Solving the Twin Paradox with Lorentz Transformation

[AdVen]

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 until 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 until 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 until 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 which is larger then zero: 0 < v. Is this correct?

 

[Ich]

If we take as our steady reference system (x,t) Stella

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.

according to formula (2) the time until the point of returning, calculated from the viewpoint of Gea (x',t'), is equal to 6.25.

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".

 

[Fredrik]

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

I'm calling the twin on Earth "A" and the twin in the rocket "B".
Blue lines: Events that are simultaneous in the rocket's frame when it's moving away from Earth.
Red lines: Events that are simultaneous in the rocket's frame when it's moving back towards Earth.
Cyan (light blue) lines: Events that are simultaneous in Earth's frame.
Dotted lines: World lines of light rays.
Vertical line in the upper half: The world line of the position (in Earth's frame) where the rocket turns around.
Green curves in the lower half: Curves of constant -t^2+x^2. Points on the two world lines that touch the same green curve have experienced the same time since the rocket left Earth.
Green curves in the upper half: Curves of constant -(t-20)^2+(x-16)^2. Points on the two world lines that touch the same green curve have experienced the same time since the rocket turned around.

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 http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/TwinParadox/twin_paradox.html about the twin paradox. Edit: That link doesn't work for me right now. Try this one instead.

 

[D H]

Another link that often gets posted in these threads (there's an absurd number of them) is to the http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/TwinParadox/twin_paradox.html about the twin paradox.

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.

 

[Dale]

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.

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.

 

[AdVen]

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:

http://kspark.kaist.ac.kr/Twin Paradox/Twin-Paradox Events and Transformations.htm

Is this explanation correct?

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

 

[Ich]

One of you wrote: "So I'm not going to try to understand your argument in its present form."

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?

 

[Fredrik]

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:

http://kspark.kaist.ac.kr/Twin Paradox/Twin-Paradox Events and Transformations.htm

Is this explanation correct?

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

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.

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. I also find it odd that you're not saying anything about the answers you got. Does that mean that you understood them perfectly or that you didn't understand them at all? By not saying anything about the answers you got, and instead asking about someone else's answer, you're making it look like you didn't even read the answers you got.

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.

 

[phyti]

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.

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

Per this explanation, the paradox vanishes due to the disparity in the time at which the received signal switches from slow to fast.

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.

 

[D H]

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

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.

There is no paradox ...

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.

 

[Fredrik]

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

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).

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

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.

 

[stevmg]

Didn't DaleSpam answer this here?

https://www.physicsforums.com/showthread.php?t=382591&highlight=stevmg&page=9

 

[Fredrik]

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

 

[AdVen]

Answer to Fredrik.

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.

 

[AdVen]

Answer to all of you.

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

Thanks for your valuable time, effort and goodwill.

Ad.

 

[Fredrik]

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.

 

[AdVen]

To Fredrik

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

 

[stevmg]

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

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?

 

[Fredrik]

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?

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:

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.

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".

 

[stevmg]

If you're specifically talking about his post #129
Also, no one has been talking about general relativity. And I had to look up "linac".

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

 

[AdVen]

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".

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 spaceship on the outbound trip with Stella on it and

3. the spaceship on the inbound trip with Stella on it.

Thanks a lot.

 

[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.

 

[Fredrik]

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.

 

[Dale]

This article uses another convention that's at least as natural.

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".

 

[starthaus]

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".

The author of this paper is less enthusiastic.

 

[Fredrik]

The author of this paper is less enthusiastic.

I have only had a quick look at it, but it seems to me that he's saying that there's nothing wrong with what Dolby and Gull did. It seems that his only real objection is that Dolby and Gull (according to him) claimed that the convention they're using is "right" and that using co-moving inertial frames is "wrong". If Dolby and Gull actually said that, then I agree with Eagle about that specific thing. These are just two different conventions. Neither of them is "right".

 

[Dale]

I have read the Eagle paper before. Eagle's section IV on the conventionality of simultaneity is especially valuable.

