# Not banal question on twins

1. Jul 7, 2006

### Born2Perform

Imagine an empty universe, just two planets exists.

If they move with relative speed, it is impossible to decide which is moving right?

But when the people on the moving planet go to meet their twins on the other planet, they are younger.

So in this way is possible to decide which is moving???? thx.

2. Jul 7, 2006

### Staff: Mentor

Which twin travels to which planet?

3. Jul 7, 2006

### Staff: Mentor

The issue is not who is moving, but who is accelerating.

4. Jul 7, 2006

### Born2Perform

isn't that indifferent?
If i call Planets A and B, and when after some journey peoples meet together, if all the people who where in B are younger than their twins in A, doesn't it mean that B moved?
thx.

5. Jul 7, 2006

### Born2Perform

i know that this type of question is like doing 1+1 for you but still i don't understand.
What about if the two planets are alredy in motion, and instead to stop and chech the age of their twins, they see it with a telescope.

6. Jul 7, 2006

### MeJennifer

If they check with a telescope the people on the other planet always appear younger.

7. Jul 7, 2006

### Staff: Mentor

In order to come together, one twin has to accelerate and move to where the other twin is. That breaks the symetry.

8. Jul 7, 2006

### kknull

so... if we erase all Doppler effects, the twin who is travelling believes that his brother is younger, but when he returns home, he magically sees that his brother is older then himself??!!

9. Jul 7, 2006

### Janus

Staff Emeritus
No. While he is traveling outward he sees his brother aging slowly.Just before he starts his return trip, he will determine that his brother is younger, As he decelerates to a stop and then accelerates for the return trip he will see his brother aging much faster. After this he will determine that his brother is older. On the return trip, he will see his brother aging more slowly again. When he returns to his brother he will find that the period where the brother aged faster dominated and his brother will be older.

10. Jul 7, 2006

### kknull

so if the brother travels for 1E100000000000000 years and then returns home, the acceleration/decelaration time dominate ALWAYS his trip?

11. Jul 7, 2006

### MeJennifer

Well this is confusing to say the least, as it reeks like "this can be all explained due to the space-time effects of the acceleration stages only".

Twin paradoxes are about shorter or longer paths traveled in space-time and not particularly about the space-time effects during the short acceleration and decelleration stages.

When the planets are in relative motion with each other each will encounter a slower progression of time on the other planet. When one of the twins decides to travel then his space-time interval will be shorter with a result that the elapsed time on his local clock is less than of the twin who does not travel. Thus when the traveler reaches the planet he is in fact younger than his twin brother. The space-time effects during the short acceleration and deceleration stages of his voyage do have a small effect on the calculations but are most certainly not an completel explanation for the difference in age between the twins.

Furthermore no one aged any faster or slower.
The younger brother simply traveled a different path in space-time which had an effect that less time elapsed on his clock than on the one that stayed behind.

Last edited: Jul 7, 2006
12. Jul 7, 2006

### kknull

Thus we return to my first question:"so... if we erase all Doppler effects, the twin who is travelling believes that his brother is younger, but when he returns home, he magically sees that his brother is older then himself??!!"

13. Jul 7, 2006

### MeJennifer

There is no magic involved in this.

14. Jul 7, 2006

### kknull

Yes, sure! My question is only a provocation. Nothing is magic in physics

15. Jul 7, 2006

### clj4

Why doon't you try reading this simple and CORRECT explanation before making broad and incorrect statements:

http://sheol.org/throopw/sr-ticks-n-bricks.html [Broken]

Last edited by a moderator: May 2, 2017
16. Jul 7, 2006

### pervect

Staff Emeritus
Huh?

To answer this question, I have to understand what you mean by "erasing all Doppler effects".

I don't understand what you mean by this question. In fact, it is the only part of your problem description that seems "magic" to me, because Doppler shifts are a part of physical law, and erasing them would take some sort of "magic" that I don't understand.

