1. Apr 30, 2010

We use the following formulas for the Lorentztransformation:

x’ = [ x / sqrt(1-v**2/c**2)] - [vt / sqrt(1-v**2/c**2)] (1)

and

t’ = - [ (vx)/c**2 / sqrt(1-v**2/c**2)] + [ t / sqrt(1-v**2/c**2)] (2)

The twin paradox reads as follows. Gea and Stella are identical twins. Stella leaves from earth for an interstallar journey with a constant velocity v, where 0 < v. According to the Lorentztransformation Stella's clock will be slower then Gea's clock. Therefore Stella will be younger then Gea when returning back to earth. However, due to the symmetry of the situation Gea's clock will also be slower then Stella's clock, which leads to a paradox.

We choose the following situation for Stella's yourney. In the beginning of the story Stella travels from the earth a distance of three light years with a constant velocity 0.6*c. Then she returns back instantneously and travels the same distance back with the same constant veocity. If we take c = 1, then the time she travels untill the point of returning is equal to 3 / 0.6 = 5. Shortly, x = 3, v = 0.6 en t = 5. The time she will need to travel back is equal to the time she needed to travel to the returning point. Therefore the total total time of her yourney, calculated from the viewpoint of Gea (x,t), is equal to 5 + 5 = 10. However, according to formula (2), the time Stella travels untill the point of returning, calculated from the viewpoint of Stella (x',t'), is equal to 4. Therefore the total total time of her yourney, calculated from the viewpoint of Stella, is equal to 4 + 4 = 8. So we have t / t' = 10 / 8 = 1.25.

If we take as our steady reference system (x,t) Stella, then taken from her viewpoint, Gea would travel towards her returning point, with a velocity -v with 0 < v, and according to formula (2) the time untill the point of returning, calculated from the viewpoint of Gea (x',t'), is equal to 6.25. The total time for Stella, calculated from the viewpoint of Stella, would be 5 + 5 = 10 and total time for Gea, calculated from the viewpoint of Gea, would be 6.25 + 6.25 = 12.5 and again the ratio t / t' = 1.25.

Therefore, actuall there is no paradox.

It seems however, that, in this way of reasoning, v should represent a number wich is larger then zero: 0 < v. Is this correct?

2. Apr 30, 2010

### Ich

You are aware that this is possible only for either the inbound or the outbound trip? There is no inertial frame in which Stella is at rest in both parts of the journey. That's actually the resolution of the paradox.

No, Gea doesn't return. It's Stella who changes velocity. That's a difference.
At the time Stella fires her thrusters (t=4 y in Stella's coordinates), Gea is 2.4 ly away, and 1.92 y older.
Directly after firing her thrusters, in her new coordinate system, Stella will calculate Gea to be 2,4 ly away and 8.08 y older.

All that happens to Gea in her proper time interval [1.92 y, 8.08 y] is neither covered in Stella's first coordinate system until t=4 nor her second coordinate system starting from t=4.

Have a look at the attached spacetime diagrams. There are many coordinate systems you can use. You can use "Stella out" or "Stella in", but there's no inertial frame "Stella".

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3. Apr 30, 2010

### Fredrik

Staff Emeritus
It's difficult to follow your argument, because when you're specifying a time or a distance, you're not specifying which coordinate system you're talking about. So I'm not going to try to understand your argument in its present form.

If we're only interested in finding out the age of the twins at the event where they meet on Earth after Stella's trip, the answer follows immediately from one of the axioms of SR: A clock measures the proper time of the curve in spacetime that represents its motion. A direct calculation shows that Stella is younger.

A naive application of the time dilation formula leads to the logically inconsistent conclusion that Stella is younger than Gea and Gea is younger than Stella. The reason is that a naive application of the time dilation formula fails to account for the fact that in the inertial frames that are co-moving with Stella, the moment just before the turnaround is simultaneous with a much earlier event in Earth's history than the moment just after the turnaround. See the spacetime diagram.

