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Lorentz transformations formulas

  1. Aug 5, 2011 #1
    I'm slowly trying to understand sp relativity. I admit I got lost in the last thread I posted :blushing:. But thanks to all who replied!

    I have a question about the Lorentz transformations formulas. This is more of a mathematical question about how the formulas are derived.


    If you have the two formulas,

    x'= γ( x- vt) and x= γ(x' + vt')

    which represent the x components for two reference frames S and S', and where γ is the Lorentz factor,

    and you combine them to solve for t:

    x'= γ[γ(x' + vt') - vt]

    how do you arrive at the formula

    t= γ(t' + vx'/c^2) ?

    I know that you simply solve for t from the other formula, but I really cannot figure out how. Sorry, I realize this is more of a math-related problem, but I'm wondering if anybody can give me some tips? :smile:
     
  2. jcsd
  3. Aug 5, 2011 #2

    tiny-tim

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    hi blueberrynerd! :smile:

    (try using the X2 icon just above the Reply box :wink:)
    x'= γ[γ(x' + vt') - vt]

    γvt = (γ2 - 1)x' + γvt'

    t = (γ2 - 1)x'/γv + t' :smile:

    (the https://www.physicsforums.com/library.php?do=view_item&itemid=19" is often easier to understand if you use the rapidity, α, defined by tanhα = v

    then coshα = 1/√(1 - v2), sinhα = v/√(1 - v2), cosh2α - sinh2α = 1 :wink:)​
     
    Last edited by a moderator: Apr 26, 2017
  4. Aug 5, 2011 #3

    Dale

    Staff: Mentor

    You also have to use the formula:
    [tex]\gamma=\frac{1}{\sqrt{1-v^2/c^2}}[/tex]
     
  5. Aug 5, 2011 #4
    Oops, just a small mistake there. It should be

    x'= γ[γ(x' + vt') - vt]

    γvt = (γ2 - 1)x' + γ^2 vt'

    t = (γ2 - 1)x'/γv + yt'

    Using y = 1 / sqrt(1 - (v/c)^2), it can then be reduced further to

    t = (y - 1 / y) x' / v + y t'

    t / y = (1 - 1 / y^2) x' / v + t'

    t / y = (1 - (1 - (v/c)^2)) x' / v + t'

    t / y = x' v / c^2 + t'

    t = y (t' + x' v / c^2)
     
  6. Aug 5, 2011 #5
    Blueberrynerd,

    This may be the easiest way. You start with this ...

    . x = γ(x' + vt')

    and since x = ct, then ...

    . x = γ(x' + vt')
    . ct = γ(x' + vt')

    and t' = x'/c, so ...

    . ct = γ(x' + vt')
    . ct = γ(x' + v(x'/c)

    and x' = ct', so ...

    . ct = γ(x' + vx'/c)
    . ct = γ(ct' + vx'/c)

    dividing thru by c ...

    . t = γ(t' + vx'/c2))

    GrayGhost
     
  7. Aug 5, 2011 #6

    DrGreg

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    Er, why should that be? The Lorentz transform applies to all values of x and t, not just the subset you consider.
     
  8. Sep 21, 2011 #7

    This reply is almost a month delayed, but THANK YOU! You helped a lot! :biggrin:
     
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