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, and in terms of p-time it would also involve delays due to the time for light signals to cross from one to the other, so I assume you're not just trying to talk about apparent visual rates here). And likewise we can get the total elapsed clock time differences along the > worldlines by integrating along them until any dt'/dt = f( ). > Total elapsed time for a clock with a known path is found by integrating clock rate as a function of time in some coordinate system, which can be defined as a function of v(t), velocity as a function of time in that coordinate system. The actual integral can be seen towards the bottom of the page at http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.htmlfor example (part of a useful larger series on different conceptual approaches to understanding the twin paradox in relativity at http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html). > > This establishes a 1:1 relationship between the clock times, t and t', at > every point along the worldlines. > If you agree with my statements above, you can see that the 1:1 relationship only exists within the context of a particular choice of coordinate system. If you disagree, and think there is a coordinate-independent way to define "how one clock time varies with respect to another clock time" (and which is also more than just a statement about apparent visual rates), then you are just expressing a basic misconception of how calculations are done in relativity. Also, as usual you ignored the direct question I asked to you in the previous post you are responding to: 'As I asked before, 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?' Can you please answer the above question yes or no? If you continue to consistently avoid answering simple questions I ask, I'm going to conclude you're not really interested in a civil discussion and analysis of concepts, but rather are behaving like a lawyer or politician who just wants to "win" rhetorically and has no interest in addressing the other's concerns in an intellectually honest way. I am happy to answer any questions you have for me, I just ask that you extend me the same courtesy, at least in cases where the questions are simple ones that don't require a lengthy write-up. Jesse > > On Sunday, February 9, 2014 12:05:32 AM UTC-5, jessem wrote: >> >> >> >> On Sat, Feb 8, 2014 at 8:07 PM, Edgar L. Owen <[email protected]> wrote: >> >> Jesse, >> >> Consider another simple example: >> >> 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. >> >> >> When you say "synchronized", do you mean they are synchronized according >> to the definition of simultaneity in their mutual rest frame? As I asked >> before, 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? If so you are going to 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...I will await a clear answer from you on this question >> before elaborating on such a scenario, though. >> >> >> >> >> Now magically they are in non-accelerated relative motion to each such >> that each sees the other's clock running half as fast as their own. >> >> >> Physics textbooks often consider examples where there are "instantaneous" >> accelerations such that the velocity abruptly changes from one value to >> another, with the objects moving inertially both before and after the >> "instantaneous" acceleration, is that the same as what you mean by >> "magically they are in non-accelerated relative motion"? >> >> >> >> >> During the duration of the relative motion whenever A reads t = n on his >> OWN clock and B reads t'=n on his OWN clock they will be at the same >> current moment of p-time. They can use this method later on to know what >> they were doing at the same present moment. >> >> >> Even if we assume instantaneous jumps in velocity, there are multiple >> ways they could change velocities such that in their new inertial rest >> frames after the acceleration, each would say the other's clock is running >> half as fast as their own. For example, in the frame where they were >> previously at rest, if each one's velocity symmetrically changed from 0 in >> this frame to 0.57735c in opposite directions in this frame, then in each >> one's new rest frame the other would be moving at 0.866c (since using the >> relativistic velocity addition formula at http://math.ucr.edu/home/baez/ >> physics/Relativity/SR/velocity.html their relative velocity would then >> be (0.57735c + 0.57735c)/(1 + 0.57735^2) which works out to 0.866c), and a >> relative velocity of 0.866c corresponds to a time dilation factor of 0.5. >> But likewise, in the frame where they were previously at rest, it could be >> that one twin would remain at rest in this frame while the other would jump >> to a velocity of 0.866c in this frame, and then it would still be true that >> in each one's new rest frame the other is moving at 0.866c. Does your >> statement above that "whenever A reads t = n on his OWN clock and B reads >> t'=n on his OWN clock they will be at the same current moment of p-time" >> apply even in the case of asymmetrical changes in velocity? Are you saying >> all that matters is that in either one's new inertial rest frame, the other >> one's clock is ticking at