Offhand, it sounds like you're probablly mired with an incorrect notion of simultaneity - i.e. that you think "at the same time" has an unambiguous meaning. Unfortunately, "at the same time" means different things to different people according to relativity.

This can be seen with a space-time diagram.

$$\] \unitlength 1mm \begin{picture}(95,80)(0,0) \linethickness{0.3mm} {\color{red}\put(40,20){\line(0,1){60}} \linethickness{0.3mm} \put(40,20){\line(1,0){5}} \linethickness{0.3mm} \put(50,20){\line(1,0){5}} \linethickness{0.3mm} \put(60,20){\line(1,0){5}} \linethickness{0.3mm} \put(70,20){\line(1,0){5}} \linethickness{0.3mm} \put(80,20){\line(1,0){5}} \linethickness{0.3mm} \put(90,20){\line(1,0){5}}} \linethickness{0.3mm} {\color{blue}\multiput(40,20)(0.12,0.36){167}{\line(0,1){0.36}} \linethickness{0.3mm} \qbezier(40,20)(40,20)(40,20) \qbezier(40,20)(40,20)(40,20) \linethickness{0.3mm} \multiput(39,20)(0.41,0.12){17}{\line(1,0){0.41}} \linethickness{0.3mm} \multiput(51,24)(0.35,0.12){17}{\line(1,0){0.35}} \linethickness{0.3mm} \multiput(61,27)(0.35,0.12){17}{\line(1,0){0.35}} \linethickness{0.3mm} \multiput(72,31)(0.41,0.12){17}{\line(1,0){0.41}} \linethickness{0.3mm} \multiput(84,35)(0.41,0.12){17}{\line(1,0){0.41}}} \end{picture} \[$$
Above is a space-time diagram. Arbitrary coordinates have been assigned to the set of events that comprise space-time. Every event in space-time is represented by one and only one location on the diagram.

In this diagram the solid red line represents the path through space-time of observer #1.

The solid blue line represents the path through space-time of observer #2.

Observer #1 and observer #2 are moving with respect to each other. In fact, as you can see, we have adopted the coordinates of observer #1 to label the points on this diagram. This was an arbitrary choice on our part, and has no physical significance - coordinates never have any physical significance, they are the map, not the territory,.

For our convenience, however, we need to use some unique labels at every point to distinguish different points in space-time from each other.

We can see that on this diagram, for obsever #1, the "time coordinate" is the vertical position on this diagram, and the "space coordinate" is the horizontal position on this diagram.

The dotted red line represents the notion of simultaneity of obsever #1 - all points on this line are regarded as simultaneous by observer #1.

The dotted blue line represents the notion of simultaneity of observer #2.

Note that the two set of events that are regarded as simultaneous are not the same - they are different sets!

Last edited: Jul 7, 2006
17. Jul 7, 2006

### DaveC426913

Hey that second graph makes a lot of sense.

How can two people, each receding relativistically from each other, see the other as aging slower? Isn't this a pardox?

No. No more than two people driving trucks, each receding from each other at a 45 degree angle, see the other's truck as being shorter (because, realtive to themselves, the other truck is foreshortend due to the angle).

And that's not just an analogy. If you see time as simply another dimension.

Last edited by a moderator: May 2, 2017
18. Jul 7, 2006

### clj4

Sigh, what can we do with the never ending generations of relativity deniers that hang onto the "paradox"

19. Jul 7, 2006

### actionintegral

The original poster was correct in his statement. Two syncronized clocks in the same frame will always stay in sync. But then give one a velocity so he can go meet the other and his proper time will slow. So yes, you can decide which one is moving. What's the controversy?

20. Jul 7, 2006

### MeJennifer

A very good question and something perhaps that many educators who visit this forum should think about.

