Speed of Light: A universal Constant?
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From: Henri Wilson on 8 Apr 2005 17:39 On Fri, 8 Apr 2005 14:48:03 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:ajnb51dd873nu01pibt29747tlc58ahk04(a)4ax.com... >> On Fri, 8 Apr 2005 00:20:38 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >> >>> >>>"Henri Wilson" <H@..> wrote in message >>>news:gvbb519mnbu1f2gkf04fe92nhkuhson1ui(a)4ax.com... >>>> On 7 Apr 2005 05:43:16 -0700, "Dishman" <george(a)briar.demon.co.uk> >>>> wrote: > ><snip all areas agreed> > >>>> I will also try to incorporate the c+v factor at the source and each >>>> mirror. >>> >>>Think carefully about that. c+v from the source >>>hitting a mirror moving at v means a relative >>>incident speed of c, hence a reflected speed of >>>c relative to the mirror or c+v in the lab frame >>>again. >> >> Not so George. >> You forget the mirrors rotate slightly as the light is traveling between >> them. >> So each mirror is moving at slightly off 90 deg wrt to the previous >> one...... >> that complicates the issue somewhat. > >Not so Henry, I have not forgotten that at all. I >should add the mirror orientations at the time of >reflection to make this clearer but if you check >my applet, you will find the angles are such that >the mirrors are not at 90 degrees except when the >source and detector are co-located, i.e. the table >is not moving. > >A subtle point: when you adjust the slider, it does >not represent the speed of the table but the angle >through which it will have turned when the light >reaches the detector. Given the angle and the fact >that the light of interest is that which reaches >the detector, simple trig will calculate the path >length. Divide that by the speed of light (whatever >you choose) and that tells you the time of flight. >Divide the angle by the time of flight and you get >the table speed. > >The situation has been simplified by a number of >assumptions we made, in particular that all the >mirrors are at equal radius, all are perpendicular >to a radius and the table turns at constant speed. >The basic rule for the mirrors is of course that >the angle of incidence equals the angle of >reflection so a little geometry will show that >each leg of the path plus the two end radii must >form similar triangles. > >That means the legs take equal time so the table >turns through equal angles, quarter of the total >per leg in the case of three mirrors. > >Knowing that, it becomes easy to plot the path >and you can then work back to speeds and times >if you want. Trying to work forward is much >harder. > >George > All right. I will take some time to look at this more slosely and get back to you later. 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 8 Apr 2005 17:57 On Fri, 08 Apr 2005 04:00:03 GMT, The Ghost In The Machine <ewill(a)sirius.athghost7038suus.net> wrote: >In sci.physics, H@..(Henri Wilson) ><H@> > wrote >>>> Empirical fact of life, Jim. >>>> >>> >>>Confirmable, as well. The SR and the BaT predict different results >>>for such things as spectroscopic binaries, even if one can't >>>measure the speed directly. >> >> You are very confused now Ghost. Getting desperate I would say. > >Am I? > >Here's a hint for you. Assume two stars traveling around a common >center at 30 km/s = 10^-4 c, although we can't tell the speed directly. >What would be the wavelengths observed as these stars orbit each other, >assuming a spectral line initially at 500 nm [*] and an approximate >distance of 10 lightyears? > >BaT: > >The star is spewing out particles at lightspeed, relative to itself. >These particles are of course 500 nm apart. However, since the >star is moving toward us, the particles in realspace will be a >tad longer apart -- namely, 500.05 nm apart. The other star >moving away from us will generate light of wavelength 499.95 nm, >as measured by us. The delta is 120.0000012 GHz between the two signals. Question, Ghost: What is this 'realspace'? Is it another name for the aether? You are definitely very confused Ghost. The wavelength is the same no matter how you look at it. Proof: let the star fire a identical rods between each particle....... S_._._._._._._._._._._._. You can see that the distance between particles is constant. > >The signals will be timeshifted relative to each other as the signal >from the star approaching us will reach us slightly more quickly. >The time delta here will be approximately 2 * 10^-4 * 10 = 31557 >seconds, or 8 hours, 45 minutes, 57 seconds. Depending on the star's >orbital period this should be easily measurable. > >SR [+]: > >The gamma is (1 - 5.0000000375 * 10^-9). If