Special relativity, simultaneity

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SUMMARY

The discussion focuses on a problem involving special relativity and simultaneity, specifically analyzing the time difference between two lightning strikes on a moving train. Using Lorentz transformations, the observer on the train calculates the time separation between the strikes to be approximately -2.4691358 × 10-13 seconds, indicating that the observer sees the front strike first. The calculations are confirmed as correct, demonstrating the effects of relative motion on simultaneity.

PREREQUISITES
  • Understanding of special relativity principles
  • Familiarity with Lorentz transformations
  • Basic knowledge of reference frames
  • Ability to convert units (km/h to m/s)
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  • Study Lorentz transformations in detail
  • Explore the concept of simultaneity in different reference frames
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Students of physics, educators teaching special relativity, and anyone interested in the implications of motion on time perception.

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Homework Statement


A train of 0.8 km (measured by an observer over the train) travels at a speed of 100 km/h. Two lightnings strike simultaneously the back and the front of the train, according to an observer on the ground. What is the time separating both strikes according to the observer on the train?

Homework Equations


Lorentz transformations.

The Attempt at a Solution


Let O be the reference frame of the observer on the ground and O' be the frame of reference of the observer on the train.
They give me the information [tex]x_B'-x_A'=800m[/tex] and [tex]t_B-t_A=0[/tex]. They ask me [tex]t_B'-t_A'[/tex].
Using Lorentz transformations, I get that [tex]t_B'+\frac{vx_B'}{c^2}-t_A'-\frac{vx_A'}{c^2}=0[/tex].
I converted the km/h to m/s and my result is that [tex]t_B'-t_A'=-2.4691358 \times 10 ^{-13}s[/tex]. It means therefore that [tex]t_A'>t_B'[/tex], thus the observer on the train sees first the strike of the lightning on the front of the train and then the one on the back of the train and their time separation is about [tex]2.4691358 \times 10 ^{-13}s[/tex].
I don't know if it's true. To me it seems a too little time, although I realize that the train is only moving at [tex]\frac{1000m}{36s}\approx \frac{30m}{s}[/tex] which is very small compared to c. The order of the strikes seems logical to me...
Can someone confirm/infirm my result?
 
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Your answer is correct.
 
Doc Al said:
Your answer is correct.

Ok thank you, good to know. Problem solved.
 

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