[PLAIN]http://web.comhem.se/~u87325397/Twins.PNG [Broken]

The diagram shows both twins' points of view...if we have defined a person's "point of view" as a description in terms of an inertial frame in which the person has velocity 0. It's actually not at all obvious that we should be doing that. This article uses another convention that's at least as natural.

Another link that often gets posted in these threads (there's an absurd number of them) is to the Usenet Physics FAQ about the twin paradox. Edit: That link doesn't work for me right now. Try this one instead.

Last edited by a moderator: May 4, 2017
4. Apr 30, 2010

### D H

Staff Emeritus
Among those I personally like the Doppler shift explanation the best. My education is in physics but my career has been in engineering. I like explanations based on what one can see/measure. The acceleration explanation, e.g., A ages 25.6 years during B's turnaround event just doesn't jibe with the hardboiled engineer in me.

The Doppler shift explanation does jibe with that hardboiled engineer. I'll use Fredrik's example of space traveler B going at 0.8 c to/from a star that is 16 light years away. First, some assumptions:
• A and B continuously transmit a signal to and receive a signal from one another. Occasionally A and B will use this transmission to send messages, view family pictures, whatever.
• These continuous transmissions include a timing signal. For example, the mission elapsed time as measured on the sender's clock will be embedded once per second as measured by the sender's clock.
• B's spacecraft is equipped with sensors that can measure the distance (in B's local frame) to the target star (outbound leg) / Sun (return leg).
• B's spacecraft can similarly measure the relative velocity to the target star (outbound leg) / Sun (return leg). For example, the spacecraft might detect the frequency of the star's hydrogen alpha line and compute the velocity from the observed blueshift.

Just after B has accelerated to 0.8 c on the outbound leg, and for some amount of time thereafter, both A and B will see, by means of the communications link, the other as aging at 1/3 the rate at which they themselves are aging. Just before B returns to Earth, and for some amount of time before, both A and B will see, by means of the communications link, the other as aging at 3 times the rate at which they themselves are aging. Somewhere in between, each twin will see a transition from that 1/3 aging rate to a factor of 3 aging rate. The resolution of the paradox is that this transition occurs at distinctively different times.

What B sees
At the moment of departure (i.e., right before accelerating), B will see the target star as being 16 light years away. Assuming a rapid acceleration event, B will see the distance to the target star shrink to 9.6 light years upon reaching 0.8 c. B calculates the time needed to reach the target star as 9.6 light years / 0.8 c, or 12 years. B will see A doing things in slow motion during this 12 year long outbound leg. Suppose A gets married and has a child four years (A's clock) after B departs. When B arrives at the target star 12 years later (by her own clock), she will have just received a picture of A's newborn child. The signal from A will indicate a mission elapsed time of 4 years.

Now B stops, takes a few pictures, and turns around, doing all rather quickly. During the brief stop, B will sense that the Sun is 16 light years distance. Upon accelerating to 0.8 c, she will once again sense that distance has shrunk to 9.6 light years. Now B sees things happening to A and the child in fast motion (3x speed, to be precise). B will see A turn into an elderly gentleman and the baby zoom through life. By the time B reaches Earth 12 years later (her clock), she will see A as having aged by another 36 years. If the voyage started when A and B were both 20 years old, B will see herself as being 44 years old upon return, A as being 60, and the child as 38 -- only six years younger than B!

What A sees
From A's perspective it is B that is doing every in slow-mo on the outbound leg. A will get married, have a child, and the child will have just graduated from high school by the time A calculates that B has reached the target star. This is just a calculation, however. As far as what A can tell based on seeing, in the signal sent by B, B has a long ways to go to reach the star. A will continue to receive a slowed-down signal for another 16 years after the calculated turn around time. It will take 36 years before A sees the pictures from the target star and receives a congrats message on the birth of the child. At this point, when A is 56 years old, B will report to A that she is 32 years old. For the next 4 years, A will see B as working in fast motion and aging rapidly. By the time B returns four years later, A will be 60 years old, B will be 44, and the child will be 38.

