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Special Relativity - Which reference frame experiences which time?
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[QUOTE="malawi_glenn, post: 6870973, member: 71734"] [USER=734390]@AronYstad[/USER] I will PM you now a page from my book, it is written in swedish so I can't post it here. It covers the time dilation as a special case of the Lorentz transformation so you will not understand the derivation of the forumla. But, there is a nice diagram at the bottom of the page where you can see what is going on which each frames clock. [ATTACH type="full" alt="1680290299821.png"]324282[/ATTACH] Note. The problem is will worded. That a person on Earth [B][I]sees [/I][/B]the travel taking 43 min is not pure time dilation. Thing is that we see things by receiving light signals. So, when the rocket has completed the journey, light needs to travel from that location to the stationary observer in the Earth frame. If the distance between the observer and the landing spot for the rocket is L according to the earths frame of reference, you need to add the time L/c. Time dilation is what is depcited in the diagram above: two stationary clocks C & D in the ##\tilde S## frame (I use tilde instead of prime not to confuse with prime for derivatives) at the origin of that frame, two stationary clocks A & B in frame ##S##, one is located at the origin and one at the coordinate ##x_1##. The frame ##\tilde S## is moving w.r.t. ##S## with a constant velocity ##v##. When the origins of the two frames coincide, clock A and C are stopped. When the origin of ##\tilde S## coincide with ##x_1## clocks B and D are stopped. The ratio of the time differences ##t_B - t_A = \Delta t## and ##\tilde t_D - \tilde t_C = \Delta \tilde t## are equal to ##\Delta t / \Delta \tilde t = \gamma ##. This is where the "asymmetry" lies, you have two clocks at the same location according to one frame (##\tilde S##), but two clocks at different locations according to the other frame (## S##). [/QUOTE]
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Special Relativity - Which reference frame experiences which time?
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