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The case for True Length = Rest Length

by rjbeery
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ghwellsjr
#109
Mar3-11, 06:41 PM
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Quote Quote by rjbeery View Post
Reciprocity exists UNTIL the return trip begins; during the acceleration involved in turning around, the reciprocity is broken and both parties will now realize that "something is amiss"
The traveling twin will observe during his turn around that the ticks from the stationary twin's clock suddenly come in faster than his own but applying Relativistic Doppler Factor, he calculates that the time dilation is identical to what it was before. The stationary twin has no knowledge of the traveling twin's turn around until long after it has occurred at which point he will also observe a similar change in tick rate but it also calculates to the same time dilation factor as before. Both twins observe exactly the same time dilation of the other twin during the entire trip (except for the insignificant time it takes for the turn around and for the take off and for the landing).
Quote Quote by rjbeery View Post
Please explain specifically how this would happen if the traveling twin kept a keen eye on his twin's clock for the entire trip; are you suggesting that "just as he lands" his Earth-bound twin's clock instantly LEAPS FORWARD in time to his amazement?
I already explained some of how this works in the first part of this post but the rest of the story is that each twin keeps track of the other twin's clock by counting the observed ticks. During the outbound half of the trip, the traveling twin counts so many ticks coming in from the stationary twin at a low rate and during the inbound half of the trip, he counts a much larger number of ticks coming in from the stationary twin at a much higher rate. In contrast, the stationary twin counts the ticks coming in at a low rate from the traveling twin for way more than half of the trip and then near the very end he counts them coming in at a high rate. You have to take into account the light travel time. Since the traveling twin counted high rate ticks from the stationary twin for a much longer percentage of the trip (one half of the trip, to be precise) than the stationary twin counted of the traveling twin, the traveling twin's total count of the stationary twin's clock is much higher than the stationary twin's count of the traveling twin's clock. Nothing happens instantly during any portion of the trip and neither twin is amazed by what happens. It's all very logical, reasonable, understandable and systematic.
rjbeery
#110
Mar3-11, 09:08 PM
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Quote Quote by ghwellsjr
Nothing happens instantly during any portion of the trip and neither twin is amazed by what happens. It's all very logical, reasonable, understandable and systematic.
I don't believe the twin paradox is anything but logical, reasonable, understandable and systematic. The contradiction arises if you try to claim that it isn't the acceleration of the traveling twin that causes the age differential.
Have you ever seen a "lines of simultaneity" analysis of the twin paradox? It looks something like this:

See that "gap" in the stationary twin's world from the traveling twin's perspective? If you study this for a bit you'll realize that ALL of their relative age differential exists because of this gap. (Note: the only reason the stationary twin appears to age "instantly" is because the graph shows, as most do, that the traveler turns around "instantly". A more reasonable rate of acceleration while turning around would produce a period of extremely fast aging for the stationary twin, yet one devoid of an unnatural gap.)
Quote Quote by ghwellsjr
Both twins observe exactly the same time dilation of the other twin during the entire trip (except for the insignificant time it takes for the turn around and for the take off and for the landing)
This analysis shows that your dismissal of the "insignificant time it takes for the turn around" is precisely what is wrong with your description and understanding.
bobc2
#111
Mar3-11, 10:07 PM
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Quote Quote by rjbeery View Post
This analysis shows that your dismissal of the "insignificant time it takes for the turn around" is precisely what is wrong with your description and understanding.
rjbeery, I think you might consider the twin example from a little different perspective. Here is a spacetime diagram that simplifies the analysis by using the symmetric spacetime diagram for the trip out. We compare times with the use of the hyperbolic calibration curves for the trip back when the speeds are different (otherwise it would defeat the ability to compare distances and times directly). Both twins experience the same proper time lapse at their respective number 9 stations. But, owing to the short cut taken by the round trip twin on the return flight (blue goes faster to catch up with red), you can see the proper times would be quite different when they get back together. Red has moved 17 proper time increments and blue has only moved through 13 proper time increments.

