Jesse, Let me try to clarify my response to your A, B, C past p-time simultaneity example, because I think I misstated it in my previous post.
Assume three observers A, B, C, with three clock times t, t' and t''. It is important to specify these are the clock time readings of their OWN clocks in their OWN frames. Case 1: Assume they are all initially in the same inertial frame with synchronized clocks. In this same inertial frame they are in the same current moment of p-time at every synchronized clock time tick of their clocks. Both their clocks and their p-times are synchronized for every clock time. In this case the same current p-time relation is transitive from A to C through B. If t = t' and t' = t'' then t = t''. Case 2: Assume A is in a gravity well that makes it's clock run at 1/2 the rate of B's clock. Now tx2 = t', t' = t'' and tx2 = t''. The p-time simultaneity relationship is still transitive. Case 3: Now C is in relative motion to B (and therefore to A as well). In this case C will see both A's and B's clocks run slow, and both A and B will see C's clock run slow but this is irrelevant to p-time because it's not a 'real' effect in the sense that 1. It is not agreed upon by all observers and 2. the effect vanishes when the relative motion ceases. So in this case also tx2 = t', t' (on B's comoving clock though not on C's clock) = t'' (on C's comoving clock though not on B's clock), and once again the p-time relation is transitive because tx2 (on A's comoving clock) = t'' (on C's comoving clock). Equals here means in the SAME current moment of p-time. B and C are in the same current moment of p-time when t' on B's clock = t'' on C's clock. Thus A, B and C are in the same current moment of p-time when tx2 (on A's clock) = t' on (B's clock) = t'' (on C's clock). So I think the current p-time relation, as defined in terms of various clock times IS transitive if we just make sure we use the clock times of the clocks in THEIR OWN FRAMES to specify their clock times. In fact the current p-time relationship MUST be transitive for there to be a universal common p-time that all observers agree upon. So it's just a matter of making sure we use the frames of each observer to denote their own clock times rather than any other observer frame. I neglected to do that in the previous post when I concluded that current p-time relationship wasn't transitive. That was incorrect. So to calculate simultaneous p-times based on clock times we must be sure we use the clock time readings of each observer's OWN clock. AND, in general, we can't do it from any single relativistic equation, but only by understanding how relativity works and compares in the frames of the particular observers in the cases. Edgar On Sunday, February 9, 2014 1:44:05 PM UTC-5, Edgar L. Owen wrote: > > 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]> 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 > > ... -- 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.
