<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=""><div style="font-size: 14px;" class="">Hello Vivian,</div><div style="font-size: 14px;" class=""><br class=""></div><div class=""><span style="font-size: 14px;" class=""> Again, thank you for your further explanations. Unfortunately, you still do not seem to have realized, and therefore do not accept, that you incorrectly derived the de Broglie wavelength Ldb = h/(gamma m v) from your electron model. Since you knew the correct de Broglie wavelength formula in advance, it is understandable that you accepted your faulty derivation when it seemed to give the “right answer”. This can be a lesson for all of us. Let me be more specific. On p 13 in the middle right-hand column of your article (attached below for others' convenience) you write:</span> </div><div class=""><br class=""></div><div class=""><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">The rotational component of the electron at rest is
unavailable for any interaction, except for exchanging
photons in an electromagnetic interaction. That leaves
only the translational component of the moving elec</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">tron’s electromagnetic field available for interaction.
</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">That is the wavelength of its kinetic energy. However
the electron is entirely composed of an electromag</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">netic wave. Its total energy is </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">E= </span><span style="font-size: 11pt; font-family: Symbol;" class=""> </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">hf</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">v</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">. Its rest energy
is hf</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">0</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">. The kinetic energy component of the electro</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">magnetic field, pc, is given by </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">E sin (theta)=</span><span style="font-size: 11pt; font-family: Symbol;" class=""> </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">h</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">f</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">v </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">sin (theta) =</span><span style="font-size: 11pt; font-family: Symbol;" class=""> </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">pc = </span><span style="font-size: 11pt; font-family: Symbol;" class=""> </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">h</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">f</span><span style="font-family: TimesNewRomanPSMT; font-size: 9px;" class="">ke .</span></div><div class=""><div class="page" title="Page 13"><div class="section"><div class="layoutArea"><div class="column"><div class=""><br class=""></div><div class=""><br class=""></div><div class=""><span style="font-size: 14px;" class="">Summarizing, you derived the “kinetic energy component of the electromagnetic field, pc" from the relativistic energy-momentum triangle as KE = pc = </span><span style="font-size: 11pt; font-family: Symbol;" class=""> </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">h</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">f</span><span style="font-family: TimesNewRomanPSMT; font-size: 9px;" class="">ke</span><span style="font-size: 14px;" class=""> (as you labeled the vertical side of the triangle in fig. 13) . Then you write: </span></div><div class=""><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class=""><br class=""></span></div><div class=""><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">Substituting c/lambda</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">KE </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class=""> for f</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">KE </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">and
simplifying yields </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">pc = </span><span style="font-size: 11pt; font-family: Symbol;" class=""> </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">hc/lambda</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">KE </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">where lambda</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">KE </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">is the component wavelength of the elec</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">tromagnetic oscillation that is the electron, in the