It is clear that shortcut taken by the twin doing the round trip accounts for the difference in age, not the turn-around acceleration. All the turnaround does is to give the round trip twin interesting variations in his view of the other twin's clock (as has already been pointed out in ealier posts). We can show the respective views each has of the other's clocks on the return trip if necessary (someone else could probably do that since I'm running out of steam).

rjbeery
#112
Mar3-11, 10:28 PM
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Quote Quote by bobc2
Here is a spacetime diagram that simplifies the analysis by using the symmetric spacetime diagram for the trip out.
Bob, with respect, I LOL'ed at this one.

Anyway, you are correct in a sense; it isn't the acceleration per se, it's the frame change. Have you ever played the game Portal? Super fun. Anyway, you have a gun that can open "portals" on any flat surface. You create two of them, and then you can travel between them instantly. Jump through one, and your momentum is carried through the other. The physics really plays with your head, especially when jump through the floor and enter through a vertical wall (and your momentum continues), or you place one directly on the floor below the other in the ceiling (so you fall "for eternity").

Anyway, my point is that if we could get our hands on one of these guns then producing an asymmetrical time dilation between two observers without either one of them accelerating would be possible. Until then, frame changing is synonymous with accelerating!
bobc2
#113
Mar3-11, 10:44 PM
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Quote Quote by rjbeery View Post
Until then, frame changing is synonymous with accelerating!

I certainly agree with you that the round trip twin did accelerate during the turnaround. However, the spacetime diagram implies a relatively insignificant increase in proper time during the turnaround. We could have shown a magnification of the turnaround to indicate that the g-levels for the blue guy would not be as high as might be inferred from my diagram. But, again, the length of the world line (curve) during turnaround for the blue guy would be relatively insignificant.

Besides, it is obvious that it is the high speed at the end of the acceleration that provides the short cut through spacetime. We're not doing anything like sending the blue guy off to the neighborhood of a black hole. In any case we keep the acceleration under control so as to keep the problem in the realm of Special Relativity.
ghwellsjr
#114
Mar4-11, 01:42 AM
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You asked me:
Quote Quote by rjbeery View Post
Please explain specifically how this would happen if the traveling twin kept a keen eye on his twin's clock for the entire trip; are you suggesting that "just as he lands" his Earth-bound twin's clock instantly LEAPS FORWARD in time to his amazement?
I explained to you what the traveling twin's keen eye would see of his twin's clock during the entire trip. I said that during the outbound half of the trip, he will see his twin's clock ticking at some rate slower than his own. Then after he turns around, he will see his twin's clock ticking at some higher rate than his own. The sum total of all the ticks is the amount of aging the stationary twin experienced.

And then I also explained what the stationary twin's keen eye sees of the traveling twin's clock. I said that for way more than half of the trip, he sees the traveling twin's clock ticking at some rate slower than his own (the same slow rate that the traveling twin sees during the first half of the trip). Then I said that near the end of the trip, he sees the traveling twin's clock ticking at some rate higher than his own (the same high rate that the traveling twin sees during the last half of the trip). The sum total of all the ticks is the amount of aging the traveling twin experience.

The fact that the stationary twin counted low rate ticks for much more than half of the trip illustrates how he sees the traveling twin as aging less than himself.

But then you responded by saying that all of the differential aging occurs during the acceleration at turn around which has nothing at all to do with what the twin's keen eye sees. Why do you ask me to explain what he sees and then complain about something that has nothing to do with what he sees?

You also asked me if I can see something in a graphic but the graphic is broken. All I can see is a framed box with an X in it. So I cannot respond to your questions but it really doesn't matter because as I already explained, you haven't shown what either twin sees which is what you asked me to explain.

And keep in mind, I explained what each twin sees without bringing Special Relativity into the picture. You can also explain the Twin Paradox, as I said before, by using any frame of reference. But you have to be careful to illustrate in that frame what each twin actually sees and it will be exactly the same as I described without using SR. It doesn't matter which frame you use to analyze a scenario, they all agree on what each observer sees.

