Could somebody help me check the solution please. (special relativity)

Click For Summary

Homework Help Overview

The discussion revolves around a problem in special relativity involving a rocket traveling at a significant fraction of the speed of light and the behavior of light pulses emitted from it. The original poster seeks to determine the time taken for a light pulse to travel from the nose to the tail of the rocket and back, as measured in different reference frames.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of time dilation and the conditions under which it can be used. Questions arise regarding the differences in measurements from the rocket's frame versus the observer's frame, particularly when multiple clocks are involved.

Discussion Status

Some participants have provided guidance on the use of time dilation and the Lorentz transformation, noting the complexities introduced when measuring time intervals involving different locations. The conversation reflects a productive exploration of the concepts without reaching a definitive consensus on all aspects.

Contextual Notes

The discussion highlights the importance of understanding the implications of special relativity, particularly regarding time measurements in different frames of reference and the role of length contraction.

phalanx123
Messages
29
Reaction score
0
A rocket of length l0 measured in its rest frame S0 is traveling away from an observer in a frame S with a velocity u = 2/3 c. A light pulse is emitted from the nose of the rocket (x' = l0) at t0 = 0 and travels to the tail (x' = 0) where it is reflected back to the nose.

In S0, when does the light pulse reach the tail and when does it get back to the nose?

How long does the light pulse take to travel from nose to tail and back again
as determined by the observer in S?



Here is my solution

the time the pulse takes to travel from nose to tail is l0/c.
The time when it get back to the nose is t=2*l0/c.

For the second part, the total time it takes is gamma*2*l0/c where gamma=1/(1-v^2/c^2)^1/2 and v=2/3c

I think my solution is too simple, have I misunderstood the question:confused: ? Could somebody help me check it please. Thanks
 
Physics news on Phys.org
phalanx123 said:
the time the pulse takes to travel from nose to tail is l0/c.
The time when it get back to the nose is t=2*l0/c.

Correct.

For the second part, the total time it takes is gamma*2*l0/c where gamma=1/(1-v^2/c^2)^1/2 and v=2/3c

Also correct, but do you understand what allows you to apply the simple time dilation formula? Since, in the rocket frame, a single clock is used to measure the round-trip time of the light pulse, you are justified in using the time dilation formula (which describes the behavior of a single moving clock).

What if they asked for the time the pulse takes to get from nose to tail according to observers in S?
 
Doc Al said:
What if they asked for the time the pulse takes to get from nose to tail according to observers in S?

Would that not be gamma*l0/c where gamma is the same as defined above?
 
phalanx123 said:
Would that not be gamma*l0/c where gamma is the same as defined above?


No, it wouldn't. That case is actually a bit more complicated to answer since two clocks are involved in the rocket frame, one at the nose and another at the tail. So you can't just apply the time dilation formula to that time interval, since no single clock directly measures it.

You can figure that one out by considering what distance the light must travel according to the S frame observers. (You'll need to take into account both length contraction and the motion of the rocket during the light's journey from nose to tail.)
 
Oh I see. so if the signal returned to the nose, only one clock is involved, so the time dilation can be used. but when measure the time interval between the emmission and reflection at the back, the events takes place at diffenrent places, so two clocks were involved, in that case the Lorentz transformation has to be used. Is my explanation/logic right?
 
Exactly right. (You could figure it out just using the known behavior of light, clocks, and measuring sticks... but that's equivalent to using the LT.)
 
Ok thank you very much. That cleared the SR up a little for me ^_^
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
13
Views
5K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 18 ·
Replies
18
Views
2K