[General] inertia

Richard Gauthier richgauthier at gmail.com
Sat Apr 2 13:53:30 PDT 2016


Hello Andrew, Chip, Albrecht and all,

   Andrew, thanks for the summary of your proposed explanation for inertia. I like the part where an incoming photon (in the process of e-p pair production in the vicinity of a nucleus to absorb excess momentum) splits into two 1/2 photons that are +e and -e charged before the two 1/2 photons curl up to form an electron and a positron. I think the equal magnitude +e and -e charges on the two  1/2 photons are produced (perhaps continuously) as the single photon splits into two 1/2 photons. I would call these two 1/2 photons  as charged spin-1/2 photons since they equally divide the spin 1 of the initial photon, and maybe this is what you meant also. So I think that we may be in agreement here. Otherwise our explanations of the origin of an electron's inertia are quite different.

   I proposed in my inertia article that the electron’s inertial mass is due to the circulating momentum Eo/c = mc of a double-looping spin 1/2 Compton-wavelength h/mc charged photon that forms a resting electron. In my calculation,  F is the time rate of change of the circling momentum mc, which is F= dp/dt = wp = wmc  where w is the zitterbewegung angular frequency w = 2mc^2/hbar of the double-looping charged Compton-wavelength photon. When this inwardly-directed F= dp/dt is divided by the inwardly-directed centripetal acceleration A-cent = w^2 R of the circling charged photon, it gives M-inertial = F/A  = (dp/dt) / A-cent = m the invariant mass Eo/c^2 of the electron.  So a circulating Compton wavelength photon of momentum Eo/c = mc gives rise to the electron’s inertial mass m. 

  I recently showed that a Compton wavelength photon of energy Eo doesn’t have to circulate to give rise to an inertial mass M=m=Eo/c^2. It it is enough that the Compton photon is reflected from a reflecting area that changes the direction of the photon by the angle rule for reflection (angle of incidence = angle of reflection, with the two angles in the same plane). Here’s the proof:

  Use the Newton’s second law formula for inertial mass M = F/A = (dp/dt) / A as in the circulating charged photon calculation above. But now dp/dt is the average force F-av acting on the photon during the reflection process, and A-av is the photon's average acceleration during the reflection process.  When a Compton photon with momentum p = Eo/c = mc  approaches a reflecting area having an angle of incidence theta to the normal, the photon's component of momentum normal (perpendicular) to the reflecting area is  p-normal =  - p cos (theta) , taking the direction of the outward normal to the area as the + direction. After the photon is reflected, the component of its reflected momentum that is normal to the surface is p-normal = +p cos(theta).  The change in momentum of the reflected Compton photon is therefore delta p = 2 p cos(theta) = 2 Eo/c  cos(theta). Take the time for the photon to reflect as delta t (it will be the same delta t for the average force calculation and the average acceleration calculation.)  Then the average force acting on the photon during the reflection process is F-av = (delta p) /(delta t) = (2 Eo/c  cos(theta) ) / delta t. There is no change in the component of the photon’s momentum parallel to the reflection area during the reflection process.

  Now calculate the average acceleration A-av of the reflected photon during the same reflection time interval delta t.  The photon approaches the reflecting area with speed c and incident angle theta, so the component of its incoming velocity in the normal direction is  v-normal = -c cos(theta) . After the photon is reflected the normal component of its velocity is v-normal = +c cos (theta). Therefore the change in the normal component of its velocity during the reflection process is delta v = 2 c cos(theta) . The photon’s average acceleration during this time is A-av = (delta v)/delta t =  (2 c cos(theta) )/ delta t . There is no change in the component of the photon’s velocity parallel to the reflection area during the reflection process.

  Putting these two above results F-av and A-av  into Newton's formula for inertial mass M=F/A = (dp/dt) / A gives M = { (2 Eo/c cos(theta) ) /delta t} / { (2c cos(theta)/delta t) } = Eo/c^2 . Since this photon is a Compton wavelength photon,  Eo/c^2 = mc^2/ c^2 = m , the invariant mass of the electron. Generalizing the above derivation for a reflecting photon of any energy E=hf , this gives any photon's inertial mass as calculated in the reflection process to be M=E/c^2 = hf/c^2.  So a photon’s inertial mass M equals hf/c^2 , as determined by the simple reflection calculation, and so is non-zero while the same photon’s invariant mass m is zero. 