Why is it that about a century after relativity theory we still teach students Newtonian concepts as true? I for one think it is an embarrassment!
And the pompousness in which it is presented, "Newton's laws", laws, presented as eternal truths engraved in the stones of nature, "invisible forces that causes objects to attract each other" and more of that.
What can you expect once that stuff is taught?

I am not talking about applying Newtons' formulas, since they are sufficient most of the time, but why teach Newtonian concepts as truths, why call it laws while they are really approximations of a false model?
Teach relativity as the correct model first, the concepts, I am not talking about GR formulas here, and then inform them about the incorrect but often practical Newtonian model and formulas.

So are we really surprised that when most high school kids get taught and are examed on Newtonian concepts that they have difficulties in understanding relativity? I am not!

Understanding relativity is not hard at all in my view. The problem is not understanding it, the problem is the difficulty in unlearning what people were taught at school as correct.

Last edited: Jul 7, 2006
21. Jul 7, 2006

### Janus

Staff Emeritus
Yes, because the amount that he determines that his brother aged more is determined by not just the acceleration but how far apart (as measured along the line along which they are accelerating) they are the further apart they are, the more he will age.

To visualise this, imagine DaveC426913's example of two trucks driving at 45° to each other. Assuming that each truck started at the same point. (our two twins on Earth). as Each truck drives along, he will note that the other truck is "behind" him as judged by his progress in the direction he is driving. (this is the equivalent of each twin seeing the other as aging slower.)

Now one driver turns his truck so that its path interesects the other truck's and once he does he turns again until he is traveling in the same diirection as the other truck. (This is the one twin turning around, heading back and returning to Earth.)

This is how this appears to the truck(truck 1) that does not turn around:

The other truck (truck 2) slowly falls behind , changes direction, continues to fall behind unitl it crosses the path of truck 1, where it no longer continues to fall behind but is still behind. (the first twin see the second twin age more slowly for almost the entire trip and is younger when they meet up again. )

This how it appears to Truck 2:

Truck one slowly falls behind unitl till truck 2 changes direction, at which point truck 1 will seem to swing in front of truck 2. (Imagine your driving a car and turn a corner. objects that were behind you and in the direction you turn will now be in fornt of you.) After the turn truck 1 will still progress more slowly than you, but will still be ahead of you once you cross its path. Truck 2 turns to follow truck 1, and truck one now maintains a constant distance ahead of truck 2. (the second twin sees the first twin age more slowly while he is on the outward leg, age more quiclky as he turns around, and more slowly again during the return leg. Upon the meeting of the twins, twin 1 will have in total, aged more thant twin 1.)

22. Jul 7, 2006

### Janus

Staff Emeritus
Well, while space-time intervals and paths through space-time are a more technically coreect way to describe SR, many times while trying to expalin Relativity to laymen, such descriptions just go right over their head. They are just not familiar enough with concepts.

Sometimes you have to try to bring it down to their level first and build from there.

23. Jul 8, 2006

### robphy

A picture or a movie might help
http://physics.syr.edu/courses/modules/LIGHTCONE/LightClock/ [Broken]

Last edited by a moderator: May 2, 2017
24. Jul 8, 2006

### RandallB

Even without formulas it would be just as dishonest to declare “relativity as the correct model first” as to Name Newton’s Laws as “LAWS”.
Especially with GR included in the relativity mix:
We are still spending big money testing to provide more conclusive confirmation of GR.
AND there is still the conflict between GR and QM (and variations) Where we cannot not say which one is correct, and we suspect that cannot both be right, so we just apply the one that works best for the task at hand.

So when you start with those high school kids, rather than declare your favorite theory be it GR or QM as THE ONE, best let them know what we do not know.
After all if we did have it right and complete – we’d have both a GUT and a TOE.

25. Jul 8, 2006

### masudr

Surely it's best to emphasise the fact that theoretical physics comes up with models to describe reality, and not reality itself. That's all we, as physicists, do.

We can compare our predictions with our experimental results, but ultimately, 'tis all a model.