Solving the Twin Paradox with Lorentz Transformation

In summary: I'll post another one.In summary, the conversation discusses the twin paradox and the Lorentz transformation formulas used to explain it. The paradox involves identical twins Gea and Stella, with Stella going on an interstellar journey and returning to Earth younger than Gea due to time dilation. However, it is also argued that Gea's clock will appear slower to Stella, creating a paradox. The conversation also explores different coordinate systems and spacetime diagrams to understand the paradox and concludes that there is no actual paradox, but rather a misunderstanding of the application of the time dilation formula.
  • #1
AdVen
71
0
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?
 
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  • #2
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".
 

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  • #3
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

Fredrik said:
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.
 
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  • #4
Fredrik said:
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.
 
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  • #5
D H said:
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.
 
  • #6
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.
 
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  • #7
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?
 
  • #8
AdVen said:
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.
 
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  • #9
D H said:
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.
 
  • #10
phyti said:
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.
 
  • #11
phyti said:
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).

phyti said:
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.
 
  • #13
DaleSpam is one of many people here who have answered all these things many, many times.
 
  • #14
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.
 
  • #15
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.
 
  • #16
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
To Fredrik

Oké I'll do. Till soon and thanks again.
 
  • #18
Fredrik said:
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?
 
  • #19
stevmg said:
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:
Fredrik said:
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".
 
  • #20
Fredrik said:
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
 
  • #21
Ich said:
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.
 
  • #22
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.

gr.gif
 
  • #23
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
Fredrik said:
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".
 
  • #25
DaleSpam said:
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.
 
  • #26
starthaus said:
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".
 
  • #27
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.
 
  • #28
Fredrik said:
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.
 
  • #29
D H said:
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.
 
  • #30
starthaus said:
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?
 
  • #31
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.
 
  • #32
Fredrick;
Ignore, my drawing isn't correct.
 
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  • #33
phyti said:
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". :smile: 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.
 
  • #34
DaleSpam said:
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).
 
  • #35
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.
 

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