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</o:shapelayout></xml><![endif]--></head><body lang=EN-US link=blue vlink=purple><div class=WordSection1><p class=MsoNormal><span style='color:black'>Hi John M<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>I concur that we have not identified all the forces. We can illustrate the existence of such forces in many different ways.<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>I also agree that space is filled with small amplitude waves, both transverse and longitudinal in form.<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>I think the emission of photons from the CO molecule is regulated by the momentum considerations I mentioned. However the CO molecule is a very complex structure, and assuming a simple dipole configuration is probably an over simplification which could result in drawing improper conclusions. The CO molecule in a simplified view, is more accurately envisioned as a collection of negative charges moving about two positive charges, which are also probably circulating about each other. But it is not a simple dipole. In my view it would take a significant amount of work to model the CO molecule in order to draw reasonable conclusions from it. But I do not think viewing it as a simple dipole is adequate.<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>Chip<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><div><div style='border:none;border-top:solid #E1E1E1 1.0pt;padding:3.0pt 0in 0in 0in'><p class=MsoNormal><b><span style='font-size:11.0pt;font-family:"Calibri",sans-serif'>From:</span></b><span style='font-size:11.0pt;font-family:"Calibri",sans-serif'> General [mailto:general-bounces+chipakins=gmail.com@lists.natureoflightandparticles.org] <b>On Behalf Of </b>John Macken<br><b>Sent:</b> Saturday, April 04, 2015 2:05 AM<br><b>To:</b> 'Nature of Light and Particles - General Discussion'<br><b>Subject:</b> Re: [General] Nature of Light and Particles<o:p></o:p></span></p></div></div><p class=MsoNormal><o:p> </o:p></p><p class=MsoNormal><span style='color:blue'>Chip,<o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'>I do not agree with your analysis contained in the email below. Suppose we move to a real example of a rotating dipole emission. A carbon monoxide molecule has charge separation so that there is a positive and negative electrical charge on this diatomic molecule. In a vacuum with no apparent forces, this molecule can only rotate at 115 GHz or integer multiples of this frequency. The “rotational line spectrum” emitted by a CO molecule implies these energy levels associated with physical rotation of the dipole molecule. Each spectral line is a single frequency with a very narrow bandwidth if there is no “pressure broadening” caused by collisions with other molecules which shortens the emission time and broadens the emission spectrum. <o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'>There is no redshifted and blue shifted emission from the linearly polarized photons emitted in the equatorial plane. If your model was correct, circularly polarized photons emitted along the rotational axis would be emitted at the true rotational frequency and linearly polarized photons emitted in the equatorial plane would have two frequencies - one higher and one lower than the actual rotational frequency. There is no difference in the emission spectrum between linearly and circularly polarized photons emitted from rotating dipole molecules. <o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'>This question reminds me of two additional interesting thoughts having to do with the example of a rotating CO molecule. I once did a calculation to see how much torque needed to be applied to the CO molecule to cause it to change its rotational frequency by 115 GHz when it emitted a photon. The CO molecule has a known distance between the carbon and oxygen nuclei and a known emission lifetime, so the force couple on the CO molecule can be calculated. I do not remember the exact force, but the force required to cause the known change in angular momentum was vastly larger than would be generated by the photon recoil force applied off center over the atom separation distance. It appears as if the transfer of angular momentum when a photon is emitted or absorbed is a different force mechanism than the 4 known forces. Now that I have written this, perhaps it should be studied further. <o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'>The second interesting point has to do with the limits of mathematical analysis. It is very easy to write an equation which gives the rotational line emission if the energy levels are separated by 115 GHz. However, what physically enforces this quantized angular momentum? It is not enough to write an equation which corresponds to experiment. I want to know why this happens. Suppose that medical doctors were satisfied with merely measuring the physical properties of a sick patient. They would measure temperature, strength, etc. However, they need to ask the “why” question before they can make progress and discover germs and virus. Similarly, I claim that physics have to ask “why” and not just “how” an effect takes place. I claim that many of the “why” questions can be answered if we assume that the vacuum is filled with a vast amount of energy in the form of small amplitude waves in spacetime. <o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'>John M.