Speed of Light: A universal Constant?
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From: The Ghost In The Machine on 8 Apr 2005 00:00 In sci.physics, H@..(Henri Wilson) <H@> wrote on Fri, 08 Apr 2005 01:09:34 GMT <v4mb515iav7rae4gj5avs8nnvnev0e8f03(a)4ax.com>: > On Thu, 07 Apr 2005 17:00:03 GMT, The Ghost In The Machine > <ewill(a)sirius.athghost7038suus.net> wrote: > >>In sci.physics, H@..(Henri Wilson) >><H@> >> wrote > >>>>> >>>>> It is completely unproven. >>>> >>>>And c' = c+v has been disproven, experimentally. >>> >>> Don't be silly Ghost. You know that is not true. >> >>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. 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? > >>> >>> That is not the same as saying light leaves the star >>> with an infinite number of speeds. >> >>Light leaves with one speed: c. >>Light enters the user's eyeball with one speed: c. >>Light can be intercepted but is always measured with one speed: c. >> >>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! > > 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. 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. :-) > >> >>[rest snipped] > > > 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. -- #191, ewill3(a)earthlink.net It's still legal to go .sigless.
From: "N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox on 8 Apr 2005 00:02 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. > 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? > 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. How how about solving the photoelectric effect using a wave model? David A. Smith
From: RP on 8 Apr 2005 01:00 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. > >>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. > > >>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. 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. 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. 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, nor does it have knowledge of the surrounding geometry, the surrounding geometry in effect is communicating with it "at the target." 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. QM was absolutely contrived, right out of the box. That is simply not the way to do physics..."hmmmm lessee, I'll bet it's *particles*...... of pure *energy*, ooh cool, and then I'll bet it's probabilities guiding them and that motion is like, just an illusion dude. Whoah what a rush." Crank yanker space cadets doing physics. Oughta be illegal. 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? 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. :) Richard Perry > > David A. Smith > >
From: Sue... on 8 Apr 2005 08:46 Richard Perry wrote: 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. :) >> I believe it might have been you that pointed our that a length of lasing material with collimating lens can be the equivalent of an end-fire array many meters in length. At lower frequecies we can't maintain the phase stability of such an array more than 13 or so elements. Radiating elements composed of stimulated atoms have the ability to constantly *tune* each other so a very long array is practical. In some cases the atmosphere may even act as a dielectric lens to reduce the divergence. There is no free lunch tho. The beam still has a spatial distribution that is Gaussian, not a stream of corpuscles. So this reversed telescope can make a good thing better. http://www.csr.utexas.edu/mlrs/pictures/lunar_ra.jpg From: http://www.csr.utexas.edu/mlrs/ Sue...
From: bz on 8 Apr 2005 08:08
RP <no_mail_no_spam(a)yahoo.com> wrote in news:WISdnXVlWoqGncvfRVn-1g(a)centurytel.net: > > > bz wrote: > >> H@..(Henri Wilson) wrote in >> news:4kcb511ci10n1g13s7q4l0kgj0u5hih6vs(a)4ax.com: >> >> >>>You cannot seem to discriminate between generated radio waves and >>>individual photons. >>> >> >> >> Other than energy/wavelength/frequency there is no difference. >> >> generated radio waves consist of large groups of large, low frequency, >> low energy photons. >> When the frequency gets high enough, the energy high enough, individual >> photons have enough energy to cause distinguishable individual events, >> like electron emission. > > 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, thus how do you explain > the emission of a single electron per 10meter photon that strikes it. Picture a photon[lets assume plane polarized in the vertical direction], as seen when looking along the axis of travel: it is going to look like a point with an E field in the vertical direction and the B field in the horizontal direction. The 1 cm photo cathode will experience the variation in E and B fields 'faster' because of the photo cathodes velocity. The photon will have sufficent energy to knock an electron off of the photo cathode. The same photon, when looked at 'from the side' from the emission frame of reference is 10 meters in length. When looked at 'from the side' from the target's frame of reference is nano meters in length. > 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. My photon's volume only stretches along the axis of travel. Its E and B fields can only induce effects for very short distances, so it only sees one of the photo cathodes. > > You might, in the face of such logical difficulties, suspect that the > photon is, or perhaps collapses to, a point particle before > absorption. No, it was already a point particle, when seen end-on. > But now you have the difficulty of explaining how this > point particle simultaneously accelerates all of the electrons in a > quarter wave whip. Hmmm. It doesn't do so by itself. At 10 meters, we need the E fields from a LOT of photons, acting together, to accelerate some of the electrons in the quarter wave whip. One photon per electron. > > There are no photons bz. BTW, glad to have you on board the "no FTL > information transfer...period" campaign. I have been in that camp for quite some time. > Your next step is to admit > that the reason for this is that there are no photons to be > correlated; Rather, if the photons are correlated, it is unimportant to the transfer of information because the information is encoded when the photons are created, not when they are received. As I said in another article: > Someone cuts a coin in half. One half has the head, the other half has > the tail. > > The half coins are sent to different locations in sealed envelopes. > You get one of the envelopes. Someone you know that lives miles away > gets the other envelope. > > Wonder of wonder, when you open your envelope, you _immediatly_ know > which half of the coin went to the other person, even though you are > miles apart. > > Faster than light 'communication' has just taken place [not!]. > the observed interactions are just interactions along > different portions of the very same spherical wave or superposed > system of spherical waves. Sorry, you just convinced me, with your 1 cm photo cathode experiment, that photons ARE 'point objects' bUT only when seen along the axis of travel. When seen from the side, they have wave properties, and naturally, as they pass near to 'resonant objects', they --like the bow of a violin-- transfer energy to the 'resonant object'. Usually ALL of their energy is transfered and the photon ceases to exist. When they pass near the edge of something, like a slit, they are absorbed and reradiated with a time delay that depends on the configuration of the atom(s) of the slit edge. The configuration is 'predictable' in that there are quantum states of electron energy and of atomic and molecular vibrational energy. Some positions are more probable than others. The time delay between absorbtion and emission means that the photons that are emitted will NOT be following the exact same path that the incoming photon was following. The result will be a 'diffraction pattern' that is visible, even in single photon experiments. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap |