OWLS is not equal to c
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From: kenseto on 8 Jun 2005 09:37 "The Ghost In The Machine" <ewill(a)sirius.athghost7038suus.net> wrote in message news:4dbin2-vjk.ln1(a)sirius.athghost7038suus.net... > In sci.physics, kenseto > <kenseto(a)erinet.com> > wrote > on Tue, 07 Jun 2005 14:09:20 GMT > <kUhpe.25608$XA6.20997(a)tornado.ohiordc.rr.com>: > > > > "The Ghost In The Machine" <ewill(a)sirius.athghost7038suus.net> wrote in > > message news:qgkbn2-fue.ln1(a)sirius.athghost7038suus.net... > >> In sci.physics, kenseto > >> <kenseto(a)erinet.com> > >> wrote > >> on Sun, 05 Jun 2005 13:53:13 GMT > >> <dtDoe.13776$iu.1148(a)tornado.ohiordc.rr.com>: > >> > > >> > "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 > >> > >> You need to tighten up your experiment. > >> > >> [1] Why 1 second intervals? Why not 1 millisecond, microsecond, or > >> nanosecond intervals? > > > > Sigh....if you make the pulse length too short the whole > > pulse will miss the detector completely. > > Well, since a light pulse of 500 nm is 600 THz, a pulse width less > than about 1.66 femtosecond wouldn't be much use, at that... :-) > > Of course since the uncertainty is on the order of hbar = > h / (2*Pi) = (6.626 * 10^-34) / (6.283) = 1.0546 * 10^-34 joule-seconds, > I'd say that the pulse width can be very short -- a few nanoseconds > wouldn't bother it at all. I have choosen 1 second intervals for each trial. If you do the experiment you can choose any time interval you want. > > > BTW this is the reason for the uncertainty > > principle.....one can't determine the momentum and position of a photon > > simutlaneously due to the absolute motion of the detector wrt the photon. > > An interesting idea, that. Its the only valid idea that explains the Uncertainty Principle. > > >> > >> [2] Why 3-20mm aperature? > > > > I have no idea whta you are talking about. The detecting surface > > is 20 cm in diameter. A cover plate (20 cm in diameter) with a > > 3 mm diameter hole in the center covers the detecting surface. > > This ism used to test for the existence of absolute motion. > > Noted. So I ask again: why those values -- 3mm and 20 cm -- in particular? Read the experiment again. The 3mm diameter hole on Cover plate #1 is an arbitrary value. The 20 cm cover plate is to cover the whole surface of the detecting surface. > > > Another cover plate (20 cm in diameter) with a 3mm radial aperture > > is cut from the center to the rim of the cover plate. This is used > > to determine the direction of absolute motion. > >> > >> [3] What are the expected values for D_25, D_50, and D_100, > >> assuming that the absolute motion of the Earth is on the > >> order of 10^-4 c? > > > > You do the calculation using equations 15 and 16. > > V_100 = D_100 / 2 * (Delta T" 1). > > I do not know D_100 (it's a measured "critical diameter" ). I do not > know Delta T"1. I *do* know V_100, as I've assumed it. > > If I substitute V_100 = 10^-4 c, I get D_100 / (2 * (Delta T" 1) = 30,000 m/s. > > That doesn't tell me an awful lot. You asked for the value estimated value of D_100 at V_100=10^-4 c. Delta T"1 is equal to=2*100/c(10^-4c). You can figure D_100 using equation 16. > > >> > >> [4] Would the wavelength of the light make any difference? > > > > NO. > >> > >> [5] Would a moving light source make any difference? > > > > The you are not detecting absolute motion. > > Maybe not of the apparatus, but there are some interesting > possibilities -- are you measuring the absolute motion of the > light source in this case? NO. >How do the absolute motion of the > light source and the detection apparatus interrelate? > > > >> > >> At least you've answered one of my objections; the slit is rotatable > >> and 2mm wide, in front of a 3-20mm aperature. > > > > .No you got it wrong....still. > > Well, OK, be that way: 0.20 m and 0.003 m. :-) I suggest that you read my experiment again....carefully this time. http://www.geocities.com/kn_seto/2005Experiment.pdf Ken Seto
From: russell on 8 Jun 2005 12:46 [I'm snipping Jerry's post... see original] Here's something else to think about. AIUI Gagnon et al. claim to have falsified their semiclassical GGT theory, and this contradicts Roberts, Zhang, etc. You claim further that LET is falsified, but I think that doesn't follow directly; you have to make additional arguments a la Roberts to reach that conclusion. A problem for you. Now let me suggest (humbly, because I'm just an amateur) that Gagnon et al. go wrong in their analysis in the following subtle way: they do not consider carefully enough how slow transport will affect phase in the waveguide that is run at extinction frequency. In essence they are assuming that *signal* propagation speed is the same in both waveguides, but they have no reason (other than standard physics, which they are trying to test) to make that assumption. Put another way, they assume that the waveguide is (in main part) driving the wave in one case, and not in the other (since it is idealized as traveling unhindered through the aether) but that makes the first waveguide essentially a clock, or rather a whole series of clocks since it is extended in space. And clocks that are slowly transported wrt each other are supposed to go out of phase in the test theory since their frequencies wrt the aether frame will differ. I think they do not address this objection in their paper. Despite its being a worthy piece of work.