That Eagle paper only says that the Dolby and Gull paper goes too far in claiming that standard resolutions are incorrect. It does not say that the Dolby and Gull resolution is itself incorrect. I agree with both, so, while I don't believe that other resolutions are incorrect, I prefer Dolby and Gull's.

 

[starthaus]

I have only had a quick look at it, but it seems to me that he's saying that there's nothing wrong with what Dolby and Gull did. It seems that his only real objection is that Dolby and Gull (according to him) claimed that the convention they're using is "right" and that using co-moving inertial frames is "wrong". If Dolby and Gull actually said that, then I agree with Eagle about that specific thing. These are just two different conventions. Neither of them is "right".

He's saying that Dolby's claim that the textbook treatment is wrong is false. In other words, he's saying that Dolby created a strawman.
Personally, I much prefer the treatment shown here. The immediately following paragraph shows the view as calculated from the perspective of the traveling twin as well. This is quite unusual. The mathematical treatment is much tighter, does not leave any room for debate. No debate about lines of simultaneity.

 

[phyti]

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.

It isn't large changes at issue, it's instantaneous changes, continuous motion vs. discontinuous.

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.

If you compare each leg of the trip separately, not if you compare the whole trip, which is the problem. It is not symmetrical or reciprocal.

When moving clocks are brought together, the difference is real, not apparent, regardless if they appeared to run slower or faster during the separation.

 

[Fredrik]

He's saying that Dolby's claim that the textbook treatment is wrong is false. In other words, he's saying that Dolby created a strawman.

Yes, that's what I thought Eagle was saying, but I have also skimmed the Dolby and Gull article, and I didn't find any such claims in there. So maybe it's Eagle who's creating a strawman?

 

[Dale]

I think the objectionable statement by Dolby and Gull is this one: "Barbara’s hypersurfaces of simultaneity, which define ‘when events happened’ according to her, have consistently been misrepresented". That is probably overstating it, but they are certainly correct that any coordinate system which assigns multiple times to a single event is asking for trouble.

 

[phyti]

Fredrick;
Ignore, my drawing isn't correct.

 

[Fredrik]

When B-clock reads 12, a signal (blue) from the A-clock would show 4.

It's true that a radio signal from Earth that reaches B at the event where B's clock reads 12 must have been sent at an event where A's clock reads 4, but you seem to have misunderstood what the diagram is about, or the meaning of the word "when". From B's point of view, A has aged 7.2 years "when" B has aged 12 years. You find this result, not by following the world line of a light ray, but by following a simultaneity line. This is because the word "when" refers to a given time, and a simultaneity line is a set of events that are all assigned the same time coordinate.

The blue (and red) lines in my diagram are simultaneity lines, not light rays. And the text in the boxes isn't about what the twins see with their eyes or video cameras, but about what coordinates their co-moving inertial frames assign to these events.

 

[Fredrik]

I think the objectionable statement by Dolby and Gull is this one: "Barbara’s hypersurfaces of simultaneity, which define ‘when events happened’ according to her, have consistently been misrepresented". That is probably overstating it, but they are certainly correct that any coordinate system which assigns multiple times to a single event is asking for trouble.

I agree that D & G are overstating it in that sentence. Regarding the coordinate system that assigns multiple times to a single event, there are two easy ways out of that: a) Don't bother trying to combine all the co-moving inertial frames into a single coordinate system, and b) if we really want to combine them into a single coordinate system, just say that it's only defined in the region where no simultaneity lines intersect. (We will have to do that anyway when we use D & G's convention).

 

[stevmg]

To anyone out there -

How do I get quotations from different sections of a post (without the whole post) or quotations from different posts into a reply? Fredrik did just that above.

I have tried the radio button "multiquote" which does nothing.

Sorry to clutter this topic with this mundane question. If there is another site for these housekeeping type questions please direct me to it.

 

[Fredrik]

How do I get quotations from different sections of a post (without the whole post)

Use the quote button. Then delete the text you don't need. (In this case "To anyone out there -"). Then type [noparse][/quote][/noparse] after the text you want to quote, and type your reply to the quoted text after that. Always preview before you post, to make sure it looks OK. (If you e.g. mistype quote as qoute or leave out the /, the result will be ugly).

