The case for True Length = Rest Length

ghwellsjr

Why do you say "restricted to SR"? Are you leaving open a loop-hole through which you can explain the Twin Paradox?

rjbeery

It's because SR effects produce measurements that are apparently contradictory and reciprocal (i.e. each party concludes the other's watch is slower), similar to mutual foreshortening. When you involve acceleration you break that reciprocity.

ghwellsjr

You can analyze the Twin Paradox from any frame of reference. 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. The reciprocity is not broken just because one of them accelerated.

You can also analyze the Twin Paradox without using any frame of reference and without using Special Relativity. You can analyze it simply from the observations of each other's clocks during the trip using Relativistic Doppler. Again, they always see each other's clock as running slow compared to their own in a reciprocal manner and yet when they re-unite, the traveling twin's clock has progressed a lesser amount of time.

SR does not make or create the way nature works, it's merely one way to describe and analyze it.

rjbeery

This is incorrect. If acceleration did not determine which twin was aging slower we could consider the "traveling" twin to be motionless and the other twin to be on a giant, Earth-shaped spaceship. You're not realizing that labeling one of them as "traveling" is equivalent to requiring that they undergo acceleration (i.e. at the very minimum to turn around and head back to their sibling).

ghwellsjr

You missed my point. I said that the acceleration of the traveling twin does not break the reciprocity of each twin seeing the other's clock as always running slower than their own. That's what you said in post #104. If after acceleration, the traveling twin saw the stationary twin's clock as running faster, then your comment might have some merit.

So I don't know what it was in my statement that you quoted that you believe is incorrect. Could you be more specific?

rjbeery

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?

ghwellsjr

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).
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

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.)
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

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

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

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

You asked me:
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

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

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

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

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

barnpole.pdf (61.8 KB, 8 views)

ghwellsjr

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: