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Special relativity: fireworks exploding

  1. Jun 20, 2014 #1
    A firecracker explodes at the origin of an inertial reference frame. Then, 2.0 microseconds later, a second firecracker explodes 300m away. Astronauts in a passing rocket measure the distance between the explosions to be 200m. According to the astronauts, how much time elapses between the two explosions?
    Okay. My textbook answers this question using spacetime interval consistency which is simple.
    I don't understand why the time dilation formula does not work:
    $$Δt = \frac {Δτ}{ \sqrt{1-\frac{v^2}{c^2}}}$$
    Where v is simply the ratio between 300m and 2 microseconds.
    Δt=2x10^-6 s
  2. jcsd
  3. Jun 20, 2014 #2

    Doc Al

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    Staff: Mentor

    The time dilation formula is a special case of the more general Lorentz transformations. It would apply if the firecrackers exploded at the same place in the 'moving' frame. But here they are 300m apart, so you cannot use it.
  4. Jun 20, 2014 #3
    The easiest and least error-prone way to do this problem is to use the Lorentz Transformation directly.

    Event 1: x=0, t=0, x'=0, t'=0

    Event 2: x = 300, t = 2x10-6, x'=200, t'=?

    So, Δx = 300, Δt = 2x10-6, Δx'=200, Δt'=?

    What does the Lorentz Transformation predict for Δt'.
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