If you want to quote more text, you copy and paste the starting quote tag that was created automatically when you clicked quote. In this post I'm copying [noparse]

[/noparse] from the top and pasting it below this line.

or quotations from different posts into a reply? Fredrik did just that above.

I have tried the radio button "multiquote" which does nothing.

And of course, I still have to type a closing quote tag before this line.

To use multiquote. Either click the multiquote buttons next to each of the posts you want to reply to and then click the "new reply" button at the bottom, or click the multiquote button next to all the posts you want to reply to except the last one and click the quote button next to the last one. Note that when you click multiquote, the button changes color. You can deselect that post by clicking the button again. The color will change back.

If there is another site for these housekeeping type questions please direct me to it.

I've seen questions like these in the feedback section.

 

[phyti]

It's easier for me to have a drawing of each frame in its own scale, rather than one with two scales.
I use light rays because there's an event at each end, and the events help simplify the transformation from one frame to another without using the hyperbolic scale.
I do use the axis of simultaneity.
The left drawing shows A's description in agreement Fredrick's post 3.

This is a good example to demonstrate how the instantaneous reversal produces misleading results. The middle drawing shows the A-clock running slower than the B-clock 14.4/24. After adding the missing 25.6 at reversal, the A-clock is running faster.
The conflict is lack of symmetry. When two clocks diverge from a common origin, the clock that returns to the other accumulates the least time. In this case that is the B-clock. The number of events 40:24, does not change with a transformation.

The right drawing shows the doppler effect as mentioned by DH, with each half of the A-events at frequencies of 1/3 and 3. It also shows why I select event A4, because it is the last event arriving at B prior to reversal, and the changeover for frequencies. The A-path extends past the vertex at B12 because A never changed course at A20. Just more distortion when trying to make frames equivalent when they are not.

[PLAIN]https://www.qdrive.net/en/home/images/?userid=2104&ky=Ktj2Xn4gX81MnzXBpCtjs7vXFLXm1X7M8rx&id=124603&fn=twin-pardx2.gif

 

[Fredrik]

I don't see how any of this demonstrates that "the instantaneous reversal produces misleading results".

Note by the way that if we consider a scenario that takes much longer, say 40 million years of Earth time instead of 40 years. Then we can do the turnaround using a constant proper acceleration for a long time, and the diagram would still look more or less the same. If you make it fit on your screen, you probably won't even see the details that reveal that the turnaround took a long time, and yet it's the details you would see that lead to the description you've been arguing against.

 

[phyti]

We know that Fredrik, so what's new?

Since the B path is discontinuous at B12, when he switches frames, B's description can be done in two parts as shown. His distance is 10.6 instead of 9.6, because he never decelerates to 0. That's the difference.

[PLAIN]https://www.qdrive.net/en/home/images/?userid=2104&ky=Mg79nCX3KJpBFgnnbxzbxtOXsLD7LXxxNMX&id=124665&fn=twin-pardx3.gif

 

[stevmg]

Fredrik -

Thanks for the input on multiquoting.

To both Fredrik and phyti, instead of thinking of this as twin B turning around, just visualize the first leg of the "moving" twin's journey as one rocket ship alread at Mach ten zillion from the Earth to the "turnaround point." This rocket ship just blasts on by.. Then have a second rocket ship at mach ten zillion which is going the opposite direction and it crosses the first rocket ship exactly at the same instant and place as the first rocket ship at what is called the "turnaround point" and goes to Earth, You will have two separate trips in exact coordination from start to finish with NO acceleration/deceleration. Adding the two times up gives you the elapsed age of the traveler.

This approach was posted by DaleSpam and JesseM to avoid the problems associated with acceleration/deceleration and any semantic questions that could arise. It is not precise but gives the general idea.

 

[Fredrik]

Phyti: I still have no idea what you're talking about. Sorry.

I'm also having a hard time trying to make sense of your B diagrams. In particular I don't see the point of representing light signals going from A to B with lines that have the same slope as the ones representing light signals going from B to A. I haven't got a clue what these 9.6, 10.6, 5.3 numbers are.