So my simple question to you is: do you deny my description of what the twin's see?
JesseM
#115
Mar4-11, 03:00 AM
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Quote Quote by ghwellsjr View Post
But then you responded by saying that all of the differential aging occurs during the acceleration at turn around which has nothing at all to do with what the twin's keen eye sees.
Did rjbeery ever say that exactly? If so, in which post? I thought rjbeery was just arguing that the acceleration explains why one twin has aged less in total when they reunite, which is not the same as saying all the differential aging occurred "during the acceleration". Neither of what either of you are saying about the twin paradox seems incorrect to me so I don't quite understand what you're disagreeing about, either I misunderstood something about your arguments or you guys are misunderstanding each other...
GrayGhost
#116
Mar4-11, 04:18 AM
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Quote Quote by rjbeery to hgwellsjr View Post
Sure...

This part is wrong. Reciprocity exists UNTIL the return trip begins; during the acceleration involved in turning around, the reciprocity is broken and both parties will now realize that "something is amiss". Otherwise, the following statement leads to a direct and obvious logical contradiction:

Please explain specifically how this would happen if the traveling twin kept a keen eye on his twin's clock for the entire trip; are you suggesting that "just as he lands" his Earth-bound twin's clock instantly LEAPS FORWARD in time to his amazement?
Well, during periods of twin B inertial motion, the reciprocity always exists and can be observed. However when twin B undergoes proper acceleration, it's another story ...

The overall experience per twin B is twin A's wildly spinning distant clock during periods of twin B's own proper acceleration. However, this "overall experience" is the superposition of 2 relativistic effects ...
(1) the reciprocity of slower ticking clocks, and

(2) the change in relative simultaneity between the 2 POVs.
So, the reciprocity of moving clocks always holds mathematically (as ghwellsjr stated), however the change in relative simultaneity counters that effect (from B's POV), twice over ... and so the reciprocity of moving slower-ticking-clocks cannot be observed, and can only be deduced as the superposition of 2 relativitic effects that concurrently concur.

That said, I see you and ghwellsjr both as correct. However, if you think that relative clock rates are illusionary effect, in this you are mistaken. Whether inertial or undergoing proper acceleration, what a clock presently reads dictates its real time and thus the proper time experienced by the clock since the 1st of 2 spacetime events.

GrayGhost
bobc2
#117
Mar4-11, 07:36 AM
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Quote Quote by ghwellsjr View Post
And keep in mind, I explained what each twin sees without bringing Special Relativity into the picture. You can also explain the Twin Paradox, as I said before, by using any frame of reference. But you have to be careful to illustrate in that frame what each twin actually sees and it will be exactly the same as I described without using SR.
ghwelsjr, I was pretty much following everything you've been saying, except I don't follow what you mean about not using special relativity to explain the twin paradox. I've always understood the twin example as following from application of the knowledge of special relativity (I did rely upon it in my spacetime diagram above--post 102).
Mentz114
#118
Mar4-11, 07:46 AM
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Quote Quote by PAllen View Post
I've made some attempts at this without ever carrying it through to a conclusion; but enough to see that what a movie shows would be quite different from the same data interpreted by removing light delays with standard conventions. Also, note that you can remove issues of interpreting light signals (at least in thought experiments) by such direct means as a hypothetical sheet of detecting tissue across each door opening (separate from the doors), attached to recording clock 'right there' so no time delay. Then, irrespective of what an observer would 'see' from any single vantage point, they could put all their data together and find it hard to avoid concluding they had momentarily trapped the 100 meter rocket in the 10 meter barn.
I had a good look at the barn-pole scenario and this is what I got. I hope I haven't made an error but I'm sure someone will tell me if I have.
Attached Files
File Type: pdf barnpole.pdf (61.8 KB, 8 views)
ghwellsjr
#119
Mar4-11, 08:50 AM
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Quote Quote by JesseM View Post
Quote Quote by ghwellsjr View Post
But then you responded by saying that all of the differential aging occurs during the acceleration at turn around which has nothing at all to do with what the twin's keen eye sees.
Did rjbeery ever say that exactly? If so, in which post? I thought rjbeery was just arguing that the acceleration explains why one twin has aged less in total when they reunite, which is not the same as saying all the differential aging occurred "during the acceleration". Neither of what either of you are saying about the twin paradox seems incorrect to me so I don't quite understand what you're disagreeing about, either I misunderstood something about your arguments or you guys are misunderstanding each other...
First off, I appreciate that you agree with me that my description of what each twin sees is correct and that my statement that each twin views the other one as experiencing time dilation during the entire trip is correct.

