Question

ques

Answer 1

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 ??

Answer 2

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.

Answer 3

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
\[t_{2}=t_{1}=t\]

Answer 4

use lorentz transformation for length contraction

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

I'm getting the wrong answer after using lorentz transformation

Answer 9

\(L_0 = x'_2-x'_1 \)

Answer 10

redo it

Answer 11

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

Answer 12

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

Answer 13

ok thanks