Thread: twin paradox
View Single Post
neopolitan
neopolitan is offline
#248
Oct10-08, 10:18 PM
P: 645
Quote Quote by DaleSpam View Post
I don't know if you are thinking the right thing and just writing it wrong, but no, what you wrote is not correct. First, proper times are frame invariant, so it doesn't make much sense to say "the proper time in a single inertial frame". Second, simultaneity is frame variant, so you need to specify which frame. What you have described here only works in the rest frame of one of the worldlines, in any other rest frame you will have to consider the distance traveled by both worldlines. Third, I don't know if this is general for arbitrary paths or if it only applies for straight worldlines. Finally, it sounds like you are considering scenarios where the twins do not start together and reunite at some other time. If so, then there are 4 events of interest, the two events of each worldline starting and the two events of each worldline ending. Those two starting events must be simultaneous with each other in the rest frame of one worldline as must the two ending events.
What I wrote was this (with emphasis added):

Quote Quote by neopolitan
Now, for the purposes of explaining to the student, I would have thought we could say that we know that the square of the proper time in a single inertial frame which is taken to be at rest is equal to the sum of the squares of the proper time and coordinate distance in another inertial frame (or set of inertial frames) - between any two simultaneous events (simultaneous as defined by Einstein).
Initially, I thought about making the red "a" the definite article "the", to keep it to the twin paradox scenario, but then I thought that in reality it doesn't have to be.

As I read my paragraph again, I see I left ambiguity. I did indeed mean four simultaneous events, in two pairs. I should have written "two pairs of simultaneous events". My error. Also, to remove less obvious ambiguity, I should point out that the coordinate distance in the other inertial frame is according to the first mentioned "taken to be at rest" frame, and the simultaneous events are simultaneous in the "taken to be at rest frame". I took these latter two to be obvious, but I shouldn't do that.

So, rephrasing:

Quote Quote by neopolitan, rephrasing
Given an inertial frame as our reference, a frame which is taken to be at rest (thus having a coordinate distance of zero), and another inertial frame, or set of contiguous inertial frames, if we consider two pairs of simultaneous events (each pair of events has one event to each frame) and if we use values according to an observer in our reference frame then: we know that the square of the proper time in former inertial frame which is taken to be at rest is equal to the sum of the squares of the proper time and coordinate distance in the other inertial frame (or set of inertial frames).
Then this would work for the twin paradox scenario, with two pairs of events which are actually just two events (departure and return) or pairs of events which are not collocated in spacetime, or a mix (like Al68 has sort of been discussing when breaking the twin paradox into two legs).

Is this less ambiguous and, being less ambiguous, correct?

If correct, could we be less exacting in order to convey the concept to the student, and then slowly build up the understanding of the conditions under which the equation is correct? (Sort of how one would explain a rainbow in steps, without at first talking about suspension of droplets of water of the correct size to refract rays of light, or the relative placement of those droplets, or how the image of the rainbow is formed in the eye and is not something external to our perceptions of it.)

cheers,

neopolitan