But, rjbeery does not agree with you or with me. I'm trying to figure out exactly what he disagrees with me about. That's why I asked him at the end of my post you referenced if he denies my description of what each twin sees.

But to answer your question about where he said that all the differential aging occurs during the acceleration at turn around:
Quote Quote by rjbeery View Post
(Note: the only reason the stationary twin appears to age "instantly" is because the graph shows, as most do, that the traveler turns around "instantly". A more reasonable rate of acceleration while turning around would produce a period of extremely fast aging for the stationary twin, yet one devoid of an unnatural gap.)
PAllen
#120
Mar4-11, 09:41 AM
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Quote Quote by Mentz114 View Post
I had a good look at the barn-pole scenario and this is what I got. I hope I haven't made an error but I'm sure someone will tell me if I have.
Didn't spot any errors. However, see my post #98 for more aspects of the situation. In your scenario, a key point is to imagine cameras on the inside of each barn door taking pictures every nanosecond, previously synchronized to extreme precision (in the barn frame). These cameras will have a period where one photographs the front of the rod and the other the back. However, the camera snapping the approaching rod will see it extending out the barn (or in my more extreme scenario, before it has even entered the barn). All the same, the other camera will simultaneously show the back of the rod. Also, the readout from my proposed tissue detectors will clearly show the rod inside the barn. As, for the cameras, it is easy to understand that the camera snapping the approaching rod is showing a very stale image, and thus concluding that the two cameras together 'prove' the rod was in the barn.

This all shows the even in one frame, you better clearly define how you measure something and how you interpret those measurements. Different, perfectly reasonable choices can lead to different results.
ghwellsjr
#121
Mar4-11, 09:48 AM
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Quote Quote by GrayGhost View Post
Well, during periods of twin B inertial motion, the reciprocity always exists and can be observed. However when twin B undergoes proper acceleration, it's another story ...

The overall experience per twin B is twin A's wildly spinning distant clock during periods of twin B's own proper acceleration. However, this "overall experience" is the superposition of 2 relativistic effects ...
(1) the reciprocity of slower ticking clocks, and

(2) the change in relative simultaneity between the 2 POVs.
So, the reciprocity of moving clocks always holds mathematically (as ghwellsjr stated), however the change in relative simultaneity counters that effect (from B's POV), twice over ... and so the reciprocity of moving slower-ticking-clocks cannot be observed, and can only be deduced as the superposition of 2 relativitic effects that concurrently concur.

That said, I see you and ghwellsjr both as correct. However, if you think that relative clock rates are illusionary effect, in this you are mistaken. Whether inertial or undergoing proper acceleration, what a clock presently reads dictates its real time and thus the proper time experienced by the clock since the 1st of 2 spacetime events.

GrayGhost
Your statement:
The overall experience per twin B is twin A's wildly spinning distant clock during periods of twin B's own proper acceleration.
is not correct.

Twin B (the traveling twin) never experiences twin A's clock wildly spinning and by that I mean twin B never observes twin A's clock wildly spinning. During the acceleration period twin B will see twin A's clock transition from a tick rate lower than his own to a tick rate higher than his own. If the acceleration is instant, the transition will be instant. If it is gradual, the transistion will be gradual. If the acceleration is in two parts where twin B decelerates to a stop and stays there for awhile and then accelerates in the back-home direction, the transition will be in two parts, first twin B will see twin A's clock transition from a tick rate lower than his own to a tick rate identical to his own and stay there for awhile and second twin B will see twin A's clock transition from the same rate as his own to a higher rate than his own.

