Liz, You keep missing my point. There is NO inertial frame in this example, neither A nor B's frame is inertial.....
Neither A nor B are in an inertial frame in this example. The specific point of the example is that they BOTH experience exactly the same NON-inertial 1g acceleration for the whole trip. So now what's your answer to my original question? Edgar On Monday, February 3, 2014 5:42:48 PM UTC-5, Liz R wrote: > > On 4 February 2014 09:29, Edgar L. Owen <[email protected] <javascript:>>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 [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
