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In the most general form of the paradox, equal speed profiles are not a requirement.

OK, I understand your notation. The above is: 1. true only if B and C have the same speed wrt A 2. false if B and C have different speeds wrt A I showed this in my writeup.

What are events AB, BC, AC?

Good, because elapsed proper time is what the twins "paradox" is all about. Anything else, you can open a different thread and discuss it to your heart content, please stop trying to hijack this thread.

Apparently, you don't know. The twins paradox is all about comparing proper times. Oh, one last thing, you should learn how to express physics in symbolic math. All this stuff about numbers reeks of not knowing how to express yourself.

It is really simple: elapsed proper time is frame invariant, if you get different results depending on the frame of reference, you have done something wrong. Like you defending an indefensible calculation.

True, it is the sleigh of hand in resetting the C clock to the B value that does the trick. The point is that you created a BOGUS scenario to give creedence to your claim that you can generate a "twin paradox without acceleration". All you have done is to create a BOGUS scenario where you fiddle with C clock midway and you hope to hide the sleigh of hand. In the beginning, I thought that you honestly did not realize what you have done but, after all these exchanges, I am beginning to doubt that. I think you fully know the extent of your sleigh of hand but you think that denying it will make it go away.

Elapsed proper time is frame invariant, so it doesn't depend on the frame of reference . To make things simple, you can choose a single frame , in this case the frame chosen was A. If, on the other hand, you get different results when calculating from different frames of reference, it means that you have the wrong result. So, you can stop defending a wrong position for the time being.

Arbitrarily advancing/regressing C's clock is , in terms of calculating elapsed time, nothing but a jump in speed, i.e......acceleration. This is precisely what happens at the meeting point, where C resets his clock to the value of B. I thought that was very clearly explained in my writeup.

Neither have I. I was talking about my calculations, not his. You need to pay attention. On the other hand, it doesn't matter if md's calculations are right, the point is that they hide the presence of acceleration, this is what the debate is all about. You need to pay attention.

The twin paradox is judged from a single, unique frame perspective (A , in the case above). I am sorry that you don't know this elementary fact. If you compare apples and oranges, don't be surprised that you get ....bananas. I know he's trying to do that. The counter-argument is that the acceleration is there, it is hidden (or he doesn't see it).

Frame swapping (jumping) is the same thing as infinite acceleration. The situation is pretty bad when u=v (v_B=v_C). It only gets much worse (see writeup) when the speeds are different.

Yes, the infinite acceleration is replaced with a frame swap at the midpoint where B and C meet. I pointed out the frame-jumping at the very beginning of the thread. The whole explanation is "according to A" in calculating the elapsed proper time for B and C. The only exception is the calculation of total elapsed proper time for A. RoS never enters into the picture, especially when calculating elapsed proper time. So, please stop making up strawmen. I am used to arguing with people like you, you will never admit to error, this doesn't mean that you are correct.

You have the math to work with, the language of physics is math, "literary composition" doesn't cut it. This problem is judged from one frame only , so it has nothing to do with RoS. Perhaps if you stopped making sweeping statements and you started writing math, you could be more believable. For the time being, not. If B anC have the same speed, B is irrelevant to the problem, the paradox reduces to the form where B turns around instantaneously, with infinite acceleration. This is the fourth time when I point this out. No matter how you three guys twist and turn, the paradox is explained only in the presence of acceleration.

You need to learn how to follow the math. It appears that md is starting to understand where he went wrong. Sometime, in the future, maybe you will understand as well.

Of course you are right, even when you are totally wrong, even when you demonstrate that you don't know basics.

You are still wrong, proper time does not depend on the observer, it is frame invariant. So you cannot have: "According to A, A ages 4 years, while C ages 2 years. According to C, A ages 4 years, while C ages 8 years.." I cannot ever agree with the stuff you just wrote, basic SR says you are wrong and you continue along the path of being wrong.

Good, you are making some progress in admitting you are wrong. The mistake runs much deeper than that: in your naive setup, B is made totally irrelevant because it has the same speed as C, B doesn't play any role, A and C move inertially towards each other and record the same proper time, nothing to do with the "twins paradox" . So, your argument in the OP falls apart. If, on the other hand B and C have different speeds, your argument falls apart for a different reason, C clock has been advanced (or retarded) by a non-null jump when it encountered B. So, your "experiment" is bogus. I know becuase I took the trouble to write the exact mathematical description instead of piling words. This is how I know.

