# Ambiguity of time dilation/twin paradox

1. Mar 24, 2013

### ShreyasR

The first thing we learn in relativity is that what we measure is relative motion, i.e., with respect to a particular frame of reference. The values of position, velocity, acceleration differs with the frame of reference chosen.
When we go on, we come across time dilation. The theory mentions that "a moving clock(1) tics slower than a clock(2) at rest". The terms "moving" and "rest" actually creates a lot of ambiguity.
Even if it is mentioned that this is explained with respect to the frame of reference in which the clock(2) is fixed at the origin of a coordinate system taken as the reference, and clock(1) is moving, clock(1) ticks slower than clock(2).

Using this, the "Twin paradox" is explained.

"Consider a pair of twins "A" and "B"(age 20). Let "B" blast off in a spaceship on Jan 1st 2013, moving at about 2.8x10^8 m/s, close to the speed of light, and return to earth on Jan 1st 2023. The observations of "A" are:
"B" is in the spaceship.
"B" and spaceship are moving away at 2.8x10^8 m/s.
"B"'s clock ticks slower.
After "B" returns to the earth, "A" would be physically 30 years old, and "B" would physically be about 21 years. In other words, To "A", it seems that "B" is 9 years younger. In this situation, obviously, it should appear to "B" that "A" is 9 years older than "B".

If the same theory is explained with a frame of reference in which the spaceship of "B" is taken as the origin (B is at rest), the observations are:
"A" is on the earth.
"A" and earth are moving away at 2.8x10^8 m/s.
"A"'s clock is ticking slower,
And hence, after "B" returns to the earth, "B" would be physically 30 years old, and "A" would physically be about 21 years. In other words, To "B", it must seem that "A" is 9 years younger. In this situation, obviously, it should appear to "A" that "B" is 9 years older than "B".

The theoretical results obtained from the first case is exactly opposite to that of the second case.

Now my question is, as the theory of relativity itself is quite ambiguous, have I not understood it properly, or is there something wrong with this twin paradox?

If there is something wrong with my explanation, please correct it line by line.

2. Mar 24, 2013

### tensor33

3. Mar 24, 2013

### Ibix

That is your mistake. There is no inertial reference frame in which B is at rest for the whole trip. There is one where he's at rest outbound, but he's moving on the return leg. There is one where he's at rest returning, but he's moving on the way out. A is moving in both.

4. Mar 24, 2013

### 1977ub

Experimentally, B will find after he turns around, if he is using radar to generate his view of the world, that suddenly it is much later on the Earth than before he turned around.

"In special relativity there is no concept of absolute present. The present from the point of view of a given observer is defined as the set of events that are simultaneous for that observer. The notion of simultaneity depends on the frame of reference, so switching between frames requires an adjustment in the definition of the present... In a sense, during the U-turn the plane of simultaneity... very quickly sweeps over a large segment of the world line of the Earth-based twin. The traveling twin reckons that there has been a jump discontinuity in the age of the Earth-based twin." - http://en.wikipedia.org/wiki/Twin_paradox

5. Mar 24, 2013

### Staff: Mentor

Not quite. This paper describes what he will see if he is using radar. There is no "jump" in time using radar.

http://arxiv.org/abs/gr-qc/0104077

6. Mar 24, 2013

### 1977ub

For any given IRF, I was under the impression that radar yields the same results as Einstein method. I'll read on.

[2:54] There still appears to be a jump for B's view of A, IRF-wise. There is a discontinuity between the before-turn IRF and the after-turn IRF. This method appears to solve "over-determination" where B ends up with multiple coordinates for some events, but that isn't our problem, right? A certain time range *disappears* on Earth when for B's IRF when he turns suddenly around. That isn't still true here? Then it sounds like radar yields something different than Einstein...

Last edited: Mar 24, 2013
7. Mar 24, 2013

### ShreyasR

How is that a mistake...? can you be a bit more clear please? Ok now if i consider B is observing the earth, holding his hand in front of him, and marking the tip of his index finger as the origin, Wont he note that the earth is moving away at 2.8x10^8 m/s? if yes, according to the equation of time dilation, he must notice that every event on the earth is taking place slower than the events on his spaceship, isn't it? in this case B is aging faster than A... Please clarify this point?!

8. Mar 24, 2013

### 1977ub

To "observe" events on earth is to receive light signals from earth and then decide *when* they were sent. Any time you turn your ship in a different direction or change your speed that is an "acceleration" and your whole perspective regarding when events took place and which events are simultaneous shifts. When you then perform calculations about  how much time has passed on Earth, you will end up with different answers than you did before your acceleration.

