Richard, That's my point exactly. He can't. See my response just posted explaining that in detail.
Edgar On Thursday, February 13, 2014 1:46:18 PM UTC-5, yanniru wrote: > > > > > On Thu, Feb 13, 2014 at 1:39 PM, Jesse Mazer <[email protected]<javascript:> > > wrote: > >> >> >> On Thu, Feb 13, 2014 at 12:55 PM, Edgar L. Owen <[email protected]<javascript:> >> > 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? >> > > How can Bob age 5 years because Alice instantly accelerated into his rest > frame? > I do not agree. > >> >> >>> >>> 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 [email protected] <javascript:>. >> To post to this group, send email to [email protected]<javascript:> >> . >> Visit this group at http://groups.google.com/group/everything-list. >> For more options, visit https://groups.google.com/groups/opt_out. >> > > -- 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.
