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X and x' in two coordinate relativistic systems

  1. Jan 20, 2009 #1
    let's say we have two coordinate systems ( S-OXYZ and S'-O'X'Y'Z') , S' moving with velocity v away from S . At t=0 O=O'.

    According to Lorentz' transformations we have :

    At t=0 x in as a function of x' :
    [tex]x=\frac{x'}{\sqrt{1-\beta^2}}=x'\gamma[/tex]

    Now, say I know x(I just found it above) , and I want to find x'
    [tex]x'=x\gamma=x'\gamma^2=>\gamma^2=1[/tex]

    Where does this come from and what have I missed ?
    Thank you !
     
  2. jcsd
  3. Jan 20, 2009 #2

    JesseM

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    Science Advisor

    Why are you just looking at cases where t=0? The general transformation would be [tex]x' = \gamma (x - vt)[/tex], so if t=0 then you do have [tex]x' = \gamma x[/tex] (you seem to have reversed x and x' though, unless you meant to write t'=0).
    I agree with [tex]x' = x \gamma[/tex] in the case of t=0, but where are you getting [tex]x \gamma = x' \gamma^2[/tex]? Are you taking the reverse Lorentz transform [tex]x = \gamma (x' + vt')[/tex] and setting t'=0 as well, thus getting [tex]x = \gamma x'[/tex]? The problem is that if you set both t=0 and t'=0, the only coordinates where both time coordinates are 0 are (x=0, t=0) and (x'=0, t'=0), so obviously if x'=0 then you can have [tex]x' = x' \gamma^2[/tex] even when [tex]\gamma[/tex] is not 0.
     
    Last edited: Jan 20, 2009
  4. Jan 21, 2009 #3
    Tnx.
     
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