A constant speed of light in all reference frames? Surely you can't be serious.
in [Relativity]
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From: Bill Hobba on 12 Mar 2010 15:59 On 13/03/2010 2:36 AM, PD wrote: > On Mar 11, 9:32 pm, Paul Stowe<theaether...(a)gmail.com> wrote: >> On Mar 10, 8:57 am, PD<thedraperfam...(a)gmail.com> wrote: >> >> >> >>> On Mar 9, 9:41 pm, PaulStowe<theaether...(a)gmail.com> wrote: >> >>>> On Mar 8, 8:05 pm, "Inertial"<relativ...(a)rest.com> wrote: >> >>>>> "PaulStowe"<theaether...(a)gmail.com> wrote in message >> >>>>> news:1132a230-92d9-484a-b0c1-d3a97532cad9(a)z10g2000prh.googlegroups.com... >> >>>>>>>>> SR explains it as having to be c due to the geometry of spacetime >> >>>>>>>> That's simply a silly idea... >> >>>>>>> That you think it is silly is your problem, not that of SR >> >>>>>> Something physical may be represented by a geometric description. >> >>>>> And our universe is represented by Minkowski geometry. >> >>>> Yes, you can descibe localized behavior with that format. BUT! to do >>>> so you must depend on finite light speed and its physical >>>> independence. Geometry neither predicts. explains, or has a basis for >>>> that. >> >>> That's incorrect, Paul. The geometric structure of spacetime imposes >>> both a finite speed of light AND makes it frame-independent. >> >>> The geometric structure of spacetime *necessarily* divides pairs of >>> events into three categories: spacelike-separated, timelike-separated, >>> and nullcone-separated. This structure also immediately leads to the >>> result that any wordline that could be traversed by something between >>> timelike-separated events will, in any other inertial reference frame, >>> still be between timelike-separated events. What this means explicitly >>> is that this object can never span two spacelike-separated events. >>> Thus, the universe of events is strictly divided into two completely >>> separated causal domains. The boundary of these domains is the null >>> cone. Since the null cone has a definite slope of space vs time, this >>> imposes a causal speed limit. (This limit does not exist in Euclidean >>> 3D+1D space -- it is a unique feature of the 4D space and its >>> geometry.) >> >>> Furthermore, while transformations between inertial frames will shift >>> the slopes between pairs of timelike events (that is, the speed of an >>> object traveling between the two events), the same transformation >>> between pairs of events on the null cone do not change slope. What >>> this means is that any object that can travel between two events on >>> null cone will have the same speed regardless of inertial reference >>> frame. >> >>> So you see, the geometric structure DOES imply both a causal speed >>> limit and the invariance of that causal speed limit with choice of >>> inertial reference frame. It just so happens that light appears to be >>> one of the candidate objects that can travel between nullcone- >>> separated events. >> >>> If you need to see how the structure does impose those limits >>> formally, I could point you to a reference book or two that derives >>> this unambiguously. >> >>> At the time that Einstein proposed special relativity, he did not >>> understand how such a geometric structure could produce those two >>> conclusions as necessary consequences. And so he just posited the >>> invariance of the speed of light as a postulate (or equivalently, >>> demanded that Maxwell's equations obey the principle of relativity). >>> It was only later that the geometric structure was uncovered and it >>> was understood how the light postulate follows directly from this >>> structure. >> >>> PD >> >> I wasn't going to bother with a reply since we have gone round& round >> on this very point. I find your argument without merit and I'm >> certain that you mind is made up. Why act like kid and continuously >> and say no it ain't, yes it is??? > > I'm not sure what you mean by "find your argument without merit". I'm > not attempting to make an argument. I'm explaining facts about the > theory and what implies what in that theory. If you do not understand > what implies what, and you were hoping that my response would make it > plainer to you, then perhaps this is what you mean by "without merit". > Perhaps something is "without merit" if you are not convinced. > >> >> In minkowski math c can be any finite value. > > Yes, and in Gauss' Law, the constant in the expression between the > field and the source charge can take any value. That value is > empirically determined. In that case, it is the constant epsilon-zero. > In this case, it is c. In the case of Gauss' Law applied to Newtonian > gravity, the constant is G. > >> As Tom Roberts would >> argue the are nearly a infinite number of variations which fit this >> form. Thus it's dependent upon c being a 'physical' constant. > > Yes, that is so. As is true for just about every physical law. > >> And, >> as GR shows, it not even global. > > I'm not sure what you mean by that. Even in GR, the slope of the local > lightcone is c always. > >> Now why might that be??? The logic >> (actually lack thereof) and thought process is 'in my opinion' silly >> and no one, not in print nor herein has provided any argument that is >> convincing that the math and geometry is NOT! a resultant of physical >> processes rather some magical geometry... > > It may well be the result of what you call "physical processes", which > I take to mean matter banging on matter in the fashion you're used to > from macroscopic physics. After all, Einstein's postulates were found > to be explainable in terms of something more fundamental, as I've > explained. It is entirely possible that there is another, more > fundamental principle or interaction that accounts for Minkowski > geometry, which in turn accounts for