we assume the star >is moving directly towards us then t_O = (t_A - v * x_A / c^2) * g. >Also t_O' = (t_A' - v * x_A / c^2) * g. The difference is >(t_A' - t_A) * g. Since x_O = (x_A - v * t_A) * g and >x_O' = (x_A - v * t_A') * g, there is a corrective factor, >even though x_A = 0, and the final difference will be > >T_1 = g * ( (t_A' - t_A) - (v/c) * (t_A' - t_A)) > >for star #1, and > >T_2 = g * ( (t_A' - t_A) + (v/c) * (t_A' - t_A)) > >for star #2. Since v = 10^-4 c and t_A = 1.667 * 10^-15 s, >we get T_1 = 1.666500008332 * 10^-15 or f_1 = 600060003000300, >and T_2 = 1.6668333416675 * 10^-15 or f_2 = 599940002999700. >Frequency difference: 120.0000006 GHz. > >The signals will approach Earth and reach here at exactly the same time. Why bother with all that circular maths, Ghost. It was all derived from the postulate that the light from both stars WILL take the same time to reach the observer. Naturally that will be your conclusion. ....but why don't you simply state the (unproven) postulate. > >I won't go into details regarding precession and period changing, >as I lack the mathematical expertise therein. However, it's >clear there *is* a difference, although not a discernable one in >this case (too much imprecision in the distance and velocity >measurements). Ghost, you have made an error in your original assumptions, therefore your whole theory is wrong. > >Mercury's orbit is probably a better example. Yes, it is s long way from us, close to the sun, very hot and dry and in a fairly elliptical orbit. There are many possible explanations for its 'anomalous' precession. > >> 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 8 Apr 2005 18:20 On Fri, 08 Apr 2005 04:00:04 GMT, The Ghost In The Machine <ewill(a)sirius.athghost7038suus.net> wrote: >In sci.physics, H@..(Henri Wilson) ><H@> >>> >>>Ah, Ok. In that case I'm sure you'll have no trouble >>>substantiating it with such things as PSR B1916+16. >> >> What is that? > >Google it and find out. No hits >But here's a hint: the cost is >astronomical to go there. :-) > >> >> >>>> This is just another way of expressing the postulate. >>>> It doesn't prove it. It is an extension of the old aether idea. >>> >>>In a way, it is. There's even a postulate that the >>>Earth is twisting it. (Gravity Probe B is checking >>>this hypothesis.) Of course it has different properties >>>than the aether the MMX disproved (the old aether didn't >>>postulate spacetime distortions). >> >> That's probably because 'spacetime' is not a physical entity >> and, unlike SR, aether theory was at least a physical theory. > >Ah, so BaT must be correct because it's not an aether theory? > >An interesting viewpoint. Got any math? don't need any. Logic will suffice. > >>> >>>The catch is that the user's eyeball might be moving, and >>>the interceptor moving as well. But one still measures lightspeed c. >> >> Ghost you are preaching something you have been taught and >> accepted without question. You are indoctrinated. > >Oh no, not that! Sorry, but you are! > >> >> There is absolutely no proof to back up your claim. >> Why do you stick to it? > >So give me something provable then. Prove, within a >reasonable doubt, that c' = c+v. My expereiment in the other thread does that. >You may include MMX >in your proof with a stationary light source, as MMX >cannot tell the difference between BaT and SR unless >it is in fact rotating at a high enough speed, in which >case it ceases to be MMX and instead becomes Sagnac's >experiment, which I for one would have to look up (though >I do have a page bookmarked). > > >>> >>>> >>>>> >>>>>I'm not sure this is proof of much, of course, beyond >>>>>the fact that the math is internally self-consistent. >>>>>However, it does show that it is possible to define a >>>>>Universe with a metric (namely, the Minkowski) which >>>>>gives OWLS=c everywhere under the Lorentz transformation. >>>> >>>> It is a maths technique to make the postulate true. >>>> It is not a proven physical reality. >>> >>>There is no way to prove physical reality anyway. All we >>>can do is take measurements. >> >> ..and one day very soon, somebody will find that OWLS from >> a moving source is not c. > >We already have. The measurements just don't show it yet. :-) You mean "they are hushed up by the establishment". 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: Jim Greenfield on 8 Apr 2005 18:58 The Ghost In The Machine <ewill(a)sirius.athghost7038suus.net> wrote in message news:<evahi2-hq5.ln1(a)sirius.athghost7038suus.net>... > In sci.physics, H@..(Henri Wilson) > <H@> > wrote > on Fri, 08 Apr 2005 02:15:33 GMT > <u9qb515cv860phceb8f9qjj5c313ju2ll2(a)4ax.com>: > > On Thu, 07 Apr 2005 17:00:04 GMT, The Ghost In The Machine > > <ewill(a)sirius.athghost7038suus.net> wrote: > > > >>In sci.physics, Sam Wormley > >><swormley1(a)mchsi.com> > >> wrote > >>on Thu, 07 Apr 2005 12:41:29 GMT > >><ZT95e.11438$g65.373(a)attbi_s52>: > >>> Jim Greenfield wrote: > >>> > >>>> > >>>> Absolutely I agree that speed=frequency x wavelength; what I > >>>> absolutely disagree, is that the "speed" is always the same. > >>>> > >>> > >>> Empirical fact of life, Jim. > >>> > >> > >>Confirmable, as well. The SR and the BaT predict different results > >>for such things as spectroscopic binaries, even if one can't > >>measure the speed directly. > > > > You are very confused now Ghost. Getting desperate I would say. > > Am I? > > Here's a hint for you. Assume two stars traveling around a common > center at 30 km/s = 10^-4 c, although we can't tell the speed directly. > What would be the wavelengths observed as these stars orbit each other, > assuming a spectral line initially at 500 nm [*] and an approximate > distance of 10 lightyears? > > BaT: > > The star is spewing out particles at lightspeed, relative to itself. > These particles are of course 500 nm apart. However, since the > star is moving toward us, the particles in realspace will be a > tad longer apart -- namely, 500.05 nm apart. Not im my BaT! The particles (photons) will have the same separation, but will arrive slightly sooner than simultaneously emitted photons from the regressing star, and will appear to be bluer (higher frequency). As both stars emit a wide range of frequencies, and differring "amounts" at each wavelength, the whole of the spectrums would need to be analysed to see which star's light was more energetic due to the motion of the source (KE of all the light). I am not convinced that some of the redshifting might cause some photons to become undetectable, so the results might still be questionable (dammit) Jim G c'=c+v
From: "N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox on 8 Apr 2005 23:12
Dear RP: "RP" <no_mail_no_spam(a)yahoo.com> wrote in message news:s-udnaukW7_fksvfRVn-ug(a)centurytel.net... > > > N:dlzc D:aol T:com (dlzc) wrote: > >> Dear RP: >> >> "RP" <no_mail_no_spam(a)yahoo.com> wrote in message >> news:WISdnXVlWoqGncvfRVn-1g(a)centurytel.net... >> ... >> >>>Thanks for reminding me of a photon counter >>>argument that I formulated a couple years ago. >>> >>>A target, say 1 centimeter square photo >>>cathode, is approaching Earth sufficiently >>>fast that a 10meter broadcast appears to be in >>>the ultraviolet region wrt it. Now wrt your frame >>>(at the transmitter), the target is much smaller >>>than the photon, >> >> Good trick! The photon has been >> experimentally determined to have zero size. > > bz understood that the complaint was about > his particular hypothesized notion of the > photon, not about yours. I will address yours > as well however, per your requested below. You claimed the target was much smaller than the photon. not bz. >>>thus how do you explain the >>>emission of a single electron per 10meter >>>photon that strikes it. >> >> >>>Once you've rationalized a solution to this, >>>then assume a second surface adjacent to >>>the first and comoving with it. By what means >>>will the photon choose between the two, >>>since both are fully immersed in the volume of >>>your photon. >>> >>>You might, in the face of such logical >>>difficulties, suspect that the photon is, or >>>perhaps collapses to, a point particle before >>>absorption. But now you have the difficulty of >>>explaining how this point particle >>>simultaneously accelerates all of the >>>electrons in a quarter wave whip. Hmmm. >> >> Electrons have charge, no? A displacement of a single >> electron will "ripple" the charge field at what speed? > > Er, that would be c. So the absorption of the photon by a single electron would disturb the entire contents of "a quarter wave whip" at c, just as if they all had a hand in absorbing it... right? >>>There are no photons bz. BTW, glad to >>>have you on board the "no FTL information >>>transfer...period" campaign. Your next step >>>is to admit that the reason for this is that >>>there are no photons to be correlated; the >>>observed interactions are just interactions >>>along different portions of the very same >>>spherical wave or superposed system of >>>spherical waves. >> >> >> Or better still, photons are neither wave nor >> particle, and quantum effects don't give a >> sh*t about either space or time... except in the statistical >> domain. > > Agreed, and agreed, and agreed....that this > is indeed standard belief. > >> How how about solving the