Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
1  
$v=dx/dt\neq dx/d\tau$ –  Michael Brown Apr 21 '13 at 23:06
    
Ah, how silly of me! Thanks, Michael! :) –  Bess Apr 21 '13 at 23:09
add comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
    
Thanks, Josh! Silly me...! –  Bess Apr 21 '13 at 23:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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