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post 185 exposes your sleigh of hand, the setup in post 1 just hides the acceleration by skipping proper elapsed time on the C clock (time has a discontinuity at event BC, introduced by your scenario). post 73 already exposed the sleigh of hand, so we could have saved more than 100 posts if you admitted to the inner error

[math]\tau_A[/math] is not necessary in proving that your conjecture is wrong. Your conceptuual error is identically exposed in both posts 73 and 185.

There is no "corrected equations", the correct equations have been there since post 73. How about you simply admit that your claim in the title of the thread is FALSE? While you are at it, feel free to reverse al the negs you have been giving me. Time to close this thread, it really promovates anti-mainstream ideas.

Why is it relevant for the exercise? The OP has been proven FALSE. This is the end of the line.

Yes, finally. Give yourself a negative for the brainfart.

A quick proof that my approach obtains the same exact results as the one using the Minkowski diagrams above (but it is much more general). The amount of coordinate time needed for B to reach BC as calculated in frame A is: [math]T=\frac{x_{BC}}{v_B}=\frac{3.75}{0.6}[/math] The amount of proper time elapsed on B's clock is: [math]\tau_B=\int_0^T{\sqrt{1-v_B^2}dt}=T*\sqrt{1-v_B^2}=0.8 \frac{3.75}{0.6}=5[/math] ....which is exactly the number that can be read from the Minkowski diagram.

No, I am not doing any of the imagined mistakes you are accusing me of. Proof is, that if I dumb down the setup for the trivial cases, I recover the results obtained in the simplistic scenarios. The claim in your thread title is false, you will need to learn to live with it, no matter how you twist and turn.

I really don't feel like helping you just to see you turning around and claiming that you found "mistakes" in my posts. I don't like hypocrites, you have al the formulas, you can do the calculations yourself. You can stop giving me negatives, I don't really care about them.

After the encounter, two things happen: - C "dumps" its accumulated proper time and gets its clock reset to [math]\tau_B=\frac{D}{v_B+v_C}\sqrt{1-v_B^2}[/math] - C continues on towards A and arrives in A at the coordinate time [math]t'=\frac{D}{v_C}[/math] accumulating an additional: [math]\tau'_C=(\frac{D}{v_C}-\frac{D}{v_B+v_C}) \sqrt{1-v_C^2}[/math] So, the total time shown on C's clock when it arrives in A is: [math]\tau^"_C=\tau_C+\tau'_C=(\frac{D}{v_C}-\frac{D}{v_B+v_C}) \sqrt{1-v_C^2}+\frac{D}{v_B+v_C} \sqrt{1-v_B^2}[/math] Obviously, four things are true: 1. [math]\tau^"_C \ne \tau_A[/math] 2. A "jump" in C's clock has been inserted by the OP at the point where it met B via artificially resetting it to B's clock value. The effect is nothing but a sleigh of hand , fully equivalent to instantaneously accelerating from [math]v_C[/math] to [math]v_B[/math] retroactively (speak about unphysical stuff happening here!) at the point where C meets B. 3. Absent the sleigh of hand with resetting the clock at the BC event, the C clock will show: [math]\tau^"_C=\frac{D\sqrt{1-v_C^2}}{v} [/math] since C measures a length contracted distance to A. 4. for the trivial case whre [math]v_C=v_B=v[/math] [math]\tau^"_C=\frac{D}{v} \sqrt{1-v^2}[/math] i.e. the exercise reduces to the "instantaneous turnaround" for B, i.e. the case where B turns around with infinite acceleration. So, the claim made in the title of the thread is bogus, you cannot get a twin "paradox" in the absence of acceleration (or, as in the case of this thread, a silly sleigh of hand that the naive would not notice).

You don't know what you are doing, that much is pretty clear.

What you did is to work out the trivial case where C's speed wrt A is equal to B's speed wrt A. This is the particular, trivial case I described in my earler post that makes C irrelevant. It is equivalent with the case where B makes an instantaneous, infinite acceleration turnaround at the midpoint. See here, for example. I have shown what happens when [math]v_B \ne v_C[/math] already. The above shows why your earlier attempt to construct the equation [math]\frac{d}{v}=\frac{d}{v}*\sqrt{1-v^2}[/math] and arrive to the conclusion [math]v=0[/math] was wrong. You were trying to equate C's proper time with A's cordinate time.

