# Twins paradox confusion

1. Sep 10, 2006

### leright

I am having trouble understanding how the explanation of the paradox solves the problem. What if people in two different frames were moving wrt one another with constant velocity. Say for instance, these people never turn around to meet up once again and are constantly moving at a constant velocity wrt one another.

Clearly, each will percieve the other's frame to be moving more slowly than their own. There is never a change in reference frame so they disagree on who is older. Who is actually older???? Are they both younger and both older simultaneously? That seems to be a weird consequence of relativity. What am I missing here?

Last edited: Sep 10, 2006
2. Sep 10, 2006

### leright

Also, I know the speed of light is constant in all inertial frames. However, it does not necessarily have to be constant in accelerated frames, correct? I believe this because the postulate of relativity states that "all laws of physics are the same in all inertial reference frames", so some laws of physics are not the same in accelerated reference frames, and therefore the speed of light is not necessarily constant in all reference frames, but it is only constant in inertial frames?

3. Sep 10, 2006

### HallsofIvy

Each observes the other as being older. Since they are never in the same frame of reference, there is no paradox there. There is no such thing as "actually older" unless they can be compared in the same frame of reference.

Remember that two events being "simultaneous" only applies for two immediately adjacent events in the same frame of reference. You cannot compare the "actual" ages because you cannot compare when each observes the other.

Yes, that is true. The fact that the speed of light is the same constant in all inertial frames of reference applies only to inertial frames. In fact, there is no good way to measure the speed of light in a non-inertial frame.

4. Sep 10, 2006

### Staff: Mentor

To clarify: replace "simultaneous" with "absolutely simultaneous." That is, if two events are at the same location, then all observers (no matter how they are moving) will agree about whether those two events are simultaneous or not.

Two events that are not at the same location can be simultaneous in at most one inertial reference frame. In all other inertial reference frames, the events are not simultaneous.

Most of the apparent paradoxes involving length contraction and time dilation can be resolved by taking this relativity of simultaneity into account.

5. Sep 10, 2006

### leright

Nobody has proposed a means of calculating what the speed of light would be in a certain non-inertial frame? Yes, there is no way to test these calculations so doing such mat would be meaniningless, but I have seen people do physics that cannot be proven (string theory?) Why is the speed of light any different?

Last edited: Sep 10, 2006
6. Sep 10, 2006

### leright

So, basically, they can think they are both the older twin when they are apart, but when they come together one will in fact be older and the other will be younger.

7. Sep 10, 2006

### leright

btw, I tried to determine the time dilation effect for accelerated frames (assuming light speed is the same for them) by taking the time dilation formula and replacing the deltas with differentials and then I replaced the v in the formula with at for constant acceleration, and simply integrated from 0 to t0. I got delta(t) = (c/a)(inversesin(a*(deltatnaught)/c)).

Clearly, this only works when the argument of the sine is between -1 and +1, so a cannot generally be larger than c, unless the acceleration is for a very small amt of time (one second), and if c=a then the acceleration can only occur for a time arbitrarily smaller than one second.

8. Sep 10, 2006

### JesseM

Integrating $$\sqrt{1 - v(t)^2/c^2} \, dt$$ will give you the amount of time elapsed on an accelerating clock with a variable velocity v(t), but as seen in a single inertial frame where the clock has this velocity function v(t), not as seen in an accelerated frame--I'm not sure if that's what you meant by "the time dilation effect for accelerated frames" or not. However "constant acceleration" in relativity usually means the G-force experienced by observers on the ship is constant, which is equivalent to saying that the acceleration measured in the inertial frame where the ship is instantaneously at rest at a given moment will be the same from one moment to another; if this is the case, then the acceleration as seen in a single fixed inertial frame will not be a*t for constant acceleration, see the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] page for the correct formulas in this case.