      Richard

> On Apr 2, 2016, at 9:57 AM, Andrew Meulenberg <mules333 at gmail.com> wrote:
> 
> Dear Richard,
> 
> It was presumptuous of me to think that, even if you had read that email, you would have remembered any details.
> 
> I don't know how to search the emails for a specific item and I don't think that the thread involved had a label that would help. I'll quickly outline the model that leads to inertia.
> 
> The electron (positron):
> is a bound 1/2 photon (I actually have come to accept your term of 'charged' photon to describe the transitory EM wave existing between the 'rectification' of a >1MwV photon by a nucleus and the condensation into an electron / positron pair).
> the 1/2 photon is self-trapped by total internal reflection (the high energy density of the compacted photon gives a higher refractive index (so that n > 1 and  v = c/n) that confines the photon.
> The photon 'wraps' about itself (maybe 1e4 - 1e8 times) as a long rubber band on a ball. The lowest energy state is that of uniform wrapping, not a planar wrap that would compress all of the field lines. The path circumference is lambda/2.
> During the traverse of the circumference, the photon twists 180 degrees so that the E-field is always pointing outward (not necessarily normal to the surface of the 'ball'). This twist is automatic in the case of circularly polarized photons, but augmented by the Imbert–Fedorov effect.. It depends on the Goos–Hänchen effect for linearly-polarized photons. Both effects shift the photon out of the planar configuration.
> The tricky part of the model comes from addressing the oppositely-directed E-Field always pointing inward. There are 3 distinct options to consider to counter the argument of the electrons' self-field:
> because the EM field has been rectified, the inward-pointing reverse polarity has zero amplitude. There is no inward pointing field.
> the inward pointing field concentration is intense enough to give an energy density great enough to give a gravitational gradient sufficient to balance the self-charge repulsion of each of the leptons.
> the rectification process separates the photon potentials (giving the spatial potential gradients seen as charges). However, the separation process leaves a connection between the leptons (thru time?) to close the field lines. This connection is a wormhole.
> The spherically 'wrapped' photon gives an angular momentum component in all directions, so the designated spin axis for a stationary lepton in a field-free region is entirely arbitrary.  The leptons equally share the photon's ang mom of 1. Thus, there is a spin 1/2 in all directions (as far as I know, there is no other model that can do this).
> Inertia:
> 
> Since the lepton has photon 'flow' in all directions at close to c (v = c/n), any motion will create a relativistic effect that seeks to reduce the added velocity of all the portions with components in the direction of motion. Since there is a net spin, the effect of velocity is unbalanced and results in a precession of the spin vectors about the velocity vector. The distance traveled during a single cycle of the precession is the deBroglie wavelength.
> Any acceleration to add velocity will create a torque from the relativistic effect that will result in a change in precession about the velocity vector.
> The net torque is composed of the effect on all the 'windings' of the photon, but moreso for those with velocity components in the direction of motion. Thus there is a 'flattening' of the lepton along the velocity vector.
> The flattening (thus concentration) of the leptons increases their total EM field energies. 
> Thus, the change of energies from the change in precession angular velocity and the distortion of the photon (EM field) distribution, gives the resistance to change in momentum and energy defined as inertia.
> Andrew
> ___________________________
> 
> On Sat, Apr 2, 2016 at 7:28 PM, Richard Gauthier <richgauthier at gmail.com <mailto:richgauthier at gmail.com>> wrote:
> Hi Andrew,
>    I hope all is well with the twins.
>    I don’t want to propose as a new idea about inertia something that I might have read of yours earlier but forgot. Can you point me to where you expressed your ideas on inertia in the forum, or at least very briefly summarize what you said about your inertia hypothesis in this forum? Thanks!
>      Richard
> 
>> On Apr 1, 2016, at 8:17 AM, Andrew Meulenberg <mules333 at gmail.com <mailto:mules333 at gmail.com>> wrote:
>> 
>> Dear Richard,
>> 
>> I think that the closest I've come to writing it up is earlier in this forum. The twins and other works (e.g., getting the anomalous solution of the Dirac equations accepted) have priority at the moment. Maybe by the end of the year.
>> 
>> Andrew
>> 
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>> On Fri, Apr 1, 2016 at 6:40 PM, Richard Gauthier <richgauthier at gmail.com <mailto:richgauthier at gmail.com>> wrote:
>> Hello Andrew,
>>      Is your own relativistic model of the electron’s inertia (electron as a spherically bound photon) written up? You can put it on academia.edu <http://academia.edu/> or another website if it’s not already. I’d like to compare it with mine at https://www.academia.edu/23184598/Origin_of_the_Electrons_Inertia_and_Relativistic_Energy_Momentum_Equation_in_the_Spin-_Charged_Photon_Electron_Model <https://www.academia.edu/23184598/Origin_of_the_Electrons_Inertia_and_Relativistic_Energy_Momentum_Equation_in_the_Spin-_Charged_Photon_Electron_Model>  .
>>       Richard
>> 
>>> On Apr 1, 2016, at 12:19 AM, Andrew Meulenberg <mules333 at gmail.com <mailto:mules333 at gmail.com>> wrote:
>>> 
>>> Dear Albrecht,
>>> 
>>> You have repeatedly based your model on lack of alternatives (with very precise results). E.g., 
>>> 
>>> Why 2 particles in the model? I say it again:
>>> 
>>> 1) to maintain the conservation of momentum in the view of oscillations
>>> 2) to have a mechanism for inertia (which has very precise results, otherwise non-existent in present physics)
>>> 
>>> I will be happy to see alternatives for both points. Up to now I have not seen any.
>>> 
>>> I'm sure that alternatives exist. Whether they have very precise results to support them may be up for debate. 
>>> 
>>> My own relativistic model for inertia depends on the electron being, in its ground (restmass) state, a spherically bound photon. Until that concept is accepted, it makes little sense to go further in a description. However, if accepted, it then also leads to understanding the inertia of a photon. 
>>> 
>>> Your two-particle model faces the same challenge. Unless you are able to shape that premise into an acceptable form, it is unlikely that anything that follows will matter. Can you (re)define your particles to be acceptable to an audience and still fulfill your assumptions and derived results?
>>> 
>>> Andrew
>>> 
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