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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: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 [mailto:general-bounces+john=macken.com@lists.natureoflightandparticles.org] <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-right:0in'><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-right:0in'><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>