# Time Dilation, Twin paradox etc.

1. Oct 20, 2014

### kmm

I know there have been a lot of other threads on this topic but I wanted to get some feedback on my thoughts on this. Time dilation is a result of Einstein's second postulate of SR, although the first postulate is what gives rise to the apparent twin "paradox". The common solution to the paradox is that the twin that travels to a star and back to earth experiences a force, but it seems like this is argued in a way to say that the twin traveling to the star is somehow the one that's "truly" in motion. It seems the more fundamental way to look at it is that since the twin who travels to the star feels a force, it's not that they are in "true" motion and therefore the twin back on earth is "truly" at rest, but that the twin that leaves earth is not in an inertial reference frame, so Einstein's first postulate doesn't apply to the twin that leaves, therefore their clock actually runs slower.

There are similar paradoxes such as an observer on the ground and another in the train. The train passes the observer on the ground calculates that the clock on the train is running slower, and vice versa. It seems the solution to this is the same though. The person had to get on the train and accelerate, therefore it's not an inertial reference frame. For both of these scenarios, it seems then that it is the frame that experienced a force that will have the slower clock when each clock from each reference frame are compared later.

One other scenario is two astronauts A & B are floating in space and say orbiting a star in opposite directions. B sees A go by him then they synchronize their clocks. This seems like a situation where it's impossible to say who's clock will actually be slower when they compare their clocks at a later time. Maybe their clocks will read the same when they meet again. Is my thinking correct on this?

2. Oct 20, 2014

### Staff: Mentor

For that kind of scenario you will have to leave SR and use GR. The twins will then just calculate $\int d\tau$ along their respective worldlines.

3. Oct 20, 2014

### A.T.

If they move on symmetrical trajectories in a symmetrical gravitational field, you have the symmetry again, and they will have experienced the same time. But you can also have cases where they both experience no proper acceleration and yet accumulate different proper time between two meetings.

4. Oct 20, 2014

### kmm

Thanks, although I will need to study GR. I'm going through SR at the moment. So is my analysis correct regarding the twin paradox and the observers with one on the train and one on the ground?

5. Oct 20, 2014

### A.T.

In flat space time, free falling (zero proper acceleration, geodesic world line) maximizes the elapsed proper time between two events.

In curved space time it's complicated. There might be several free fall worldliness that connect two events, but have different proper times. I even think (not sure) that you can even have non-geodesic world lines accumulating more proper time than some of the geodesic world lines, between two events.

6. Oct 20, 2014

### PAllen

Definitely true, for a very mundane case. Consider the world line of rocket holding a fixed position relative to the sun. Compare an orbit geodesic that contains this same fixed point. Between two coincidences of these world lines, the static (non-geodesic) world line will accumulate more proper time than the orbital geodesic. The geodesic that maximizes proper time between these same events is the one that shoots radially outward such that it falls back to the fixed point in the duration of one orbit.

7. Oct 20, 2014

### kmm

Yeah, I just don't know enough GR to relate it to my question. I haven't dealt with geodesics. Although I thought my concern regarding the twins and the train, was really mostly a matter of SR.

8. Oct 20, 2014

### Staff: Mentor

If gravity is involved, as in your last example involving the orbiting astronauts, then you have to use general relativity. If there's no gravity involved, as in your other exampls, then SR is sufficient.

You may want to take a look at the FAQ at http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html to be sure that you really have the twin paradox down cold.

9. Oct 20, 2014

### Simon Bridge

You are thinking maybe that if the twins passed each other (in a third observers, call him O, reference frame) going in opposite directions at the same speed (as measured by O), turn around and come back ... in such a way that their accelerations in O's frame are equal (if opposite direction) and so forth. As they pass each other for the second time, they compare clocks. Now the system is exactly symmetrical, both twins get the same forces and everything. This how you are thinking?

A very useful tool to sort this stuff out is the space-time diagram. Since you want to think about these things I will strongly urge you to learn to use it.
A crash course that I find lots of people seem to relate to is:
http://www.physicsguy.com/ftl/html/FTL_intro.html
... ch2 deals with space-time diagrams and various popular paradoxes in such a way that you can start using them yourself.

10. Oct 20, 2014

### kmm

Indeed, I'll need some GR for my last example. I will definitely look at that link, however I wanted feedback on whether my thinking regarding the first two examples I gave is correct as of now.

Yes I've been using the space-time diagrams to help sort these things out. For my last example I wasn't really thinking about the forces involved or of a third observer, but I see I'll need to study more GR for that scenario. However, I was basing my thinking on the first two examples which I'm not 100% sure my thinking is correct about. In those first two examples that I brought up, if we looked at the world lines of each observer in those scenarios, the twin that left earth and the observer on the train would have world lines that showed that they accelerated and are therefore not in inertial reference frames. My thinking then is that what solves the "paradox" is the fact that we can identify an observer that isn't in an inertial frame. So with my astronaut example, my thought was that ultimately we can't say which one is in an inertial frame. Although I see the error I made that because they are both in orbit, they are not in inertial frames and gravity matters in this example.

A better scenario that would get to the heart of my question is if we had two astronauts in space that are moving in opposite directions and passed each other, although each would calculate that the others clock is running slower, can't we say that their clocks would actually be ticking at the same rate because they are both in inertial frames?

I'm at work and trying to hurry here, so I hope I was clear in my questions.

11. Oct 20, 2014

### Staff: Mentor

There is no such thing as the rate at which a clock is "actually" ticking. Any time that I speak of the rate at which a clock is ticking, I am either implicitly or explicitly comparing it against some other clock that is at rest relative to me so measures my sense of passing time.

12. Oct 20, 2014

### kmm

I agree. That's also what I mean when I talk about the rates that the clocks are ticking.

13. Oct 20, 2014

### Staff: Mentor

Ok, so we have two astronauts moving past each other in opposite directions. (This is exactly the same situation as either astronaut at rest while the other is moving).

Each astronaut will correctly conclude that the other's clock is ticking more slowly than his own. There is no meaningful way in which we can say that the two clocks are "actually ticking" at the same rate.

14. Oct 20, 2014

### Simon Bridge

No - this is because it makes no sense to talk about an "actual" ticking rate ... only the rate with respect to some observer. You have not said which observer you are talking about - like you say, the only two you mentioned would each "see" the other's clock as actually ticking slower... so which is the actual "actual" you refer to?

Note: The frame is which it is correct to say "two astronauts in space ... are moving in opposite directions and pass ... each other", is a frame where both astronauts are moving. This implies a third observer. It is important to use explicit observers to avoid confusion.

15. Oct 20, 2014

### kmm

That's true, and so it would seem then that there's also no meaningful way to say if either clock is ticking faster or slower than the other. We can only say what each astronaut observes.

I see. I should clarify that. So each astronaut observes the other fly past them. We can't say which one is "actually" moving, so we can't make any claims that either clock is "actually" moving faster than the other or even at the same rate. Although with the twin scenario, we can say that the clock of the twin that leaves earth actually is ticking slower. Right?

16. Oct 21, 2014

### Simon Bridge

No.

Both twins will agree that one of the twin's comes back younger than the other. Each twin will disagree about how that happened. We can only repeat what the twin's say because we weren't there.

http://www.physicsguy.com/ftl/html/FTL_part2.html#sec:twin
... but you need to have space-time diagrams down first.

17. Oct 21, 2014

### Staff: Mentor

No, we cannot say that (and I pointed you at the Twin Paradox FAQ because I thought that might be your misunderstanding).

The amount of aging you experience has nothing to do with the rate at which your clock ticks; as far as you are concerned, your clock runs at a rate of one second per second, so you age one second for every second that your clock counts. Sure, some other person somewhere else in the universe is looking at a different clock, but what do you care?

The amount you age does depend on how much time your clock measures on your journey. If it counts off ten years you'll age ten years, just as surely as your car will experience 10,000 kilometers of wear during a journey in which the odometer counts off 10,000 kilometers.

In the twin paradox thought experiment, at every step of the journey the traveler
finds that stay-at-home is moving relative to him, so correctly concludes that stay-at-home's clock is running more slowly than his own. Stay-at-home finds that traveler is moving relative to him, so just as correctly concludes that traveler's clock is running more slowly.

They are both right.
However, when they meet up at the end and compare notes, they will find that traveler's clock ticked less often and counted off less time. Therefore traveler aged less.

That sounds impossible, right? Go read that FAQ....

It's all special relativity, no GR involved.

18. Oct 21, 2014

### kmm

OK, so I read both of those links (Only the twin paradox section that you posted, Simon) and that helped a lot (I should have read them sooner). I do understand that in your own reference frame, regardless of what an observer in another reference frame observes about my clock, I will not see my clock change the rate at which it ticks.

So with the twin paradox, at the end each twin finds that the twin who left earth has aged less so there really is no paradox. They just each have a different story describing the outcome. But it was the acceleration undergone by one of the twins that breaks the symmetry between them and allows us to determine who has aged less. So in a situation where two observers appear to be moving to from the others perspective, as long as neither experiences acceleration and the symmetry between them remains, there's no absolute answer to the question of who has aged less.

Although, with the twin scenario, we still have to say time dilation really occurred and in total, less time passed for the twin who left earth.

19. Oct 21, 2014

### harrylin

No doubt you meant that the twin that leaves earth is not all the time at rest in an inertial reference frame. That is correct: "feeling a force" has nothing to do with time dilation in SR. Ignoring possible effects of gravitation on time keeping, the original "twin" example was even given for a swing shot around a star so that the astronaut feels no force at any time during the voyage [1].
And while my GR knowledge is very limited, what I do know is that at least for small gravitational fields the time dilation effects from gravitation and velocity can simply be added (this fact was applied to the first GR test on Earth, gravity probe A [2]). Thus for a sufficiently long journey at sufficiently high speed the gravitational time dilation effects from the Earth and from the star can be neglected for the calculation.
[1] p.50 and further in https://en.wikisource.org/wiki/Translation:The_Evolution_of_Space_and_Time
[2] http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.45.2081

Last edited: Oct 21, 2014