And twin A will see identical transitions in the tick rates of twin B's clock, except that instead of them occuring at the half-way point in the trip, they occur closer to the end of the trip. It's only this lack of symmetry in when the transitions occur that accounts for the twin's observations of the difference in aging when they finally reunite.

Keep in mind that there are similar transitions that occur during twin B's initial acceleration and final deceleration but all these acceleration periods only complicate the issue which we are discussing which is the reciprocal time dilation that each twin observes of the other one's clock. If we make the accelerations be instant, then we can say that each one always observes the same time dilation of the other one's clock during the entire trip. If we insist on making the accelerations take time, then we have to put in a little caveat that the time dilation will not be reciprocal nor be constant during the entire trip but the effect is quite minor and really doesn't impact the point of the discussion.
ghwellsjr
#122
Mar4-11, 09:58 AM
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Quote Quote by bobc2 View Post
Quote Quote by ghwellsjr View Post
And keep in mind, I explained what each twin sees without bringing Special Relativity into the picture. You can also explain the Twin Paradox, as I said before, by using any frame of reference. But you have to be careful to illustrate in that frame what each twin actually sees and it will be exactly the same as I described without using SR. It doesn't matter which frame you use to analyze a scenario, they all agree on what each observer sees.
ghwelsjr, I was pretty much following everything you've been saying, except I don't follow what you mean about not using special relativity to explain the twin paradox. I've always understood the twin example as following from application of the knowledge of special relativity (I did rely upon it in my spacetime diagram above--post 102).
I'm not saying that my knowledge of Special Relativity hasn't influenced my thinking about how to explain the twin paradox without using SR, in fact, it is from SR that we learn how to properly understand Relativistic Doppler which is what I have been using. It's just that I have not established a frame of reference to explain what each observer sees of the other one's clock during each part of the trip. Do you think if you were the traveling observer, you would need to think about a frame of reference when you watch something? I don't think you go around all day long giving any thought to what frame of reference you should be considering your observations to be made in. And the point is, when you do invoke SR and establish a frame of reference, it does not in any way change what the observers see, it merely adds another level of complexity to the issue at hand which is: what do the observers see?
rjbeery
#123
Mar4-11, 10:32 AM
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ghwellsjr, I'd like to apologize only for the reason that our discussion seems to have taken on a small bit of defensive posturing. Our mutual goal should be to further our understanding of Physics rather than to play a game of "speech and debate", parsing each others' words for errors.

That being said, since you're asking explicitly, I'll explain my position one last time. First of all, if you can't see my graphic of lines of simultaneity, let me try again, because it's important...

Quote Quote by RJBeery
See that "gap" in the stationary twin's world from the traveling twin's perspective? If you study this for a bit you'll realize that ALL of their relative age differential exists because of this gap. (Note: the only reason the stationary twin appears to age "instantly" is because the graph shows, as most do, that the traveler turns around "instantly". A more reasonable rate of acceleration while turning around would produce a period of extremely fast aging for the stationary twin, yet one devoid of an unnatural gap.)
The picture is from Wikipedia, but you can read more about this analysis here:
http://chaos.swarthmore.edu/courses/PDG/AJP000384.pdf

That's my perspective. All age differential is due in toto to the frame change caused directly by acceleration of one of the twins. Now, on to what you're saying...
Quote Quote by ghwellsjr
They all agree that each party views the other party's clock as running slower than their own during the entire trip and yet when they re-unite, the traveling twin's clock has progressed a lesser amount of time.
First of all, this isn't right and doesn't even coincide with the very lengthy explanations you have subsequently given. I suspect it was a mental lapse that you corrected when you did a thorough walk-through.

