Why physicists refuse to measure the one-way speed of light directly?
in [Relativity]
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From: Tom Roberts on 23 Jan 2010 11:05 glird wrote: > On Jan 19, 3:22 pm, PD <thedraperfam...(a)gmail.com> wrote: >> http://edu-observatory.org/physics-faq/Relativity/SR/experiments.html... > > I looked and found this in it: "M�ssbauer experiments show that the > rate of a clock is independent of acceleration (~1016 g) and depends > only upon velocity." Bailey et al extends this to ~ 10^18 g. > How can an acceleration, a = dv/dt, have no affect on something that > varies with v? Experiments show that the INTRINSIC rate of a clock does not depend on its velocity v, up to at least v ~ 0.9999 c; it does not depend on a (= dv/dt), up to at least a ~ 10^18 g. Both are of course measured relative to a locally inertial frame. An inertial observer will MEASURE the rate of a moving clock as slowed down by the gamma factor of SR, but this does not mean the clock ITSELF is affected; indeed the clock itself CANNOT be affected by either v or a in order for the theory to agree with experiments. Some clocks are broken by sufficiently large a. This only applies to clocks that are not broken. Fortunately both atomic transitions and muon decays are highly robust clocks. You REALLY need to learn what SR actually says. Your head is so full of nonsense that you never seem to get anything right. Tom Roberts
From: PD on 23 Jan 2010 15:33 On Jan 23, 8:09 am, kenseto <kens...(a)erinet.com> wrote: > 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. No, they do not. If step 6 shows an inequality, this indicates the clocks were not synchronized. Suppose that t3-t2 is 38 usec and t2-t1 is 36 usec. Then these two clocks are not synchronized. But the inequality tells you what you have to do to make the correction. Clock B is slow by 1 usec. If you set it back by 1 usec and repeat the procedure, then you will find that t3-t2 is 37 usec and t2-t1 is 37 usec. Now step 6 shows an equality. Congratulations! You have synchronized the clocks. Do you understand now? Has it really taken 12 years for you to have this procedure explained to you in a way that you can understand it? > 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? Because the resolution and experimental accuracy obtained by using two synchronized clocks does not beat the resolution and experimental accuracy obtained by the combination of TWLS and anisotropy experiments. The reasons have to do with the detailed analysis of sources of experimental uncertainty, which you can only get by reading the full experimental papers and not just the abstracts. > > 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: eric gisse on 23 Jan 2010 16:22 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? One way speed of light tests have been done, Ken. Directly, and indirectly through anisotropy measurements of light speed in addition to two way light speed tests. Do you see how limits on anisotropy and knowledge of two way light speed limits one way light speed, or do you need another 15 years to parse the concept? [...]
From: Sam Wormley on 23 Jan 2010 19:17 On 1/23/10 4:13 PM, Henry Wilson DSc wrote: > On Sat, 23 Jan 2010 09:42:23 -0600, Sam Wormley<swormley1(a)gmail.com> wrote: > >> 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! > > > ......and ken has pointed out that the same distance, measured with light > speed, is used to measure light speed. > > Hahahahahhahaaha! Wormey can't even see the joke.... > hahahhaaha! > > > Henry Wilson... > I've never understood your need, Henri, to play the fool. But if you must... you must!
