Why physicists refuse to measure the one-way speed of light directly?
in [Relativity]
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From: PD on 25 Jan 2010 10:43 On Jan 24, 2:39 pm, glird <gl...(a)aol.com> wrote: > 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. What makes you think they will have different readings? > > glird
From: kenseto on 25 Jan 2010 11:04 On Jan 25, 10:42 am, PD <thedraperfam...(a)gmail.com> wrote: > 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. Yes they will if you bring them next to each other with the same velocity they will have different readings. > > > > > > > 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 -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: PD on 26 Jan 2010 12:07 On Jan 25, 10:04 am, kenseto <kens...(a)erinet.com> wrote: > On Jan 25, 10:42 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > 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. > > Yes they will if you bring them next to each other with the same > velocity they will have different readings. No, they won't, Ken. What on earth gave you the idea that they would? > > > > > > 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 -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -
From: glird on 31 Jan 2010 14:16 On Jan 25, 10:43 am, PD <thedraperfam...(a)gmail.com> wrote: > OnJan24,2:39 pm, glird <gl...(a)aol.com> wrote: > > OnJan24, 11:24 am, kenseto wrote: > > > ... 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. > > What makes you think they will have different readings? Did you read Einstein's demo re "the relativity of simultaneity"? If so, you know that he let the clocks of the moving system keep the same time per clock as those of a stationary one. That's why, as he wrote, the moving system will NOT measure c as a constant. Now reset those clocks, via Einstein's defined method, so they DO measure c as a constant and you will see that they WILL have different readings than each other, as per Lorentz's Voigtian "local time" equation in which x, t and t' are co-ordinates of the same system and v is its velocity in Einstein's "empty space". glird
From: Peter Webb on 31 Jan 2010 18:51
<SNIP> and > v is its velocity in Einstein's "empty space". How do you measure velocity relative to empty space? |