From: russell on 8 Jun 2005 12:53 russ...(a)mdli.com wrote: [snip to a point I want to clarify] > Put another way, they assume that the waveguide is (in > main part) driving the wave in one case, and not in the > other (since it is idealized as traveling unhindered > through the aether) By "it", I meant the wave in the second case. That is, the first waveguide drives its wave, but the second waveguide has no effect on *its* wave. (In the idealization.)
From: George Dishman on 8 Jun 2005 17:21 "Henri Wilson" <H@..> wrote in message news:9lm9a1l02h33pl4fqg4cf7aurvk5etiho7(a)4ax.com... > On Mon, 6 Jun 2005 20:03:15 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >>... a few weeks ago Henri posted a >>Visual Basic program illustration how the >>Sagnac experiment falsified the Ritzian >>(ballistic light) model. While there were >>some details to be resolved, he showed it >>produced an error of "a factor of about 2". > > Don't lie George. > My program obvioously surprised you because it proved what I wa saying all > along. Not really, you were working towards something that matched my own but you said it would take some time and we left the conversation there. I expect that when you complete it you will find you get the same as me but you need to correct the angles so that the beams both hit the same point and then we can look at the speeds you are using. > The sagnac effect occurs no matter what light speed is used. > It certainly does not prove the BaT wrong. > >> >>Henri, how is your program development going? >>Have you got the beams to coincide on the >>detector yet? > > They only coincide when there is no rotation. > That's the main principle behind the sagnac effect. I thought you had grasped how an interferometer worked during our chat. The intensity at any point on the screen (or whatever other detector is used) depends on the amplitudes and relative phase of the two beams falling on that point. It should be obvious that a ray landing elsewhere cannot change the intensity at that point. It's quite simple to work out the numbers because the legs are equal so each reflection point is 90 degrees plus 1/4 of the overall angle moved by the detector between emission and reception. This applet shows the beam paths: http://www.briar.demon.co.uk/Henri/SagnacAngles.html If I get some time, I intend to animate this to match your format but it won't be soon, sorry. George
From: Henri Wilson on 9 Jun 2005 05:26
On Wed, 8 Jun 2005 22:21:16 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:9lm9a1l02h33pl4fqg4cf7aurvk5etiho7(a)4ax.com... >> On Mon, 6 Jun 2005 20:03:15 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >> >>>... a few weeks ago Henri posted a >>>Visual Basic program illustration how the >>>Sagnac experiment falsified the Ritzian >>>(ballistic light) model. While there were >>>some details to be resolved, he showed it >>>produced an error of "a factor of about 2". >> >> Don't lie George. >> My program obvioously surprised you because it proved what I wa saying all >> along. > >Not really, you were working towards something >that matched my own but you said it would take >some time and we left the conversation there. >I expect that when you complete it you will >find you get the same as me but you need to >correct the angles so that the beams both hit >the same point and then we can look at the >speeds you are using. My beams do hit the same point when there is no rotation. > >> The sagnac effect occurs no matter what light speed is used. >> It certainly does not prove the BaT wrong. >> >>> >>>Henri, how is your program development going? >>>Have you got the beams to coincide on the >>>detector yet? >> >> They only coincide when there is no rotation. >> That's the main principle behind the sagnac effect. > >I thought you had grasped how an interferometer >worked during our chat. The intensity at any >point on the screen (or whatever other detector >is used) depends on the amplitudes and relative >phase of the two beams falling on that point. >It should be obvious that a ray landing elsewhere >cannot change the intensity at that point. I dont think that is the whole story at all. I reckon the divergence of the beams has as much to do with it than path length difference. Huygens principle has no QM basis. > >It's quite simple to work out the numbers because >the legs are equal so each reflection point is >90 degrees plus 1/4 of the overall angle moved by >the detector between emission and reception. This >applet shows the beam paths: > >http://www.briar.demon.co.uk/Henri/SagnacAngles.html I don't agree that they meet when the thing is rotating, according to either theory... > >If I get some time, I intend to animate this to >match your format but it won't be soon, sorry. Time is becoming quite a problem, I'm afraid. > >George > HW. www.users.bigpond.com/hewn/index.htm Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong. |