Relativity of simultaneity (astronaut and simultaneous explosions.)

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SUMMARY

The discussion centers on a physics problem involving an astronaut traveling at 0.17c between two planets and observing simultaneous nuclear explosions. The astronaut calculates the time difference in the arrival of light from the explosions using the Lorentz transformation equation, specifically t = (gamma)(t' + (vx'/c^2)). The calculated gamma factor is approximately 1.01477, leading to a time difference of about 20.687 minutes. Participants suggest simplifying the problem by rewriting it in the astronaut's frame and recommend using a space-time diagram for clarity.

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  • Understanding of special relativity concepts, particularly simultaneity.
  • Familiarity with Lorentz transformations and the gamma factor.
  • Basic knowledge of light speed and its implications in relativistic physics.
  • Ability to interpret space-time diagrams.
NEXT STEPS
  • Study the derivation and application of the Lorentz transformation equations.
  • Learn how to construct and interpret space-time diagrams in special relativity.
  • Explore the implications of simultaneity in different inertial frames.
  • Investigate the effects of relativistic speeds on time dilation and length contraction.
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Students of physics, particularly those studying special relativity, educators teaching these concepts, and anyone interested in the implications of relativistic motion on simultaneity.

Gator93
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Homework Statement



An astronaut is space-traveling from planet X to planet Y at a speed of 0.17c. When he is precisely halfway between the planets, a distance of 1 light-hour from each, nuclear devices are detonated. The explosions are simultaneous in the astronaut's frame. What is the difference in time of arrival of the flashes from the explosions as observed by the astronaut?

v = 0.17c = 51000000 m/s
x = 2(1.079E12)
t' = 0 is simultaneous?
t = ?

Homework Equations



t = (gamma)(t' + (vx'/c^2))

gamma = 1/(sqrt (1-B^2))

The Attempt at a Solution



1/(sqrt (1-0.17^2))
1/sqrt(0.9711)
= 1.014770943

(1.014770943)(0 + (51000000*2.158505698E12)/(9E16)
= 20.687 min?

Now I'm not seeing how the units follow through on this without taking it from sec to min, but I've arrived at this answer a few other ways... namely multple equations and setting them equal to see when the first flash would encounter the ship and also when the second flash would catch up to the ship, and taking the difference.

So, I was giving this as a multiple choice problem.
A) 14 min B) 0 min C) 56 min D) 28 min

There is not an E) other... but wouldn't put it past him... then again I'm not confident I'm doing it correctly.
 
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Hi Gator93! Welcome to PF! :smile:
Gator93 said:
An astronaut is space-traveling from planet X to planet Y at a speed of 0.17c. When he is precisely halfway between the planets, a distance of 1 light-hour from each, nuclear devices are detonated. The explosions are simultaneous in the astronaut's frame. What is the difference in time of arrival of the flashes from the explosions as observed by the astronaut?

You're making this too complicated :redface:

first rewrite the question in the astronaut's frame …

what does it say? :smile:
 
You might try drawing a space-time diagram...
 

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