OWLS is not equal to c
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From: Jerry on 5 Jun 2005 07:39 Jerry wrote: > The idea is that since the phase velocity of the RF beams > traveling through the two waveguides is different, then > the two beams would be differentially sensitive to the > effects of motion through any hypothetical aether. Uh, I guess I shouldn't have used the word "any". A Lorentz aether, for example, would not be detectable by this or any other experiment. Jerry
From: kenseto on 5 Jun 2005 08:38 Here's my point: SR says: 1. Two touching and synchronized clocks will remain synchronized after moving in the opposite directions at the same speed and came to rest again. 2. It is impossible to determine the value of OWLS because OWLS is dependent on the synchronization procedure choosen. The question is: Why can't we use the synchronized clocks described in item #1 to measure OWLS? 3. Numerous experiments were performed to confirm that OWLS is isotropic and thus from that OWLS is equal to TWLS and equal to c because TWLS is measured to be isotropic c. The question is: Why were the values of OWLS for these experiments not reported? Why did they have to use the isotropy of OWLS to conclude that it is equal to c? Is it because the measured value of OWLS is not equal to c even though that OWLS is isotropic? I think so. What do you think? Here's what a correct ether theory would say: 1. Two touching and synchronized clocks will remain synchronized after moving in the opposite directions at the same speed and came to rest again. These two clocks can be used to measure the value and isotropy of OWLS. 2. The OWLS is measured to be isotropic using the clocks described in item #1. 3. The measured value of OWLS is not c. The measured value of OWLS is distance dependent. 4. The paper in the following link describes a correct ether theory: http://www.geocities.com/kn_seto/2005Unification.pdf Ken Seto <rotchm(a)gmail.com> wrote in message news:1117898353.467089.225430(a)g43g2000cwa.googlegroups.com... > >When two clocks are synchronized they will read the same if you can compare > >them directly. The procedure use for such synchronization is the same as I > >described before. > > What do you meand by "directly"? > The procedure you proposed is a synch procedure. But how do you define > the concept of synchronized clocks? In other words, when the two clocks > are far away, how do you define and how do you verify that the clocks > are (still) syncronized? > > >Assertion is not an arguement. SR says that the clocks remains synchronized. > > Correct. SR *says* that they remain synch. The verification procedure > is to send a light signal from one clock (Ta on A) to the other. If > the other receives it a time > Tb then Tb must = Ta + 2L/c to conclude that they are in synch. That is > what i am askng you above, what synch verification procedure you are > refering to when you say that they remain in synch? > > >All the OWLS experiments done so far only tested for OWLS isotropy. > >2L/(Tb-Ta) will not have the value of 299,792,458m/second. > > That I would like to see. Have you any references on that (or > equivalent) experiments? I have not seen any and perhaps they hide the > results as you claim... > > >Both theories do not predict the same. > > Assertion is not an argument. > > >The ether theory does not predict > >OWLS to have the value of 299,792,458m/sec if the clocks are synchronized > >the way I described it and the length between the two clocks is measured > >with a physical ruler. > > Yes it does predict again the value of 288972458. I would like to see > your math where it does not predict that value. > I can show you my math where it predict that value. But then again, > what ether theories are we refering too...? > On the next rainy days, I will do the math and send it to you. >
From: Jerry on 5 Jun 2005 09:11 kenseto wrote: > 3. Numerous experiments were performed to confirm that > OWLS is isotropic and thus from that OWLS is equal to TWLS > and equal to c because TWLS is measured to be isotropic c. > The question is: Why were the values of OWLS for these > experiments not reported? Because an experiment designed to test for OWLS anisotropy is not necessarily capable of providing a figure for OWLS itself. > Why did they have to use the isotropy of OWLS to conclude > that it is equal to c? Is it because the measured value of > OWLS is not equal to c even though that OWLS is isotropic? > I think so. What do you think? I think you have never bothered to familiarize yourself with the details of the experiments that have verified OWLS isotropy. I recommend that you download, read, and try to understand the three papers that I posted on the subject at http://imaginary_nematode.home.comcast.net/LightSpeed.htm You are lacking in basic logic skills, if you think there is any way around the fact that isotropic OWLS implies OWLS=TWLS. Jerry
