On Sun, Feb 9, 2014 at 1:44 PM, Edgar L. Owen <[email protected]> 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. > Rereading doesn't help, I just don't understand what you mean since I can't think of another way to interpret "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." Perhaps another question would help. Say it's true in some sense that a meteor impact on Mars happens at the "same p-time" as a lightning strike on Earth. Does either of these capture your view on how p-time works? 1. The fact that these events are "simultaneous in p-time" is an objective truth by itself, it requires no context of a particular reference frame (though there may still be a rule for determining this objective truth that refers to reference frames, like "two events happening to objects that share the same rest frame are objectively simultaneous in p-time if and only if they are simultaneous in the time coordinate of their mutual rest frame). 2. It is an objective truth that these events are "simultaneous in p-time" in the context of one frame, and "not simultaneous in p-time" in the context of another frame, but there is no frame-independent objective truth about which events are simultaneous in p-time. If either of these does capture your view, please point out which one...if neither does, then perhaps in trying to explain your view to me you could keep in mind that these are the only two options *I* can imagine at the moment, so perhaps you could explain how your third alternative would differ from each of these two in turn. > > > As for your last example, establishing past p-time simultaneity across > multiple frames is NOT transitive > So does "p-time simultaneity across multiple frames" mean that p-time simultaneity is frame-dependent, as suggested in option 2 above? If not I don't know how to interpret that phrase. > (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. > Relativity doesn't talk about p-time at all, so not sure what you mean. Perhaps you would take my option 1 above, but you would just say that although there are objective truths about p-time simultaneity, these truths aren't transitive? In other words you could be saying that there is a frame-independent objective truth that events A and B are simultaneous in p-time, and a frame-independent objective truth that events B and C are simultaneous in p-time, but this does not imply that A and C are simultaneous in p-time. Is that right or am I still misunderstanding your view? Jesse > > 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 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]. > 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. 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