OWLS is not equal to c
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From: Paul Stowe on 1 Jun 2005 23:57 On 1 Jun 2005 19:44:11 -0700, russell(a)mdli.com wrote: >Paul Stowe wrote: > >[snip] > >> If you wish to orient distances based upon OWLS it's pretty >> straight forward. The Earth rotates and there is a directional >> in the CMBR. Pick a point on the Earth that is most parallel >> to the orientation of the CMBR. Next, Set up opposing tracks >> that are the distance you want. Then, when one of the track >> direction aligns with the CMBR have the receiver move outward >> from a repeating pulse transmitter until the desired delay in >> the reception is achieved. This is a OWL pulse moving from the >> Transmitter to the receiver. The receiver (having a high >> precision clock) is computing the difference in reception times >> to get the increasing delay. >> >> Now, wait until the Earth rotates 180ý and the other track aligns >> in the same direction & repeat. This will assure that BOTH! >> distance are equal based upon an OWLS measurement. > > What does CMBR have to do with this? If the aether exists, the CMBR illuminates the rest frame. Any motion wrt this will be detectable by a directionalized doppler shift. Thus, using the CMBR dipole allows you to orient wrt that possible aether rest frame. We know the the entire solar system is moving at ~ 370 kps wrt to the CMBR zero. Thus, sending a signal along (up or down) that direction will maximize any OWLS asymmetry. > You are using slow transport as your synchronization method; I'm not synchronizing anything (or trying to). We're moving a predefined distance based upon an assume ct where t is a predefined delta (or delay). Starting with the transmitter and receiver together (no delay) and separating while just listening for the pulses from the transmitter. When the lag between pulses have increase the defined amount the receiver stops. The Light pulses all move in one direction only. Thus, if there is any c +/- v effects they will be 'in there'. > the lengths are equal if and only if slow transport in > the two directions induces the same time difference per > unit distance. That's the point of waiting for the Earth to rotate 180ý. The second rail then aligns in space in exactly the same direction as the earlier one. Same c +/- v, same distance. EVEN IF, there is an anisotropic OWLS, Right? Look at my first sentence, "If you wish to orient distances based upon OWLS it's pretty straight forward." To clarify, I should have said, "If you wish to orient distances of equal length based upon OWLS it's pretty straight forward." > Now, if indeed you found that, having done this, the > wires when brought together are different lengths, you > would have a significant result, one in conflict with > Michelson-Morley etc. No, it wouldn't, even IF so. MMX is TWLS not OWLS. Isn't that the point? Paul Stowe
From: Jerry on 2 Jun 2005 01:36 Paul Stowe wrote: > If the aether exists, the CMBR illuminates the rest frame. Why do you think this? What makes you so sure that the aether constitutes an absolutely rigid backdrop for the propagation of EM waves? Can there not be streams and rivers flowing in the aether? Can not the aether rotate and swirl, form vortices? Can the aether not be entrained by the passage of matter? As you know, many of the early aether theories postulated such properties for the aether. So... What in heck is so special about the CMBR, such that you postulate that it "illuminates" the universal rest frame, when many aether theories implicitly deny such a role by postulating a dynamic, flowing, entrainable aether? Jerry
From: The Ghost In The Machine on 2 Jun 2005 03:00 In sci.physics, Jerry <Cephalobus_alienus(a)comcast.net> wrote on 1 Jun 2005 17:27:25 -0700 <1117672045.562429.47700(a)f14g2000cwb.googlegroups.com>: > kenseto wrote: > >> Sigh....all these are two-way experiments. The one-way value with two >> spatially separated and synchronized clocks was never determined. They >> performed experiments that confirms the one-way isotropy. But the one-way >> value for those experiments were not reported. Why? Because the one-way >> value for those experiments was not c. > > If OWLS is isotropic, then OWLS must be equal to TWLS. > OWLS is observed to be isotropic. > Therefore, OWLS is equal to TWLS. In any event, if OWLS is *an*isotropic one runs into a variant of the "headwind/tailwind" problem. Briefly, the photon takes time d / (c+v) to go there, and time d / (c-v) to come back. Total time: (d/(c+v) + d/(c-v)) = d( (c+v) + (c-v) )/(c^2-v^2) = 2dc/(c^2-v^2) != 2d/c, assuming Newtonian physics (which admittedly isn't quite right). > > Jerry > -- #191, ewill3(a)earthlink.net It's still legal to go .sigless.
