Calculating Time and Distance in Lorentz Transforms: A Homework Example

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SUMMARY

The discussion focuses on calculating time and distance using Lorentz transforms in the context of special relativity. The problem involves a spaceship moving at a velocity of 0.9995c, where the captain measures a clock tick of 1.0 second. The calculations confirm that the time measured by stationary observers on Earth is t = 0.01 seconds, and the position at which the second tick occurs is x' = 0.9995c. The proper time on Earth is expressed as t_{earth} = γ t_{spaceship}, establishing the relationship between the two frames of reference.

PREREQUISITES
  • Understanding of Lorentz transformations
  • Familiarity with the concept of proper time
  • Basic knowledge of special relativity
  • Ability to perform calculations involving gamma (γ) factor
NEXT STEPS
  • Study the derivation of Lorentz transformations in detail
  • Learn about the implications of time dilation in special relativity
  • Explore the concept of simultaneity in different reference frames
  • Investigate practical applications of Lorentz transforms in modern physics
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Students of physics, educators teaching special relativity, and anyone interested in understanding the implications of time dilation and Lorentz transformations in high-speed scenarios.

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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.
 
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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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