Summary
Per this explanation, the paradox vanishes due to the disparity in the time at which the received signal switches from slow to fast. For B this happens right at the turnaround event. A has spent half of B's 24 year journey aging slowly and the other half aging quickly. From B's perspective, she has aged 24 years while A has aged 12/3 + 12*3 = 40 years. For A the transition occurs when the turnaround signal comes back to Earth. From A's perspective, A has aged 40 years while B has aged 36/3 + 4*3 = 24 years.

Last edited: Apr 30, 2010
5. Apr 30, 2010

### Staff: Mentor

I like that one too because it really shows that the symmetry argument is bogus without relying on any concepts that come in a coordinate-dependent and a coordinate-independent flavor.

6. May 4, 2010

One of you wrote: "So I'm not going to try to understand your argument in its present form." Okay with me. Please, would you then answer the following question:

What is your opinion about the explanation of the twin paradox given by Dr. Kyung ('Ken') S. Park on the site underneath:

Is this explanation correct?

As far as I could understand he does not make use of general relativity, nor of the Doppler shift explanation.

Last edited by a moderator: Apr 25, 2017
7. May 4, 2010

### Ich

Well, and three of us gave you good answers.
Did you read them? If not, why were you asking? If yes, you understood everything and have no more questions?

8. May 4, 2010

### Fredrik

Staff Emeritus
It looks good. I don't like that he says "is in S" when he should be saying something like "has velocity 0 in S", but the explanation is fine.

No one here has said anything about general relativity. If you meant my comment about proper time, that's definitely SR.

If you still want to find out what's wrong with your original argument, then you should edit it to include the information I requested and post the new version, or, if you think Ich understood what you meant, you could just continue the discussion with him.

Last edited by a moderator: Apr 25, 2017
9. May 4, 2010

### phyti

That's only because the turnaround is instantaneous, a violation of the rules of physics, and the consequent strange results.

There is no paradox because B records 40 new years for A, during B's 24 years. A is aging faster than B.

It is as you say, relativistic doppler shift.

10. May 4, 2010

### D H

Staff Emeritus
Acceleration doesn't have to be instantaneous to result in a large *calculated* change in A's age over a short period of time. People use instantaneous turnaround primarily because it makes the math easier. An instantaneous acceleration is not essential.

Sure there is. You are interpreting paradox to mean "a self-contradictory statement." Paradox has other meanings. In the case of the many paradoxes in relativity, paradox means "a statement that is seemingly contradictory or absurd and yet is true." The twin paradox is seemingly absurd, and yet it is true. In other words, it is paradox.

11. May 4, 2010

### Fredrik

Staff Emeritus
As D H said, that effect is still there even if the turnaround takes a while. If anything is to blame for this effect, it's the standard way to associate an inertial frame with an object with constant velocity. (And of course the fact that we're talking about Minkowski spacetime).

In inertial frames comoving with A, yes. But B is aging faster than A in any inertial frame that's comoving with B at any point on B's world line where B's velocity is constant. That's a paradox in the sense D H defined.

12. May 4, 2010

### stevmg

13. May 4, 2010

### Fredrik

Staff Emeritus
DaleSpam is one of many people here who have answered all these things many, many times.

14. May 5, 2010

You wrote:

"I find it odd that you went through the trouble of writing up an argument, and then abandoned it immediately when I said I can't follow it because you kept specifying coordinates of events without saying which coordinate system you were using."

Your are quite correct I 'abandoned it immediately' because reading your reply I became aware I was probably completely wrong and decided to think it all over again.

You further wrote:

"I also find it odd that you're not saying anything about the answers you got."

