Simultaneity question: light in a moving spaceship

In summary, we are presented with a scenario of a spaceship with a proper length of 100m moving at 0.2c in the +x direction, with light sources at the back and front of the spaceship. The question asks how long it would take for the light to reach the front and back of the spaceship, measured by the frame of the spaceship. The answer is 100m/c for both scenarios, as the light is moving at c in all reference frames. There is no time dilation or length dilation as the light is moving at the same speed as the spaceship.
  • #1
senorhosh
3
0

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.
 
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  • #2
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.
 
  • #3
Doc Al said:
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?
 
  • #4
senorhosh said:
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.)
 
  • #5


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.
 

1. What is the simultaneity question in relation to light in a moving spaceship?

The simultaneity question in this scenario refers to the concept of whether or not two events that occur at the same time in one frame of reference will also occur at the same time in another frame of reference. In the case of light in a moving spaceship, this question arises because the speed of light is constant in all frames of reference according to the theory of relativity.

2. How does the theory of relativity impact the simultaneity question in a moving spaceship?

The theory of relativity states that the speed of light is constant in all frames of reference, meaning that it will always travel at the same speed regardless of the velocity of the source or observer. This means that the concept of simultaneity is relative and can differ between different frames of reference.

3. Can two events in a moving spaceship occur at the same time in one frame of reference but not in another?

Yes, according to the theory of relativity, two events that are simultaneous in one frame of reference may not be simultaneous in another frame of reference. This is because the speed of light is constant and can affect the perception of simultaneity in different frames of reference.

4. How does the speed of the spaceship impact the simultaneity question?

The speed of the spaceship is a crucial factor in the simultaneity question. As the speed of the spaceship increases, the perception of simultaneity between two events may differ between different frames of reference. This is because the speed of light remains constant, but the speed of the observer or source (the moving spaceship) is changing.

5. What implications does the simultaneity question have on our understanding of time?

The simultaneity question challenges our traditional understanding of time as a universal concept. It suggests that the perception of time can differ between different frames of reference and is relative to the observer's speed. This can have significant implications for our understanding of causality and the way we experience events in the universe.

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