<o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><div><div style='border:none;border-top:solid #E1E1E1 1.0pt;padding:3.0pt 0in 0in 0in'><p class=MsoNormal><b><span style='font-size:11.0pt;font-family:"Calibri",sans-serif'>From:</span></b><span style='font-size:11.0pt;font-family:"Calibri",sans-serif'> General [<a href="mailto:general-bounces+john=macken.com@lists.natureoflightandparticles.org">mailto:general-bounces+john=macken.com@lists.natureoflightandparticles.org</a>] <b>On Behalf Of </b>Chip Akins<br><b>Sent:</b> Friday, April 03, 2015 5:55 PM<br><b>To:</b> 'Nature of Light and Particles - General Discussion'<br><b>Subject:</b> Re: [General] Nature of Light and Particles<o:p></o:p></span></p></div></div><p class=MsoNormal><o:p> </o:p></p><p class=MsoNormal><span style='color:black'>Hi John Macken<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>If we suppose the dipole in your though experiment is a one wavelength dipole with the half wave point at the center, (at the frequency of photon emission), the equatorial photons emitted would be blue shifted on the advancing side and redshifted on the retreating side of the spinning dipole. The blue shifted emission region would contain more energy and therefore more momentum regardless of spin, and the spinning dipole would be slowed due to this effect. This effect becomes even easier to understand if the photon wavelength is shorter related to the dipole length. In the equatorial case, for the slowing of the spinning dipole, longitudinal momentum is imparted to the photons (tangential with respect to the dipole) in lieu of spin momentum. I am thinking that in your though experiment example, if any spin momentum is imparted to the photons, it will be equal parts of left and right spin, yielding a plane polarized photon. But since it would have the blue shifted and redshifted regions during emission, even equal amounts of left and right spin angular momentum imparted to the photons would slow the dipole.<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>Thoughts?<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>Chip<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><div><div style='border:none;border-top:solid #E1E1E1 1.0pt;padding:3.0pt 0in 0in 0in'><p class=MsoNormal><b><span style='font-size:11.0pt;font-family:"Calibri",sans-serif'>From:</span></b><span style='font-size:11.0pt;font-family:"Calibri",sans-serif'> General [<a href="mailto:general-bounces+chipakins=gmail.com@lists.natureoflightandparticles.org">mailto:general-bounces+chipakins=gmail.com@lists.natureoflightandparticles.org</a>] <b>On Behalf Of </b>John Macken<br><b>Sent:</b> Friday, April 03, 2015 5:49 PM<br><b>To:</b> 'Nature of Light and Particles - General Discussion'<br><b>Subject:</b> Re: [General] Nature of Light and Particles<o:p></o:p></span></p></div></div><p class=MsoNormal><o:p> </o:p></p><p class=MsoNormal><span style='color:blue'>Chip,<o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'>I am responding to your questions about the spin of a linearly polarized photon because I intend to include this subject in my paper if my abstract is approved for inclusion in the Nature of Light conference. I am going to first present a thought experiment. <o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'>Suppose that we have a rotating electrical dipole which physically consists of two opposite polarity electrical charges at opposite ends of a rotating rod. The rotation is around the center of the rod and the rotation axis is perpendicular to the rod length dimension. The rotating dipole will emit electromagnetic radiation into a classical rotating dipole emission pattern. The photons emitted along the axis of rotation will be circularly polarized with the rotation direction the same as the rotation direction of the rod. If the rotating dipole is visualized in a vacuum and an inertial frame of reference, then it can be shown that the angular momentum being carried away by the circularly polarized photons emitted along the axis slows down the rotation speed of the dipole by the exact amount that corresponds to the energy being carried away by the circularly polarized photons. So far there are no surprises.<o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'>Now suppose that we look at the photons being emitted in the equatorial plane of the rotation. The well-known emission pattern of a rotating dipole emits linearly polarized photons in this plane. If these photons are carrying away equal amounts of the two opposite spin rotational directions, then the rotating dipole is experiencing no net loss of angular momentum which implies that the rotating dipole does not lose any energy when it emits equal amounts of photons with opposite spins. A perpetual motion machine could be made if a special reflector was made which only allowed light emitted in the equatorial plane escape. <o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'>This obviously must be wrong. The implication is that linearly polarized light is carrying away angular momentum also and the angular momentum always is such that it slows down the rotating dipole. The proposed answer is that linearly polarized photons are carrying away orbital angular momentum (my laser background) and the rotation axis is perpendicular to the photon’s propagation direction. This should be experimentally provable, but a practical experiment will be difficult devise.<o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:blue'>John M. <o:p></o:p></span></p><p class=MsoNormal><span style='color:blue'><o:p> </o:p></span></p><p class=MsoNormal><b><span style='font-size:11.0pt;font-family:"Calibri",sans-serif'>From:</span></b><span style='font-size:11.0pt;font-family:"Calibri",sans-serif'> General [<a href="mailto:general-bounces+john=macken.com@lists.natureoflightandparticles.org">mailto:general-bounces+john=macken.com@lists.natureoflightandparticles.org</a>] <b>On Behalf Of </b>Andrew Meulenberg<br><b>Sent:</b> Friday, April 03, 2015 10:27 AM<br><b>To:</b> Nature of Light and Particles - General Discussion; Andrew Meulenberg; Jean-Luc Pierre P.