Speed of Light: A universal Constant?
in [Relativity]
|
Prev: Joan-Claude van Dirk Helps to Trivialize Special Relativity
Next: GOD=G_uv Measure your IQ in 30 seconds
From: The Ghost In The Machine on 20 Apr 2005 11:00 In sci.physics, H@..(Henri Wilson) <H@> wrote on Tue, 19 Apr 2005 22:21:55 GMT <trua61hacub6blube26g1halciji3qs8hb(a)4ax.com>: > On Tue, 19 Apr 2005 11:09:52 +0200, "Paul B. Andersen" > <paul.b.andersen(a)deletethishia.no> wrote: > >>Henri Wilson wrote: >>> >>> Paul, here is a simple question, Please answer. >>> >>> S________r________c----------------------------------------------A >>> >>> S is a light source connected to a clock by a long rod r, which point towards >>> Andromeda. >>> >>> S emits a light pulse towards A. >>> >>> Clock c intercepts the pulse and indicates ONE single travel time from S. It >>> also observes that the length of the pulse is the same as when emitted by S. >> >>All what can be read off clock c when the pulse is >>intercepted is the reading of the clock at that instant. >>To find the travel time, you must read off the clock c >>when the pulse is emitted. >>Please specify exactly how that's done. >>No hand-waving. > > Paul, I anticipated such a 'last resort' response from you. > > THE TRAVEL TIME IS QUITE UNIMPORTANT. > > What matters is that there is ONLY ONE TIME. Ah, but what does that reading represent? The clock, after all, is in a different reference frame from the stars of Andromeda. It conducts the measurement in that frame. > > So I will ask again: how can the pulse be traveling at precisely c > relative to every object in Andromeda, when the experiment > conclusively shows it has only one speed? Yes. You don't understand the Lorentz. Let (x_O,y_O,z_O,t_O) be in space O, where x_O^2+y_O^2+z_O^2-c^2*t_O^2 = 0. Basically, this is a hypercone conic surface (or a lightcone surface, if one prefers). An alternative interpretation is that it's the locus of a photon which originated at the origin. The Lorentz transformation between coordinate systems A and O is of course x_A = (x_O - v * t_O) / sqrt(1-v^2/c^2) y_A = y_O z_A = z_O t_A = (t_O - v * x_O/c^2) / sqrt(1-v^2/c^2) It turns out that, if x_O^2+y_O^2+z_O^2-c^2*t_O^2 = 0, then x_A^2+y_A^2+z_A^2-c^2*t_A^2 = 0 as well. I'd have to find my post proving the above but it's not that difficult to prove, mathematically. The corollary, of course, is that lightspeed under this transform is c everywhere. Is the Universe required to be Euclidean? :-) But the real proof is in the experimentation, which makes this worth diddly-squat without at least some experimental evidence to back it up. However, there's a number of experiments that suggest that lightspeed *is* in fact c everywhere, regardless of motion or gravitation. Your dual-probe moon experiment might corroborate SR and GR, but that's about it. [snip for brevity] > Here are a few you might look at: > > R Aquilae > R Andromedae > R Arietis > R Aur > X Aur > R Boo > S Boo > U Boo* > V Boo* > V CVn** > R Cam > V Cam" > X Cam > Z Cam > R Cas* > S Cas** > t Cas** > W Cas > S Cep* > T Cep* > Omicron Ceti > R Com > R Crb*** > S Crb > V Crb > W Crb > R Cyg > S Cyg > V Cyg > W Cyg > AF Cyg*** > CH Cyg----- > Cyg---- > Chi Cyg > R Dra > R Gem > S Her* > RU Her** > SS Her > AH her > R Hya > SU Lac > X Oph > U ori > RU Peg--- > GK Per--- > R Scuti** > R Ser > V Tau > R Uma > S Uma > T Uma > CH Uma*** > S Umi > R Vul > V Vul* > What, precisely, are we looking for again? All these are are luminosity vs. time curves, AFAICT. [.sigsnip] -- #191, ewill3(a)earthlink.net It's still legal to go .sigless.
From: Henri Wilson on 20 Apr 2005 17:02 On Tue, 19 Apr 2005 19:21:21 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message George, I am making progress. See: http://www.users.bigpond.com/hewn/sagnac.exe It only goes half way round but already shows how the red line is going to reach the startpoint much earlier than the blue one. It represents your standard model. How am I going? Does it work on your computer? 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.
From: Henri Wilson on 20 Apr 2005 17:11 On Wed, 20 Apr 2005 16:45:21 +0200, "Paul B. Andersen" <paul.b.andersen(a)deletethishia.no> wrote: >Henri Wilson wrote: >> On Tue, 19 Apr 2005 10:46:00 +0200, "Paul B. Andersen" >> <paul.b.andersen(a)deletethishia.no> wrote: >> >> >>>>>Sagnac rings and ring lasers both falsifies the ballistic theory. >>>>>Obviously. >>>>> >>>>>Paul >>>> >>>> >>>>You are not being at all helpful for the purpose of this discussion. >>> >>>You are claiming that the fringes in a Sagnac interferometer >>>are moving when the interferometer is rotating at a constant rate. >>>That is wrong. >>> >>>So you don't know what is happening in the Sagnac experiment. >>> >>>I think it would be very helpful for the purpose of this >>>discussion to know what is actually happening in the real world. >>> >>>But maybe you prefer your fantasy world were what >>>actually happens surely isn't what happens? >> >> >> Paul, please read the whole thread. >> I pointed out that according to both SR and the BaT, fringes would not move >> during constant rotation. >> >> I merely queried whether that was the case or or not. > >Henri, you wrote: >"The pattern would remain fixed. > And that is not what happens, surely." > >That's not a query, it is an assertion. >A wrong one. It was intended as a query! > > > It apparently is and so a >> continuous integration must be carried out (wrt time) in order to establish >> total rotation angle moved. > >OK. >So you have realized that your assertion was wrong. >Enough about that. I was a query. I didn't know the principle so how could I make an assertion? > >And of course you are right when you say that >you have to integrate the angular velocity to get >the angle. > >> I pointed out that there are inherent inaccuracies in such a method...but the >> use of a long fibre obviously helps. > >Of course there are inherent inaccuracies, >but the integration isn't the problem, this >is done very precisely with DSPs these days. >The precision with which the angular velocity is >measured is what determines the precision. > >And that's why ring laser gyros rather that fibre optic gyros >are used in inertial navigational systems, they are by their >very nature much more precise. > >Compare this fibre optic gyro (reference given by Dishman): >http://www.kvh.com/pdf/DSP3000_5.04.pdf >to this ring laser gyro: >http://content.honeywell.com/dses/assets/datasheets/ds13_gg1320_an.pdf > fibre optic ring laser >Bias: +/-20deg/h ? very small by nature >Bias stability: 1deg/h 0.0035deg/h >Angular random walk: 0.667deg/sqr(h) 0.0035deg/sqr(h) > >The angular velocity of the Earth is ca. 15deg/h. This >means that for the fibre optic laser, the rotation of >the Earth is smaller than the precision of the gyro, even >when the gyro is parallel to the equatorial plane. >The ring laser gyro is several order of magnitudes more precise. >(I suspect this is reflected in the prices!) > >The reason for this big difference is easy to understand. > >In the fibre optic gyro, the phase difference is static when >the angular velocity is constant. This means that the phase >difference must be compared to the phase difference when >the gyro is not rotating. This must be remembered, it can >not be measured while the gyro is rotating. And it will >probably drift! > >The ring laser gyro is very different. >If the phase difference is static, the gyro is not rotating. >When the gyro is rotating, the phase difference is changing >at a rate proportional to the angular velocity. >If the phase difference is changing with 360deg/s, >(equivalent to a fringe speed of one inter fringe length >per second - only there are hardly any fringes in these >instruments) >then the angular velocity is (360deg/2N)/s, where N is >the number of wavelengths around the ring. With a ring >length of - say 25cm - and a wavelength of 0.5um we >get that a rotation of 0.000360deg/s or 1.3deg/h will >give a phase difference change of 360deg/s. >Easily measurable, and the beauty of it is that this >only depend on geometric parameters and not on >any calibration done at some earlier time. > >I know that the inertial navigational systems in >commercial and military aeroplanes all use ring lasers. So what is the basic difference between a 'fibre optic' and a ring gyro? I didn't know there was any. > >(The maintenance manuals for the MD-80 family >and Boeing 737 happens to occupy a couple of metres >of my shelf.) > >Paul 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.
