<html><head><meta http-equiv="Content-Type" content="text/html charset=utf-8"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space;" class="">Hi Chip,<div class=""> I don’t know if an electric charge radiates under gravitational acceleration. But I googled the question and got this: <a href="http://physics.stackexchange.com/questions/21830/does-a-charged-particle-accelerating-in-a-gravitational-field-radiate" class="">http://physics.stackexchange.com/questions/21830/does-a-charged-particle-accelerating-in-a-gravitational-field-radiate</a> . After reading this I still don’t know if the issue is resolved. </div><div class=""> But I’ve recently thought about this question (non-radiation of a circling (accelerating) charge in a resting or a constant speed charged-photon electron) in another way. This is the first time I’ve mentioned this. In a double looping charged-photon electron model (whose double-looping is needed to get the electron's spin 1/2 hbar), the photon as you know makes 2 loops for each Compton wavelength h/mc in a resting electron. Each circumference is therefore 1/2 of a Compton wavelength around. So that after 1 round the photon is out of phase by 180 degrees with the photon on its path on the following and preceding round. If the photon (being charged) IS radiating with the wavelength of the Compton wavelength while traveling along this zitterbewegung double-looping path, then each set of E-M waves radiated outward by the electric charge on one round will cancel the E-M waves waves radiated outward by the charge on the following round, so the total E-M wave being radiated outward will cancel to zero and the resting electron (composed of a circulating therefore accelerating charged photon) won’t radiate EM-radiation. However, when the electron is moving at a constant speed (and its charged photon follows a double-looping helical light-speed trajectory as in my article, rather than a double-looping light-speed circle in the resting electron) these radiated E-M waves from the charged photon do not cancel out in the longitudinal/forward direction in which the electron is moving, but rather combine in the forward direction, and generate the de Broglie wavelength/de Broglie waves in this longitudinal direction as described in my article “electrons are spin 1/2 charged photons generating the de Broglie wavelength”. These de Broglie waves allow the electron to be diffracted as waves by crystals and also to go through the double-slit apparatus as waves, before being detected later as particles.</div><div class=""> What do you think?</div><div class=""> Richard</div><div class=""><br class=""><div><blockquote type="cite" class=""><div class="">On Feb 26, 2016, at 4:07 AM, Chip Akins <<a href="mailto:chipakins@gmail.com" class="">chipakins@gmail.com</a>> wrote:</div><br class="Apple-interchange-newline"><div class=""><div class="WordSection1" style="page: WordSection1; font-family: Helvetica; font-size: 12px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: auto; word-spacing: 0px; -webkit-text-stroke-width: 0px;"><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">Hi Richard<o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">Yes. I see your point. Good one.<o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">But we know that accelerating a complete charge unit (elementary charge) causes it to radiate, and that the charge in the electron does not radiate when the electron is at rest. So I am imagining that the confinement force of the energy for the electron topology plays a part in creating the charge unit, which also prevents that self-confined charge unit from radiating away its energy.<span class="Apple-converted-space"> </span><o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">So the (wave) energy in the electron is simply following its zero net force geodesic path and does not feel any acceleration.<o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">This would imply that an electric charge only radiates when accelerated by the addition of energy. Which in turn would imply that an electric charge might not radiate under gravitational acceleration??<o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">Thoughts?<o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">Chip<o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div class=""><div style="border-style: solid none none; border-top-color: rgb(225, 225, 225); border-top-width: 1pt; padding: 3pt 0in 0in;" class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><b class=""><span style="font-size: 11pt; font-family: Calibri, sans-serif;" class="">From:</span></b><span style="font-size: 11pt; font-family: Calibri, sans-serif;" class=""><span class="Apple-converted-space"> </span>Richard Gauthier [<a href="mailto:richgauthier@gmail.com" class="">mailto:richgauthier@gmail.com</a>]<span class="Apple-converted-space"> </span><br class=""><b class="">Sent:</b><span class="Apple-converted-space"> </span>Thursday, February 25, 2016 10:22 