Last edited by a moderator: May 2, 2017
9. Sep 11, 2006

### yogi

Not necessarily - you have to define the experiment so there is an assymetry - if two separated stationary clocks are synchronized and each quickly accelerated to the same velocity so they travel at a contant speed until they meet each other at some point, each will conclude that the other guys clock is running slow during the flight - but when they meet and stop - both clocks will read the same - now compare this with the case where only one clock moves - when it reaches the other clock, the one that has been put in motion will have accumulated less time - what is the difference - the clock put in motion will have undergone both a temporal and spatial increment in his world line - the clock that has not moved will only have experienced a temporal change during the experiment (a straight world line) - in both cases the spacetime interval is the same - but the moving clock has both a temporal and space interval that must be taken into account - so the temporal increment must of necessity be less for the clock that is put in motion

Last edited: Sep 11, 2006
10. Sep 11, 2006

### HallsofIvy

And the point is that, in order for the twins, who have been moving apart, to come back together, one or both must accelerate- so the laws of special relativity no longer apply.

11. Sep 11, 2006

### M1keh

Okay. So Bill stays on the Earth and Bob flies away at 0.6c (180,000kms)for 1 Earth hour. Bob's watch shows a time 12 minutes slower than Bill's after this hour ? Time dilation factor of 0.8 ?

If Bob then turns back and travels at 0.6c for another Earth hour, his watch would be 24 mins behind Bill's when he returns. Is this correct ?

Thanks.

12. Sep 11, 2006

### HallsofIvy

Not necessarily. What happens to his clock during acceleration and decelleration? You would have to specify exactly how the acceleration-decelleration work.

13. Sep 11, 2006

### yogi

You do not have to turn around and return to the starting point to experience time dilation - in fact most of the experiments are one way - e.g. pion created in the lab accelerated to a high velocity will exhibit time dilation as it travels a straight path - if the twin stops at the end of his outbound journey and compares his watch to one on earth, it will be found to have logged less time - the traveling clock can be stopped at any point and compared to a local clock at that point and it will be found to be out of sync therewith - every clock at rest wrt the earth will read the same as an earth clock - it is in the earth rest frame - the twin exercise is best treated as two one way experiments - half the time is lost on the outward journey - the other half is lost on the inward journey - and while you can get the correct result by considering the way each twin observes things during the three phases of the trip, it is easier to comprehend the trip by breaking it into separate events

In Einsteins original publication, he first considers one way motion - then a round trip ... later in 1918 he did a 180 and decided that the twin paradox could best be explained in terms of a pseudo G field that one or the other twin experiences during turn around - this has led several prominent authors to claim general relativity is necessary to explain the twins difference in ages - its of little wonder there are so many different views on the subject

Last edited: Sep 11, 2006
14. Sep 12, 2006

### M1keh

15. Sep 12, 2006

### Staff: Mentor

For the sake of discussion, let Bob do a simple straight-line out-and-back trip, with very short periods of acceleration at the beginning, turnaround and end, and constant-velocity cruising periods in between

From the relativistic point of view, the essential difference between Bill and Bob is that Bill remains at rest in a single inertial reference frame throughout, whereas Bob is at rest in two different inertial reference frames during the trip: one inertial reference frame for the outbound trip and another one for the inbound trip. To put it another way, if you view Bob's rocket ship as a single reference frame for the entire trip, it cannot be an inertial reference frame.

The difference between inertial and non-inertial reference frames is crucial in relativity. From a relativistic point of view, inertial reference frames are indeed "favored".

The difference between inertial and non-inertial reference frames is not merely mathematical or semantic. There are concrete physical differences which can be perceived by Bill and Bob. If they close their eyes, or equivalently are locked up in a windowless room and spaceship respectively, each one can easily tell whether he remains in an inertial reference frame or not. Bill, in his stationary room, does not perceive anything special. Bob, on the other hand, knows that he is accelerating during takeoff, turnaround, and landing, because he can feel those events taking place, the same way that you can feel it when your car accelerates and your seat pushes against you.

