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anonymous
 one year ago
ques
anonymous
 one year ago
ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Length 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 \[L=x_{2}\primex_{1}\prime\] 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
 one year ago
Best ResponseYou've already chosen the best response.0I don't get at : \(L_0 = x_2x_1\) is the function w.r.t t while \(L = x_2'x_1'\) is the function w.r.t.t' you apply \(x_2'= \gamma (x_2vt)\) 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
 one year ago
Best ResponseYou've already chosen the best response.0The 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 \[t_{2}=t_{1}=t\]

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1use lorentz transformation for length contraction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's what I did I used the transformation \[x'=\gamma(xvt)\]

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1\(\gamma = \huge \frac{1}{\sqrt{1\frac{v^2}{c^2}}} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep and that's exactly what I've used...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm getting the wrong answer after using lorentz transformation

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1\(L_0 = x'_2x'_1 \)

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1i am at work now so can't really focus much, but I found this for you http://hyperphysics.phyastr.gsu.edu/hbase/relativ/tdil.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[L_{o}=x_{2}'x_{1}'\] \[L=x_{2}x_{1}\] \[L_{o}=x_{2}'x_{1}'=\gamma L\] \[\implies L=\frac{L_{o}}{\gamma}\] Why is it now working after switching the coordinates?
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