d\tau}=\gamma=\frac{1}{\sqrt{1-v^2}},$$ taking $c=1$.

Problem: I don't understand why...

In the Minkowski metric, using the $(-+++)$ signature and taking $c=1$, $$ds^2=-dt^2+d\vec x^2\\ d\tau^2=-ds^2\\ \implies d\tau^2=dt^2-d\vec x^2\\ \implies 1=\left(\frac{dt}{d\tau}\right)^2-v^2\\ \implies \frac{dt}{d\tau}=\sqrt{1+v^2}\neq \frac{1}{\sqrt{1-v^2}}$$

What has gone wrong? with my reasoning?

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$v=dx/dt\neq dx/d\tau$ –  Michael Brown Apr 21 '13 at 23:06
Ah, how silly of me! Thanks, Michael! :) –  Bess Apr 21 '13 at 23:09

You made a simple error; $dx/d\tau\neq v$!. Start from your equation $$d\tau^2 = dt^2 - d\vec x^2$$ Now, divide both sides by $dt$ not $d\tau$ to get $$\left(\frac{d\tau}{ dt}\right)^2 = 1-v^2$$ which gives $$\frac{d\tau}{dt} = \frac{1}{\gamma}$$ as desired.