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Lorentz Transformation - Clock

  1. Jan 8, 2015 #1

    CGM

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    1. The problem statement, all variables and given/known data
    Use the Lorentz Transformation equations to derive the formula relating the time period of a moving clock to that of a stationary clock
    2. Relevant equations
    X'=y(X-vt)
    Y'=Y
    Z'=Z
    t'=y(t-vx/c^2)
    3. The attempt at a solution
    t'=1/sqrt(1-(v/c)^2) . (t-vx/c^2)
     
  2. jcsd
  3. Jan 8, 2015 #2

    stevendaryl

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    You're almost there. If the clock is at rest in the primed coordinate system, that means that its location as a function of time can be chosen to be [itex]X' = 0[/itex]. So using the first equation from your "Relevant equations", you can solve for [itex]x[/itex] in terms of [itex]t[/itex]. Plug that into your equation.
     
  4. Jan 8, 2015 #3

    CGM

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    So, t'=yt?
     
  5. Jan 8, 2015 #4

    stevendaryl

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    How did you get that? Take your equation, [itex]t' = \gamma (t - \frac{vx}{c^2})[/itex] and replace [itex]x[/itex] by [itex]v t[/itex].
     
  6. Jan 8, 2015 #5

    CGM

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    Never mind I did something silly.
    So 0=y(x-vt) => x=vt

    t'=y(t-(v/t)^2 . t)
     
  7. Jan 8, 2015 #6

    stevendaryl

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    I'm afraid you did something silly again. If you start with [itex]t' = \gamma (t - \frac{vx}{c^2})[/itex] and substitute [itex]x=vt[/itex], you get: [itex]t' = \gamma (t - \frac{v^2 t}{c^2})[/itex]. Now factor out a factor of [itex]t[/itex] from the right-hand side, and remember what [itex]\gamma[/itex] is.
     
  8. Jan 9, 2015 #7

    CGM

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    Yes that's what I said, maybe my notation is confusing.

    t'=yt-((v/c)^2)t
    t'=t(y-(v/c)^2)

    Is this the answer or is there more simplification to be done?
     
  9. Jan 9, 2015 #8

    stevendaryl

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    No, you have the expression

    [itex]t' = \gamma (t - (v/c)^2 t)[/itex]

    Factor out [itex]t[/itex] to get:

    [itex]t' = \gamma t (1 - (v/c)^2)[/itex]

    Now, if you remember what the definition of [itex]\gamma[/itex] is, you can write [itex]1-(v/c)^2 = 1/\gamma^2[/itex]
     
  10. Jan 9, 2015 #9

    CGM

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    Ah yes, I see my mistake. Thank you.

    t'=t/y
     
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