Why physicists refuse to measure the one-way speed of light directly?
in [Relativity]
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From: PD on 1 Feb 2010 12:35 On Jan 31, 1:16 pm, glird <gl...(a)aol.com> wrote: > 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. Ah, so what you are saying is that clocks can be synchronized in frame A, and clocks can be synchronized in frame B, but frame B's clocks will not be synchronized in frame A and frame A's clocks will not be synchronized in frame B. And that's right. > That's why, as he > wrote, the moving system will NOT measure c as a constant. Oh, yes it will. It is not necessary for frame B's clocks to be synchronized in frame A for frame B to measure the speed of light to be still c. > 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: glird on 1 Feb 2010 15:00 On Feb 1, 12:35 pm, PD wrote: > On Jan 31, 1:16 pm, glird wrote: > > On Jan 25, 10:43 am, PD wrote: > > > OnJan24,2:39 pm, glird wrote: > > > > OnJan24, 11:24 am, kenseto wrote: > > > > > ... after these procedures the clocks will have different readings so they are not synchronized. > gl: 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. > pd: What makes you think they will have different readings? > > gl: > > 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. > PD: Ah, so what you are saying is that clocks can be synchronized in frame A, and clocks can be synchronized in frame B, but frame B's clocks will not be synchronized in frame A and frame A's clocks will not be synchronized in frame B. And that's right. > > > That's why, as he wrote, the moving system will NOT measure c as a constant. > pd: Oh, yes it will. It is not necessary for frame B's clocks to be synchronized in frame A for frame B to measure the speed of light to be still c. > That's right; but its irrelevant wrt this discussion, in which -- s Einstein said, clocks of frame B ARE set identically to those of frame A. Why didn't you do what I suggested next? (Here is that suggestion, which you omitted) > > 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: PD on 1 Feb 2010 15:18 On Feb 1, 2:00 pm, glird <gl...(a)aol.com> wrote: > On Feb 1, 12:35 pm, PD wrote:> On Jan 31, 1:16 pm, glird wrote: > > > On Jan 25, 10:43 am, PD wrote: > > > > OnJan24,2:39 pm, glird wrote: > > > > > OnJan24, 11:24 am, kenseto wrote: > > > > > > ... after these procedures the clocks will have different readings so they are not synchronized. > > > gl: 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. > > pd: What makes you think they will have different readings? > > > gl: > > 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. > > PD: Ah, so what you are saying is that clocks can be synchronized in > frame A, and clocks can be synchronized in frame B, but frame B's > clocks will not be synchronized in frame A and frame A's clocks will > not be synchronized in frame B. And that's right. > > > > That's why, as he wrote, the moving system will NOT measure c as a constant. > > pd: Oh, yes it will. It is not necessary for frame B's clocks to be > synchronized in frame A for frame B to measure the speed of light to > be still c. > > > That's right; but its irrelevant wrt this discussion, in which -- s > Einstein said, clocks of frame B ARE set identically to those of frame > A. That means the same procedure is used. It does not mean they are set to read the same readings. It's a shame that such an elementary misunderstanding has gotten in your way. > > Why didn't you do what I suggested next? > (Here is that suggestion, which you omitted) > > > > 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: glird on 2 Feb 2010 13:43 On Feb 1, 3:18 pm, PD wrote: > On Feb 1, 2:00 pm, glird wrote: > > > gl: 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. > pd: What makes you think they will have different readings? > > gl: > > 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. > > > PD: Ah, so what you are saying is that clocks can be synchronized in frame A, and clocks can be synchronized in frame B, but frame B's clocks will not be synchronized in frame A and frame A's clocks will not be synchronized in frame B. And that's right. > > > > > That's why, as he wrote, the moving system will NOT measure c as a constant. > > > pd: Oh, yes it will. It is not necessary for frame B's clocks to be synchronized in frame A for frame B to measure the speed of light to still be c. > > > > gl: That's right; but its irrelevant wrt this discussion, in which -- as Einstein said, clocks of frame B ARE set identically to those of frame A. > > That means the same procedure is used. It does not mean they are set to read the same readings. > You either didn't read or didn't understand Einstein's paper. Or perhaps you don't understand simple English, as in "clocks of frame B ARE set identically to those of frame A"? (If you do, then how come you don't know the difference between "are set identically TO those of frame A" and (are set identically AS those of frame A), as in "the same procedure is used"? It's a shame that such an elementary misunderstanding has gotten in your way. > > gl: Why didn't you do what I suggested next? (Here is that suggestion, which you omitted) 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: PD on 2 Feb 2010 14:02
On Feb 2, 12:43 pm, glird <gl...(a)aol.com> wrote: > On Feb 1, 3:18 pm, PD wrote:> On Feb 1, 2:00 pm, glird wrote: > > > > gl: 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. > > pd: What makes you think they will have different readings? > > > gl: > > 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. > > > > PD: Ah, so what you are saying is that clocks can be synchronized in > > frame A, and clocks can be synchronized in frame B, but frame B's > clocks will not be synchronized in frame A and frame A's clocks will > not be synchronized in frame B. And that's right. > > > > > > That's why, as he wrote, the moving system will NOT measure c as a constant. > > > > pd: Oh, yes it will. It is not necessary for frame B's clocks to be > > synchronized in frame A for frame B to measure the speed of light to > still be c. > > > > > gl: That's right; but its irrelevant wrt this discussion, in which -- as Einstein said, clocks of frame B ARE set identically to those of frame A. > > > That means the same procedure is used. It does not mean they are set > > to read the same readings. > > > You either didn't read or didn't understand Einstein's paper. Or > perhaps you don't understand simple English, as in "clocks of frame B > ARE set identically to those of frame A"? > (If you do, then how come you don't know the difference between > "are set identically TO those of frame A" and (are set identically AS > those of frame A), as in "the same procedure is used"? > It's a shame that such an elementary misunderstanding has gotten in > your way. He wrote in German. What are the prepositions used in the original? > > > > gl: Why didn't you do what I suggested next? (Here is that suggestion, which you omitted) > > 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 |