From: russell on 2 Jun 2005 03:24 Paul Stowe wrote: > On 1 Jun 2005 19:44:11 -0700, russell(a)mdli.com wrote: > > >Paul Stowe wrote: > > > >[snip] > > > >> If you wish to orient distances based upon OWLS it's pretty > >> straight forward. The Earth rotates and there is a directional > >> in the CMBR. Pick a point on the Earth that is most parallel > >> to the orientation of the CMBR. Next, Set up opposing tracks > >> that are the distance you want. Then, when one of the track > >> direction aligns with the CMBR have the receiver move outward > >> from a repeating pulse transmitter until the desired delay in > >> the reception is achieved. This is a OWL pulse moving from the > >> Transmitter to the receiver. The receiver (having a high > >> precision clock) is computing the difference in reception times > >> to get the increasing delay. > >> > >> Now, wait until the Earth rotates 180° and the other track aligns > >> in the same direction & repeat. This will assure that BOTH! > >> distance are equal based upon an OWLS measurement. > > > > What does CMBR have to do with this? > > If the aether exists, the CMBR illuminates the rest frame. What does it mean for a *frame* to "be illuminated"? Why don't you just say what you mean -- that you assume that the CMBR is isotropic in the rest frame of the aether. That might be a reasonable assumption based on some cosmological model you are attempting to verify. But it *is* an assumption and cannot be said to be true until verified. Any > motion wrt this will be detectable by a directionalized doppler > shift. Thus, using the CMBR dipole allows you to orient wrt that > possible aether rest frame. We know the the entire solar system > is moving at ~ 370 kps wrt to the CMBR zero. Thus, sending a > signal along (up or down) that direction will maximize any OWLS > asymmetry. So you think. Perhaps a reasonable place to look first for asymmetry, but you really don't know. > > > You are using slow transport as your synchronization method; > > I'm not synchronizing anything (or trying to). I can believe you're not trying to. We're moving > a predefined distance based upon an assume ct where t is > a predefined delta (or delay). Starting with the transmitter > and receiver together (no delay) and separating while just > listening for the pulses from the transmitter. Understood; and you have a high precision clock at the receiver. You assume (or posit) that it stays synchronized with the pulsing source as it moves. This is what we call "slow transport". It is a synchronization convention. However, this only matters if you want to know the exact distance; I concede that it's not important if your intent is merely to set equal distances as you describe below. When the lag > between pulses have increase the defined amount the receiver > stops. The Light pulses all move in one direction only. Thus, > if there is any c +/- v effects they will be 'in there'. > > > the lengths are equal if and only if slow transport in > > the two directions induces the same time difference per > > unit distance. > > That's the point of waiting for the Earth to rotate 180°. > The second rail then aligns in space in exactly the same > direction as the earlier one. Same c +/- v, same distance. > EVEN IF, there is an anisotropic OWLS, Right? Ok, I get your point this time. There's a problem, though. Even though you can say that the two tracks are exactly the same length (i.e. you can in effect use your method as a definition of length) you still don't know how the aether affects propagation of signals down wires laid along the respective tracks. One signal may well propagate faster than the other, exactly counteracting the anisotropy in LS that you are attempting to measure. Or perhaps partially. You can never know by how much, except by measuring it with a pair of clocks -- and that requires you to specify a synchronization convention. That scotches any hope of getting a OWLS measurement independent of convention. > > Look at my first sentence, > > "If you wish to orient distances based upon OWLS it's > pretty straight forward." > > To clarify, I should have said, > > "If you wish to orient distances of equal length based > upon OWLS it's pretty straight forward." > > > Now, if indeed you found that, having done this, the > > wires when brought together are different lengths, you > > would have a significant result, one in conflict with > > Michelson-Morley etc. My details here are wrong, based as they were on my misunderstanding of your setup. Rather, a groundbreaking result (overthrowing SR) would be any difference in the respective arrival times of a light pulse emitted from the central point. > > No, it wouldn't, even IF so. MMX is TWLS not OWLS. Isn't > that the point? Sorry, but your method is TWLS too, if you use signals traveling down the tracks that you have measured.
From: kenseto on 2 Jun 2005 10:21
"Jerry" <Cephalobus_alienus(a)comcast.net> wrote in message news:1117672045.562429.47700(a)f14g2000cwb.googlegroups.com... > kenseto wrote: > > > Sigh....all these are two-way experiments. The one-way value with two > > spatially separated and synchronized clocks was never determined. They > > performed experiments that confirms the one-way isotropy. But the one-way > > value for those experiments were not reported. Why? Because the one-way > > value for those experiments was not c. > > If OWLS is isotropic, then OWLS must be equal to TWLS. > OWLS is observed to be isotropic. > Therefore, OWLS is equal to TWLS. That's an assumption. Why didn't they report the value for OWLS when they tested OWLS for isotopy?? Why did they have to assume that OWLS is equal to TWLS when the value of OWLS is readily available from those experiments?? BTW, the assumption that OWLS is equal to TWLS because OWLS is isotropic is bogus. Why? Because in TWLS the mirror did not reflect the return signal instantaneously. In a true OWLS test with two spatially separated and synchronized clocks there is no such delay. Ken Seto |