Simultaneity question: light in a moving spaceship

[senorhosh]

Homework Statement

Spaceship with proper length 100m is moving at 0.2c in +x direction. In the back and front of the spaceship is a light source.
a. When the light in the back is turned on, how long will it take for the light to reach the front of the spaceship?
b. How about the light in the front? (to reach the back)

Homework Equations

The Attempt at a Solution

a. I'm guessing because light always moves at c in all reference frames, t = d/c or 100/c? There would be no time dilation or length dilation because light is moving the same frame as the space ship
b. Same answer as a. for the same reasons.

For some reason, I don't think this is correct. I'm not sure why... it just seems off..
Any help? thanks.

 

[Doc Al]

I suspect they want the time it takes for the light to travel according to the frame that views the rocket as moving at 0.2c.

 

[senorhosh]

I suspect they want the time it takes for the light to travel according to the frame that views the rocket as moving at 0.2c.

Whoops I forgot to add this in.
"Measured by the frame of the spaceship". So no..
sorry about that. Anyway would my answer be correct then?

 

[Doc Al]

Whoops I forgot to add this in.
"Measured by the frame of the spaceship". So no..
sorry about that. Anyway would my answer be correct then?

Yes, in that case your answers are correct: the time of travel is simply 100m/c in each direction. (From the frame of the rocket, the rocket is at rest.)

 

[paranormal]

Your intuition is correct. The answer to both parts of the question is not simply d/c, where d is the distance between the light source and the front/back of the spaceship. This is because of the phenomenon of time dilation and length contraction in special relativity.

In this scenario, we can use the Lorentz transformation equations to calculate the time and distance in the reference frame of the spaceship. The equations are:

t' = γ(t - vx/c^2)
x' = γ(x - vt)

Where t' and x' are the time and distance measured in the reference frame of the spaceship, t and x are the time and distance measured in the stationary reference frame, v is the velocity of the spaceship (0.2c in this case), and γ is the Lorentz factor given by γ = 1/√(1-v^2/c^2).

a. When the light in the back is turned on, it will take a time of t' = γ(100/c - 0.2*100/c) = 0.98*100/c for the light to reach the front of the spaceship. This is slightly longer than the time it would take in the stationary reference frame, which would be 100/c.

b. Similarly, when the light in the front is turned on, it will take a time of t' = γ(100/c + 0.2*100/c) = 1.02*100/c for the light to reach the back of the spaceship. This is slightly shorter than the time it would take in the stationary reference frame, which would be 100/c.

The reason for this difference is that in the reference frame of the spaceship, the distance between the light source and the front/back of the spaceship is contracted due to length contraction. This means that the light has a shorter distance to travel, but the speed of light is still c in this reference frame, so it takes slightly less time to reach the end of the spaceship.

I hope this helps clarify your understanding. If you have any further questions, don't hesitate to ask.