Last edited: Mar 24, 2013
9. Mar 24, 2013

### ShreyasR

Yes i agree to your point. But when when you divide the "main" event, that is the trip of "B", into many intervals/parts... That is,
1) Leaving the earth by spaceship with an initial acceleration to reach a speed of 2.8x10^8 m/s (wrt earth): time taken = t1,
2) Once the speed is reached, the spaceship moves with a constant speed of 2.8x10^8 m/s for a time interval t2.
3) after that the spaceship decelerates to rest in time t3,
4) The spaceship now accelerates back to 2.8x10^8 m/s in the reverse direction in time t3,
5) the spaceship moves with the constant speed of 2.8x10^8 m/s for time t2,
6) and decelerates in time t1 to land back on earth.
(all time measured by B)

total time spent = 2(t1+t2+t3)

Can you give me a mathematical explanation about the time in each interval measured by A and the total time.
Please explain it like in a tabular column, like in the Left side column, every interval,what A sees, what time he measures, and on the right hand side of the column, what B sees and measures... (According to the concept of time dilation, time measured by B should not be equal to the time measured by A) But i feel if the problem is mathematically illustrated in the form of a tabular column, total time measured by both will be the same... Please try it and reply?!

10. Mar 24, 2013

### ShreyasR

Also, u can include a 3rd person C, who moves such that A and B are equidistant from him, So he'll measure the speeds and accelerations of A and B to be exactly the same wrt himself (but in opposite directions), throughout the whole trip. This should mean that The calculations of C should result in A and B being the same age after the trip isn't it?

11. Mar 24, 2013

### 1977ub

Person C might be able to position himself so that he can calculate accelerations in his coordinate system that are the same for A & B. However the gorilla in the room is the "source of inertia" which is simply not understood at the current time. Real world measurements will find that A will experience no accelerations and so all of his measurements make sense in an inertial frame. B *will* experience them and so this will throw off the measurements of simultaneity which he might attempt to make. No assignment of coordinates by C to A will cause A to *experience* acceleration, and it is this - the change between different inertial frames - which leads to real differences in elapsed time once B comes back to A. http://www.calphysics.org/haisch/science.html

12. Mar 24, 2013

### 1977ub

Let's see.

It's easier if you consider the acceleration period to be negligibly rapid and short in duration. And why not.

B leaves point X.

1) acceleration = negligible.

2) once B moving, the landscape which is at rest to X appears contracted to him, and so he feels he arrives at point Y much more quickly than he anticipated. meanwhile it appears to take the expected amount of time for for B to reach Y as measured by observer A at point X, who does however find B's ship to be contracted.

3) deceleration = negligible.

4) negligible as #1

5) as #2, landscape is contracted, trip takes less time than expected.

6) negligible.

So I reckon B's time to be 2*t2 shrunken by the lorenz factor.

So what accounts for the difference? B "Changed inertial frames" aka "accelerated," unlike A.

There are multiple descriptions elsewhere in this site and also here.

I honestly think people become so distracted that each observer has their own IRF and is entitled to declare the other person's clock as moving slowly that they become distracted by this effect and lose sight of the fact that it is something related to the change in inertial frame which gives rise to the objectively agreeable difference in clock rates.

13. Mar 24, 2013

### ShreyasR

But then isn't earth itself supposed to be a non inertial frame when it comes to observing B? I mean even the earth is rotating and revolving around the sun so we must consider the spaceship of B also revolving around the sun, and moving radially outwards away from the line joining the earth and sun. So all 3 frames A,B,C are revolving around the sun, and hence my argument is that All three frames are non inertial...

and this quote: "B *will* experience them and so this will throw off the measurements of simultaneity which he might attempt to make. No assignment of coordinates by C to A will cause A to *experience* acceleration" contradicts the statement "ALL MOTION IS RELATIVE" doesn't it?

14. Mar 24, 2013

### 1977ub

Generally for these SR discussions we find the Earth's non-inertial aspects to be negligible. Think of someone sitting on the rock in the middle of space and calling it "Earth". Just as you can't draw a perfect circle, no observer is in a perfectly inertial frame. To isolate the effects of SR, you draw up scenarios where other effects can be disregarded as very small.

All motion is relative, but not all acceleration is. That's what you might not realize until you get a little more up to your elbows in this stuff. Any why? Nobody really knows. It takes empirical observation to determine who is accelerating and who is not.

