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Derivation of an old-time formula

  1. Jul 3, 2010 #1
    In certain physics textbooks, one starts with the assumption (in a one linear and one time dimension) that

    1) x2 - c2t2 = x'2 - c2t'2

    I don't want to go into that. Let us start from there.

    Now, if you assume that

    2) x = vt, then

    3) x' = 0 always because the origin is moving at v along the x-axis so that x' is always zero.

    Using those three bits of information one can derive the Lorentz equations:

    x' = [tex]\gamma[/tex](x - vt)
    t' = [tex]\gamma[/tex](t - xv/c2)

    where [tex]\gamma[/tex] = SQRT[1 - v2/c2] (I can't get the square root to come out in LATEX)

    I have tried, tried, tried to do that but I cannot.

    Anyone can help?

    If I substitute the Lorentz equations back into the above three bits of information, it will work out but that's backwards. I want to do it forwards.
     
  2. jcsd
  3. Jul 3, 2010 #2
    [tex]c^2\text{d}\tau^2=c^2\text{d}t^2-\text{d}x^2[/tex]

    This is a usual way to express the minowski metric, or SR metric. [tex]\tau[/tex] is proper time, measured by an observer at rest relative to the frame being measured.

    From here:

    [tex]\left(\frac{\text{d}\tau}{\text{d}t}\right)^2=1-\frac{1}{c^2}\left(\frac{\text{d}x}{\text{d}t}\right)^2[/tex]

    We use the symbol [tex]v[/tex] for [tex]\frac{\text{d}x}{\text{d}t}[/tex].

    [tex]\left(\frac{\text{d}\tau}{\text{d}t}\right)^2=1-\frac{v^2}{c^2}[/tex]

    [tex]\frac{\text{d}\tau}{\text{d}t}=\sqrt{1-\frac{v^2}{c^2}}[/tex]

    In the case of constant velocity, the right hand side is a constant. We can integrate wrt. [tex]t[/tex], letting [tex]\tau=t=0[/tex] initially and obtain

    [tex]\tau=t\sqrt{1-\frac{v^2}{c^2}}[/tex]

    or the more familiar

    [tex]t=\frac{\tau}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
     
    Last edited: Jul 4, 2010
  4. Jul 3, 2010 #3
    Perfect!

    Actually, that makes it clearer than the text I was reading from.

    Thanks,

    Steve G
    Melbourne, FL
     
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