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/ > baez/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" or "same point in > spacetime". If I bounce some light off you, surely you agree that the event > of it reflecting off you occurred at a slightly earlier point in p-time > that the event of reaching my eyes (or instruments)? Likewise if I feel our > palms meet in a handshake, I don't actually begin to feel it until a > slightly later moment of p-time than the moment our palms first made > physical contact, and likewise for any shift or movement you might make > with your hands. If you want to talk in a non-approximate way, all our > experiences are slightly delayed impressions of events that occured in the > past, regardless of whether we're talking about p-time or coordinate time. > > On this subject, could you address the question I asked in another post > about whether you think there's any empirical way to determine whether two > events in the past occurred at the same p-time, or whether the assumption > of p-time simultaneity is a purely metaphysical one and that there's no way > of knowing whether a specific pair of events we have records of actually > happened simultaneously in p-time? > > > > and that same point in time is obviously not anything that relativity > predicts, because no matter what set of coordinates you choose, relativity > always gives 2 different real answers for the ages of the twins. > > > > I don't know what part of this you're not understanding, "same point in > time" in relativity just MEANS that two events are assigned the same time > coordinate, relativity doesn't deal with any absolute notion of > simultaneity of distant events whatsoever. And relativity definitely does > predict situations where clocks show different readings at the same > coordinate time--do you deny this? > > Like I said earlier, there is a direct spatial analogy here that makes > perfect sense if you don't assume p-time from the start. If two different > measuring tapes cross, and the point where they cross is at the 30 cm mark > on one tape and the 40 cm mark on the other, and there's a Cartesian > coordinate grid on the surface under them which says this point has an > x-coordinate of 50, wouldn't you say that the measuring tapes DO cross at > the "same point in space"? Would the fact that the tapes themselves show > two different readings at that point negate this? > > As for your last paragraph you seem to agree that both our operational > definitions DO support the notion of a same present moment, just not that > time flows. > > > How do you figure? My last paragraph was just clarifying what I meant by > arguments "dependent on conscious experience" vs. arguments defined in > terms of straightforward experiments whose results we can all observe and > agree on. Nowhere did I say anything in support of an absolute "same > present moment". > > > Jesse > > > > On Friday, February 7, 2014 8:49:32 AM UTC-5, jessem wrote: > > > On Fri, Feb 7, 2014 at 7:57 AM, Edgar L. Owen <[email protected]> wrote: > > Jesse, > > OK, here's the detailed analysis of how I see the current state of this > issue that I promised: > > > A few points: > > 1. Since you asked let me repeat my 'operational definition' of the > present moment that I used before. The twins meet, shake hands and compare > watches. That is the operation definition. > > That is essentially the same as your reflected light operational > definition with which I have no problem. > > 2. However it is important to note that that works not just for the twins > together, but for every single twin by himself. Because any twin or > observer can shake his own hand, look at his own watch, or note that the > light reflected from a mirror in his hand takes minimal time to return. > > Therefo > > ... -- 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]. 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