Lorentz Transformation - Clock

  • #1
CGM
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Homework Statement


Use the Lorentz Transformation equations to derive the formula relating the time period of a moving clock to that of a stationary clock

Homework Equations


X'=y(X-vt)
Y'=Y
Z'=Z
t'=y(t-vx/c^2)

The Attempt at a Solution


t'=1/sqrt(1-(v/c)^2) . (t-vx/c^2)
 

Answers and Replies

  • #2
stevendaryl
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Homework Statement


Use the Lorentz Transformation equations to derive the formula relating the time period of a moving clock to that of a stationary clock

Homework Equations


X'=y(X-vt)
Y'=Y
Z'=Z
t'=y(t-vx/c^2)

The Attempt at a Solution


t'=1/sqrt(1-(v/c)^2) . (t-vx/c^2)
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.
 
  • #3
CGM
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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.
So, t'=yt?
 
  • #4
stevendaryl
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So, t'=yt?
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].
 
  • #5
CGM
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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].
Never mind I did something silly.
So 0=y(x-vt) => x=vt

t'=y(t-(v/t)^2 . t)
 
  • #6
stevendaryl
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Never mind I did something silly.
So 0=y(x-vt) => x=vt

t'=y(t-(v/t)^2 . t)
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.
 
  • #7
CGM
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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.
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?
 
  • #8
stevendaryl
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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?
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]
 
  • #9
CGM
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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]
Ah yes, I see my mistake. Thank you.

t'=t/y
 

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