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System S’ has velocity +0.45c relative to S. The clocks of S and S’ are synchronized at t=t’=0s when origins O and O’ coincide. An event is observed in both systems. The event takes place at x=600m and at a time t=1.9µs, as measured by an observer in S. Calculate the x’-coordinate of the event and the time t’ of the event, measured by an observer in S’ (15pts).

Answer :

Answer:

  Δt ’= 2,128 10⁻⁶ s   and Δx ’= 535.8 m

Explanation:

In modern physics the speed in vacuum is constant measured in any inertial system, so Newton's formulation has two assumptions

* the laws of physics are the same in all inertial systems

* the speed of light is constant in all systems

Based on this Einstein wrote the relationships for the change of position and time in two systems that move relative to each other.

      Δt ’ = γ Δt

      Δx ’= ΔL / γ

     γ = 1 /√(1- v²/c²)

where the system without premium is fixed with respect to the event and the prima system is easy.

For our case

the speed between the two systems is   v = 0.45c

The position in the fixed system is            x = 600 m

The time in the fixed system is                  t = 1.9 μs = 1.9 10⁻⁶ s

let's use the relationships to find the position and time in the mobile system

               γ  = 1 /√(1 - (0.45c / c)²)

               γ  = 1,119785

we calculate time o and position

            Δt ’= 1,119785 1.9 10⁻⁶

            Δt ’= 2,128 10⁻⁶ s

            Δx ’= 600 / 1,119785

            Δx ’= 535.8 m

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