15. Mar 24, 2013

### ShreyasR

OH! The last point has caught my attention... The thing is, the second time i read the basics of general relativity, i read that experimental results have shown that the length of a moving plane is found to be shorter than the plane when it was at rest. this proves that the theory of relativity is valid. Personally, i partially don't believe in this theory. I immediately felt that the difference/error found in the measurement of the length of the plane must be because of the reason that light has got finite speed, and we start and stop a clock when the light reflected from the body under observation to our eyes, creates the error in the time measured, as light has to travel a longer distance before reaching our eyes when the plane is moving away... Who knows? I never made any calculations by imagining a hypothetical experiment, and stressing on the light reflected by the body under observation and so on...

16. Mar 24, 2013

### Staff: Mentor

Nope. Time dilation and length contraction are what's left over AFTER we've corrected for the light-travel time.

17. Mar 24, 2013

### ShreyasR

Oh! Okay... Thats awesome. In the books that i have read, they haven't mentioned about the correction in light travel time. This is the reason why i was driven nuts! :P

18. Mar 24, 2013

### Ibix

Why not try it yourself? Say there is a star a distance d away from Earth, as measured in the Earth's frame. A ship travels to it at speed v, turns round and comes back. How long does the trip take according to a stay-at-home observer, and according to an observer on the rocket?

You'll need the Lorentz transforms, which relate time, t, and distance-from-Earth, x, as measured by someone on Earth to the time, t', and distance-from-Earth, x', as measured by someone doing speed v. These are:
$$\begin{eqnarray} x'&=&\gamma(x-vt)\\ t'&=&\gamma(t-vx/c^2) \end{eqnarray}$$
where
$$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$
Post what you can do if you get stuck, and we'll help.

19. Mar 24, 2013

### ShreyasR

@Ibix, I have understood the calculations given in the book, using the equations mentioned... I was only confused as I was only aware that all motion is relative but I was not aware that all acceleration is NOT relative... Now the equations make lots of sense... Thank you all! :)

20. Mar 24, 2013

### 1977ub

Only true for someone who is not moving with the plane, who is measuring the length of a moving plane. It is intimately related to the fact that for someone who is not moving relative to the plane, they will find the clocks at different ends of the plane to show different times while someone on the plane would find them to be in synch.

If you are on the plane, you will not find these results.

21. Mar 24, 2013

### Staff: Mentor

For an inertial observer radar yields the same result as Einstein's method for the frame where the inertial observer is at rest at the origin. But B is not inertial. See figure 9 of the paper I referenced for a diagram of B's frame using radar coordinates.

Could you point out what you are referring to here? There is no discontinuity in figure 9, so I am unsure what you mean.

Correct, on both counts (for a non inertial observer).

22. Mar 24, 2013

### Staff: Mentor

Yes, he will notice that the earth is moving away, but the equation of time dilation is specifically for an inertial frame, so it doesn't apply to any frame where B is at rest the whole time.

23. Mar 24, 2013

### 1977ub

B is initially outward, in IRF1. Then at the turnaround point, B is suddenly traveling exactly coincident with C, who was always in IRF2 earthward and then they both return to Earth together in IRF2. So we are suddenly shifting from IRF1 outward to IRF2 inward. In IRF1, certain Earth times were in the future. In IRF2, many of them are suddenly in the past. If we define IRF1 and IRF2 using the Einstein method, which should be the same as the radar method for these IRFs, there must indeed be a discontinuity. [ this method appears to be different than plain old radar, since they are defining coordinates differently in the different regions? ]

Last edited: Mar 24, 2013
24. Mar 24, 2013

### 1977ub

[corrected url]
http://arxiv.org/abs/physics/0411008

"Title: A note on Dolby and Gull on radar time and the twin 'paradox'

Abstract: Recently a suggestion has been made that standard textbook representations of hypersurfaces of simultaneity for the travelling twin in the twin 'paradox' are incorrect. This suggestion is false: the standard textbooks are in agreement with a proper understanding of the relativity of simultaneity."

Last edited by a moderator: Mar 24, 2013
25. Mar 24, 2013

### Staff: Mentor

Suppose that A, B, and C are each using radar to generate their coordinate systems. Then A's radar coordinates will be the same as Einstein's method for the frame where A is at rest at the origin. Also, C's radar coordinates will be the same as Einstein's method for the frame where C is at rest at the origin. Now, B's coordinates never match with A's coordinates, but if you look in figure 5 in region F you can see that there is a region where B's coordinates do match C's coordinates.

The text below the figure describes it in some detail. Basically, in any region where the radar user is inertial between sending and receiving the radar pulse, radar coordinates give the same result as Einstein's method. However, where there is some acceleration between sending and receiving then you do get a difference.