Discussion of Note 385(3)

In the Note, omega sub 0 = m c squared / (2 pi h bar), (Eq. 34), and not zero. Assuming B = 0 and partial A / partial t = 0, gives Eq. ( 5), which is developed as Eq. (12). Using Eq. (34) gives curl A = 0. Using Eqs. (26) and (34) gives the vector potential (27) in spherical polar coordinates, whose curl is zero. It would be interesting to graph this A. It is related to the linear momentum of the aether by p = e A. So the vacuum or aether electric vector potential (27) imparts a momentum p to a charge e in material matter, for example one electron. The electric field strength induced in a circuit is directly proportional to the spacetime or aether or vacuum A. This shows how the circuit in UFT311 picks up an electric field strength. The solution omega sub 0 = – c / r is also valid mathematically and could also be used with the dipole electric field. The computer alebra you sent over this morning shows that omega sub 0 = – c / r introduces a constraint in order to have curl E = 0.

To: EMyrone@aol.com
Sent: 08/08/2017 10:22:48 GMT Daylight Time
Subj: Re: Computer Check of curl E = 0 for the Electric Dipole Field

Assuming omega_0 = 0 solves the equations but I wonder why a dipole should be that different from a point charge where we have omega_0 = -c/r. The choice

omega_0 = -c/r * cos(theta)

is much more plausible for me and fulfills

curl (omega_0 * A) = 0.

Is there any equation that requires

curl(A) = 0

for the electric dipole potential?
Horst

Am 08.08.2017 um 09:57 schrieb EMyrone:

Many thanks! This is an important demonstration of the self consistency of ECE2 electrostatics. The quadrupole, octopole and hexadecapole electric field strengths will also be irrotational, and in general the n pole electric field strength must be irrotational. In Note 385(3) the time like spin connection was defined as the de Broglie rest frequency in hertz of the vacuum particle. It follows that gradient omega sub 0 = 0, and that curl A = 0. So this new universal definition of omega sub 0 can be used from now one and improves understanding.

To: EMyrone
Sent: 07/08/2017 21:22:46 GMT Daylight Time
Subj: Re: 385(3): Complete Solution for the Electric Dipole Field

I checked that the electic dipole field irrotational. This is true in cartesian as well as spherical coordinates. The calculation for cartesian coordinates is highly complicated but works.
I also applied the scalar spin connection

omega_0 = -c/r

and tested the resulting vector potential

bold A = -1/omega_0 bold E.

The question is if A is also rotation-free. When assuming the above omega_0 or even a general omega(r), it is not possible to achieve this. When assuming a more general omega(r, theta), a condition comes out when A becomes irrotational:
= 0

which can further be simplified. However the question is if this required for electrostatics. I think it is not. Eq. (12) has to be valid.
It can be shown that the spin connection
omega_0 =

Fulfills eq.(12) for the dipiole field.

I will prepare some graphics the next days.

Horst

Am 05.08.2017 um 15:50 schrieb EMyrone:

This note shows that the electric dipole field is irrotational, curl E = 0, and that its vector potential is Eq. (27) where omega sub 0 is a fundamental constant of nature, the rest frequency in hertz of the vacuum particle discovered in PECE2 (UFT366). Therefore omega sub 0 is given by the well known de Broglie equation (34), where the mass of the vacuum particle can be calculated from the missing mass of the universe as in PECE2. Louis de Broglie first devised this equation for photon rest mass, which is incorrectly zero in the standard model. The calculation and concepts in this note conserve antisymmetry provded that the vector spin connections are correctly calculated from Eq. (27). This can be done in any coordinate system. In general any static electric field conserves antisymmetry, a major step forward for ECE2 theory. The static electric field is defined by curl E = 0, and partial B / partial t = 0 from the Faraday Law of Induction. A particular solution of the latter equation is B = 0, the solution used in this note. Note carefully that in the standard model, a static electric field does not have a vector potential, so the standard model violates the fundamental antisymmetry of the electromagnetic field tensor. The standard model violates the first antisymmetry law of ECE2, Eq. (1) of this note This was first shown in UFT131 ff. and the conclusion has been accepted by the vast readership of ECE and ECE2. Physics is divided therefore into two major schools of thought.

Advertisements
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: