Why physicists refuse to measure the one-way speed of light directly?
in [Relativity]
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From: PD on 22 Jan 2010 16:37 On Jan 22, 1:41 pm, Da Do Ron Ron <ron_ai...(a)hotmail.com> wrote: > T Roberts asked: > > >I have no idea what you mean. The propagation of light just occurs > >however it happens, and clocks are simply synchronized according to > >some prescribed method. > >What "physical process" do you mean? > > It is the process that you mentioned just prior to your query, namely, > the > "prescribed method" of clock synchronization; however, since you have > not yet properly applied this method to more than one frame, you > cannot > fully understand it. > > Yes, I know that you believe that you have done this, because you > said > the following: > > >As no mention of which inertial frame was used, this applies in any > >inertial frame, thus ensuring invariance of the one-way speed of light. > > But this is wrong, as my prior (simple) diagram should have shown. > As I have tried to get you to see, merely repeating the same frame > over and over (as you just did, and as the Einsteinian version of his > method does) does _not_ convey the full story. > > There is only one way to properly show Einstein's convention of > synchronization in more than one frame, and that, as I have tried > to get across, is by letting the frames share the light source. > > Giving each frame its own source is to merely and uselessly repeat > the same frame over and over and over. > > Here, _again_, is a picture of the start of the physical process > about > which you asked above: > > Frame A > [0]---------x----------[?] > Source S~~>light > [0]---------x----------[?] -->v > Frame B > > Notice the very careful and very necessary usage of a single light > source. > > Notice the equally necessary usage of at least two frames. > > Since this is NOT done in any relativity text, no one has yet seen > the > full version of Einstein's definition of clock synchronization. > > Therefore, no one has yet seen the full truth re Einstein's > definition. > > To repeat, the _ONLY_ way to see the full truth of the definition > is by carrying the above picture to completion. > > You, or PD, or Android, or Dirk, or Seto, or Gisse, or _some_ person > MUST fill in the blanks to complete the diagram. Sorry, but no. If you do not understand what Einstein's procedure is, look at what I wrote to you earlier, where I explained it simply. No one is obligated to follow YOUR boondoggles, just because you claim it's what Einstein really meant (which he did not). > Only then will the > full physical process of which we are speaking be made perfectly > clear. > > Have I made myself perfectly clear? > > I will even bend over backward to carry the picture one step further: > > Frame A > [0]---------x----------[?] > Source S---------------->light > ----------[0]---------x----------[?] -->v > Frame B > > WHAT, pray tell, is the reading NOW on A's right-hand clock per > Einstein's definition of clock synchronization? > > Can anyone tell us? > > ~~RA~~ > (as was given, x is the ruler-measured distance > given a ruler at rest wrt the frame in which the > measurement is made)
From: Tom Roberts on 22 Jan 2010 20:50 Da Do Ron Ron wrote: > Tom Roberts wrote: >> I have no idea what you mean. The propagation of light just occurs >> however it happens, and clocks are simply synchronized according to >> some prescribed method. >> What "physical process" do you mean? > > It is the process that you mentioned just prior to your query, namely, > the "prescribed method" of clock synchronization; Calling that a "physical process" is quite unusual. It is a procedure carried out by humans. To the rest of us, "physical process" implies something performed by some natural phenomenon. > however, since you have > not yet properly applied this method to more than one frame, you > cannot fully understand it. I described techniques applicable to ANY inertial frame. That is infinitely more than one. > Yes, I know that you believe that you have done this, because you > said > the following: >> As no mention of which inertial frame was used, this applies in any >> inertial frame, thus ensuring invariance of the one-way speed of light. > > But this is wrong, as my prior (simple) diagram should have shown. Your diagram is impenetrable. And what I said is not wrong -- it does indeed apply to ANY inertial frame. > There is only one way to properly show Einstein's convention of > synchronization in more than one frame, and that, as I have tried > to get across, is by letting the frames share the light source. No. One can apply either method I described in however many frames you choose to use. This requires pairs of clocks and a light source in each of the inertial frames you are using, plus light detectors to start each clock. Sharing the light source is not necessary at all. While it is in principle possible to share the light source among frames, arranging to do so is quite complicated and COMPLETELY unnecessary. > Giving each frame its own source is to merely and uselessly repeat > the same frame over and over and over. Not true. Performing synchronization in a lab on earth, and in a lab on a moving train, is not at all "using the same frame". They are MOVING relative to each other, and thus are DIFFERENT inertial frames. > Have I made myself perfectly clear? No. Your diagram is impenetrable. You need to describe what it purports to show. Tom Roberts
From: glird on 23 Jan 2010 00:27 On Jan 19, 3:22 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Jan 19, 2:13 pm, Da Do Ron Ron wrote: > > > One-way isotropy was the subject at hand. We all know that experiment > > has shown round-trip isotropy, so your ref. is useless. > > http://edu-observatory.org/physics-faq/Relativity/SR/experiments.html... I looked and found this in it: "Mössbauer experiments show that the rate of a clock is independent of acceleration (~1016 g) and depends only upon velocity." How can an acceleration, a = dv/dt, have no affect on something that varies with v? Does it mean that since dt is a function of v, dv/dt is a constant? If so, suppose the value of a is steadily changing? > > PD incorrectly stated: > > > Synchronization is a condition that is only > > >satisfied in one frame anyway. > > >< You cannot have one-way light speed invariance unless observers in _all_ frames obtain the same speed for light's one-way speed, and this means that clocks in _all_ frames must be set per Einstein's definition. > > >< No, that is incorrect. Each frame has its OWN set of clocks, and in EACH frame THAT set is synchronized according to the Einstein procedure. In EACH frame, then, the speed of light will have the same value. This does NOT mean that the clocks in one frame are synchronized along with the clocks in a different frame, and in general this will NOT be the case. > A frame of reference is a conceptual device. As such, it extends infinitely in all directions; as do all other differently moving frames. each such frame is anchored to a visible referent; taken as the origin of an attached co-ordinate system ("cs"). At every hypothetical point of such an imaginary cs there is an infinitely small mythical observer with an equally small mythical clock. Since all of these frames are everywhere, thus overlap each other, if we arbitrarily stipulate that at t=t'=t^x=0 their origins coincide, then so do all their observers and clocks at that point. If we now let clocks of a given frame be set via Einstein's method, that identical process can also be used to set the differently moving clocks of any other frame. Inded, as Einstein sort of showed in his 1905 paper, that was the reason for his relativity of simultaneity. > >< And by refusing to complete the given task, you have blocked yourself from understanding Einstein's definition of clock synchronization. > Yes! > No, I'm sorry, but Einstein was quite clear. > > >< Here, again, are the rules: 1. At least two frames must be used (for invariance). 2. Only one light source must be used (to separate the frames). 3. The proper version of the definition must be used. (This is the one that can be applied to two or more frames using a single light source.)> 1. One frame can be used even though an infinity of other frames coexist with it. Invariance simply means that AFTER clocks have been esynched (set by Einstein's method) the speed of light will be c as measured in each such frame. 2. ANY light source can be used to set the clocks of any and all inertially moving systems. None of them are or need to be separate from each other, even though their origin points are, as time passes. 3. That's what Einstein actually did in his demo of the rel of simul. In that demo, though, he did NOT esynch the clocks of cs 2; he let them have the same settings as esynched cs 1. > >< Anyone who ignores any one of these bedrock rules will not be able to grasp the full physical significance of Einstein's definition. > Perhaps you are talking about his "relativity of simultaneity". As to his definition of "synchrony"; it means that if a beam of light (from any source) travels from clock A to clock B of any given system, those clocks must be hand set to MEASURE its speed as a constant each way. > >< Here, again, is the task that you must complete in order to fully comprehend that definition: Frame A [0]---------x----------[?] Source S~~>light [0]---------x----------[?] -->v Frame B Why are you afraid to fill in the blanks? There aren't any blanks. > > Forget about everything else, and do this now. If you mean, "Put numbers in place of the ??s"; let x = x = .5 and ? = ? = 1. > > Only then will you see the truth. > > Guaranteed. > > ~~RA~~ The "truth" is that AFTER clocks have been set by Einstein's method, they will measure 0?/c = 1/c = c in frame A, and 0?/(c-v) = 1/(c-v) = c in frame B. In EINSTEIN's paper, the diagram would have been like this: Stationary Frame K [A at x=0]---------[B at x=1] light ~~> [A at xi=0]------[B at xi=1] -->v Frame k He would then have said, "We have so far defined only an `A time' and a `B time'. We have not defined a common `time' for A and B, for the latter cannot be defined at all unless we establish by definition that the `time' required by light to travel from A to B equals the `time' it requires to travel from B to A. Let a ray of light start at the `A time' from A towards B, let it at the `B time' be reflected at B in the direction of A, and arrive again at A at the `A time'. "In accordance with definition the two clocks synchronize if tB - tA = tA' - tB." A bit later he said, "In agreement with experience we further assume the quantity 2AB/(tA'-tA) = c to be a universal constant--the velocity of light in empty space." Had he left his initial demo in place, it might have continued thus: To be continued tomo... later today. g-nite now... glird