From: kenseto on 5 Jun 2005 09:53 "Jerry" <Cephalobus_alienus(a)comcast.net> wrote in message news:1117977112.464092.305300(a)g44g2000cwa.googlegroups.com... > kenseto wrote: > > > 3. Numerous experiments were performed to confirm that > > OWLS is isotropic and thus from that OWLS is equal to TWLS > > and equal to c because TWLS is measured to be isotropic c. > > The question is: Why were the values of OWLS for these > > experiments not reported? > > Because an experiment designed to test for OWLS anisotropy > is not necessarily capable of providing a figure for OWLS > itself. This is bull. The true test for isotropy is by determining the flight times of light between the two synchronized clocks A and B in both directions (A--->B and B--->A). If the flight time in both directions is the same then you have isotropy. The value of of OWLS can be determined simply by measuring the distance between A and B using a physical ruler. I notice that you snipped out my items #1 and 2. Here it is again: "SR says:" 1. Two touching and synchronized clocks will remain synchronized after moving in the opposite directions at the same speed and came to rest again. 2. It is impossible to determine the value of OWLS because OWLS is dependent on the synchronization procedure choosen. The question is: Why can't we use the synchronized clocks described in item #1 to measure OWLS? > > > Why did they have to use the isotropy of OWLS to conclude > > that it is equal to c? Is it because the measured value of > > OWLS is not equal to c even though that OWLS is isotropic? > > I think so. What do you think? > > I think you have never bothered to familiarize yourself with > the details of the experiments that have verified OWLS > isotropy. I recommend that you download, read, and try to > understand the three papers that I posted on the subject at > http://imaginary_nematode.home.comcast.net/LightSpeed.htm > > You are lacking in basic logic skills, if you think there is > any way around the fact that isotropic OWLS implies OWLS=TWLS. No it is you who lack logic skills. On earth OWLS can be isotropic and yet have a different value than TWLS. The following link will explain why: http://www.geocities.com/kn_seto/2005Experiment.pdf Ken Seto
From: Jerry on 5 Jun 2005 10:15
kenseto wrote: > "Jerry" <Cephalobus_alienus(a)comcast.net> wrote in message > news:1117977112.464092.305300(a)g44g2000cwa.googlegroups.com... > > kenseto wrote: > > > > > 3. Numerous experiments were performed to confirm that > > > OWLS is isotropic and thus from that OWLS is equal to TWLS > > > and equal to c because TWLS is measured to be isotropic c. > > > The question is: Why were the values of OWLS for these > > > experiments not reported? > > > > Because an experiment designed to test for OWLS anisotropy > > is not necessarily capable of providing a figure for OWLS > > itself. > > This is bull. Consider a photofinish camera used at horse races, track events, etc. The design of the camera allows it to assess very accurately the winner of a race. But it is useless for assessing the time it took for a horse to run the track. An instrument designed to test OWHS anisotropy (one way horse speed anisotropy) is not necessarily capable of measuring OWHS. > > You are lacking in basic logic skills, if you think there is > > any way around the fact that isotropic OWLS implies OWLS=TWLS. > > No it is you who lack logic skills. On earth OWLS can be isotropic > and yet have a different value than TWLS. The following link > will explain why: > http://www.geocities.com/kn_seto/2005Experiment.pdf You write: "Michelson-Morley failed to ask the relevant question: What is the direction of absolute motion of the apparatus with respect to the defined horizontal plane of the light rays that will produce a null result for all the orientations of the horizontal arms? The answer to this question is: If the apparatus is moving vertically then a null result will be obtained for all the orientations of the horizontal arms." Michelson-Morley took measurements six hours apart during separate experimental runs spaced months apart. Are you saying that the apparatus was moving vertically on all occasions? Are you saying that -everybody- who has repeated the MMX experiment with ever increasing sensitivity over the years has had their apparatus oriented vertically with respect to Earth's absolute motion through space? Are you stating that the Earth spirals through space in a direction perpendicular to whatever MMX experiment is being run at the moment? Oh, yes. Your incomprehensible E-matrix garbage... Jerry |