PM<br class=""><b class="">To:</b><span class="Apple-converted-space"> </span>Chip Akins <<a href="mailto:chipakins@gmail.com" class="">chipakins@gmail.com</a>><br class=""><b class="">Cc:</b><span class="Apple-converted-space"> </span><a href="mailto:davidmathes8@yahoo.com" class="">davidmathes8@yahoo.com</a><br class=""><b class="">Subject:</b><span class="Apple-converted-space"> </span>Re: Sarfatti discussion<o:p class=""></o:p></span></div></div></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">Chip,<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> But the electron internally generates the electron’s magnetic moment, which requires at least a light-speed charge circulating. So I think there’s no way to avoid a circulating electric charge that is accelerating (which an electric charged even moving at a constant speed in a circle is doing with its centripetal acceleration.<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> Richard<o:p class=""></o:p></div></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div><div class=""><blockquote style="margin-top: 5pt; margin-bottom: 5pt;" class="" type="cite"><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">On Feb 25, 2016, at 4:17 AM, Chip Akins <<a href="mailto:chipakins@gmail.com" style="color: purple; text-decoration: underline;" class="">chipakins@gmail.com</a>> wrote:<o:p class=""></o:p></div></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div><div class=""><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">Hi Richard<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> <o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">This lack of radiation of the “charged photon” when it is accelerated to form the closed double loop, is the principal reason that I feel the term “charged photon” is incorrect. I we assume charge to be created by the divergence of the E field as a result of the double loop confinement, then the photon is not charged, but the electron is charged due to its topology.<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> <o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">Chip<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> <o:p class=""></o:p></div></div><div class=""><div style="border-style: solid none none; border-top-color: rgb(225, 225, 225); border-top-width: 1pt; padding: 3pt 0in 0in;" class=""><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><b class=""><span style="font-size: 11pt; font-family: Calibri, sans-serif;" class="">From:</span></b><span class="apple-converted-space"><span style="font-size: 11pt; font-family: Calibri, sans-serif;" class=""> </span></span><span style="font-size: 11pt; font-family: Calibri, sans-serif;" class="">Richard Gauthier [<a href="mailto:richgauthier@gmail.com" style="color: purple; text-decoration: underline;" class="">mailto:richgauthier@gmail.com</a>]<span class="apple-converted-space"> </span><br class=""><b class="">Sent:</b><span class="apple-converted-space"> </span>Wednesday, February 24, 2016 10:52 PM<br class=""><b class="">To:</b><span class="apple-converted-space"> </span><a href="mailto:davidmathes8@yahoo.com" style="color: purple; text-decoration: underline;" class="">davidmathes8@yahoo.com</a><br class=""><b class="">Cc:</b><span class="apple-converted-space"> </span>Chip Akins <<a href="mailto:chipakins@gmail.com" style="color: purple; text-decoration: underline;" class="">chipakins@gmail.com</a>><br class=""><b class="">Subject:</b><span class="apple-converted-space"> </span>Re: Sarfatti discussion</span><o:p class=""></o:p></div></div></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> <o:p class=""></o:p></div></div><div class=""><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">David,<o:p class=""></o:p></div></div></div><div class=""><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> The charged photon is moving in a circular (in a resting electron) or helical (in a moving electron) path while the uncharged photon moves in a straight line. But the charged photon doesn’t radiate when moving in this circular or helical path (although its electric charge is being accelerated in this helical trajectory), so this path is like a geodesic for the charged electron — its natural path. The charged photon (electron) only radiates when it accelerates off of this helical trajectory by an externally applied force produced by for example an electric or magnetic field. I think that the electron’s charge and its helical ‘geodesic’ trajectory are closely related — the charge produces the helical trajectory that the uncharged photon doesn’t have (superluminal motion in both the uncharged and charged photon complicates this picture but the genera idea is the same.)<o:p class=""></o:p></div></div></div><div class=""><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> Richard<o:p class=""></o:p></div></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> </div></div></div></blockquote></div></div></div></blockquote></div></div></body></html>