From: Paul B. Andersen on 20 Apr 2005 17:25 Henri Wilson wrote: > On Tue, 19 Apr 2005 11:09:52 +0200, "Paul B. Andersen" > <paul.b.andersen(a)deletethishia.no> wrote: > > >>Henri Wilson wrote: >> >>>Paul, here is a simple question, Please answer. >>> >>>S________r________c----------------------------------------------A >>> >>>S is a light source connected to a clock by a long rod r, which point towards >>>Andromeda. >>> >>>S emits a light pulse towards A. >>> >>>Clock c intercepts the pulse and indicates ONE single travel time from S. It >>>also observes that the length of the pulse is the same as when emitted by S. >> >>All what can be read off clock c when the pulse is >>intercepted is the reading of the clock at that instant. >>To find the travel time, you must read off the clock c >>when the pulse is emitted. >>Please specify exactly how that's done. >>No hand-waving. > > > Paul, I anticipated such a 'last resort' response from you. > > THE TRAVEL TIME IS QUITE UNIMPORTANT. > > What matters is that there is ONLY ONE TIME. You mean that the fact that the clock c is showing a specific number when the pulse is intercepted is all that matters? Why is this obvious triviality so important? > So I will ask again: how can the pulse be traveling at precisely c relative to > every object in Andromeda, when the experiment conclusively shows it has only > one speed? How can you conclude anything about the speed of anything with only one reading of a single clock? What is your point? >>Specify a method which actually is possible to use. You evaded this one, didn't you? Without it, there is no way you can say anything about the speed of the light. >>I think we can agree that c would measure the length (duration) >>of the pulse to be the same as measured by a clock at S. >> >> >>>How can you claim that the pulse is traveling at 'c' wrt all the moving objects >>>in Andromeda? It clearly has only one speed, not an infinite number. >> >>So let's introduce a "moving object" - a second clock d. >>Let's suppose that this clock is adjacent to c when the pulse >>is intercepted by both clocks at the same instant. (Coinciding events.) >> >> d -> v >>S________r________c >> >>All what can be read off clock d when the pulse is >>intercepted is the reading of the clock at that instant. >>To find the travel time, you must read off the clock d >>when the pulse is emitted. >>Please specify exactly how that's done. >>No hand-waving. >>Specify a method which actually is possible to use. > > > Paul, my experiment doesn't require any second clocks that are moving. It > proves the point as it is. I can't imagine which point a single reading of a single clock can prove. >>I think we can agree that d would measure the length (duration) >>of the pulse to be different from what is measured by a clock at S. > > > It would still get only one reading and prove that the pulse could NOT be > traveling at c wrt every object in Andromeda. That's two readings. > The value of that reading is unimportant. You mean the duration is unimportant? What is then your point? >>>Note: the ballistic theory, as applied to brightness curves, is in no way >>>dependent on how the speed of light might be measured by any observers. >>>As we know, the aether (and pseudo-aether) concept is that all observers will >>>measure OWLS as being 'c' because their clocks and rods will miraculously >>>change in order to make it so. Even if that were true, (haha) it would not >>>affect the BaT's predictions concerning variable stars. >> >>You sure are right about the latter. :-) >>Every time we - you and I - have checked what >>the BaT predicts for concrete binaries, the predictions >>have proven to be wrong. > > > Here are a few you might look at: > > R Aquilae > R Andromedae > R Arietis > R Aur > X Aur > R Boo > S Boo > U Boo* > V Boo* > V CVn** > R Cam > V Cam" > X Cam > Z Cam > R Cas* > S Cas** > t Cas** > W Cas > S Cep* > T Cep* > Omicron Ceti > R Com > R Crb*** > S Crb > V Crb > W Crb > R Cyg > S Cyg > V Cyg > W Cyg > AF Cyg*** > CH Cyg----- > Cyg---- > Chi Cyg > R Dra > R Gem > S Her* > RU Her** > SS Her > AH her > R Hya > SU Lac > X Oph > U ori > RU Peg--- > GK Per--- > R Scuti** > R Ser > V Tau > R Uma > S Uma > T Uma > CH Uma*** > S Umi > R Vul > V Vul* Yea, Right. :-) I won't bother to repeat what everybody knows. Paul
From: George Dishman on 20 Apr 2005 17:39
"Henri Wilson" <H@..> wrote in message news:ikgd61966v4b9r1gd195nkr8ddk6vjllk3(a)4ax.com... > On Tue, 19 Apr 2005 19:21:21 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <H@..> wrote in message > George, I am making progress. > > See: http://www.users.bigpond.com/hewn/sagnac.exe > > It only goes half way round but already shows how the red line is going to > reach the startpoint much earlier than the blue one. It represents your > standard model. > How am I going? Does it work on your computer? It works and gives me this at the end of the first leg http://www.briar.demon.co.uk/Henri/henri_01.gif and this at the end of the second http://www.briar.demon.co.uk/Henri/henri_02.gif The breaks in the mirrors seem odd and using shorter lengths would avoid the overlaps. Clicking the "second leg" button while the first leg is still running is interesting. The broadened left hand side is because the mirror line is drawn continuously from when the red beam hits until the blue beam arrives. Assuming you have the angles right, you are going in the right direction. I don't have time tonight and will be out tomorrow but I'll try to check on Friday. I have a backlog of other emails though. George |