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latentcorpse
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I'd just like someone to clarify that my argument is correct here.
Assume twin A stays on Earth and B goes 2 light years away and then back to Earth at a velocity of 0.8c
From Earth frame of reference:
Twin A makes the calculation that 2/0.8=2.5 years will be needed for the outward journey (5 years in total). However, we know that Twin A sees twin B moving at 0.8c and using his knowledge of Special Relativity, expects B's clock to be running slow (due to time dilation). He calculates this effect using Lorentz transformations to be
[itex]2.5 = \frac{\tau_0}{0.6} \Rightarrow \tau_0 = 0.6 \times 2.5 = 1.5[/itex] years will elapse for B on outward journey (3 years in total).
This means when they are reunited, A will be 5 years old and B only 3. Fair enough!
From spaceship frame of reference:
To begin with we look at the outward part of the journey. B will see the distance to the turnaround point undergo length contraction (since he knows he is moving at 0.8c) and so he must only travel 0.6 x 2 = 1.2 light years
Since B is going at a velocity of 0.8c, he predicts that this will take him 1.2/0.8=1.5 years
Since B must turn his ship around to come back, the return part of the journey will take place in a different inertial frame. However, exactly the same analysis as above tells us that B will see the return journey take another 1.5 years. This means, when B gets back to Earth, he will be 3 years older (in agreement with what A predicted).
What about how B sees A?
Well on the outward part of the journey, A will be moving away from B at a negative velocity - because of this, B predicts A's clock will be running slow. Indeed this is true and in fact, when B turns around after 1.5 years in his own frame, he will see A having only aged by 0.6x1.5=0.9 years.
A symmetric analysis can be used to show that when B is returning to Earth, A will age by another 0.9 years in the eyes of B (who will of course age by 1.5 years).
Therefore, in order for our two arguments to match up, the act of turning the ship around (switching inertial frames for the return journey), must cause A's clock to "jump" 5-2(0.9)=5-1.8=3.2 years.
Hopefully this is correct! I have a few outstanding questions however:
(i) Did I miss anything crucial out?
(ii) My analysis assumes that B is the twin who turns around i.e. changes inertial frames. How do we know this? Surely, from B's frame of reference, he sees A turn around after the midpoint and change frames? Of course, this can't be true otherwise A would be the younger twin and this would contradict the above. So how do we know which twin does the turning around (frame change)? Is this by considering the situation from the perspective of a third observer stationed at infinity?
(iii) From the reading I've been doing on this, I keep coming across the fact that A's clock must "jump". What does this mean? What happens to the missing years? Is this just an artefact of simultaneity not being absolute in Special Relativity? In other words, B's return frame was always chronologically ahead of B's outgoing frame and so when he had to turn around (or switch frames), he was always going to lose some time?
(iv) People keep saying that the acceleration is unimportant in Special Relativity (unless gravitational). I accept this since, the same argument could be applied if B went out to the turnaround point where he instantaneously synchronised his clock with an observer C who was returning to Earth. This wouldn't necessitate a turnaround (i.e. acceleration) but would still yield the same result. Furthermore, the Lorentz factor only depends on v not a and so time dilation and lorentz contraction will be unaffected by accelerations. However, I cannot find any examples of Special Relativity succesfully dealing with acceleration - can anyone provide some?
Thanks a lot for your help. I have put a lot of time into trying to get my head round this successfully so hopefully it has paid off!
Assume twin A stays on Earth and B goes 2 light years away and then back to Earth at a velocity of 0.8c
From Earth frame of reference:
Twin A makes the calculation that 2/0.8=2.5 years will be needed for the outward journey (5 years in total). However, we know that Twin A sees twin B moving at 0.8c and using his knowledge of Special Relativity, expects B's clock to be running slow (due to time dilation). He calculates this effect using Lorentz transformations to be
[itex]2.5 = \frac{\tau_0}{0.6} \Rightarrow \tau_0 = 0.6 \times 2.5 = 1.5[/itex] years will elapse for B on outward journey (3 years in total).
This means when they are reunited, A will be 5 years old and B only 3. Fair enough!
From spaceship frame of reference:
To begin with we look at the outward part of the journey. B will see the distance to the turnaround point undergo length contraction (since he knows he is moving at 0.8c) and so he must only travel 0.6 x 2 = 1.2 light years
Since B is going at a velocity of 0.8c, he predicts that this will take him 1.2/0.8=1.5 years
Since B must turn his ship around to come back, the return part of the journey will take place in a different inertial frame. However, exactly the same analysis as above tells us that B will see the return journey take another 1.5 years. This means, when B gets back to Earth, he will be 3 years older (in agreement with what A predicted).
What about how B sees A?
Well on the outward part of the journey, A will be moving away from B at a negative velocity - because of this, B predicts A's clock will be running slow. Indeed this is true and in fact, when B turns around after 1.5 years in his own frame, he will see A having only aged by 0.6x1.5=0.9 years.
A symmetric analysis can be used to show that when B is returning to Earth, A will age by another 0.9 years in the eyes of B (who will of course age by 1.5 years).
Therefore, in order for our two arguments to match up, the act of turning the ship around (switching inertial frames for the return journey), must cause A's clock to "jump" 5-2(0.9)=5-1.8=3.2 years.
Hopefully this is correct! I have a few outstanding questions however:
(i) Did I miss anything crucial out?
(ii) My analysis assumes that B is the twin who turns around i.e. changes inertial frames. How do we know this? Surely, from B's frame of reference, he sees A turn around after the midpoint and change frames? Of course, this can't be true otherwise A would be the younger twin and this would contradict the above. So how do we know which twin does the turning around (frame change)? Is this by considering the situation from the perspective of a third observer stationed at infinity?
(iii) From the reading I've been doing on this, I keep coming across the fact that A's clock must "jump". What does this mean? What happens to the missing years? Is this just an artefact of simultaneity not being absolute in Special Relativity? In other words, B's return frame was always chronologically ahead of B's outgoing frame and so when he had to turn around (or switch frames), he was always going to lose some time?
(iv) People keep saying that the acceleration is unimportant in Special Relativity (unless gravitational). I accept this since, the same argument could be applied if B went out to the turnaround point where he instantaneously synchronised his clock with an observer C who was returning to Earth. This wouldn't necessitate a turnaround (i.e. acceleration) but would still yield the same result. Furthermore, the Lorentz factor only depends on v not a and so time dilation and lorentz contraction will be unaffected by accelerations. However, I cannot find any examples of Special Relativity succesfully dealing with acceleration - can anyone provide some?
Thanks a lot for your help. I have put a lot of time into trying to get my head round this successfully so hopefully it has paid off!