That is false, the time on C is CHANGED to the time on B when the two meet. Pay attention before you post such mistakes. "Start with 2 passing clocks, A and B, which are each set to zero at passing. Let B travel some distance at velocity v, where it passes clock C traveling in the opposite direction at the same speed, and have C set its clock to match B's as they pass. When C and A meet they'll record the same difference in proper time as if the experiment was run with clock B instantly turning around as it passes clock C."

Nope, it is the same "experiment": "Start with 2 passing clocks, A and B, which are each set to zero at passing. Let B travel some distance at velocity v, where it passes clock C traveling in the opposite direction at the same speed, and have C set its clock to match B's as they pass. When C and A meet they'll record the same difference in proper time as if the experiment was run with clock B instantly turning around as it passes clock C." The only difference is that you set the speed of C to be equal to the speed of B. According to my explanation, this makes B totally irrelevant (I told you this several times already). What I did is to generalize to "u \ne v" and to show the absurdities your scenario leads to. Instead of rushing your answers in trying to defend your position, how about you read the counter-argument I wrote. It dissects your scenario with math, not with empty words. It is hard to write any math on this forum, this is why people don't post any, they just talk arond the subject.

LOL, the C clock shows one time when it arrives in P and a DIFFERENT time when it leaves P. I agree with you, this thread belongs in Speculation. It belonged from the very first post.

It is physically possible but it produces BOGUS results. Try following the counter-argument instead of pretending that it isn't there. The math is pretty elementary.

This is the whole point, in the BOGUS scenario that you describe in the first post, the clocks DO exhibit a jump via the fact that B assigns its time to C. You need to follow the argument, it is pretty basic.

This is the whole point, in the BOGUS scenario that you describe in the first post, they DO.

the reason that you keep getting unphysical answers is that you do not realize that the exercise, as set up, hides acceleration. Not only that it hides acceleration but, in certain instances, it produces contradictory results. At the root of all the problems is the "jump" in elapsed proper time on C's clock. This forum is horrible for math, so you will have to bear with what is to follow. In the description, we have A and B at one end of a segment of length d (as measured by A, at rest wrt the segment) and C at the other end. In frame F_A, B and C start simultaneously from the ends of the segment with speeds u and v, respectively. They meet at t=d/(u+v) as measured by A in point P. At that point, the proper times shown on their clocks are: tau_BP=d sqrt(1-(u/c)^2)/(u+v) tau_CP=d sqrt(1-(v/c)^2)/(u+v) C "dumps" its time and "assumes" the time shown by B. C continues towards A accumulating an additional: tau_PA=sqrt(1-(v/c)^2)(d/v-d/(u+v)) So, the total time shown by C when it meets A is: tau_CA=tau_CP+tau_PA=d/(u+v)(sqrt(1-(u/c)^2-sqrt(1-(v/c)^2)+d/v*sqrt(1-(v/c)^2) On the other hand, the proper time accumulated by A can be calculated by a frame comoving with C, F_C as : tau_A=d/v*sqrt(1-(v/c)^2) (all frames agree on proper time so we might as well use F_C) It means that the difference between the times showed by C and A is: tau_CA-tau_A=d/(u+v)(sqrt(1-(u/c)^2-sqrt(1-(v/c)^2) We have three possibilities; 1. u=v . This is the case usually shown in textbooks. B plays no role (at the P "exchange" B and C show the same elapsed time). So: tau_CA=tau_A (there is no difference in total elapsed time , so the whole "paradox" is NONEXISTENT) 2. u>v tau_CA< tau_A 3. u<v tau_CA>tau_A case 3 is particularly disturbing because it CONTRADICTS the spirit of the "twin paradox", the "traveling twin" shows MORE elapsed proper time. The reason why the results are BOGUS (sorry for the nonscientific term but it applies) is that the whole setup is BOGUS, C's clock is forced to "jump" (forward or backward, depending on u<v, u>v) at point P. A BOGUS scenario results into BOGUS results. So, the OP was false from the get go.