Proper time is frame invariant, all frames agree on proper time, please stop asking these type of questions, it shows that you aren't learning. No, it won't, you (and Delta) need to learn how to use symbolic calculations.

This is exactly as in the muon exercise: the muons ( C ) are located at altitude d , as measured by the Earh (A) and they move at speed v (wrt the Earth as well). It takes the muons [math]\frac{d \sqrt{1-(v/c)^2}}{v}[/math] to hit the Earth (A). Where you make the mistake is that, indeed, A measures the coordinate time [math]T=\frac{d}{v}[/math] for C to reach him (A). But what A wants, is the proper time, [math]\tau_A=T\sqrt{1-(v/c)^2}[/math]

The calculation I gave you is correct, here is your obvious mistake, right off the bat: "A measures it's proper time as d/v." No , he doesn't.

[math]\tau_A=\frac{d \sqrt{1-(v/c)^2}}{v}[/math] in the post where I did all the calculations, you need to go a few pages back.

This exercise is not about clock rate, it is about elapsed proper time. I have shown you how to calculate the elapsed proper time as a function of coordinate time, textbook stuff.

I don't need any lessons so it looks like you pulled the formula out of your butt and you are peddalling very fast trying to cover up that you have no derivation. Thanks for playing.

I don't have to use the frame for C, it was just a convenient way. Did you get the line: "all frames agree on proper time so we might as well use F_C" ? When I do the calculations, I replace [math]d[/math] with [math]d \sqrt{1-(v/c)^2}[/math], did you notice that?

Frame C has no impact on the solution, all proper times are calculated in one frame, frame A. You (and Iggy) need to come with grips with this fact.

Please do so, there is no derivation for the second term that you are adding in. <quote>You think I don't know this?</quote> Then stop claiming that I am switching frames, would you?

You have just showed two disjoined formulas without being able to connect them into your final formula that you are advocating. How do you derive this: If B measures [math]\tau_B[/math] between AC and BC then he will calculate [math]\tau_A[/math] between those events as follows, [math]\tau_A = \tau_B \sqrt{1 - {v_B}^2 } + \tau_B \frac{{v_B}^2}{\sqrt{1-{v_B}^2}}[/math] There is no "coordinate switching" in my approach, all calculations are done in A frame. This is the third time I am telling you this.

How about you showed the derivation of your formulas? This is the second time I ask you. I have shown you the way it is done in standard textbooks, so how about you explained how you arrived to the formulas that you posted?

I simply pointed out how I do the calculations, complete with the derivation from scratch. You haven't answered my question, how do you arrive to your formulas? Where is the derivation?

I don't need the basic stuff on time dilation. The calculations are very simple, in A frame B and C meet at [math]t=\frac{D}{v_B+v_C}[/math] The total elapsed proper time in frame A for: 1. B to reach the BC event is [math]\tau_B=\int_0^t{\sqrt{1-v_B^2}dt}=\frac{D}{v_B+v_C}\sqrt{1-v_B^2}[/math] 2. C to reach the BC event is [math]\tau_C=\int_0^t{\sqrt{1-v_c^2}dt}=\frac{D}{v_B+v_C}\sqrt{1-v_c^2}[/math] Very simple.

This seems to make no sense: according to "Delta" AC is the event of C passing through A and BC is the event of B passing through C so [math]\tau_B[/math] seems to have no sense for the problem. Nor does your attempt at applying corrections through RoS, the standard way of calculating proper time is to integrate [math]\sqrt{1-v^2}dt[/math], nor RoS "corrections" needed . This doesn't look right, proper time is frame invariant, so A and B need to measure the time between the same two events (AB,BC) as being the same: [math](cd \tau)^2=(cdt)^2-dx^2=(cdt')^2-dx'^2=.....[/math] You are applying a "gamma correction" when going from a frame comoving with A to a frame comoving with B, this is not correct. The times measured by A are proper times, not coordinate times , so no "gamma correction" is allowed.