16. Sep 12, 2006

### gijeqkeij

Space and time are just 1 thing in SR and GR. If both can say "you are more distant from me" and being both right why both can't say "you are older than me" and being both right?

gijeqkeij

Universe it's so simple that it's almost impossible for us to understand it

17. Sep 12, 2006

### JesseM

That's a little like saying that since all three spatial axes are "1 thing", then being 5 meters away from me on the y-axis is equivalent to being 5 meters away from me on the z-axis, which is not correct. Space and time are both part of a single entity, "spacetime", but they are different directions in spacetime, so an event happening at a later time is not the same as it happening further in some spatial direction. And the time dimension is somewhat different than the spatial dimensions for other reasons--for example, you can't use the regular pythagorean theorem to calculate "distance" in spacetime, instead you must add the squares of the spatial distances and subtract the square of the temporal distance (multiplied by c^2) between two events to get their spacetime separation.

18. Sep 13, 2006

### M1keh

Good answer ...

But if they compare watches, Bob's watch should now be 24 mintues (approx) behind Bill's ?

If as Bob approaches, Bill accelerates to 0.6c (instantaneously-ish), matching Bob's velocity, his watch wont then be 24 minutes behind Bob's even though he's now in Bob's f.o.r and that's what Bob would expect ?

So why is Bill's f.o.r so 'special' ? From Bob's perspective, Bill raced away at 0.6c, stopped, turned back and returned at 0.6c and when he arrived he stopped. Bill's watch must now be 24 minutes behind Bob's ???

If not, which element is wrong ? At the final second of the journey the times shown on the watches can't surely shift over a range of 48 minutes from -24 to +24 depending on which f.o.r you jump to ? There's no (or close to no) elapsed time in either f.o.r. for this to happen ?

19. Sep 13, 2006

### Janus

Staff Emeritus
No, Bob will expect his clock to be 24 min behind, more as to why latter.
No, Bob can not say that. As he watches Bill, he can see that Bill never experiences any acceleration, while during the time that Bill apparently stops and turns around, Bob does experience the effects of acceleration; evidence that is is he who is doing the turning around.
For Bill, Bob's clock always runs slow.
For Bob, Bill's clock runs slow on the outward leg and run slows on the inward leg, but as he (Bob) stops and turns around, form his perspective, Bill's clock will run fast. How fast depends on how hard he acclerates and how far apart Bill and Bob are. It will always work out that the amount of time that Bill's clock gains according to Bob during this period will more than make up for the outgoing and incoming legs and then some. The 'then some', will account for the fact that He will expect his Clock to Be 24 min behind Bill's when they meet up.

The distance factor also explains why that, in your scenerio where Bill accelerates to match Bob's frame of reference when they come back together, Bob's clock is still 24 min slow. Bill doesn't see Bob's clock run fast, becuase there is no distance( or almost no distance) between them at that time.

20. Sep 13, 2006

### M1keh

Thanks. You're right of course. The acceleration can be felt by Bob, so his view of things is different. However, nobody seems to be able to explain why this makes a difference. The fact that Bob accelerates can't reverse the time dilation under Relativity as this would require a negative difference in speed ?

In fact, what happens if Bill accelerates to Bob's intial speed in the direction Bob was travelling, stops immediately and then returns at the same speed as Bob, timed so that they both return together ? That eliminates the acceleration completely but the watches still show inconsistent times ?

ie. Bob accelerates in 1 sec to 0.6c and travels for 1hr Earth's time. He stops taking another 1 sec, turns & accelerates to 0.6c, taking another 1sec (lets say turning takes no time at all. could just reverse ). 3 seconds before Bob returns, Bill accelerates towards Bob to 0.6c in 1 sec, stops, taking another 1 sec & accelerates back to Earth, reaching 0.6c in 1 sec. Both experience the same acceleration / deceleration and are in the same place, yet Bill's watch is behind Bob's ?

If instead of Bob being the one to 'travel', both Bill and Bob travel in opposite directions from Earth and then return and compare watches, both have experienced the same acceleration (albeit in opposite directions). I'm assuming there's nothing to suggest that he direction of the acceleration makes a difference ?

So Bob travels one way at 0.6c, Bill travels the other at 0.6c and Suzy stays and watches the clock. When the two adventurers return, they both have to have watches that show 24 mins earlier than Suzy's ? So how are their watches showing the same time ?

Last edited: Sep 13, 2006
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