Mordred Posted November 12, 2019 Posted November 12, 2019 (edited) the end result once finished being latexed will give you the hyperbolic angle of rapidity I already provided that but am now typing in the proof. I will do the case of a rotating acceleration. anyways that part is done. That was the classic case when the twin does a turnaround. The turnaround will fall onto that hyperbolic rotation under constant acceleration for the duration of the turnaround. It isn't the case you presented but I wanted to show a case where acceleration affects the Lorentz transformations. Hopefully tomorrow I will have time to show how an emitter orbiting an observer around the z axis leads to [math]d\tau=\sqrt{1-\frac{wr^2}{c^2}}dt[/math] yes I was a little loose with some of the grammar distracted by wife while trying to latex all that in is distracting lol. PS I don't have any issue with the solution presented earlier this thread. Like I said wanted to show acceleration effects for the classic case as well. Edited November 12, 2019 by Mordred
md65536 Posted November 12, 2019 Author Posted November 12, 2019 (edited) I don't understand why you would write out all those equations when they don't even relate to the case described. Even the classic twin paradox is typically presented with negligible acceleration time, or equivalently considering only 2 legs of inertial motion for the "traveling" twin. Adding in an acceleration phase makes it more complicated, and doesn't help in figuring out the simpler version of the paradox. Why complicate it if a simpler version is specified? Edited November 12, 2019 by md65536
Mordred Posted November 12, 2019 Posted November 12, 2019 Some solutions prefer the added detail such as the time taken specifically during the turnaround. Anyways I added it to show how it can be handled by using the four momentum. I had the time so figured some might learn from it.
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