the 1905 postulates. The only > problem is, nothing of the sort has been successfully produced yet. > Since you feel very strongly that this is the only kind of fundamental > explanation that is worth anything, you are invited to produce one > that works. > > You may be interested in investigating spin-foam models of quantum > gravity, which offer the attractive feature of being "backgroundless". > That is, they do not presume a pre-existing spacetime. Rather, space > and time *emerge* from the spin-foam. I have no idea whether you > consider such models (http://math.ucr.edu/home/baez/foam/ for some > introductory pointers) to be "physical processes" according to your > expectations, but they do have a feature I would guess would be > attractive to you -- that spacetime is a artifact of the explanation, > not the basis of the explanation. Note that, despite your > protestations that no one is working on a deeper explanation, loop > quantum gravity, spin-foams, and spin-networks are very much an active > and hot area of research. > > PD > Thanks for making me aware of those models. Much appreciated. Thanks Bill
From: Inertial on 12 Mar 2010 18:44 "Bruce Richmond" <bsr3997(a)my-deja.com> wrote in message news:bf49f7ee-126a-4e44-bdfd-cf323e83cdc6(a)c16g2000yqd.googlegroups.com... > On Mar 11, 10:54 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "Paul Stowe" <theaether...(a)gmail.com> wrote in message >> >> news:722fe1d3-ba1d-4439-bffe-eda2ca668f82(a)p3g2000pra.googlegroups.com... >> >> >> >> >> >> > On Mar 10, 8:57 am, PD <thedraperfam...(a)gmail.com> wrote: >> >> On Mar 9, 9:41 pm, PaulStowe<theaether...(a)gmail.com> wrote: >> >> >> > On Mar 8, 8:05 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> > > "PaulStowe" <theaether...(a)gmail.com> wrote in message >> >> >> > >news:1132a230-92d9-484a-b0c1-d3a97532cad9(a)z10g2000prh.googlegroups.com... >> >> >> > > >> >> SR explains it as having to be c due to the geometry of >> >> > > >> >> spacetime >> >> >> > > >> > That's simply a silly idea... >> >> >> > > >> That you think it is silly is your problem, not that of SR >> >> >> > > > Something physical may be represented by a geometric >> >> > > > description. >> >> >> > > And our universe is represented by Minkowski geometry. >> >> >> > Yes, you can descibe localized behavior with that format. BUT! to >> >> > do >> >> > so you must depend on finite light speed and its physical >> >> > independence. Geometry neither predicts. explains, or has a basis >> >> > for >> >> > that. >> >> >> That's incorrect, Paul. The geometric structure of spacetime imposes >> >> both a finite speed of light AND makes it frame-independent. >> >> >> The geometric structure of spacetime *necessarily* divides pairs of >> >> events into three categories: spacelike-separated, timelike-separated, >> >> and nullcone-separated. This structure also immediately leads to the >> >> result that any wordline that could be traversed by something between >> >> timelike-separated events will, in any other inertial reference frame, >> >> still be between timelike-separated events. What this means explicitly >> >> is that this object can never span two spacelike-separated events. >> >> Thus, the universe of events is strictly divided into two completely >> >> separated causal domains. The boundary of these domains is the null >> >> cone. Since the null cone has a definite slope of space vs time, this >> >> imposes a causal speed limit. (This limit does not exist in Euclidean >> >> 3D+1D space -- it is a unique feature of the 4D space and its >> >> geometry.) >> >> >> Furthermore, while transformations between inertial frames will shift >> >> the slopes between pairs of timelike events (that is, the speed of an >> >> object traveling between the two events), the same transformation >> >> between pairs of events on the null cone do not change slope. What >> >> this means is that any object that can travel between two events on >> >> null cone will have the same speed regardless of inertial reference >> >> frame. >> >> >> So you see, the geometric structure DOES imply both a causal speed >> >> limit and the invariance of that causal speed limit with choice of >> >> inertial reference frame. It just so happens that light appears to be >> >> one of the candidate objects that can travel between nullcone- >> >> separated events. >> >> >> If you need to see how the structure does impose those limits >> >> formally, I could point you to a reference book or two that derives >> >> this unambiguously. >> >> >> At the time that Einstein proposed special relativity, he did not >> >> understand how such a geometric structure could produce those two >> >> conclusions as necessary consequences. And so he just posited the >> >> invariance of the speed of light as a postulate (or equivalently, >> >> demanded that Maxwell's equations obey the principle of relativity). >> >> It was only later that the geometric structure was uncovered and it >> >> was understood how the light postulate follows directly from this >> >> structure. >> >> >> PD >> >> > I wasn't going to bother with a reply since we have gone round & round >> > on this very point. I find your argument without merit and I'm >> > certain that you mind is made up. Why act like kid and continuously >> > and say no it ain't, yes it is??? >> >> > In minkowski math c can be any finite value. >> >> Indeed it can. But we observe it to have a particular value in our >> universe. > > Would that be the value in meters per second, miles per second, miles > per hour.... I could go on. I'm sure you would carry on with such pedantic nonsense. It is still the same value .. just expressed with a different numerical value in different units as every value with units is. So yours is really no argument at all. [snip rest] And that's all you had to say?
