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Simplified Twin Paradox Resolution.
On Jan 7, 10:02 am, Absolutely Vertical wrote:
On 1/5/2013 10:24 AM, Vilas Tamhane wrote: It is perfectly symmetrical. Note that SR does not seek to find who actually fired the rocket. Between the two spaceships A and B, A can accelerate or B can accelerate or both can accelerate. SR deals with uniform motion after acceleration. that's not so. That was what PD said years ago. Stupid PD, an exprofessor of physics at a university in Texas. shrug if both accelerate, there is no time difference. After both have done their acceleration, they continue to move away from each other. What is their relative speed? Does the Lorentz transform not say time dilation? At this moment, who is actually moving, and who is not? If time dilation is building up, how does it evaporate? shrug the fact that one accelerates and the other doesn't is the reason there is a difference. Actually not according to the Lorentz transform. You cannot make up your own laws of physics. You are no god. shrug sr accounts for the difference in that case. if you thought that sr just ignores acceleration then you thought wrong and the twin example was designed to elicit that mistake. In this case, both accelerate with a coasting period to allow for mutual time dilation building up. Shouldn’t the magic effect of acceleration cancel out? If not, why not? Just what part of this simple scenario do you not understand, PD? shrug Let’s recap the mathemagic trick Einstein dingleberries like to pitch when one accelerates and the other does not. ** dt1 = dt2 / sqrt(1 – B^2) And ** dt2 = dt1 sqrt(1 – B^2) Where ** B c = Relative speed between 1 and 2 When both accelerate, well they will probably say the following. ** dt1 = dt2 And ** dt2 = dt1 It is indeed interesting what type of mathemagic trick they are going to pull out when both 1 and 2 are coasting away or towards each other. shrug For reference, the Lorentz transform always says the following regardless who is accelerating or no: ** dt1 = dt2 sqrt(1 – B^2) And ** dt2 = dt1 sqrt(1 – B^2) The only time when there is no contradiction is when (B^2 = 0). shrug 
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