From: kenseto on 23 Jan 2010 09:09 On Jan 22, 4:37 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Jan 22, 1:41 pm, Da Do Ron Ron <ron_ai...(a)hotmail.com> wrote: > > > > > > > T Roberts asked: > > > >I have no idea what you mean. The propagation of light just occurs > > >however it happens, and clocks are simply synchronized according to > > >some prescribed method. > > >What "physical process" do you mean? > > > It is the process that you mentioned just prior to your query, namely, > > the > > "prescribed method" of clock synchronization; however, since you have > > not yet properly applied this method to more than one frame, you > > cannot > > fully understand it. > > > Yes, I know that you believe that you have done this, because you > > said > > the following: > > > >As no mention of which inertial frame was used, this applies in any > > >inertial frame, thus ensuring invariance of the one-way speed of light.. > > > But this is wrong, as my prior (simple) diagram should have shown. > > As I have tried to get you to see, merely repeating the same frame > > over and over (as you just did, and as the Einsteinian version of his > > method does) does _not_ convey the full story. > > > There is only one way to properly show Einstein's convention of > > synchronization in more than one frame, and that, as I have tried > > to get across, is by letting the frames share the light source. > > > Giving each frame its own source is to merely and uselessly repeat > > the same frame over and over and over. > > > Here, _again_, is a picture of the start of the physical process > > about > > which you asked above: > > > Frame A > > [0]---------x----------[?] > > Source S~~>light > > [0]---------x----------[?] -->v > > Frame B > > > Notice the very careful and very necessary usage of a single light > > source. > > > Notice the equally necessary usage of at least two frames. > > > Since this is NOT done in any relativity text, no one has yet seen > > the > > full version of Einstein's definition of clock synchronization. > > > Therefore, no one has yet seen the full truth re Einstein's > > definition. > > > To repeat, the _ONLY_ way to see the full truth of the definition > > is by carrying the above picture to completion. > > > You, or PD, or Android, or Dirk, or Seto, or Gisse, or _some_ person > > MUST fill in the blanks to complete the diagram. > > Sorry, but no. If you do not understand what Einstein's procedure is, > look at what I wrote to you earlier, where I explained it simply. No > one is obligated to follow YOUR boondoggles, just because you claim > it's what Einstein really meant (which he did not). Einstein's synchronization procedure is circular: The procedure is as follows: 1. Note the time t1 at clock A. 2. Send a signal from clock A to clock B at a speed v. 3. When the signal arrives, note the time t2 at clock B. 4. Send a signal from clock B to clock A at a speed v. 5. When the signal arrives, note the time t3 at clock A. 6. If t3-t2 = t2-t1, then the clocks are synchronized. The clocks A and B must be pre-synchronized to begin with before Einstein's procedure is valid. If you already know the clocks are synchronized why do you need to synchronize them again using Einstein's procedure? Also if you have two spatially separated synchronized clocks why don't you use them to determine the one way speed of light? Ken Seto > > > > > Only then will the > > full physical process of which we are speaking be made perfectly > > clear. > > > Have I made myself perfectly clear? > > > I will even bend over backward to carry the picture one step further: > > > Frame A > > [0]---------x----------[?] > > Source S---------------->light > > ----------[0]---------x----------[?] -->v > > Frame B > > > WHAT, pray tell, is the reading NOW on A's right-hand clock per > > Einstein's definition of clock synchronization? > > > Can anyone tell us? > > > ~~RA~~ > > (as was given, x is the ruler-measured distance > > given a ruler at rest wrt the frame in which the > > measurement is made)- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: Sam Wormley on 23 Jan 2010 10:42
On 1/23/10 8:09 AM, kenseto wrote: > Also if you have two spatially separated synchronized clocks why don't > you use them to determine the one way speed of light? > > Ken Seto The speed of light is constant, Seto. We use light speed define distance measure for more than a quarter of a century now. Where have you been! |