Speed of Light: A universal Constant?
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From: Henri Wilson on 7 Apr 2005 06:13 On 6 Apr 2005 14:47:26 -0700, "PD" <pdraper(a)yahoo.com> wrote: > >Henri Wilson wrote: >> On Wed, 6 Apr 2005 06:13:06 +0000 (UTC), bz ><bz+sp(a)ch100-5.chem.lsu.edu> wrote: >> >> >> So if a photon that travels through space for eons suddenly comes >across an >> observer, it somehow 'knows' how much it should doppler shift? > >You have it in your mind that the wavelength (or frequency) is an >inherent property of the photon, independent of the observer. This is >not the case. > >> >> Does it carry information about its original source's speed wrt the >ultimate >> random observer? > >It carries its wavepacket with it. The wavelength (or frequency) is not >information that is carried inherently with it, any more than length is >carried along with a physical rod. However, when the observer measures >the wavepacket, that measurement is made in the observer's rest frame. You seem to know all about the structure of a photon. Please tell the world. > >> >> Photons must be pretty clever little things if they can do that. >> >> I think they just move at c wrt their source. >> >> >> >> > >> >Emission lines, even from the gas in the neon sign in the bar's >window, >> >will show the doppler effect, as the atoms/molecules (the same thing >in >> >neon's case) are in motion. >> > >> >Absorption by another molecule of neon can only occur when the >receiver is >> >going the right velocity so that the wavelengths match. >> >> OK answer this. >> >> Consider two photons that were emitted in a neon transition from >quite separate >> and relatively moving sources. By chance, they end up traveling >together >> through space. >> They eventually strike the same observer. >> >> How do they carry the information required to cause the correct >doppler shift >> to take place? > >Because they are different wavepackets. The wavepackets do not have to >do anything to change at the moment of observation. Bilgey reckons all photons are identical. He is one of your lot. So why do some exhibit doppler shift and not others? 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 7 Apr 2005 06:21 On Thu, 07 Apr 2005 06:00:10 GMT, The Ghost In The Machine <ewill(a)sirius.athghost7038suus.net> wrote: >In sci.physics, H@..(Henri Wilson) ><H@> > wrote >> >> That's just aether theory in which the one absolute frame >> is replaced by an arbitrary one. >> >> It is completely unproven. > >And c' = c+v has been disproven, experimentally. Don't be silly Ghost. You know that is not true. >> >>> >>>[rest snipped] >> >> Ghost, why did you snip my question? Was it too hard to answer. >> Here it is again: >> >> Do you claim that light emitted from a remote star is initially >> traveling at c relative to every object in the universe? > >Initially, terminally, and at every point in between. > >> Does that require an infinite number of discreet speeds? > >No. The Lorentz spacetime twist takes care of that. > >Herewith the math. Very hairy grody grumpy weedy math. >Bring your own machete. :-) > >The Lorentz, of course, can be expressed > >x_A = (x_O - v * t_O) * g >y_A = y_O >z_A = z_O >t_A = (t_O - v * x_O / c^2) * g > >where g = 1/sqrt(1 - v^2/c^2), or g^2 = 1/(1 - v^2/c^2). This will >be an important observation later on. > >Let a dual lightcone [*] erupt at a certain point (x_o, y_o, z_o, t_o) >somewhere in O-space, satisfying the equation > >0 = (x_O - x_o)^2 + (y_O - y_o)^2 + (z_O - z_o)^2 - c^2 * (t_O - t_o)^2 > >If we do the substitution and then factor out g^2, we get: > >0 = ((x_A - v * t_A) * g - x_o)^2 > + (y_A - y_o)^2 + (z_A - z_o)^2 > - c^2 * ((t_A - v * x_A/c^2)*g - t_o)^2 > > = ((x_A - v * t_A) - x_o/g)^2 * g^2 > + (y_A - y_o)^2 + (z_A - z_o)^2 > - c^2 * ((t_A - v * x_A/c^2) - t_o/g)^2 * g^2 > > = (x_A^2 + v^2 * t_A^2 + x_o^2/g^2 - 2*x_A*v*t_A - 2*x_A*x_o/g > + v*t_A*x_o/g) * g^2 > + (y_A - y_o)^2 + (z_A - z_o)^2 > + (- t_A^2 * c^2 - v^2 * x_A^2/c^2 - c^2 * t_o^2/g^2 + 2 * t_A *v * x_A > + 2 * t_A * t_o * c^2/g - 2 * v * x_A * t_o / g ) * g^2 > > = (x_A^2 * (1 - v^2/c^2) - 2 * x_A * (x_o - v * t_o) /g > + (x_o - v * t_o)^2) * g^2 > + (y_A - y_o)^2 + (z_A - z_o)^2 > + (- c^2 * t_A^2 * (1 - v^2/c^2) + 2 * t_A * (t_o - v * x_o/c^2) * c^2/g > - c^2 * (t_o - v * x_o/c^2)^2) * g^2 > + (x_o^2/g^2 - 2*x_A*v*t_A - c^2 * t_o^2/g^2 + 2 * t_A *v * x_A) * g^2 > > = (x_A^2 - 2 * x_A * (x_o - v * t_o)*g + (x_o - v * t_o)^2 * g^2) > + (y_A - y_o)^2 + (z_A - z_o)^2 > - c^2 * (t_A^2 - 2 * t_A * (t_o - v * x_o/c^2) *g > + (t_o - v * x_o/c^2)^2 * g^2) > + (x_o^2/g^2 - 2*x_A*v*t_A - c^2 * t_o^2/g^2 + 2 * t_A *v * x_A > - (x_o - v * t_o)^2 + c^2 * (t_o - v * x_o/c^2)^2 ) * g^2 > > = (x_A - (x_o - v * t_o)*g))^2 > + (y_A - y_o)^2 + (z_A - z_o)^2 > - c^2 * (t_A - (t_o - v*x_o/c^2)*g) > + (x_o^2/g^2 - c^2 * t_o^2/g^2 > - (x_o - v * t_o)^2 + c^2 * (t_o - v * x_o/c^2)^2 ) * g^2 > > = (x_A - x_a)^2 + (y_A - y_o)^2 + (z_A - z_o)^2 - c^2 * (t_A - t_a)^2 > + (x_o^2/g^2 - c^2*t_o^2/g^2 - (x_o - v * t_o)^2 > + c^2 * (t_o - v * x_o/c^2)^2 ) * g^2 > > = (x_A - x_a)^2 + (y_A - y_o)^2 + (z_A - z_o)^2 - c^2 * (t_A - t_a)^2 > + (x_o^2/g^2 - c^2 * t_o^2/g^2 > - x_o^2 - v^2*t_o^2 + 2*x_o*v*t_o > + c^2 * t_o^2 + v^2 * x_o^2/c^2 - 2 * t_o * v * x_o) ) * g^2 > > = (x_A - x_a)^2 + (y_A - y_o)^2 + (z_A - z_o)^2 - c^2 * (t_A - t_a)^2 > + (x_o^2*(1 - v^2/c^2) - c^2 * t_o^2 * (1 - v^2/c^2) > - x_o^2*(1 - v^2/c^2) - v^2*t_o^2 + 2*x_o*v*t_o > + c^2 * t_o^2 + 2 * t_o * v * x_o) ) * g^2 > > = (x_A - x_a)^2 + (y_A - y_o)^2 + (z_A - z_o)^2 - c^2 * (t_A - t_a)^2 > + 0 > >where I've written x_a = (x_o - v * t_o) * g >and t_a = (t_o - v * x_o/c^2) * g , as that >part of the equation stopped contributing to >the final result. If one wants to get >pedantic one can define y_a = y_o and z_a = z_o, >and rewrite the final result accordingly. > >The dual lightcone is invariant (except for a in retrospect >very obvious translation) under the Lorentz. Therefore >lightspeed is c everywhere under the Lorentz. Ghost, you don't have to go through all that maths every time. It is obvous what you are saying....the rods and clocks of all the observers aree supposed to contract so that they will all MEASURE OWLS as 'c'. This is just another way of expressing the postulate. It doesn't prove it. It is an extension of the old aether idea. That is not the same as saying light leaves the star with an infinite number of speeds. > >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. > 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: bz on 7 Apr 2005 08:19 jgreen(a)seol.net.au (Jim Greenfield) wrote in news:e7b5cc5d.0504062211.6957dcb5(a)posting.google.com: > ertain crystals emmit light of a fixed frequency. What is the > chemical reaction within the crystal, which causes it to alter its > emmitted wavelength, according as to how it is observed??? > Hint: the wavelength emmitted by the crystal does NOT alter from its > point of view; the Doppler shift noted by the observer is due to the > change in VELOCITY. > The relative velocity between source and observer. NOT the velocity of the photons. -- 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: Sam Wormley on 7 Apr 2005 08:41 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.
