# A variation of Twin paradox

1. Dec 26, 2011

### Snip3r

If in the twin paradox instead of the twin travelling to the distant star lets say he stays and the earth along with that star moves (imagining a rod joining earth and star moves) and the star reaching this twin goes back at the same speed(along with earth) now i guess the twin whos stationary age lesser. correct?

2. Dec 26, 2011

### ghwellsjr

No, the twin who traveled with the earth would age less than the stationary twin who was reached by the star.

3. Dec 26, 2011

### Snip3r

yeah probably i got it wrong. Now lets do some numbers star is 10 lyrs away, acceleration is instantaneous for the rod. Now immediately after the acceleration stationary twin would measure 5 lyrs to star isn't it?so total trip would take 10 yrs for him. Now what happens in the frame of rod?

4. Dec 26, 2011

### ghwellsjr

Just like we can't talk about the frame of the traveling twin in the normal twin paradox because he does not remain inertial due to his changing speed, we can't talk about the frame of the rod in your modified scenario because it changes speed.

5. Dec 26, 2011

### Snip3r

Why cant we talk about travelling twin in the normal paradox? cant we say star is coming at 0.86c and is 5 lyrs away so it would take 10/√3 yrs for the trip forward same for back hence a total of 20/√3 yrs for ship but for earth its 20/√3 + 20/√3 hence 40/√3 yrs. In a similar way can you explain what happens here from both frames of refernces?

6. Dec 27, 2011

### ghwellsjr

There are three frames of reference to consider: one for the stationary twin, one for the traveling twin during the outbound half of the trip and one for the traveling twin during the inbound half of the trip. When you ask about both frame of reference, I don't know which two of the three you are referring to. Keep in mind that any one of the three FoRs covers the entire scenario from start to finish. It's just that the traveling twin is never at rest in any FoR for the entire scenario.

But in any case, having a 10 ly long rod connecting earth to a star that instantly accelerates only complicates the analysis and in an unnecessary way. All that matters is that you have one twin that does not accelerate and one that does. And you can easily analyze the scenario from the FoR for the non-accelerating twin, as you just did, or you can more complicatedly analyze the entire scenario from the FoR in which the accelerating twin is at rest during the outbound half of the trip, or you can just as complicatedly analyze the entire scenario from the FoR in which the accelerating twin is a rest during the inbound half of the trip, or you can pick any other FoR. All will give the same conclusion that you got in your analysis that the traveling twin aged by 20/√3 years while the stationary twin aged by 40/√3 years.

7. Dec 27, 2011

### Snip3r

yes i accept but i am still not having the feel of SR so i am just coming up with random questions to test my understanding. In this rod example i thought as soon as it accelerates to 0.86c it shrinks to 5 ly to stationary observer so to and fro of the trip for the stationary observer would be 20/√3 years but you seem to say it is 40/√3 years. Just when i thought i understood normal paradox and came up with some random question i am unable to solve it :(

8. Dec 27, 2011

### ghwellsjr

I was ignoring the 10 ly long rod because, as I say, it complicates things. I was treating the problem in the following way:

There are two twins on earth. There is a star 10 ly away from earth. In a frame in which all four of these objects are at rest, suddenly all of them except one of the twins instantly accelerate to .86c in the direction of the star towards the earth. When the star reaches the stationary twin, the other three objects instantly accelerate to .86c in the opposite direction until the earth and the twin on it reach the stationary twin at which point the other three objects decelerate and come to rest. Note that we are defining everything in a single FoR. How much have the two twins aged?

You will note that since the earth and the star have identical accelerations, they remain 10 ly apart in the single FoR that we are using. We note that the star travels at .86c for 10 ly and so it takes it 11.6 years to reach the stationary twin. During this time the earth and the twin on it also have traveled at .86c for 11.6 years. We can calculate how much this twin aged by dividing the coordinate time of 11.6 years by gamma (which calculates to 2) so the earth twin has aged by 5.8 years. Now the earth/twin and star change directions and travel back to their original locations at the same speed, so the times and agings are the same again for all objects on the return trip. When they come to rest in their original locations, the earth twin has aged 11.6 years while the stationary twin has aged 23.2 years. You expressed these numbers as 20/√3 years and 40/√3 years.

Now if you want to put a rod between the earth and the star and say that it accelerates instantly, you create a new problem because you haven't specified how it accelerates. If you are thinking that every point along its length accelerates identically (what else could instant acceleration mean?) then it doesn't change anything in my previous analysis which means the rod has to physically stretch so that it remains with a length of 10 ly.

But you were thinking that it will experience length contraction but if it does, then you have to define what you mean by instant acceleration. Do you want the end connected to the star to start accelerating toward the earth and to push on the rod so that it will compress and end up with a length of 5 ly? Well, if you do, then you should be aware that it will take at least 10 years for this propagation to take effect and the earth and its twin will not even start moving until just before it's time for the star end of the rod to change directions and go back to its original position and length.

I hope you can see the problem. If you still want to have the rod shrink to half its original length, then you have to tell me precisely how you want to have it instantly accelerate.

