1. Dec 7, 2011

### latentcorpse

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

$2.5 = \frac{\tau_0}{0.6} \Rightarrow \tau_0 = 0.6 \times 2.5 = 1.5$ 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!

2. Dec 7, 2011

### BruceW

I get that you've put a lot of time into this, but I don't think you fully understand it yet. I had similar problems when I was learning this stuff. (not so long ago).

I think you need to understand world-lines and the proper time which passes for a clock, and how that depends on its path through space-time, before you can fully understand the 'twin paradox'. Try wikipedia, they give a good explanation.

Some general advice: only use one reference frame for the calculation, don't swap them mid-calculation. Also, remember that only inertial frames of reference are allowed in special relativity, so you cannot consider the frame of reference of the spaceship.

I hope I've been of some help. You could also do a search on physics forums for old threads on the 'twin paradox'.

3. Dec 7, 2011

### latentcorpse

The spaceship is an inertial frame (at least if we assume the necessary accelerations were instantaneous), since when travelling at uniform velocity (in this case 0.8c), Newton's first law will indeed hold.
Also, I'm fairly sure that we do need to swap frames half way through when considering the spaceship perspective. The reason being that otherwise, we include the non-inertial part of the motion i.e. the turning round.

4. Dec 7, 2011

### BruceW

I looked on wikipedia, in hope of making a better explanation. And it says that it would be possible to have one ship on the outward journey, who gives the information of its clock time to an earth-bound ship, and that earth-bound ship keeps counting on from there until it gets back to earth.

So I guess it is possible to explain the 'twin paradox' by using a 'frame swap' method, and with zero acceleration.

It looks more complicated than an argument about the proper time of a clock which undergoes acceleration, though.

5. Dec 7, 2011

### D H

Staff Emeritus
The Lorentz factor $\gamma = 1/\sqrt{1-(v/c)^2}$ is a function of velocity only; acceleration is not a part of the equation. This does not mean that acceleration is unimportant. If an object is accelerating, the velocity changes over time, so that means that gamma also changes over time.

The standard presentations of the twin paradox use an instantaneous change in velocity at the turnaround point because doing so makes the math come out easier. One could also use a finite acceleration, but the math would be a bit messier.

This is a widely promulgated misconception. It isn't true. The only thing that special relativity can't handle is gravitation. The relativistic rocket equations are an example of where acceleration is used in special relativity. There's nothing wrong with thinking in terms of the rest frame of the accelerating rocket.

6. Dec 7, 2011

### vela

Staff Emeritus
While B is turning around, he will, in principle, be able to detect fictitious forces in his frame. In contrast, A won't see any such forces. So it's clear that B is the one who accelerated, not A.
Wikipedia's article on the twin paradox has a good explanation on the resolution using the Minkowski-space viewpoint. It might help you to think of B turning around smoothly, rather than discontinuously. B's plane of simultaneity, rather than switching instantaneously and causing A's age to jump, will instead sweep quickly through a long portion of A's world line. From B's perspective, he sees A aging a lot during a short time for himself.

7. Dec 7, 2011

### latentcorpse

Excellent! Thanks to both of you.

Can you confirm that I didn't miss anything from my original argument?

And also, B can detect fictitious forces because he will get pressed up against the side of his rocket (let's say he's pushed to the right of the rocket) when he turns his spaceship around (assuming he was initially going right and turns it round to face left).
(i) These forces are fictitious because someone in B's frame of reference would need to account for them in order to see B remain stationary. Right?
(ii) Surely, from B's frame of reference, he will see objects on Earth pushed to the left when the Earth "turns around" and comes back to him - won't he? i.e. I can't picture why the Earth doesn't have fictitious forces from B's frame!!!

Thank you

8. Dec 7, 2011

### vela

Staff Emeritus
B has to introduce the fictitious forces in order to maintain the illusion that Newton's laws still hold for him. When the rocket turns, the wall presses against him. If there were no force to counter that force and F=ma still held, he'd accelerate away from the wall. So either he has to discard the validity of F=ma or he has to posit there's a force pushing him into the wall, the centrifugal force, so that the net force on him and thus his acceleration remain 0.

Similarly, when B sees the Earth changing direction, he has to assume there's a force on Earth causing it to accelerate or he has to discard Newton's laws.

An inertial observer seeing B would instead say the rockets exert a force on the ship, causing it to turn. The wall of the ship then pushes against B to cause B to turn along with the ship. Similarly, to the observer, the Earth doesn't change direction, so there's no force needed there either. The inertial observer doesn't see any fictitious forces. Only B experiences them.

Last edited: Dec 7, 2011
9. Dec 8, 2011

### D H

Staff Emeritus
You meant real forces, not fictitious. Fictitious forces aren't detectable (locally measurable). They're fictitious. Accelerometers measure acceleration due to real forces.

To be a true experiment of special relativity, it is probably better to have the journey start from a space station well outside a gravitating body rather than from the Earth. The traveler departs from the space station, quickly accelerating to near c relative to the space station. The traveler then coasts to the target star, accelerates briefly to turn around at the target star, and accelerates one last time to once on approach to the space station. Now the use of an accelerometer to test whether the rest frame is inertial works quite nicely. Per those accelerometers, the space station doesn't accelerate but the spaceship does.

Note that one can get away with using the Earth as the base because gravitational time dilation on the surface of the Earth is very, very small.

B does not "see" A aging a lot. That rapid change in age as calculated using planes of simultaneity is an unobservable effect. The observed effect is that B "sees" A change from aging slowly to aging quickly. In the example at hand, a traveler changing from moving at 0.8c away from A to moving at 0.8c toward A, B sees A change from aging at 1/3 the rate at which B is aging to aging at 3 times the rate at which B is aging. B does not "see" a sudden jump in A's age; she just sees a change in rate.

10. Dec 8, 2011

### BruceW

I thought that in special relativity, you have to use an inertial frame of reference to do the calculation. Of course, you can calculate what happens to an accelerating spaceship, but you must do this using an inertial frame of reference. Was I wrong?