On 4 February 2014 09:29, Edgar L. Owen <edgaro...@att.net> wrote: > John, > > A couple of points in response. > > Yes, I agree that both A and B see each other's clocks running slower than > their own DURING the trip. This is standard relativity theory mostly > Lorentz transform if we just take non-accelerated relative motion. Also > note that, contrary to your statement, in this case both A and B DO AGREE > on "when the race starts and stops" because they both begin and end at the > same present moment point in actual spacetime back on earth. > > I know that, and presumably we agree on it, but that was NOT the question > I asked. The question is why when A gets to the center of the galaxy and > stops relative to B that then his clock shows only 20 years passage, but > B's clock shows 30,000+? > > Perhaps you didn't see my similar questions to Brent, to which he has > either been unwilling or unable to reply, about this case. > > He says it's a matter of geometry, but neglects to point out that geometry > must have its origin at B's earth bound frame. The question is why this > geometry rather than the equal and opposite frame based in A's origin > creates not only the transitory effect you reference above but also a real > permanent effect. It seems we have to choose the correct geometry but what > is the criterion for the correct geometry? It almost seems as if there must > be some absolute real geometry centered at B's origin on the earth for this > to work. > > The situation is only equal and opposite while A and B remain in inertia frames. However, in order that they can get together to compare clocks, at least one of them can't remain in an inertial frame throughout the duration of the trip. For convenience assume that B remains unaccelerated throughout, while A accelerates, then coasts, then decelerates, then repeats the process to return. If we assume the periods of acceleration are negligible compared to the tie spent coasting at (say) 0.9c, this simplifies the problem slightly and lets us treat A and B's paths through space-time as the sides of a triangle. This shows A's path through space-time is longer than B's, no matter how you look at it (i.e. which reference frame you use), because you can't construct a triangle with two sides adding up to be shorter than the base.
-- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
