anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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 ??
Loser66
  • 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.
anonymous
  • 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 \[t_{2}=t_{1}=t\]

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nincompoop
  • nincompoop
use lorentz transformation for length contraction
anonymous
  • anonymous
That's what I did I used the transformation \[x'=\gamma(x-vt)\]
nincompoop
  • nincompoop
\(\gamma = \huge \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \)
anonymous
  • anonymous
Yep and that's exactly what I've used...
anonymous
  • anonymous
I'm getting the wrong answer after using lorentz transformation
nincompoop
  • nincompoop
\(L_0 = x'_2-x'_1 \)
nincompoop
  • nincompoop
redo it
nincompoop
  • 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
anonymous
  • 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?
anonymous
  • anonymous
ok thanks

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