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Tuesday, March 17, 2009

Quantum's Photon Interactions

In this conception of the photon, the Poynting Vector would be proportional to the cross product of ExB, and thereby would be expanding and contracting in the E-field planes perpendicular to the direction of motion. In accordance with the Maxwell Equations, the curl of E is proportional to the time rate of change of the B-field, and since a photon has no charges (photons emitt no electic fields) it experiences an oscillatory spin and thereby the angular momentum is not conserved for the photon and that leads to symmetry breaking on the quantum realm of things. This is only true for objects that don't have a charge or rest masses, but for objects that do have charges (and thereby a current density) or rest masses, the angular momentum is conserved.

UPDATE:  The above image of E and B fields for a photon is incorrect due to the law of relativity.  There can be no change in time for a lightspeed object relative to a "blind observer", and hence no -dB/dt or dE/dt, a "blind observer" is a device which doesn't interact with the entity being observed, in this case a photon.  This is the reason why photons in quantum physics don't change their orientations unless they interact with either another photon, thereby exhibiting wave diffraction patterns, or with another particle thereby being bounced, deflected, absorbed or transmitted, since it's only when a photon traverses through a medium with a refractive/absorptive index greater than 1 that it begins to oscillate relative to an "interacting observer".  Hence this quantum depiction of a static photon spin for an unobtruded photon below is actually correct, and hence angular momentum is conserved for a photon.