On Thu, Feb 13, 2014 at 12:55 PM, Edgar L. Owen <edgaro...@att.net> wrote:
> Jesse, > > See my proximate response to Liz who asked the same question. Basically > relativity theory gives you the equations for both frames for any > relativistic situation. So all you have to do is do the calculations like > I've explained to you with nearly a dozen examples. > > To the question in your last paragraph. Yes, of course we assume > originally synchronized clocks. Remember this is a thought experiment, and > that is clearly possible if we assume it's done at rest relative to each > other and then magically without acceleration (your instantaneous > acceleration, which is also physically impossible, but has the exact same > thought effect). > > So do you agree that given synchronized clocks, A and B in relative motion > will still have synchronized clocks in their own frames to each other? > I.e., that A will have the same reading on his own clock that B does on his > own clock? > "Same reading" using what definition of simultaneity? If you're talking about p-time simultaneity, then I don't agree, because I don't believe in p-time in the first place. If you're talking about the "same reading" using the definition of simultaneity assumed in each one's own rest frame, then I still don't agree. Say that two observers Alice and Bob have their clocks set to zero when they are at the same point in spacetime (i.e. if I use A to represent the event of Alice's clock reading 0, and B to represent the event of Bob's clock reading 0, then all frames will assign exactly the same space and time coordinates to B that they assign to A), and from that meeting at a common spatial location they move away from each other inertially at 0.6c, so in each one's frame the other has a time dilation factor of sqrt(1 - 0.6^2) = 0.8. Then in Alice's rest frame, the event of her clock reading 25 would be simultaneous with the event of Bob's clock reading 20. In Bob's rest frame, the event of his clock reading 25 would be simultaneous with the event of Alice's clock reading 20 (and in his frame the event of his clock reading 20 would be simultaneous with the event of Alice's clock reading 16). Do you disagree with these conclusions about frame-dependent simultaneity in SR? > > And do you also agree that when the relative motion magically stops, their > clocks will still read the same as each other's, AND they will both be the > same age because of that? > No, I don't agree. Using the numbers above, if Bob instantaneously accelerates to come to rest relative to Alice when his clock reads 20, then he will now be at rest in Alice's rest frame, and it'll still be true in this frame that the event of Alice's clock reading 25 is simultaneous with Bob's clock reading 20. Likewise, if Alice instantaneously accelerates to come to rest relative to Bob when her clock reads 20, she will now be at rest in Bob's rest frame, and it'll still be true in this frame that the event of Bob's clock reading 25 is simultaneous with Alice's clock reading 20. Do you disagree with these conclusions? > > This is just elementary relativity theory, nothing to do with p-time at > all... > Yes, it is elementary, and if you disagree with any of my statements about SR above then you need to go back and learn the basics of how SR math actually works. Jesse -- 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.