Again you are quite all right. For the time being I decided to stick to the text of Dr. Kyung ('Ken') S. Park to keep thinks not too complicated. But I will read your comment in due time. Be aware I am a mathematical psychologist interested in SR and not a physicist. My homepage is http://www.socsci.ru.nl/~advdv/

I am very grateful for your honesty and your patience with me and for your comments on the text of Dr. Kyung ('Ken') S. Park. If you do not mind I might still ask a question about this text in the near future. In the mean time I make my apologies to you and hope you will accept them.

15. May 5, 2010

I am going to read very carefully everething you wrote and I hope to reply soon.

Thanks for your valuable time, effort and goodwill.

16. May 5, 2010

### Fredrik

Staff Emeritus
That's OK. I think that if you try to fix your argument by making sure that you always specify what coordinate system you're talking about when you specify coordinates of an event, you would probably see what's wrong with it without our help. If you still don't see a mistake after you've done that, just post the new version of the argument and wait for someone to either tell you what mistakes you made, or tell you that you got it right.

17. May 5, 2010

To Fredrik

Oké I'll do. Till soon and thanks again.

18. May 5, 2010

### stevmg

I know, Fredrik and you DID answer it for me, too but this answer by DaleSpam really hit it on the head and really illustrated the relationship between time and distance and why the the "moving" twin is younger on the return - even without using General Relativity.

You know, this stuff is really fascinating. And, I don't even have a linac. "Linac" - does that date me?

19. May 5, 2010

### Fredrik

Staff Emeritus
If you're specifically talking about his post #129, what he's showing you there is just the easiest way to do the calculation I've been talking about:
We would of course have to add a few more words and some math to rigorously prove that what DaleSpam calculated really is the proper times of the two curves.

This (i.e. what I said in the quote above, plus a calculation like the one DaleSpam did in #129 in the other thread) is in my opinion the best answer to the question of what SR says the ages of the twins will be when they meet again.

If we're also interested in what's wrong with the argument that seems to lead to a logical contradiction, we need something more, like my spacetime diagram and a discussion about how and why we associate specific inertial frames with moving objects and why we call a description in terms of those coordinates "the object's point of view".

Also, no one has been talking about general relativity. And I had to look up "linac".

20. May 6, 2010

### stevmg

Thanks, Fredrik, I really needed that!

You'd be the type of person who would hold the door for me and say. "Age before beauty."

I heard that from my troops all the time and that was 10 - 20 years ago. How do you think I feel now?

By the way, I was giving you a compliment, not a put down.

Steve G

21. May 6, 2010

One thing is very clear to me now from: "There is no inertial frame in which Stella is at rest in both parts of the journey." Evidently, if a frame undergoes a change in velocity (speed and/or direction), then it is not an inertial frame and SR does not apply anymore. So, I quite well understand, that in this particuilar case you need three frames:

1. Earth with Gea on it,

2. the space ship on the outbound trip with Stella on it and

3. the space ship on the inbound trip with Stella on it.

Thanks a lot.

22. May 7, 2010

### Aaron_Shaw

I find that, when it comes to trying to understand the turnaround accounting for the apparent paradox, it helps to use a simultaneity diagram with a more realistic turnaround than an instantaneous one. It kinda shows how simultaneous events appear close together, then space out, and then close together again when looking at the stationary person on the diagram, when those same events are evenly spread when looking at the moving part of the diagram. Makes it clearer.

23. May 7, 2010

### Fredrik

Staff Emeritus
I like the diagram, but it should be mentioned (again) that these simultaneity lines are a consequence of the convention to define an observer's point of view at each point on his world line in terms of a global inertial frame in which the observer is at rest. This isn't something we have to do. See e.g. the article I linked to in #3.

24. May 7, 2010

### Staff: Mentor

I like that one. I think it is probably the most natural way to define coordinates in 1+1 dimensions that really represent a particular observer's "point of view".

25. May 7, 2010

### starthaus

The author of this paper is less enthusiastic.