From: Inertial on 12 Mar 2010 18:56 "ben6993" <ben6993(a)hotmail.com> wrote in message news:716c1760-a5db-4ce8-b116-7a739eaae397(a)o30g2000yqb.googlegroups.com... > On Mar 12, 6:16 pm, PD <thedraperfam...(a)gmail.com> wrote: >> On Mar 12, 12:00 pm, Ste <ste_ro...(a)hotmail.com> wrote: >> >> >> >> >> >> > On 11 Mar, 20:57, PD <thedraperfam...(a)gmail.com> wrote: >> >> > > On Mar 11, 2:15 pm, Ste <ste_ro...(a)hotmail.com> wrote: >> >> > > > On 11 Mar, 15:12, PD <thedraperfam...(a)gmail.com> wrote: >> >> > > > > On Mar 11, 6:43 am, Ste <ste_ro...(a)hotmail.com> wrote: >> >> > > > > > On 11 Mar, 01:51, "Peter Webb" >> > > > > > <webbfam...(a)DIESPAMDIEoptusnet.com.au> >> > > > > > wrote: >> >> > > > > > > No, perhaps you didn't understand. As I say, this is *not* >> > > > > > > the twins >> > > > > > > paradox, because in the twins paradox only *one* twin leaves >> > > > > > > Earth. >> >> > > > > > > ________________________ >> > > > > > > Its functionally the same. It is exactly the twins paradox, >> > > > > > > but with two >> > > > > > > twins apparently doing exactly the same thing. >> >> > > > > > > Even if you cannot see that, the explanation on the Wikipedia >> > > > > > > page of the >> > > > > > > Twins Paradox is trivially adapted for two twins. >> >> > > > > > > I assume that you do not understand the Wikipedia twins >> > > > > > > paradox page, or >> > > > > > > else you would know the answers to your questions already. >> > > > > > > Which parts don't >> > > > > > > you understand? >> >> > > > > > Let's just go through it step by step Peter, as we have been >> > > > > > doing. >> > > > > > It's pointless spending 10 more postings arguing about how the >> > > > > > Wikipedia page does or does not answer the question, or how it >> > > > > > is or >> > > > > > is not relevant. As I've just said in a post to Inertial, the >> > > > > > only >> > > > > > analogy between my scenario and the twins paradox is that, in >> > > > > > my >> > > > > > scenario, both twins leave Earth, and both return the same age >> > > > > > as each >> > > > > > other - hence no paradox, and hence bearing no resemblance at >> > > > > > all to >> > > > > > the twins paradox. >> >> > > > > First of all, let's establish what you think is paradoxical at >> > > > > all >> > > > > about the description of the twins in the twin puzzle. Then let's >> > > > > see >> > > > > whether this paradox is present in the case you mention. >> >> > > > As I understand it, the supposed "paradox" in the twins paradox was >> > > > that one returned younger than the other. It was, of course, not a >> > > > paradox at all, but that's besides the point. >> >> > > No, then you do not understand the paradox, because there is nothing >> > > contradictory in that statement at all. It may be surprising, but >> > > it's >> > > not contradictory, not paradoxical. Disagreement of clocks is not a >> > > paradox. >> >> > > The paradox, which is what is perceived (normally) by freshmen when >> > > first introduced to this statement, is embodied in their immediate >> > > classroom question: "But in the frame of the traveling twin, it is >> > > the >> > > earth twin that is moving away and returning. Since this is symmetric >> > > to the case of the traveling twin moving away and returning, then >> > > shouldn't the traveling twin expect the earth twin to be younger when >> > > they meet again?" Now perhaps the paradox is more apparent to you. >> >> > > However, the puzzle is specifically designed to emphasize the danger >> > > of oversimplifying. In fact, the two twins are NOT symmetric, because >> > > one unambiguously experiences acceleration and the other >> > > unambiguously >> > > experiences no acceleration. This then leads to a discussion of what >> > > produces the asymmetry in the time. >> >> > I know Paul. I know. >> >> You can imagine my surprise, since what you said explicitly above was >> that the paradox was that the twins aged differently. >> >> >> >> > > Perhaps if you had started out by asking, "Since I don't see any >> > > obvious paradox here at all, perhaps someone could illuminate me as >> > > to >> > > why this is called the twin paradox?" Then at least you would have >> > > been on square one. >> >> > Really I just wanted to avoid going off on a long tangent about the >> > twins paradox. As I said, the scenario that were were addressing is >> > different from the twins paradox, in that we have three clocks, and >> > the two clocks with which we are now concerned (B and C) both return >> > to the origin point *synchronised* (albeit both lagging