direction of its travel, namely lambda</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">KE </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">= h/p </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">which is the expression for the de Broglie wavelength.</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: 4pt;" class=""> </span></div><div class=""><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: 4pt;" class=""><br class=""></span></div><div class=""><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: 4pt;" class=""><br class=""></span></div><div class=""><span style="font-size: 14px;" class="">But the correct expression for the relativistic kinetic energy KE of a particle is KE= (gamma-1)mc^2 which is NOT equal to pc , which equals pc=(gamma mv)c or pc=gamma mvc . It is OK to set KE=</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">h</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">f</span><span style="font-family: TimesNewRomanPSMT; font-size: 9px;" class="">ke </span><span style="font-size: 14px;" class="">but this means that </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">f</span><span style="font-family: TimesNewRomanPSMT; font-size: 9px;" class="">ke </span><span style="font-size: 14px;" class="">=</span><span style="font-size: 14px;" class="">KE/h = [ (gamma-1)(mc^2) ] /h . Combining this expression for </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">f</span><span style="font-family: TimesNewRomanPSMT; font-size: 9px;" class="">ke</span><span style="font-size: 14px;" class=""> with </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">c/(lambda</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">KE </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">) = f</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">KE </span><span style="font-size: 14px;" class="">or </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">lambda</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">KE </span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class=""> = c/ f</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">KE </span><span style="font-size: 14px;" class=""> gives</span></div><div class=""><br class=""></div><div class=""><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class=""><span style="font-size: 11pt;" class="">lambda</span><span style="font-size: 7pt; vertical-align: -4pt;" class="">KE </span><span style="font-size: 11pt;" class=""> = c/ f</span><span style="font-size: 7pt; vertical-align: -4pt;" class="">KE </span></span><span style="font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class=""><span style="font-size: 11pt;" class="">= c</span><span style="font-size: 14px;" class="">/(KE/h) = c/ [(gamma-1)(mc^2)] / h</span></span></div><div class=""><span style="font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class=""><span style="font-size: 14px;" class=""><br class=""></span></span></div><div class=""><span style="font-family: TimesNewRomanPSMT; font-size: 11pt;" class="">lambda</span><span style="font-family: TimesNewRomanPSMT; font-size: 7pt; vertical-align: -4pt;" class="">KE </span><span style="font-size: 14px;" class="">= </span><span style="font-family: TimesNewRomanPSMT; font-size: 14px;" class=""> h/(gamma-1)(mc)</span><span style="font-size: 14px;" class=""> </span></div><div class=""><span style="font-size: 14px;" class=""><br class=""></span></div><div class=""><span style="font-size: 14px;" class="">which is not at all what you obtained: </span><span style="font-family: TimesNewRomanPSMT; font-size: 11pt;" class="">lambda</span><span style="font-family: TimesNewRomanPSMT; font-size: 7pt; vertical-align: -4pt;" class="">KE </span><span style="font-size: 14px;" class="">= h/p , when you incorrectly set KE = h</span><span style="font-size: 11pt; font-family: TimesNewRomanPSMT;" class="">f</span><span style="font-size: 7pt; font-family: TimesNewRomanPSMT; vertical-align: -4pt;" class="">KE</span><span style="font-size: 14px;" class=""> equal to pc . So whatever </span><span style="font-family: TimesNewRomanPSMT; font-size: 11pt;" class="">lambda</span><span style="font-family: TimesNewRomanPSMT; font-size: 7pt; vertical-align: -4pt;" class="">KE</span><span style="font-size: 14px;" class=""> is </span><span style="font-size: 14px;" class="">(you call it the wavelength of the electron’s kinetic energy)</span><span style="font-size: 14px;" class="">, it is NOT the de Broglie wavelength.</span></div><div class=""><span style="font-size: 14px;" class=""><br class=""></span></div><div class=""><span style="font-size: 14px;" class=""><br class=""></span></div><div class=""><span style="font-size: 14px;" class=""> So your electron model unfortunately does not give a correct derivation of the electron's de Broglie wavelength formula, due to a mis-interpretation of the relativistic energy-momentum equation’s “right triangle”. This led you to incorrectly equate two different quantities: pc and KE of a moving electron, resulting in your incorrect derivation of the de Broglie wavelength. One can reasonably conclude that the other relativistic mathematical derivations in your model, which you claim are consistent with known experimental properties of the electron, should be carefully checked as well; and should not be accepted by you unless they pass such checking.</span></div><div class=""><span style="font-size: 14px;" class=""><br class=""></span></div><div class=""><span style="font-size: 14px;" class="">with best regards, </span></div><div class=""><span style="font-size: 14px;" class=""> Richard</span></div><div class=""><span style="font-size: 14px;" class=""><br class=""></span></div><div class=""><span style="font-size: 14px;" class="">p.s. you might say that I’m just trying to find fault with your electron model because I have a different one. I agree, and hope you will return the favor. Everyone benefits in this way, and it hopefully speeds up the process of finding better models for the electron, photon and other fundamental particles. </span></div>
</div>
</div>
</div>
</div></div><div class=""><br class=""></div><div class=""></div></body></html>