Why physicists refuse to measure the one-way speed of light directly?
in [Relativity]
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From: kenseto on 24 Jan 2010 11:24 On Jan 23, 3:33 pm, PD <thedraperfam...(a)gmail.com> wrote: > 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. No it could mean that you have detected absolute motion. >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. No....after these procedures the clocks will have different readings so they are not synchronized. Ken Seto > > 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 -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: glird on 24 Jan 2010 14:11 On Jan 11, 4:48 pm, Da Do Ron Ron <ron_ai...(a)hotmail.com> wrote: > Tom R wrote: > >It should be clear that using Einstein's synchronization > >method ... will GUARANTEE that the one-way speed of light > >will be measured to be c. > > How is this guaranteed? (That is, what exactly did > Einstein do to make it happen?) > > ~~RA~~ He postulated that clocks be set to MEASURE the speed of light as c. It is obvious that if wer do set our clocks to measure the speed of light as c, they WILL do that regardless of what the speed might be wrt those clocks. glird
From: glird on 24 Jan 2010 14:57 On Jan 11, 4:55 pm, "Androcles" <Headmas...(a)Hogwarts.physics_r> wrote: > "Da Do Ron Ron" <ron_ai...(a)hotmail.com> wrote in messagenews:f5decbd3-4e3c-4581-a010-9fd911ad1396(a)j14g2000yqm.googlegroups.com... > > > Tom R wrote: > >>It should be clear that using Einstein's synchronization method ... will GUARANTEE that the one-way speed of light will be measured to be c. > >> > > > How is this guaranteed? (That is, what > > exactly did Einstein do to make it happen?) > > ~~RA~~ > > He said this: > http://www.androcles01.pwp.blueyonder.co.uk/Shapiro/Crapiro.htm I looked and found this on that page: "We assume that this definition of synchronism is free from contradictions- Albert Einstein. "We don't need to assume Albert Einstein was a ranting lunatic, clearly his 'definition' is absurd, obviously it takes more time for light to reach Mars from Earth than it does to return." Obviously that was written by Androcles, who seems congenitally incapable of understanding precisely what Einstein's statements AND equations actually mean. Though Einstein's definition IS absurd, it is NOT for the reason given by Androiy. Here is what Einstein said: "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'' t_A from A towards B, let it at the ``B time'' t_B be reflected at B in the direction of A, and arrive again at A at the ``A time'' t'_A. In accordance with definition the two clocks synchronize if t_B - t_A = t'_A - t_B." In that definition, clocks A and B, which are the targets of the light beam, are at rest wrt each other. That is totally unrelated to Andy's objection, "obviously it takes more time for light to reach Mars from Earth than it does to return"; in which the target Earth is NOT at rest wrt target Mars.
From: glird on 24 Jan 2010 15:39 On Jan 24, 11:24 am, kenseto <kens...(a)erinet.com> wrote: > On Jan 23, 3:33 pm, PD wrote: > > > > > 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. > > > > No it could mean that you have detected absolute motion. > > >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. > > > No....after these procedures the clocks will have > different readings so they are not synchronized. You are right, Ken, but you are also wrong. Although the clocks WILL have different readings, they ARE "synchronized" as per Einstein's rather silly definition. glird
From: PD on 25 Jan 2010 10:42
On Jan 24, 10:24 am, kenseto <kens...(a)erinet.com> wrote: > On Jan 23, 3:33 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > 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. > > No it could mean that you have detected absolute motion. Or it could mean that 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. > > No....after these procedures the clocks will have different readings > so they are not synchronized. I'm sorry, but no they won't. It's a pity that after 12 years, you STILL do not understand this very simple procedure, when freshman students by the hundreds get it after just a few minutes. > > Ken Seto > > > > > 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 -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - |