ghwellsjr said:
The relative part of SR states that we can analyze any scenario in any frame we choose, none is preferred over any other, including frames in which different observers are at rest. So if I choose a frame in which the planet is at rest (because I like simple solutions) you cannot claim that I'm not done and need to do something more complicated.
OK, I made an unstated assumption that each participant in this example would draw up a space-time diagram in which he is at rest, and relate the other's clocks to his diagram.
ghwellsjr said:
There is nothing wrong with correctly drawing up multiple space-time diagrams, as long as you realize that they are adding nothing to the analysis, especially that they are not associated with any "actual observations" of any observer. It's nothing more than a fun exercise for you (not for me) and is in no way a requirement of relativity.
I don't understand the "not associated with any actual observations" bit. But we normally choose space-time doiagrams to suit the problem and simplify the mathematics. We use a heliocentric frame to work out Mercury's precession, and the results do match with "actual observations". We use a geocentric system when discussing time dilation on Earth relative to satellites, and again get figures that are associated with actual observations. So I don't understand what you are getting at with that comment.
Going back to the original problem of two triplets A and B going on symmetrical journeys in opposite directions while C stays at home, C will see their clocks agreeing all the time, but retarded relative to his. A and B in turn will both see C's clock retarded, and each other's clocks even more retarded as their relative velocity is greater (Relativistic addition of velocities applies, so their relative velocity is less than double their velocity relative to C).
For simplicity, we assume they accelerate, travel at constant velocity for a long time, turn around and return at the same velocity. Assume a velocity that gives a 50% time dilation - I think the figure is around 0.84c. then after 4 years traveling on C's clock, A and B turn around. But their clocks read 2 years (they have traveled half the distand C sees, because of Lorentz contraction of the distance), and they see C's clock reading 1 year - because the velocity and time dilation is relative. So on the return journey they should see C's clock advance another year, and read 2, but when they get there it reads 8! And this is the SR twin paradox.
The paradox is resolved because A and B change direction, and as thery do so they see C's clock accelerate and advance from 1 to 7 years. As ibriggs444 says, the time change is based on acceleration times distance. But the calculation for acceleratiion is a bit too complicated for me, and I like to use Einstein's equivalence principle and look at it from a gravity point of view.
Picture A in his spaceship, now turned to point back to C. He switches his rocket engines on to decelerate and then accelerate back to C. From his point of view, he is motionless and experiences a gravitation acceleration pulling him down to his rockets floorboards, and he sees this same gravity slow down C's recession from him and C starts falling towards him.
Because he and C are now in a gravitational field, and C is much higher, see sees C's clock run faster than his own, just as we on Earth see clocks in orbit above us run faster.
Please note that in all this discussion I have ignored the time for light signals to pass between them, so what they "see" is what they would calculate allowing for this time, or what their space-time diagram would indicate if they drew a straight line across it to denote their "present".
And please accept that my responses on this forum might be time-retarded becauase I live in a different time-frame - Australia!
Mike
