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Derivation of Lorentz transformation (time)

  1. Aug 24, 2014 #1
    Hi all,

    I'm trying to derive that t=δ(t'+vx'/c^2)

    Using x'=δ(x-vt) then substituting for x=δ(x'+vt') I should be able to isolate t and solve the problem but I am getting to the following point after simplification and can't figure out where to go next...?

    x'= δ[δ(x'+vt')-vt] (isolate "vt")

    vt = -x'/δ + δ(x'+vt') (divide by "v")

    t = -x'/δv + (δ/v)(x'+vt')
     
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  3. Aug 24, 2014 #2

    Fredrik

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    If you're trying to solve the system of equations
    \begin{align}
    &t'=\gamma(-vx/c^2+t)\\
    &x'=\gamma(x-vt)
    \end{align} for t, it's really weird to use the result for x in terms of t' and x', because that's no easier to find than the result for t.

    I recommend that you start by finding a linear combination of x' and t' that doesn't involve x. (I would also use units such that c=1, to avoid having to keep track of factors of c in every step, but that's optional).
     
  4. Aug 24, 2014 #3
    I agree but the assumption is that I do not know the equation for t or t', only x and x'. I see this approach on numerous websites including the wikipedia derivation but everyone seems to get to where I am the just write that this becomes the familiar t=(gamma)(t'+vx'/c^2)
     
  5. Aug 24, 2014 #4

    Fredrik

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    OK, in that case, the trick is to rewrite ##\gamma^2-1## in a clever way. Can you see it? Use the definition of ##\gamma##.
    $$t=-\frac{x'}{\gamma v}+\frac{\gamma}{v}(x'+vt') =\frac{x'}{\gamma v}(\gamma^2-1)+\gamma t'.$$
     
  6. Aug 24, 2014 #5
    Yes, it took me min but now I see it! Very sneaky. I don't see how this becomes t=δ(t'+vx'/c^2)....

    Apparently δ(vx'/c^2) = [x'/δ(v)](δ^2-1) but there must be some other trick to reduce to the standard formula
     
  7. Aug 24, 2014 #6

    Fredrik

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    Not sure why there's an x' in your result for ##\gamma^2-1##, but you can cancel it from that equation. Then solve for ##\gamma^2-1##, and insert the result into the right-hand side of the calculation in post #4.

    Why are you using the symbol δ instead of γ. Is it just because the latter looks too much like a y?

    Edit: OK, now I see what you're doing. The reason you didn't solve for ##\gamma^2-1## is that you wanted to rewrite the first term on the right-hand side in post #4. You found that it's equal to ##\gamma v x'/c^2##. This turns the result in post #4 into
    $$t=\frac{\gamma v x'}{c^2}+\gamma t' =\gamma\left(t'+\frac{vx'}{c^2}\right).$$
     
    Last edited: Aug 24, 2014
  8. Aug 24, 2014 #7
    Got it! Sorry i didn't have my reading glasses with me but now i can see now that i've been using delta instead of gamma.

    Thanks for your help!!!!
     
  9. Aug 24, 2014 #8

    Fredrik

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    OK, cool. I think the easiest way to do this calculation is like this (in units such that c=1):

    \begin{align}
    &x'=\gamma(x-vt) =\gamma\big(\gamma(x'+vt')-vt\big) =\gamma^2(x'+vt')-\gamma vt\\
    &\gamma vt =\gamma^2(x'+vt')-x' =(\gamma^2-1)x'+\gamma^2vt'\\
    &\hspace{20mm}\bigg[\gamma^2-1=\frac{1}{1-v^2}-1 =\frac{1-(1-v^2)}{1-v^2}=\gamma^2v^2\bigg]\\
    &\gamma vt =\gamma^2v^2x'+\gamma^2vt'\\
    &t=\gamma v x'+\gamma t' =\gamma(vx'+t')
    \end{align}
     
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