 

[phyti]

Fredrik -

Thanks for the input on multiquoting.

To both Fredrik and phyti, instead of thinking of this as twin B turning around, just visualize the first leg of the "moving" twin's journey as one rocket ship alread at Mach ten zillion from the Earth to the "turnaround point." This rocket ship just blasts on by.. Then have a second rocket ship at mach ten zillion which is going the opposite direction and it crosses the first rocket ship exactly at the same instant and place as the first rocket ship at what is called the "turnaround point" and goes to Earth, You will have two separate trips in exact coordination from start to finish with NO acceleration/deceleration. Adding the two times up gives you the elapsed age of the traveler.

This approach was posted by DaleSpam and JesseM to avoid the problems associated with acceleration/deceleration and any semantic questions that could arise. It is not precise but gives the general idea.

That was done in my post, but it doesn't include the 'turnaround point' because it never happens with switching frames on the fly. The essence of this case is that it's not symmetrical. The closed path decides one twin aged more than the other, and you can't straighten out B's path or bend A's, to make it reciprocal.

 

[phyti]

Phyti: I still have no idea what you're talking about. Sorry.

I'm also having a hard time trying to make sense of your B diagrams. In particular I don't see the point of representing light signals going from A to B with lines that have the same slope as the ones representing light signals going from B to A. I haven't got a clue what these 9.6, 10.6, 5.3 numbers are.

I agree, they are not clear enough, so here is a new and improved one.
When B switches frames at B12, event A4 gets shifted from 5.3 ly distant to 48 ly,
because of Einstein's simultaneity convention.
The light paths are cyan, and red denotes the portions of A-events observed by B.
I wanted to show the asymmetry of B's description vs. that of A.

[PLAIN]https://www.qdrive.net/en/home/images/?userid=2104&ky=N9XnI2MOt3NCB8C84XKpPxXO32x6M6mNXrr&id=125014&fn=twin-pardx4.gif

 

[stevmg]

Fredrik -

The reason why there are an "absurd" number of posts on this subject is that this is one of the most elementary areas of relativity and simpletons like me can actually delve into it. To wit, I know what Doppler is and how it works with sound, but the relativistic argument about it requires me to state "If you say so."

To me when DaleSpam used the 4-space depiction of the space-time coordinates this illustrated the inversion (for want of a better word) of the Pythagorean theorem in which the longer pathway is shorter in proper time. Below, AC > AB + BC

t
C
...\
...\
...\
...B (turnaround point)
.../
.../
.../
A...x

I works out that way when you use the norm of vectors but it is surely counter intuitive. The diagrams shown above make sense in a weird way if I remember that counter-intuitive relationship.

Kind of intereseting from basic arithmetic: Take 12 and divide it by 2. This equals 6 + 6
Take 12² and 6² + 6²
We get 144 and 36 +36
Taking the square root of both sides,
we get 12 and 8.45. That's kind of the way it works in this paradox problem.

Maybe God did make the universe follow mathematical rules.

 

[AJ Bentley]

Here's an explanation of the famous 'twin paradox' of relativity. WITHOUT ANY MATH.

Relativity says time passes more slowly for a moving body than a stationary one. At the same time it says you can't tell which of the two is moving.
The paradox revolves around a pair of twins. If you don't know which one was moving, how can you tell which finishes up the older if one of them takes a round trip?

Here you go.

Both twins agree that Joe (the 'moving' twin) passed Barnard's star at the age of 45 and his hair fell out the same day.

Both twins agree that at the age of 75 Moe lost his teeth.

However, they cannot agree that these two events were simultaneous - Moe says yes, they did happen at the same time and what's more, Moe is much younger than he is - a consequence of his motion.
Joe says, on the contrary, he was over a hundred by the time Moe reached 75 and when he passed Barnard's star Moe was celebrating his 30th - not his 75th - and anyway, Moe is the younger - because he is moving!.

That is the ONLY point of disagreement between them - the issue of which past events happened at the same time. The disagreement is quite profound. Each looking into his telescope can see the other twin and how old he is. And allowing for the distance can calculate how old the twin is 'now'. Each sees the other is moving and ageing very slowly and that even allowing for the time taken for the light image to arrive, the other twin must be far younger.

