Jesse, No, "the definition of p-time simultaneity itself depends on the arbitrary "choice of coordinate system" is NOT true. I clearly stated otherwise and explained why. Please reread if it isn't clear.
As for your last example, establishing past p-time simultaneity across multiple frames is NOT transitive (in your sense of using the same intermediate frame t value). You can only establish it between any two frames (at a time) in general because the relativistic differences between multiple frame relationships as in your example are not transitive. However take clocks A, B and C. You can always determine same past p-times between A and B, and between B and C IN TERMS OF their clock time relationships as calculated by standard relativity theory. However you cannot in general say that because B's t' = A's t, and B's t' = C's t'' that t and t'' were at the same p-time. Relativity doesn't work like that as I'm sure you know. You'd want to calculate the relativity equations between A and C to determine which t and t'' occurred at the same past p-time. So you will be able to compute which t and which t'' are at the same p-time but in general the observer B t' t values those t and t'' values correspond to will be different. Thus the same p-time clock time points of A and C will not be transitive through B on t' t values but they will be computable. Edgar On Sunday, February 9, 2014 12:47:49 PM UTC-5, jessem wrote: > > > > > On Sun, Feb 9, 2014 at 11:19 AM, Edgar L. Owen <[email protected]<javascript:> > > wrote: > > Jesse, > > Same thing as I'm saying. My other clock time is just a clock centered in > your coordinate system. It's the same idea. If you look at the equations of > relativistic clock time they are always of the general form dt'/dt = f( ). > I just note that the dt with respect to which dt' is calculated is another > clock. You simply note that other clock is some coordinate system. Exactly > the same. MY clock is the clock at the origin of YOUR coordinate system. > The equations are exactly the same. The concept is exactly the same. You > are talking about the exact same thing as I am. > > Yes, the PARTICULAR 1:1 relationship only exists with respect to some > arbitrary coordinate system (which I stated as just some other clock). The > choice of that coordinate system is of course arbitrary. That's irrelevant > because with EVERY choice of a coordinate system there will be some such > 1:1 relationship on the basis of which clock times can be used to determine > the same points in p-time. Depending on the choice of coordinate system > those clock times will of course be different but there will be such a > relationship that defines the clock times in ANY two relativistic systems > such that a same point in p-time can be defined in terms of a 1:1 relation > between those clock times. > > > Are you saying that the definition of p-time simultaneity itself depends > on the arbitrary "choice of coordinate system"? I thought p-time > simultaneity was supposed to be an objective matter, so the question of > whether any two past events were simultaneous in p-time could have only one > TRUE answer. Is that not correct? > > > > > > > Yes is the answer to your question "if two clocks are at rest relative to > one another and "synchronized" according to the definition of simultaneity > in their mutual rest frame, do you automatically assume this implies they > are synchronized in p-time?" > > I already stated that several times in my posts of yesterday and even gave > concrete examples in which it was true, so I'm surprised you accuse me of > not answering it. > > > > Thanks for giving a clear answer. I understand that you think that some of > your statements in previous posts were answering it, like "A and B in deep > space. No gravity. Their clocks, t and t', are synchronized. They are in > the same current p-time moment and whenever t = t', which is always their > clock times confirm they are the same current p-time as well as the same > clock time." But I think there is still potential for ambiguity in that > statement, because "synchronized" could mean synchronized in p-time which > might not agree with relativistic synchronization in their rest > frame--that's why in my own question I said 'synchronized according to the > definition of simultaneity in their rest frame. If you can just quote a > question I ask and respond directly to the quote, as you did above, it'd be > appreciated, since this would avoid any possible ambiguities that might > occur to me but wouldn't occur to you. > > In any case, now that I understand your answer, let me elaborate on what I > meant when I said earlier that such a rule for p-time simultaneity will > "run into major problems if you consider multiple pairs of clocks where > each member of a pair is at rest relative to the other member of the same > pair, but different pairs are in motion relative to another". Suppose we > have two pairs of observers, with each member of a pair being at rest > relative to the other member of the same pair, but the two pairs in > relative motion. Call the first pair Alice and Bob, and the second pair > Arlene and Bart. Assume that in the Alice/Bob rest frame, Alice and Bob's > clocks are synchronized, and likewise assume that in the Arlene/Bart rest > frame, Arlene and Bart's clocks are synchronized. > > Start by considering their initial positions, velocities and clock times > in a coordinate system where Alice and Bob are at rest. At coordinate time > t=0 in this frame, Alice is at position x=0 light-years, Bob is at position > x=25 light years, and their clock readings are T(Alice)=0 years, T(Bob)=0 > years. Meanwhile at the same coordinate time t=0, Arlene is at position x=0 > light years--her position coincides with that of Alice--and her clock reads > T(Arlene)=0 years, and Bart is at position x=9 light years and his clock > reads T(Bart)=-12 years. In this frame, Arlene and Bart are both moving in > the +x direction at 0.8c. So 20 years later in this frame, they both will > have moved forward by 20*0.8=16 light-years, so at t=20 Arlene is at > position x=16 light-years while Bart is at position x=25 light years. Their > clocks are running slow by a factor of 0.6 in this frame, so in a span of > 20 years they tick forward by 12 years, meaning at t=20 Arelene's clock > reads T(Arlene)=12 years and Bart's clock reads T(Bart)=0 years, so this > event on Bart's worldline is simultaneous in his own frame with the event > on Arlene's worldline where her clock read T(Arlene)=0 years and her > position coincided with that of Alice (the fact that these events are > simultaneous in the Arlene/Bart rest frame is easily proven using the > Lorentz transformation, I can supply the details if needed). But since Bart > is at x=25 light years at this moment, his position coincides with that of > Bob who has remained at rest at x=25 light years, and whose clock is > keeping pace with coordinate time so his clock reads T(Bob)=20 years. > > Summing it all up, if we use BOTH the rule that a pair of clocks at rest > relative to one another and sychronized in their rest frame must also be > synchronized in p-time, AND the rule that events which coincide at the same > point in spacetime must happen at the same p-time, we get the following > conclusions: > > 1. The event of Bob's clock reading T(Bob)=0 and the event of Alice's > clock reading T(Alice)=0 must be simultaneous in p-time, since they are > simultaneous in the Alice/Bob rest frame. > > 2. The event of Alice's clock reading T(Alice)=0 and the event of Arlene's > clock reading T(Arlene)=0 must be simultaneous in p-time, since they happen > at the same point in spacetime. > > 3. The event of Arlene's clock reading T(Arlene)=0 and the event of Bart's > clock reading T(Bart)=0 must be simultaneous in p-time, since they are > simultaneous in the Arlene/Bart rest frame. > > 4. The event of Bart's clock reading T(Bart)=0 and the event of Bob's > clock reading T(Bob)=20 years must be simultaneous in p-time, since they > happen at the same point in spacetime. > > Thus if simultaneity in p-time is transitive, we can put these all > together and arrive at the conclusion that the event of Bob's clock reading > T(Bob)=0 is simultaneous in p-time with the event of Bob's clock reading > T(Bob)=20! I am sure you would not accept such a conclusion, but it is an > unavoidable consequence of the two rules for p-time simultaneity I listed > above, so if you want to avoid the conclusion you have to either ditch one > of the rules or say that p-time simultaneity is not transitive (which I > guess would be possible if you don't think there's a single objective truth > about p-time simultaneity and that it depends on the context of the frame > we are using, see my question about this above). > > Jesse > > > > > > > Edgar > > > > > > > On Sunday, February 9, 2014 10:51:32 AM UTC-5, jessem wrote: > > > On Sun, Feb 9, 2014 at 9:49 AM, Edgar L. Owen <[email protected]> wrote: > > Jesse, et al, > > A Propros of our discussion of determining same past moments of P-time let > me now try to present a much deeper insight into P-time, that illustrates > and explains that, and see if it makes sense. I will show how relativity > itself implicitly assumes and absolutely requires P-time to make sense. > > > Every relativistic calculation of clock times consists of some equation > describing how one clock time varies with respect to another clock time. > > > No, every relativistic calculation of clock times consists of an equation > describing how one clock time varies with coordinate time in some > coordinate system. There is no coordinate-independent way of defining how > "one clock time varies with respect to another time" when they are at > different points in space (aside from apparent visual rates, but that > involves things like the Doppler effect, a > > ... -- 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]. 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