Speed of Light: A universal Constant?
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From: Jim Greenfield on 7 Apr 2005 22:18 bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote in message news:<Xns96314B27BA908WQAHBGMXSZHVspammote(a)130.39.198.139>... > jgreen(a)seol.net.au (Jim Greenfield) wrote in > news:e7b5cc5d.0504062211.6957dcb5(a)posting.google.com: > > > Well I'll take my crystal, which we KNOW, and by definition, emmits a > > signal of known fixed frequency / wavelength in the lab, and YOU > > explain how it does NOT exhibit that colour on the film, when I bring > > the film and crystal beam together. > > No magic; film which we trust, and a crystal also > > > > > > What is your magic crystal? > The crystal of a laser diode or LED? That is a crystal and it emits a > specific frequency/wavelength/color. Any that produces a "fixed" signal. I assume that you agree with Paul D, in that film records the rate of impingement of light waves (sic photons). We agree that the wavelength emitted by the crystal doesn't alter. By the simplist algebra, that leaves only ONE possibility for the motion of the crystal ref the source showing a different colour on the film-- "c" in c=fu has CHANGED > > What makes you think it will not exhibit the color that would be > represented by the doppler shifted emission? Said doppler shift being due > to relative velocity of source and film. Nope! As above, this should read "relative velocity of PHOTONS and film." Jim G c'=c+v
From: bz on 7 Apr 2005 22:51 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. -- 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
From: bz on 7 Apr 2005 23:05 jgreen(a)seol.net.au (Jim Greenfield) wrote in news:e7b5cc5d.0504071818.32282c6b(a)posting.google.com: > bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote in message > news:<Xns96314B27BA908WQAHBGMXSZHVspammote(a)130.39.198.139>... >> jgreen(a)seol.net.au (Jim Greenfield) wrote in >> news:e7b5cc5d.0504062211.6957dcb5(a)posting.google.com: >> >> > Well I'll take my crystal, which we KNOW, and by definition, emmits a >> > signal of known fixed frequency / wavelength in the lab, and YOU >> > explain how it does NOT exhibit that colour on the film, when I bring >> > the film and crystal beam together. >> > No magic; film which we trust, and a crystal also >> > >> > >> >> What is your magic crystal? >> The crystal of a laser diode or LED? That is a crystal and it emits a >> specific frequency/wavelength/color. > > Any that produces a "fixed" signal. > I assume that you agree with Paul D, in that film records the rate of > impingement of light waves (sic photons). Nope. The film responds to the ENERGY of the photon. > We agree that the wavelength emitted by the crystal doesn't alter. agreed. > By the simplist algebra, that leaves only ONE possibility for the > motion of the crystal ref the source showing a different colour on the > film-- > "c" in c=fu has CHANGED nope. the relative velocity has added changed the frequency/energy/wavelength that the film sees. >> >> What makes you think it will not exhibit the color that would be >> represented by the doppler shifted emission? Said doppler shift being >> due to relative velocity of source and film. > > Nope! As above, this should read "relative velocity of PHOTONS and > film." the relative velocity of photons and film is c. Do you think the dopplar shift in sound is due to changes in the speed of sound? -- 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
From: RP on 7 Apr 2005 23:55 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. 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. 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. Richard Perry
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 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. 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. 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. 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). Mercury's orbit is probably a better example. > > > 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. [*] this is not an actual spectral line but it makes the math slightly easier. [+] GR introduces an additional corrective factor, though for these particular stars it's the same for both. -- #191, ewill3(a)earthlink.net It's still legal to go .sigless. |