9. Dec 27, 2011

### Snip3r

this is exactly how i thought but i dint know the later part that it has to physically stretch can you tell me why?

10. Dec 28, 2011

### ghwellsjr

Whether or not it stretches is up to you and how you say you want it to accelerate. If you take away the stipulation that it has to accelerate instantly and you allow many years for the process to take place so that normal length contraction occurs, then in the rest frame of the twin that does not accelerate, it will contract to half its original length. Of course in the rest frame of the twin that accelerated along with the earth-rod-star, it will appear to be the same 10 ly length but it will take tens of years for that twin to actually measure the length of the rod which messes up your original scenario because you had it turn around only 11.6 years.

Since we are doing a thought experiment in Special Relativity, we are ignoring the effects of gravity but in actuality, it is not possible to have a rod anywhere near 10 ly long connecting two massive bodies together so that it would be like some gigantic dumbbell. Things that large always reshape themselves into spheres.

So when you ask about what physically happens, since we are doing a thought experiment where we ignore certain physics, it's really up to your thoughts how you want it to happen.

11. Dec 28, 2011

### Snip3r

i accept but what i want to know is positions of a rod after attaining a velocity. Consider this simple scenario take a 1 feet rod along x-axis now left end is at [0,0](x,t)and right end is at [1,0] now i assume the rod instantaneously accelerates to 0.86c (i dont know if this is possible but i assume since the length is small please notify me otherwise)now just after the time Δt in the rest frame i try to mark the positions of the rod. What will the co-ordinates be?will it be [0.5,Δt] and [1,Δt]?

12. Dec 28, 2011

### Tea Jay

It will be very far away by then...or it didn't accelerate.

13. Dec 28, 2011

### Snip3r

yes but i was talking about the case Δt→0.

14. Dec 28, 2011

### ghwellsjr

It doesn't matter if the rod is one foot long or 10 light years long, if you're going to accelerate it, at any rate, you have to say how it is going to accelerate. For example, you could say that the right end is where the acceleration is applied and then some time after the final velocity has been achieved and the left end has had a chance to stabilize, it will have moved closer to the right end, according to the original rest frame.

15. Dec 28, 2011

### Snip3r

i assume every particle in the rod is accelerated instantaneously(acceleration is applied to all the particles simultaneously) . Now at t=1 ns if i mark the positions of the rod what will it be?

16. Dec 29, 2011

### ghwellsjr

You said the coordinates ([x,t] with x in feet and t in nanoseconds, assuming the speed of light is 1 foot per nanosecond) at the moment of acceleration for the left end is [0,0] and for the right end is [1,0] and that the speed instantaneously became 0.86c so at t=1 ns, the coordinates will be [0.86,1] for the left end and [1.86,1] for the right end.

17. Dec 29, 2011

### Snip3r

what about length contraction shouldn't the length be 0.5 f?

18. Dec 29, 2011

### ghwellsjr

You said, "every particle in the rod is accelerated instantaneously(acceleration is applied to all the particles simultaneously)" which means it has to stretch. If you want the length to contract, then you need to accelerate just one part of the rod--the left end, the right end, the middle, take your pick.

19. Dec 29, 2011

### Snip3r

ok let me make it this time clear
a)when every particle accelerates instantaneously it will stretch why/how?
b)when normal acceleration is allowed it will shrink why/how?

20. Dec 29, 2011

### ghwellsjr

It will stretch because there are forces applied all along the length of the rod. This stretching will be measurable in the final rest frame of the rod (after acceleration). It is not a result of the acceleration being large and instantaneous but will happen no matter how small or how long the forces of acceleration are applied as long as they are applied simultaneously along the entire length of the rod.
It will contract (according to its original rest frame) because the force is applied at only one place along the length of the rod and normal length contraction is allowed to occur. This contraction is not measurable in the final rest frame of the rod (after acceleration).

Special Relativity does not explain the mechanism of length contraction nor is it concerned with how it happens. It is a necessary result of the principle of relativity.

21. Dec 30, 2011

### Snip3r

why should forces applied all along the length lead to stretching?

22. Dec 30, 2011

### ghwellsjr

Well, if you forced one end of the rod to follow a particular acceleration, which means a particular velocity versus time, which means a particular position profile versus time and left the rest of the rod free and it contracted (according to the original rest frame), then don't you think that if you forced the other end of the rod to follow exactly the same acceleration, velocity and position versus time, then it would maintain the same distance apart from the first position (according to the original rest frame)?

23. Dec 30, 2011

### Snip3r

by this do you mean length contraction as lessening in distance between the particles of the rod or decrease in volume?

24. Dec 30, 2011

### ghwellsjr

What's the difference? Why wouldn't it be both? (In the original rest frame, of course.)

25. Dec 30, 2011

### Snip3r

yes they are the same i just re-framed the question by saying decrease in volume
if i get you correct you will accept this(of course this statement is very informal):
When 2 particles A and B form an object the object can shrink only as long as A and B dont touch each other isn't it?