Quote Quote by ghwellsjr
The reciprocity is not broken just because one of them accelerated.
Yet...
Quote Quote by ghwellsjr
The stationary twin has no knowledge of the traveling twin's turn around until long after it has occurred at which point he will also observe a similar change in tick rate but it also calculates to the same time dilation factor as before.
What you're saying is that Twin B sees an effect that Twin A does not "until long after it has occurred"...that's the very definition of a break in symmetry!

There are many ways to analyze the Twin Paradox, but trying to do so using only SR concepts devoid of acceleration will ultimately fail. The "cause" of the age differential is acceleration, period. That is and continues to be my point, and if you or anyone else has an interpretation of the Twin Paradox that does not involve a frame change based caused by acceleration I would be fascinated to learn of it. Thanks
PAllen
#124
Mar4-11, 10:54 AM
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Quote Quote by PAllen View Post
Didn't spot any errors. However, see my post #98 for more aspects of the situation. In your scenario, a key point is to imagine cameras on the inside of each barn door taking pictures every nanosecond, previously synchronized to extreme precision (in the barn frame). These cameras will have a period where one photographs the front of the rod and the other the back. However, the camera snapping the approaching rod will see it extending out the barn (or in my more extreme scenario, before it has even entered the barn). All the same, the other camera will simultaneously show the back of the rod. Also, the readout from my proposed tissue detectors will clearly show the rod inside the barn. As, for the cameras, it is easy to understand that the camera snapping the approaching rod is showing a very stale image, and thus concluding that the two cameras together 'prove' the rod was in the barn.

This all shows the even in one frame, you better clearly define how you measure something and how you interpret those measurements. Different, perfectly reasonable choices can lead to different results.
Addendum: There will be a point in time where both head on and tail on cameras see the rod inside the barn. In may be only when the rod just about the smash the camera it is approaching.
bobc2
#125
Mar4-11, 12:52 PM
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Quote Quote by rjbeery View Post
That's my perspective. All age differential is due in toto to the frame change caused directly by acceleration of one of the twins. Now, on to what you're saying...

The "cause" of the age differential is acceleration, period.
rjbeery, with due respect I really don't see how that comes out of our analysis. Certainly the change in velocity of the returning twin is a result of acceleration. But, it is the final velocity that is to be associated with the subsequent shortcut to the final destination.

Quote Quote by rjbeery View Post
That is and continues to be my point, and if you or anyone else has an interpretation of the Twin Paradox that does not involve a frame change based caused by acceleration I would be fascinated to learn of it. Thanks
But, I really have already presented the spacetime diagram that shows clearly the effect of the short cut through spacetime accounting for the difference in twin ages (post #111 pg 7 above). Your spacetime diagram does not include the hyperbolic calibration curves necessary to follow the progression of proper time for the return trip. You can easily come to the wrong conclusions without correctly representing the proper time. If you will add in the two sets of hyperbolic calibration curves onto your spacetime sketch, then the effect of shortcuts thru spacetime will become clear.
bobc2
#126
Mar4-11, 01:04 PM
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Quote Quote by PAllen View Post
Didn't spot any errors. However, see my post #98 for more aspects of the situation. In your scenario, a key point is to imagine cameras on the inside of each barn door taking pictures every nanosecond, previously synchronized to extreme precision (in the barn frame). These cameras will have a period where one photographs the front of the rod and the other the back. However, the camera snapping the approaching rod will see it extending out the barn (or in my more extreme scenario, before it has even entered the barn). All the same, the other camera will simultaneously show the back of the rod. Also, the readout from my proposed tissue detectors will clearly show the rod inside the barn. As, for the cameras, it is easy to understand that the camera snapping the approaching rod is showing a very stale image, and thus concluding that the two cameras together 'prove' the rod was in the barn.

This all shows the even in one frame, you better clearly define how you measure something and how you interpret those measurements. Different, perfectly reasonable choices can lead to different results.
PAllen & Mentz114, did you guys find something wrong with my post #102 pg 7, or were you just interested in understanding the example strictly from the standpoint of observer measurements?


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