<br><b>Subject:</b> Re: [General] Nature of Light and Particles - Request<o:p></o:p></span></p><p class=MsoNormal><o:p> </o:p></p><div><div><div><div><p class=MsoNormal style='margin-bottom:12.0pt'>Dear Chip,<o:p></o:p></p></div><p class=MsoNormal style='margin-bottom:12.0pt'>Since I have been having this discussion with someone (Jean-Luc Paillet) in a different context, I thought that I would take the time to try and find a paper that contained a statement that I had interpreted to mean that a linear-polarized photon still had a spin of 1. <br><br>I found what I think may be what I had seen (attached). However, now that I look more closely, I am not sure that it is referring to a photon or a collection. Perhaps someone more mathematically sophisticated can look at sections 6.7 (for circular-polarization) and 6.8 (for linear-polarization) of the attached and let me know if it can refer to single photons as well as collections. "We recover the classical result derived in Section 6.7: the spin is in the direction of propagation of the wave."<o:p></o:p></p></div><p class=MsoNormal style='margin-bottom:12.0pt'>Jean-Luc referred to the 3rd from last paragraph of <a href="http://mathpages.com/rr/s9-04/9-04.htm">http://mathpages.com/rr/s9-04/9-04.htm</a> , which states that linear-polarized light is only balanced circular-polarized light. However, it further states that individual photons will register as +/- hbar. Thus, it is a superposition of 2 states, rather than a 3rd state. If this is the case, does the E = n h nu relation come into play? If so, then I assume that spectrometers could respond differently to linear- and circular-polarized light of the same energy (with n = 2 and 1 respectively). On the other hand, since w = w1+/- w2, a spectrometer might see only the sum of the two coherent photons (a thermally stable BEC?). It is an interesting problem that I see no convincing solution to.<o:p></o:p></p></div><p class=MsoNormal>Andrew<o:p></o:p></p><div><div><div><div><div><p class=MsoNormal style='margin-bottom:12.0pt'>______________________--<o:p></o:p></p><div><p class=MsoNormal>On Fri, Apr 3, 2015 at 7:26 PM, Chip Akins <<a href="mailto:chipakins@gmail.com" target="_blank">chipakins@gmail.com</a>> wrote:<o:p></o:p></p><blockquote style='border:none;border-left:solid #CCCCCC 1.0pt;padding:0in 0in 0in 6.0pt;margin-left:4.8pt;margin-top:5.0pt;margin-right:0in;margin-bottom:5.0pt'><div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'>Hi John W</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'>The intent of this line of discussion is to probe more deeply into the structure of the photon and to address polarization entanglement experiments.</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'>A thought and some questions for you John.</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'>First some background. As I understand it Quantum physics posits a superposition of spin states as a cause for planar polarization. In order to reach a more causal explanation, can we then envision two fields within the photon, spinning opposite directions, and constructively interfering only in a plane, which is dependent on their spin phase? </span><o:p></o:p></p></div></div></blockquote><div><p class=MsoNormal> <o:p></o:p></p></div><blockquote style='border:none;border-left:solid #CCCCCC 1.0pt;padding:0in 0in 0in 6.0pt;margin-left:4.8pt;margin-top:5.0pt;margin-right:0in;margin-bottom:5.0pt'><div><div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'>Are you familiar with Joy Christian’s work? He writes that two non-commuting rotations (spin operators) as local variables, exactly duplicate the predictions of Quantum mechanics and satisfy Bell’s inequalities in precisely the same way. I have checked some of the math and so far it seems to be quite accurate. In both of these approaches, two oppositely rotating fields would apparently satisfy these physical aspects of the theories… ???</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'>Christian uses a Clifford algebra to illustrate his theory. Have you had the chance to compare that with the work you are doing using Clifford algebra to in your new theory of light and matter? Specifically have you had any opportunity to check to see if two opposite, (non-commuting local) spins caused by your framework would also satisfy Bell’s inequalities? Or CHSH inequalities?</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'>Of course you can see the underlying reasons for these questions. One underlying reason is to discover if two equal and oppositely spinning fields, confined within the photon, can explain polarization. In both, quantum physics, and Christian’s theories, it seems that two opposite spins are required, hinting that we would need those two opposite physical spins to be possible in a physical model of the photon.</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'>The other underlying reason is to discover if non-commuting (rotation) local variables can potentially be the cause for the appearance of entanglement.</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'>Thoughts?</span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'> </span><o:p></o:p></p><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span style='color:black'>Chip</span><o:p></o:p></p></div></div><p class=MsoNormal><o:p> </o:p></p></blockquote></div><p class=MsoNormal><o:p> </o:p></p></div></div></div></div></div></div></div></body></html>