From: glird on 23 Jan 2010 19:44
On Jan 23 glird wrote: >On Jan 19, 2:13 pm, Da Do Ron Ron wrote: Da Do: And by refusing to complete the given task, you have blocked yourself from understanding Einstein's definition of clock synchronization. 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.) > GL: 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. > Da Do: Anyone who ignores any one of these bedrock rules will not be able to grasp the full physical significance of Einstein's definition. > > GL: 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. > Da Do: 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? Forget about everything else, and do this now. Only then will you see the truth. Guaranteed. ~~RA~~ THANK You, RA, for your better image than the one I presented in my this-morning's reply: "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" Noting that the 0 in the left hand box is the origin of the given system, i will now fill in the "blanks" in YOUR diagram: Frame A [0]---------x----------[1] Source S~~>light [0]---------x----------[1] -->v Frame B Notice that x is not a point, in the image, it represents the value x of ANY point on the X axis, which is the line it's on. Accordingly, in order to pursue your objective, which is to understand the PHYSICS in Einstein's definition of synchronous clocks, we will proceed thus: In frame A let clock A be at x = 0 and clock B at x = 1. Let a ray of light emit from source S when it is far to the left of x = 0; and let the speed of light be a universal constant in empty space regardless of the motion of its source. Let~~> indicate the direction of the emitted ray relative to frame A Having agreed that (in PHYSICS) "time" is the indications of the hands of someone's clock, he soon continued thus: "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 = t'A - tB." I will now resurrect the demonstration he deleted. Let frame A and is cs and observers and clocks be at rest in his "stationary" empty space. Let the ray emit from A at tA = 0 and travel to B at c = unit/second in that space. It will therefore take tB = AB/c = 1/1 = 1 second for the ray to get from A to B. Now let there be a mirror at B which reflects the ray back to A. It will take the ray BA/c = 1 second for the return trip. The total time of the round-trip will thus be t'A = 2AB/c = 1 seconds. Placing these values into his stipulate equation, we get, (tB - tA) = 1 - 0 = 1 = (t'A - tB) = 2 - 1 = 1. Given that his definition holds good for a stationary system, we will now take the step indicated by RA's diagram. Let frame A with its clocks and cs and observers now be given an inertial velocity of .6c in the X direction. Other than that it is now moving, it remains the same system. (HUSH, puppies! Remember, this was June of 1905, BEFORE STR existed!!) Noting that the emitted light is still moving to the right at c in the same EMPTY SPACE in which frame A now moves to the right at .6c, we will now re-check the validity of his stipulated equation. Step 1. We change the NAME of frame A to "Frame B". Step 2. In order to accommodate an STR-dissenter on these groups, we will let the ray continue to travel at c on X while that frame's referent, and the attached cs with its mythical observers and their clocks, travels at .6c in the same direction. (Having done this exercise decades ago, I already know the following values, which I will not derive here.) At tA = 0 let the light be at x = 0. In 2.5 seconds it will reach clock B at x = 1. Accordingly, tB = (AB + vt)/c = (1 + 1.5)/c = 2.5 seconds. (Though a trifle different, Einstein's 1905 procedure yields the identical value, tB = AB/(c-v) = 1/.4 = 2.5 seconds.) Having been reflected from B the ray returns to A at t'A = tA + tB + [(AB - vt)/c = AB/(c+v)] = 0 + 2.5 + [(1-.6x.625)/1 = 1/1.6 = 5/8] = 0 + 2.5 + 5/8 = 3.125 seconds. It is obvious that this contradicts his second postulate. THAT is because it doesn't obey his third postulate, "We ASSUME {!!!} that this definition of synchronism is free contradiction, and possible for any number of points; and tHat the following relations are UNIVERSALLY VALID."! !! Had he left his initial demo in place, it might have continued thus: In order for the "times" of clocks of an inertially moving system {such as frame B} to obey our definition, we have to change the setting of clock B. In obedience to the "local time" equation Lorentz got from Voigt's 1887 paper on the Doppler shift, we turn back clock B by .9375 seconds, so (tB-tA) = (2.5-.9375) - 0 = 1.5625 = (t'A-tB) = 3.125 - 1.5625 = .5t'A. (Ok, puppies; you can howl now. "AROOOOooo.. that maaaay satisfy POOstulate 3; but it dOOOnt fit postuuulate 2!oo!u!ooo!" Here's a bone for you, pup. (Maybe it will sut him up..) In his 1905 paper, Einstein recognized this problem and took care of it. "HOOOOOWWWWlll", howled the pups. Like this: [Sorry; but I gotta go now. I'll look it up in one of my ancient books, such as What it all is and Why {W2}, or The Theory of Reality {Tor}; or maybe a more recent one such as "A Flower for Einstein". I COULD just type it in from his paper itself, but since I'm quitting for now anyway, I might as well look it up tomorrow and copy and paste it here instead. That will be MUSH eesier ... DOWN dogiieees!!) |