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Relativity, reference frames and Lorentz transformations

  1. Aug 22, 2010 #1
    1. The problem statement, all variables and given/known data

    The system [tex]S'[/tex] moves in relation to the system [tex]S[/tex] with velocity [tex]\upsilon[/tex] along the -[tex]x[/tex]- axis. At the time when the beginnings of the coordinate system are in the same point, clocks in both system shows [tex]t=t'=0[/tex]. Which coordinates will have a reference point during the motion in every of these systems, which has the property that in some next moment clocks in systems [tex]S[/tex], [tex]S'[/tex] shows the same time [tex]t=t'[/tex]. Determine the law of motion of motion of this point.



    2. Relevant equations
    Lorentz transformation

    [tex]x'=\frac{x-\upsilon t}{\sqrt{1-\frac{{\upsilon}^2}{c^2}}}[/tex]

    [tex]y'=y[/tex]

    [tex]z'=z[/tex]

    [tex]t'=\frac{t-\frac{\upsilon}{c^2}x}{\sqrt{1-\frac{{\upsilon}^2}{c^2}}}[/tex]



    3. The attempt at a solution

    I tried like this

    [tex]t'=\frac{t-\frac{\upsilon}{c^2}x}{\sqrt{1-\frac{{\upsilon}^2}{c^2}}}[/tex]

    [tex]t=\frac{t'+\frac{\upsilon}{c^2}x'}{\sqrt{1-\frac{{\upsilon}^2}{c^2}}}[/tex]

    [tex]t=t'[/tex]

    [tex]t-\frac{\upsilon}{c^2}x=t'+\frac{\upsilon}{c^2}x'[/tex]

    [tex]t-t'=\frac{\upsilon}{c^2}(x+x')[/tex]

    [tex]0=\frac{\upsilon}{c^2}(x+x')[/tex]

    and get

    [tex]x=\frac{\upsilon t}{\sqrt{1-\frac{\upsilon^2}{c^2}}+1}[/tex]

    Is this correct?
     
  2. jcsd
  3. Aug 29, 2010 #2
    Re: Relativity

    Any answer?
     
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