photoelectric >> effect using a wave model? > > I already have explained it to bz in another > thread, but lets have another go at it. > > Supposing a true orbital system of charges, > with positive charges moving counter in > direction to negative charges, an incident > em wave (B field) whose wavelength is > large compared to the atom will introduce > forces on these particles that are more or > less in the same direction, that is, since > the direction of force on electrons and protons (or positrons) > that are moving > counter in direction in a B field are > forced in the same direction. Now reduce > wavelength until the forces in any instant > diverge from each other and the orbital paths are forced into > greater elipticity, > i.e. they become unstable. Resonance doesn't work. Multiple incident waves do not add up to releasing photo-electrons. Only individual photons with sufficient energy can release photo-electrons. > In the > pumped state (of the photo cathode that > we discussed) the orbits are all semi- > stable, a small change in direction of an > electron will increase its elipticity such > that the electron leaves the atom and binds to another. > > The ambient thermal energy is nothing > to scoff at, electrons are moving around > at about 1/100 speed of light and the > thermal radiation exchange taking place > is incomprehensibly complex. All of the > waves emitted from every source (in the > visible universe) at a time such that the > cathode falls on their light cones, are > impinging on that surface along with > your controlled radiation (in the case of > the bz/RP argument it was green light > from a heated filament). The green "waves" > have sufficient gradient to turn the electron > away from the nucleus more so than it > was, enough to allow escape, but only in > an atom that had an electron that was in > a phase of its motion such that it was receptive to the change. > Some electrons > will have been forced into a lesser ellipse > by the same wave, and some barely > changed in their elipticity. This is where > the probabilities enter in, not of the wave to impinge on the > atom, but of the atom > to respond to the wave in such a way > that an electron escapes its orbit. There > are probabilities of interference, of all > sources to superpose either net > constructively or net destructively. Nodes > are your point photons. Multiple lower energy photons don't add up to release photo-electrons. Therefore your model fails. > From the above general description it's > easy to see why longer wavelengths don't > induce photo-emission, while shorter > wavelengths do, i.e. why there is a threshold. But not why multiple long-wavelength photons don't boost electrons anyway. Because your very mechanical model would allow such. > In turn, Feynman's sum-over-histories > approach works simply because the > absorption of radiation at some point is > every bit as determined by the > surrounding matter as it is by the source. > The light doesn't take every path, Can you prove this? Certainly a wave model requires that it does. Aren't you selling waves today? > nor > does it have knowledge of the surrounding geometry, the > surrounding geometry in > effect is communicating with it "at the > target." Or a quantum particle doens't care about goemetry, so the argument is moot. > You simply cannot omit such an obvious > contribution to the behavior of an atom as > thermal exchange, and then expect to get anything resembling > truth. Are you referring to self-interference? Thermal effects at the slits serve to spread the pattern out, as if the slits were all different geometires. This is not a good argument on your part (if I've read you mind correctly). > QM was absolutely contrived, right out of the box. What science isn't? Only religion (and philosophy) starts with "known" underlying causes. > Now here's another thought for the day. Suppose > I have a very tight laser beam, 850nm. If a > spaceship is moving away from the source sufficiently fast that > wrt it the wavelength is > 10 meters, then how is it that I see the > beam incident on an area of the ship of, oh, > about 1mm, while he sees it as completely > enveloping him? Let's see, what about a repeating pattern in a stream of 10^n photons, makes you think it applies to the width of the "illuminated" spot? You can't be this naive, can you? > Hmmm. How in the hell does a 10meter beam focus to a spot > 1mm diameter? Cool indeed, so cool that > I think I'll have to pass on it if you don't > mind. :) Detection is your problem. Were you to map intensity, you would indeed find such a spot with 10m incident irradiation. David A. Smith |