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The whole thread is about pretending that "acceleration is not important in the twin paradox" while trying to hide it all along.

No, it wasn't understood. MD has admited it at times only to recant later, followed by admitting it just a few posts later.

Yes, for the past 190 posts.

The acceleration is hidden through a trick. You jump from observer B to observer C at event BC. If you add up the proper times of B (for the outboard leg) with the time of C (for the return leg) it is as if you accelearted. You may not admit it but this is what you did. Yet, the acceleration is there, right under your nose.

There is no coordinate transformation in the explanation. So, no. Like I said, it is very simple: you either - dump C's time and reset the clock to B's time (see citation from MD's claim in the previous post, just in case you missed it), meaning that you are generating a discontinuity in C's clock reading - make up the journey from two DIFFERENT parts, one whereby B moves at [math]v_B[/math] and accumulates proper time [math]\tau_B[/math] and the second leg whereby C moves at [math]v_C[/math] and accumulates proper time [math]\tau_C[/math]. This means that at event BC the speed jumped from [math]v_B[/math] to [math]v_C[/math] It reflects your tone. Correct. Correct, the acceleration is "hidden" through this trick that has been exposed since post 73. In other words, the acceleration has always been there, it just takes an experienced debunker to expose it.

Not according to MD. From post 74: One way to describe the discrepancy in MEASUREMENT while observing the same thing, is to consider the differing distances that the light signal travels from AB to BC. Remember each observer measures light traveling at a speed of c, and if you accept that, then relativity of simultaneity has these unintuitive effects. Using gamma=2, v=.866, proper time T from AB to BC = 1 year (measured by B)... For A, the rest distance between itself and BC (imagine a marker in space at that location, or consider that events occur in all frames, they don't have an inertial frame and can be considered at rest relative to anyone) is the length-contracted distance traveled by B, converted to A's frame: v*T*gamma = 1.732 LY. Observer A ages 2 years while B ages the one, and t = 2-1.732 = 0.268, the age A is when it must send a signal to reach event BC. For B: If A sent a signal at time t, the time at B is t*gamma = 0.536, and the light spends the rest of proper time T incoming, ie. T-t*gamma = 0.464 years. For B, the signal travels 0.464 LY, during which A ages 0.464/gamma = 0.232. Add t, the age when A sent the signal, and you get that A aged 0.5 years when B reaches C, in agreement with the Lorentz transformation. For C: Observer C is closing on B extremely quickly. Using composition of velocities, its relative velocity is 0.990c, Lorentz factor gammaB=7. While B ages 1 year, C ages 7 before they pass. Taking 7 years to reach BC, it measures that A ages 7/gamma (it is also moving at -v relative to A) = 3.5 years. If A sends its signal at 0.268 years old, it will age 3.5-0.268 = 3.232 years while the light signal approaches BC, during which C ages 3.232*gamma = 6.464 years. So to recap: Due to relativity of simultaneity, and light traveling the same relative speed according to all the observers, A observes that the signal from AB to BC travels 1.732 LY, B measures 0.464 LY, and C measures 6.464 years. You need to keep on reading, IF you do that , this is equivalent to accelerating from [math]v_B[/math] to [math]v_C[/math]. And no, this is not a coordinate transformation. It is very simple really.

This is false: "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." Clock C accumulated a proper time [math]\tau_C[/math] between start and BC. At BC, clock C was reset to mathch clock B value, [math]\tau_B[/math] where [math]\tau_B \ne \tau_C[/math]. Even if you claimed that C is simply set to [math]\tau_B[/math] at BC ignoring its history, the above is STILL equivalent to executing a step in speed from [math]v_B[/math] to [math]v_C[/math], i.e. accelerating from [math]v_B[/math] to [math]v_C[/math]. This was explained twice, once in post 73 and a second time in post 177.

They will be younger. How much younger depends on the length of the trip and how fast they were going. Do not confuse frequency shift up/down with elapsed proper time, they are different entities.

It is very simple , really. If you reset the C clock to the value of the B clock, there is a jump in elapsed time because at the event BC , when the clocks met, they have accumulated different amounts of proper time. This is what post 177 demonstrates mathematically. Post 73 showed the same calculations (not in Latex).

post 177 explains your sleigh of hand quite clearly.

Because, at the turn around point the line of simultaneity makes a "jump". Half of the value of the jump can be attributed to the "before" the turnarounf and the other half to the "after" the turnaround. The situation you describe is depicted here.

Science is not a democracy, it is not settled by vote. Out of curiosity, who do you think is on your side?

The jumps are explained in post 177, they are due to your sleigh of hand in resttingC to the time of B at the BC encounter. This has been explained multiple times to you, the mathematical explanation is in post 177 (a repeat of post 73).

No, it isn't, this is a common mistake. The twins paradox is about elapsed proper time, not about time dilation.

Read posts 73 and 177. They expose where your sleigh of hand is. Post 2, by Markus exposes a different , more fundamental , flaw in your post 1. I suggest that you read all these posts.

It is clear that he would not admit to error, even if you clearly outlined it in post 2 and I exposed the sleigh of hand in the scenario in both posts 73 and 177

But in the scenario you contrived, they DO. Look at post 177 (or 73). At BC you made C dump its accumulated time and pick up the time shown by B. This results into a step, jump i, discontinuity in C's reading. This is the sleigh of hand that I have been referring to.

It means your claim is false, there is no wording you claim in my post 73. There is no "retraction", you just made a false claim.

I don't think you understand what post 73 shows, it shows that there is a skip in the C clock at the BC event. The skip is explained again, in Latex at post 185. Did you read the math at post 73? There is no such wording as you claim in post 73, there is just math.

I am sorry, I did not mean to insult. I just have a very low tolerance to cheating and lying. I have proven my point at post 73, yet , it took another 140 posts to conclude that the claims in the OP are indeed false. I hope that I will not encounter another thread like this, and, if I do, I will stay away or simply report it , as I have successfully done with the two threads that you moved to "Speculation". This is not correct, the correct statement is : "Acceleration IS required to physically measure the time dilation effect demonstrated in the twin paradox." This is not correct, the correct statement is : "Acceleration IS required to physically measure the time dilation effect demonstrated in the twin paradox." . This was known since my post 73. I have long proved it, post 73.

Good, finally. Please change the title to: "Acceleration plays a key role in the twin paradox"

I did a lot more than that, I have proven that the claim in the title is false. So, all you did now is that you made B and C to have the same speed, thus they meet midway, so you just dumbed down the scenario such that it would fit your claim. But, as I already pointed out, when you claim: "Both B and C age at the same rate according to A, because v and -v give the same value of gamma. B and C meet halfway through the experiment" you simply reduced the scenario to the case where B has no role whatsoever. We have been over this several times already. So, I did the reverse, I found the proper time for B knowing the data from A. The point I made is that the two methods, yours and mine are equivalent. Yours is much more complicated <shrug>

Why not do the honorable thing and admit that the claim in the title is false?

In other words, I proved you wrong and you surreptitiously edited the scenario? What post post did you post edit? So, to fix the ideas, the stuff in post 1, the one that gives the name to the thread has been proven to be false.

So, you continue to persist in the lie you have created? Even after it has been shown how your scenario is nothing but a manipulation of C's clock at the encounter with B?