- #1

PASta95

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- Homework Statement
- A clock is built using two mirrors (A and B) mounted horizontally (x-direction), between which a pulse of light is repeatedly reﬂected. The proper length between the two mirrors is 5.00 m. The spacecraft’s velocity (0.60c relative to our galaxy; γ = 1.25) is in the positive x direction.

1. According to an observer in the frame of our galaxy, what is the distance between the mirrors?

2. One period of this clock is the time taken for light to travel from A to B and back to A. According to an observer in the frame of our galaxy, what is the period of the clock?

- Relevant Equations
- t=t0*γ (time dilation)

L0=L*γ (length contraction)

1. I'm fine with this one, I simply calculated L = L0/γ = 5/1.2 5= 4m.

2. This is the one I'm having problems with.

My approach was to say that both observers would measure the light beam to have velocity of ‘c’. Therefore, if 4m is the distance between the mirrors as observed in the frame of the galaxy, then the light beam traverses a distance of L=8m in one period, and the period T as observed in the frame of the galaxy should be T=L/c=8/(3E8)=26ns.

However the correct answer is 42 ns. This was determined by calculating the proper time as observed by someone in the spacecraft as t0 = 2L/c = 10/(3E8), then using the time dilation formula to calculate the dilated time, i.e. t=t0*gamma = (3.33E-8)*1.25 = 42ns.

So i) I don’t understand why my answer is incorrect, and ii) I also don’t understand how 42ns could be correct when, for L=8m, the observer would measure v = 8/(42E-9) = 2E8 m/s (which doesn’t equal c).

Appreciate any assistance you could provide, thank you.

2. This is the one I'm having problems with.

My approach was to say that both observers would measure the light beam to have velocity of ‘c’. Therefore, if 4m is the distance between the mirrors as observed in the frame of the galaxy, then the light beam traverses a distance of L=8m in one period, and the period T as observed in the frame of the galaxy should be T=L/c=8/(3E8)=26ns.

However the correct answer is 42 ns. This was determined by calculating the proper time as observed by someone in the spacecraft as t0 = 2L/c = 10/(3E8), then using the time dilation formula to calculate the dilated time, i.e. t=t0*gamma = (3.33E-8)*1.25 = 42ns.

So i) I don’t understand why my answer is incorrect, and ii) I also don’t understand how 42ns could be correct when, for L=8m, the observer would measure v = 8/(42E-9) = 2E8 m/s (which doesn’t equal c).

Appreciate any assistance you could provide, thank you.