half the rate of their own? >> >> >> Jesse >> >> >> >> On Saturday, February 8, 2014 5:28:08 PM UTC-5, jessem wrote: >> >> >> >> >> On Sat, Feb 8, 2014 at 4:01 PM, Edgar L. Owen <[email protected]> wrote: >> >> Jesse, >> >> Yes, I think there is always a way to determine if any two events happen >> at the same point in p-time or not, provided you know everything about >> their relativistic conditions. >> >> You do this by essentially computing their relativistic cases BACKWARDS >> to determine which point in each of their worldlines occurred at the same >> p-time. >> >> Take 2 observers, A and B. >> >> 1. If there is no relative motion or gravitational/acceleration >> differences you know that every point t in A's CLOCK time was in the same >> present moment as every point t' in B's CLOCK TIME when t=t'. >> >> >> And what if there *are* gravitational differences, if there are sources >> of gravity nearby and they are at different points in space? Gravity is >> dealt with using general relativity, and in general relativity there is no >> coordinate-indepedent way to define the "relative motion" of observers at >> different points in space (see discussion at http://math.ucr.edu/home/ba >> ez/einstein/node2.html for details). And the only coordinate-independent >> definition of "acceleration" is proper acceleration (what an observer would >> measure with an accelerometer that shows the G-forces they are >> experiencing), but all observers in freefall have zero proper acceleration, >> so if you think there is a "gravitational/acceleration difference" between >> an observer orbiting far from a black hole and one falling towards it close >> to the event horizon, you can't quantify it using proper time. >> >> >> >> >> 3. In the case of twins DURING the trip in relative motion we can always >> back calculate the relativistic effects to make a statement of the form >> "the twins were in the same current moment of p-time when A read his own >> clock as A-t and B's clock as B-t, AND B read his own clock as B-t' and >> read A's clock as A-t'. In this case A-t will NOT = A-t', and B-t will NOT >> = B-t', but they will have specific back calculable t values for every >> current p-time during the trip. Thus if we have all the details of that >> trip's motion we should always be able to back calculate to determine which >> clock times of any two observers occurred in the same current p-time >> SIMULTANEITY even when those observers cannot agree on CLOCK time >> simultaneity among themselves. >> >> >> HOW would you "back calculate" it though? Even if we set aside my >> questions about gravity above and just look at a case involving flat SR >> spacetime, your answer gives no details. If you have any procedure in mind, >> could you apply it to a simple example? Let's say Alice is sent on a ship >> that moves away from Bob on Earth on the day they are both born, and the >> ship moves with speed of 0.8c relative to the Earth, towards a planet 12 >> light-years away in the Earth's frame. Alice arrives at that planet when >> she is 9 years old, and at that point the ship immediately turns around and >> heads back towards Earth with a relative speed of 0.6c. Alice experiences >> the return journey to take 16 more years, so when she returns to Earth she >> is 25 years old, but Bob is 35 years old when they meet. Can you show me >> how to back-calculate how old Bob was when he was in the same moment of >> p-time as Alice turning 9 and her ship reaching the planet and turning >> around? >> >> >> >> So since p-time has no metric itself you can't just compare p-time t >> values because there are none. You have to back calculate clock times to >> determine in what current p-times they occurred. >> >> So that's how we determine whether any two events occurred a the same >> p-times or not. You should always be able to determine that even though you >> can assign a p-time t value because there are none because p-time doesn't >> have a metric. >> >> >> I have never asked you for a p-time "value", I'm only interested in the >> question of which events are simultaneous in p-time. I don't think your >> answers so far have made it clear that you have any well-defined procedure >> for determining this, see my questions above. >> >> Jesse >> >> >> >> >> Edgar >> >> >> >> On Friday, February 7, 2014 12:51:32 PM UTC-5, jessem wrote: >> >> >> >> >> On Fri, Feb 7, 2014 at 12:27 PM, Edgar L. Owen <[email protected]> wrote: >> >> Jesse, >> >> Well you just avoid most of my points and logic. >> >> >> Can you itemize the specific points you think I'm avoiding? >> >> >> >> But yes, I agree with your operational definition analysis. That is >> EXACTLY my point. That what our agreed operational definitions define is a >> COMMON PRESENT MOMENT, and NOT a same point in spacetime, because the logic >> of it does not support it being in the same point in space, only in the >> same point of time >> >> >> Huh? Even if one accepts p-time, that "operational definition" still must >> be seen as a merely *approximate* way of defining the same point of p-time, >> not exact, just like with "same point in space" >> >> ... > > -- > 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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