behind A), >> > whereas the twins' ages are not synchronised on the return of the >> > astronaut twin. >> >> Well, yes, it is a different result from the application of the same >> principle. >> >> If I calculate the angle that I can tip a TV tray before the coffee >> cup on it starts to slide, I find that I'm using the same principle >> (equilibrium of forces) that I would use to determine the tension in >> picture-hanger wire when mounting a photo on the wall. >> >> Different situation. Very same principle. >> >> Fine example of losing the forest for the trees. As you've done here. >> >> >> >> >> >> > So let me say again. The twins paradox would be applicable if we were >> > talking about A and B, or A and C. In the event, we are talking about >> > what B and C observe of each other from their own reference frames. >> > There is, therefore, no correspondence with the twins paradox, because >> > unlike the twins, B and C return synchronised with each other.- Hide >> > quoted text - >> >> - Show quoted text -- Hide quoted text - >> >> - Show quoted text - > > I have written below a twin plank paradox of length. I know the > explanation must really be very straightforward, but I don't see it > yet. This is not a time paradox, but it looks to be a similar issue. > > A plank has four units of length (----) in its own frame and a garage > has two units of length (--) in its own frame. > There are two similar planks and garages in relative motion such that > plank P and garage G are moving fast towards the other plank p and > garage > > g. The relative speed is such that lengths are approximately halved > in relativistic contraction. > > <--- direction of motion > p ---- G' - > g -- P' -- > > Where ' sign indicates motion in the other twins' frame. > > Here, p does not fit within G', but P' fits approx. within g. > > > Looked at from the other frame, below, a similar result occurs: p' > fits approx. within G, but P does not fit within g'. > ----> > p' -- G -- > g' - P ---- > > > As the same two events are looked at twice, i.e. from two frameworks, > there are four outcomes in total. The shed doors are destroyed in two > outcomes and the planks fit into the sheds in the other two outcomes. > So it looks like one of the two sheds is damaged and the other is > safe. But that cannot be true as the motion is relative not absolute, > and the planks and garages are twins, and so the two outcomes should > be identical. They are identical if you do identical things . If you do the 'close both garage doors simultaneously in the garage's rest frame of reference when the pole is fully within' for both garages, both are safe. If you close the door at different times in their frame, then they may not be safe (depending on when you do)
From: BURT on 12 Mar 2010 19:23 On Mar 8, 6:35 am, Ste <ste_ro...(a)hotmail.com> wrote: > On 7 Mar, 02:51, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > wrote: > > > > > > > "Ste" <ste_ro...(a)hotmail.com> wrote in message > > >news:651a713d-7ae4-4048-bafb-f1b3219ee4fc(a)v20g2000yqv.googlegroups.com.... > > > > On 6 Mar, 12:47, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > > > wrote: > > >> > This should make perfect sense to you. If a clock is running 2% > > >> > slower, then it is running 2% slower regardless of distance. But if, > > >> > as a result of running 2% slower, it falls behind 6 minutes after > > >> > running a certain amount of time, then it will fall behind 12 minutes > > >> > after running for twice as long. > > > >> Agreed. > > > >> The question now is, if we agree that both clocks suffer time dilation > > >> in this way, then when they return to the start point, how do they > > >> each reconcile the fact that (after accounting for the effects of > > >> acceleration) it ought to be the other clock which is slow, when in > > >> fact one clock (the one that went furthest from the start point) will > > >> be slower than the other? And a third clock, left at the start point, > > >> will be running ahead of both? > > > >> _________________________ > > >> They know that the operations were not symmetric. Only one clock remained > > >> in > > >> the same inertial reference frame throughout. The other two clocks spent > > >> different amounts of time in at least 3 different inertial reference > > >> frames. > > >> Everybody can see this is true, and so nobody expects that the clocks > > >> will > > >> remain synchronised. > > > > Yes, but the important question here is whether they agree *after* the > > > effects of acceleration are taken into account. I mean, if we said > > > that each travelling clock slows by 2% when moving away from the start > > > point at a certain speed, then by rights both travelling clocks should > > > slow equally. Yes? > > > As I understand your thought experiment, no. > > > In SR, time dilation is a function of relative speed and the time for which > > they are moving at the speed. It is not a function of accleration. > > > A doesn't move. B moves at speed v for time t, and its clock will read x > > behind A. C moves at speed v for time 2t, and its clock will read 2x behind > > A. > > The question is this. We'll deal with only the outbound trip (in other > words, the clocks are on the move, but time 't' has not yet elapsed, > so there has been no further acceleration). I agree with your answer > above, as it concerns A's frame. > > The question is, from the frame of B, what will the slowdown be on > clock C, *after* having accounted for the increased distances between > them (i.e. having accounted for the increased propagation delays). It > seems to me that the natural answer is to say "4%".- Hide quoted text - > > - Show quoted text - The clocks change when accelerating and decelerating in space. Time decelerates and accelerates when there is a change in speed in space. Mitch Raemsch
From: mpc755 on 12 Mar 2010 20:01
On Mar 12, 7:23 pm, BURT <macromi...(a)yahoo.com> wrote: > On Mar 8, 6:35 am, Ste <ste_ro...(a)hotmail.com> wrote: > > > > > On 7 Mar, 02:51, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > > wrote: > > > > "Ste" <ste_ro...(a)hotmail.com> wrote in message > > > >news:651a713d-7ae4-4048-bafb-f1b3219ee4fc(a)v20g2000yqv.googlegroups.com.... > > > > > On 6 Mar, 12:47, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> > > > > wrote: > > > >> > This should make perfect sense to you. If a clock is running 2% > > > >> > slower, then it is running 2% slower regardless of distance. But if, > > > >> > as a result of running 2% slower, it falls behind 6 minutes after > > > >> > running a certain amount of time, then it will fall behind 12 minutes > > > >> > after running for twice as long. > > > > >> Agreed. > > > > >> The question now is, if we agree that both clocks suffer time dilation > > > >> in this way, then when they return to the start point, how do they > > > >> each reconcile the fact that (after accounting for the effects of > > > >> acceleration) it ought to be the other clock which is slow, when in > > > >> fact one clock (the one that went furthest from the start point) will > > > >> be slower than the other? And a third clock, left at the start point, > > > >> will be running ahead of both? > > > > >> _________________________ > > > >> They know that the operations were not symmetric. Only one clock remained > > > >> in > > > >> the same inertial reference frame throughout. The other two clocks spent > > > >> different amounts of time in at least 3 different inertial reference > > > >> frames. > > > >> Everybody can see this is true, and so nobody expects that the clocks > > > >> will > > > >> remain synchronised. > > > > > Yes, but the important question here is whether they agree *after* the > > > > effects of acceleration are taken into account. I mean, if we said > > > > that each travelling clock slows by 2% when moving away from the start > > > > point at a certain speed, then by rights both travelling clocks should > > > > slow equally. Yes? > > > > As I understand your thought experiment, no. > > > > In SR, time dilation is a function of relative speed and the time for which > > > they are moving at the speed. It is not a function of accleration. > > > > A doesn't move. B moves at speed v for time t, and its clock will read x > > > behind A. C moves at speed v for time 2t, and its clock will read 2x behind > > > A. > > > The question is this. We'll deal with only the outbound trip (in other > > words, the clocks are on the move, but time 't' has not yet elapsed, > > so there has been no further acceleration). I agree with your answer > > above, as it concerns A's frame. > > > The question is, from the frame of B, what will the slowdown be on > > clock C, *after* having accounted for the increased distances between > > them (i.e. having accounted for the increased propagation delays). It > > seems to me that the natural answer is to say "4%".- Hide quoted text - > > > - Show quoted text - > > The clocks change when accelerating and decelerating in space. Time > decelerates and accelerates when there is a change in speed in space. > > Mitch Raemsch Due to the change in the pressure associated with the aether. |