## anonymous one year ago ques

1. anonymous

Length contraction: Suppose a rod has initial length $L_{o}$ and coordinates of it's end points with respect to frame S at time t are given by $x_{1},x_{2}$ $\therefore L_{o}=x_{2}-x_{1}$ then coordinates of it's end points with respect to frame S' moving with velocity v with respect to frame S are given by $x_{1}\prime,x_{2}\prime$ Then It's length with respect to frame S' at time $t'$ will be $L=x_{2}\prime-x_{1}\prime$ But $x_{2}'=\gamma(x_{2}-vt)$$x_{1}'=\gamma(x_{1}-vt)$ So, $L=\gamma(x_{2}-x_{1})=\gamma L_{o}$ But in fact I should get $L=\frac{L_{o}}{\gamma}$ What am I doing wrong ??

2. Loser66

I don't get at : $$L_0 = x_2-x_1$$ is the function w.r.t t while $$L = x_2'-x_1'$$ is the function w.r.t.t' you apply $$x_2'= \gamma (x_2-vt)$$ but I didn't see the relationship of t and t'. Can we apply directly like that? because $$x_2'$$ is the coordinate of the end at the time t' not t.

3. anonymous

The relation between t' and t is $t'=\gamma(t-\frac{xv}{c^2})$ and $t=\gamma(t'+\frac{xv}{c^2})$ Also Something that is assumed here is $t_{2}=t_{1}=t$

4. nincompoop

use lorentz transformation for length contraction

5. anonymous

That's what I did I used the transformation $x'=\gamma(x-vt)$

6. nincompoop

$$\gamma = \huge \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

7. anonymous

Yep and that's exactly what I've used...

8. anonymous

I'm getting the wrong answer after using lorentz transformation

9. nincompoop

$$L_0 = x'_2-x'_1$$

10. nincompoop

redo it

11. nincompoop

i am at work now so can't really focus much, but I found this for you http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html

12. anonymous

$L_{o}=x_{2}'-x_{1}'$ $L=x_{2}-x_{1}$ $L_{o}=x_{2}'-x_{1}'=\gamma L$ $\implies L=\frac{L_{o}}{\gamma}$ Why is it now working after switching the coordinates?

13. anonymous

ok thanks