Please check my Lorentz Transforms

  • Thread starter mattst88
  • Start date
  • #1
29
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Homework Statement



At x = x' = 0 and t = t' = 0, a clock ticks on a fast spaceship (gamma = 100). The captain of the ship heads it tick again 1.0 s later. Where and when do we (the stationary observers) measure the second tick to occur?

Homework Equations



[tex]t = \frac{t'}{\gamma}[/tex]
[tex]x' = \gamma(x - vt)[/tex]
[tex]\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

The Attempt at a Solution



First, I solve for velocity.
[tex]v = c\sqrt{1 - \frac{1}{100}^2} = 0.9995c[/tex]

Next, solve for t when t' = 1.
[tex]t = \frac{t'}{\gamma} = \frac{1}{100} sec[/tex]

Finally, solve for x'.
[tex]x' = \gamma(x - vt) = 100(0 - (0.9995c)(\frac{1}{100})) = 0.9995c [/tex]

Have I answered this correctly? Thanks.
 

Answers and Replies

  • #2
176
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The spaceship is defined as the moving object, so proper time is measured on the Earth. Time appears to be going slower for the spaceship. So we could write that:

[tex]t_{earth} = \gamma t_{spaceship}[/tex]

Sam :smile:
 

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