From: Dishman on 7 Apr 2005 08:43
Henri Wilson wrote: > On Wed, 6 Apr 2005 23:36:36 +0100, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > > > > >"Henri Wilson" <H@..> wrote in message > >news:65g851dn7asmetb67jjv5cnue56cbsjlai(a)4ax.com... > > > >>>It should demonstrate that a change of angle > >>>as illustrated in the case of the elliptical > >>>motion has no effect on the brightness as long > >>>as the path length of both beams remains the > >>>same while path length does have an effect even > >>>if the angle remains constant. > >> > >> In the elliptical case, the intensity of the beam will remain constant > >> only in > >> a plane perpendicular to the beam direction. > > > >Note I am describing the intensity only at a > >single point. In other situations that may > >extend to other points by symmetry but not > >in this case. > > Yes, the intensity on a 'line' will be constant... I am only discussing intensity at a point, not along a line. How the intensity varies spatially also depends on other factors (like the shape of th source). That is just for clarification, it doesn't impact your next point: > ...but the fact that one beam is > coming in at a different angle as you move the mirror around the ellipse will > surely affect the way the two beams interfere. No. The general formula for a wave propagating along the x axis is f(x, t) = A * sin(w*(t - x/c) + p) A is the amplitude w is the angular frequency c is the speed p is the phase t is time x is the distance from the source At any given point x is a constant so we can define p' = w*x/c+p as a local phase which is again fixed and simplify as: f(t) = A * sin(w*t + p') When you take two such waveforms and combine them, the sum of two sine waves is also a sine wave but with an amplitude that depends only on the product of the values of A and difference between the values of p', the phase difference. Any detector averages the power over many cycles so the relative intensity depends solely on the phase difference as the amplitude is a common factor. > It isn't just a matter of > intensity. It is just a matter of relative phase. > >> In your example, a cosine > >> correction needs to be applied to compare intensities AT ANY POINT on the > >> target. > >> I'm not sure if tat is relevant. It might only affect the contrast of the > >> fringes. > > > >What is significant is that it shows that > >the angle at which the light arrives at > >point D is unimportant, hence any possible > >angle change in the Sagnac case is also > >unimportant, except insofar as it might > >lead to a path length change. > > I wouldn't agree. The angle between the two beams is moreimportant than the > intensity. Both experimental and theoretically, at any given point the angle of incidence has absolutely no effect, only the relative phase matters. As I have acknowledged several times, when you consider the spatial behaviour, then you need to know how x varies over the screen since it affects p' hence the angle would matter. > >>>You said "In my sagnac model, there is both > >>>sideways displacement AND angular change." but, > >>>if you agree, you should now understand that a > >>>change of angle cannot cause any effect at the > >>>detector. > >>> > >> But there is also a path length difference. > > > >RThat's the key and here I agree with you, but > >my point is that the effect must come from > >the length change only, not the angle change, > >hence that makes it a lot easier to calculate. > > Well, until we have a better idea of what a photon really is, I cannot say I > would agree on that either. If you want to talk about probabilities of individual photons then we would need to use QED, but for our discussion a purely classical approach is adequate. > >> Put it another way, when the > >> apparatus is rotating, the sections of the two beams that arrive > >> simultaneously > >> at any point did not leave the source at the same instant. > > > >I agree, that is what my other simulation > >will try to illustrate. > > I still cannot see why fringe pattern should change at all while the apparatus > is rotating at constant speed. I'll work on the sim tomorrow, I'm out doing more interesting things tonight. I should have it by the weekend. By then I hope you will have your Java VM working. > I have version 1.1.4 That is still the old Microsoft version which only implemented an early subset of Java. I think the main deficiency is that it doesn't have the "Swing" components that produce the GUI. If you click the download button, it should upgrade you to version 1.5.0_02 of the Sun VM. Here's the link again http://www.java.com/en/download/ > >>>I just followed the defaults and it > >>>worked fine. > >>> > >>>>>If it works, switch on one beam at a time > >>>>>using the check boxes and move the location > >>>>>of the detector using the slider. Wiggle it > >>>>>about and see if you agree with the beam > >>>>>paths. Then switch both on and look at the > >>>>>angle between the beams at the detector. Let > >>>>>me know what you think. > >True, I found the display elements tedious > >but Swing helps and the EJS program does all > >that for you. Anyway, let's get the simulation > >running so we can address the Sagnac case, the > >software is only a means to that end. > > I'll run one up on Vbasic too. Cool, then we can compare ideas :-) George |