Space traveler and time dilation

1. Jun 8, 2010

andyp2010

Maybe some one can help clear up a problem. According to Wikipedias article on time dilation

“In the case that the observers are in relative uniform motion, and far away from any gravitational mass, the point of view of each will be that the other's (moving) clock is ticking at a slower rate than the local clock.”

Which I can accept but this does not seem consistent with the “time slowing down for a space traveler” situation. If the space traveler is moving fast relative to the earth the people on earth will see that time is passing more slowly for traveler. The traveler will think that the time on the earth is moving more slowly. So what happens when he gets back to earth he’s been observing that earth time has been moving more slowly so he would be older that earth time would suggest. But also for the earth people the travelers time would have been passing more slowly so he should be younger. Which is a contradiction.

2. Jun 8, 2010

3. Jun 8, 2010

4. Jun 8, 2010

George Jones

Staff Emeritus
Yes as the traveler moves away from Earth, he sees visually time on the Earth passing more slowly, but when the traveler is on the trip back to Earth, he sees visually time on the Earth passing more quickly. The net result is that upon arrival back on Earth, the traveler finds that the Earth has aged more than he has.

See

5. Jun 8, 2010

andyp2010

thanks everyone. I'll get reading then.

6. Jun 8, 2010

JDługosz

What I'd like to know is: if you went "all the way around the Universe", i.e. it was a 4-D sphere, and passed your starting point again without ever having undergone acceleration, how would the paradox be resolved? A wormhole would have the same problem in GR.

7. Jun 8, 2010

ExecNight

Ah "the twins".. I don't know about the time dilation or the paradox.

I don't know their acceleration or anything but this topic sure never gets old =)

8. Jun 9, 2010

Dmitry67

As Universe is curved, then that can be answered in SR framework, only n GR framework. But the situation is not symmetric in GR because the world line of such observer is more curved.

9. Jun 9, 2010

yogi

You dont really need to go around the universe to resolve or simulate the problem - there is one round trip voyage where both twins remain in their own inertial frame for the entire round trip - yet their clocks show different lapsed times when the traveling twin returns - no general relativity involved and no curvature - a polar orbiting satellite will do - simple construct a 100 mile high tower on the South Pole and put a satellite in polar orbit at an elevation of 100 miles - a clock on top of the tower remains fixed in the non rotating earth centered inertial reference frame and the clock on board the satellite remains in the inertial frame of the orbiting satellite - start the clocks as the satellite passes the tower and stop them when it passes by after completing one orbit - the two clocks will not have logged the same amount of time.

10. Jun 9, 2010

matheinste

In SR if the twins part and reunite then at least one of them has been moving non-inertially at some time.

What sort of motion a round trip of the universe entails I do not know. It depends on the global geometry of the universe.

Matheinste.

11. Jun 9, 2010

MikeLizzi

Gorge Jones,
You have posted this comment several times in this forum, now. It is absolutely wrong. If you believe it, you have a serious misunderstanding of Special Relativity.

I am shocked that the other members of this forum are letting you get away with this.

12. Jun 9, 2010

Ich

George Jones' comment is not wrong. Maybe you are confusing relativistic doppler effect and time dilation? George explicitly talks about the former.

13. Jun 9, 2010

phyzguy

George Jones and Ich are correct. Attached is a picture I downloaded from among the plethora of explanations on the web. You can see that on the way out, the twin on the rocket sees two years pass on Earth, while four years pass for him. On the way back, he sees 8 years pass on Earth while four years pass for him. So when he returns, 10 years have passed on Earth while 8 years have passed for him, so the twin on the rocket is younger.

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14. Jun 9, 2010

Aaron_Shaw

If you do calculations to compensate for doppler effect, then you'll see that time passes slower in both directions (outbound and inbound). It's just that at the turnaround point at half way and decelleration to stop back on earth that perceptions of simultaneity will shift significantly in a small amount of time. This accounts for a huge 'jump' forwards in time on earth, as seen by the traveller, as he turns around.

Thus, even though earths clocks have been ticking slower on both the outbound AND inbound journey, the 'jump' forwards in time as simultaneity shifts means that the earthlings will have still aged more.

15. Jun 9, 2010

George Jones

Staff Emeritus
How is it wrong?
A huge jump in coordinates is not actually seen visually by the traveler. There is a big difference between what is seen visually, and what happens to coordinates.

16. Jun 9, 2010

phyzguy

This is absolutely correct. If the twin on the rocket had a telescope trained on Earth, he would see no jump. He would simply see things start to happen faster as he turned around, because he is now starting to "catch up" with the outward propagating wave fronts coming from Earth. Look again at the space-time diagram I posted a few posts ago.

17. Jun 9, 2010

DrGreg

Re: Space traveller and time dilation

This analogy doesn't work in either special or general relativity.

In special relativity, ignoring gravity, the south pole pole would be inertial and the "satellite" would be accelerating.

In general relativity, the south pole pole is accelerating and the satellite is inertial.

18. Jun 9, 2010

Aaron_Shaw

I'm basing this on a model of what happened after calculating to remove the doppler effect and time it takes for light to travel, as i stated originally (probably not very well). I find that when people are trying to understand, the doppler effect and all that confuses the issue. People typically want to know what is 'really' happening, rather than what it looks like by the time the light has reached you.

So i guess what i mean is that if you plotted space and time coordinates on a graph for the whole journey then what the traveller would see is the earth taking a large jump forwards in time at the turnaround point (assuming its instantaneous).

19. Jun 9, 2010

zetafunction

my question is, can a Wick rotation be made 'physical' ??

i mean you are in a metric $$t^{2}-x^{2}$$ space and time

and you 'rotate' your reference system to get a new metric $$t^{2}+x^{2}$$ which is purely Euclidean and there is no distinction between space and time

20. Jun 9, 2010

JesseM

But that isn't what's "really happening" in any meaningful sense, it's just what's happening in one particular non-inertial rest frame for the traveling twin--specifically one constructed in such a way that the definition of simultaneity in this non-inertial frame always matches up with the definition of simultaneity that would be used in the traveler's instantaneously co-moving inertial rest frame at that instant. Unlike with inertial frames, though, there isn't any one "correct" way to construct a non-inertial rest frame for a non-inertial observer, there are an infinite number of different coordinate systems you could construct for such an observer and none of them would be considered physically "preferred".