In order to bring simultaneity back into line, one twin or the other (or both) has to perform an action on his world view.
Joe can 'stop' thereby moving back into Moe's frame. Or Moe can start after Joe, entering his frame.

For the twin that performs this action (acceleration) the catalogue of simultaneous events completely changes.
As he changes his velocity, all the events he was yet to see but which have already happened for the other twin sweep past him in a great flood. He sees the other twin who he thought was much younger, suddenly grow old. Of course, this is the key to the 'paradox'. If you ignore this consequence of changing velocity, you get the paradox. Why does it happen? because it HAS to happen. For the two to ever meet over dinner again, one or the other has to go through this experience. In Treckie terms, they are on opposite sides of a 'simultaneity rift'. One or the other must cross to the other side.


This is because simultaneity as a concept has literally no meaning when applied between frames. Joe and Moe are unable to agree which events were simultaneous , neither is wrong.

 

[stevmg]

Somehow, I think the math explanation by DaleSpam or JesseM is easier to understand...

 

[MikeLizzi]

This is how to resolve the “Twins Paradox”. You must recognize the following.

1. To an inertial observer the clock of an inertially moving or accelerating body is always running SLOWER than his/her own.

2. To an accelerating observer the clock of an inertially moving body is always running FASTER than his/her own.

3. The earthbound twin is assumed to be an inertial observer throughout the course of the astronaut’s trip.

4. The astronaut is an inertial observer during some portions of the trip and is an accelerating observer during other portions.

So, the earthbound observer always calculates the astronaut’s clock as running slower. Therefore the earthbound observer must conclude that the astronaut returns younger.

But the astronaut sometimes calculates the earthbound observer’s clock to be running slower and sometimes faster. If the astronaut actually makes a calculation using numbers it turns out that the acceleration/faster clock rate portion of his trip overwhelms the inertial/slower clock rate portion and the astronaut concludes he/she will return younger.

 

[stevmg]

You're making it hard

Take Twin A...B

If twin B moves to the right then he has to move to the left to get back. On that portion his relative speed to twin A is much faster than his outbound speed from A. That's the asymmetry. Twin B never is the "center of the universe" as he has to change direction. Look up JesseM's solution for the math. Twin A never changes direction.

 

[Austin0]

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.

Hi DH With your engiuneering perspective you may be the one to answer a question I have had.

With a slight alteration of your scenario.

Earth and an inertial frame at some distance. COnsider instant accerelation for simplicity.

The ship makes a round trip and both the ship and Earth send out the timed signals.
At the end of the trip they will of course agree that the number of signal sent and received was proportianal to gamma factor.

During the trip both frames will of course have a count of received signals that is behind the current number by the time interval of the propagation time of the last received signal.
The rest being spread out through space.
Could not the ship [or earth] calculate that propagation time from D/c aand then look back through their log , subtract that dt and see how many signals they had sent at that point??
Compare that number with the number of the last received signal and get a number that was accurate at the time that last signal was sent?
Getting a comparison which would be out of date but meaningful.
That because it is based simply on their own measurement of proper time and distance ,this would still apply with acceleration and that Doppler or turn around etc. would not make any difference??

If this concept is not completely off the wall it would seem to mean that relative dilation would be apparent and quantifiable along the way albeit with the qualification that it would not be completely current.

SO what do you think?

Thanks

 

[AJ Bentley]

Doppler shift has nothing to do with it. That's a Newtonian concept that won't help you to understand SR.

The twin paradox is based on a naive view of simultaneity, which was the gist of Einstein's original point.

In order to understand relativity at all, you have to have a clear grasp of the twin paradox in such a way that you see it isn't a paradox at all.

Mathematics won't help, you can push symbols around all day and still have no idea what they ultimately mean. Especially when you elect to use mathematics that don't relate to your problem.

The ONLY thing that the twins disagree on is which events for one twin were simultaneous with those for the other. That's hardly a paradox - it's simply an interesting, mildly surprising fact.