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Many thanks again. I just sent over the general proof of conservation of antisymmetry in magnetostatics. The proof is very simple, it uses curl alpha = omega x A, thereby defining a novel vacuum current density via eq, (20) or (21). Given this new vacuum current density, del omega x A is always zero because del dot curl alpha is zero by definition. I give a couple of simple examples of vector potentials. So this is significant progress in solving the ECE2 field equations.

To: EMyrone@aol.com
Sent: 25/08/2017 11:31:02 GMT Daylight Time
Subj: Re: Discussion of Note 386(3)

del dot (omega x A) is not zero. Perhaps it is possible to determine one of the undefined omega components in a way that this condition is fulfilled, but it seems that this produces a new differential equation. Using a simple omega_y did not work in my tests.

Anyhow we have to find a suitable vector potential for the magnetic dipole field. Will try some approaches later.

Horst

Am 25.08.2017 um 09:50 schrieb EMyrone:

OK many thanks for all the work, this is very interesting, I will think about this problem and develop some ideas in the next note. There is another equation available, del dot (omega x A) = 0. In fact this equation must always be solved simultaneously with the antisymmetry equations. It comes from del B = 0. It is always possible to look for a simpler vector potential, anyone will do because this work aims to develop a general methodology and there is no need to use a vector potential that turns out to be too complicated.

To: EMyrone
Sent: 24/08/2017 18:44:35 GMT Daylight Time
Subj: Re: 386(3): Violation of Conservation of Antisymmetry by the Standard Model

I tested a lot with this potential. The good message is: The partial derivatives are handable. The bad message: solutions of the antisymm. eqs. for omega are difficile. The first and second eq. lead to the same result for omega_z, see o19 and o20. Therefore only one equation is left for omega_x and omega_y, they are underdetermined. I testes several arbitrary values for omega_y. In the protocol I used omega_y=0. (see o23). In all cases the same results for omega x A come out for the X and Y components of B, but the Z component depends on the choice of omega_y. The difference

B – B1
with
B=curl A and
B1 = omega x A

shows that the X and Y component are equal. Obviously the problem is to find a suitable omega_y so that also the Z components are equal too. Then A would be antisymmetric. Otherwise we have to change A.

Horst

Am 24.08.2017 um 13:23 schrieb EMyrone:

This note evaluates the magnetic dipole potential (7) in Cartesian coordinates at a point far from a current loop. The spin connection vector can be found from Eq (7) and Eqs. (9) to (11). Finally the ECE2 magnetic flux density B is found from Eq. (13). It is seen immediately from Eqs. (14) to (16) that the standard model violates antisymmetry, a disaster for standard (Maxwell Heaviside) electrodynamics first pointed out in UFT131 ff. Conservation of antisymmetry requires the existence of a spin connection and the existence of torsion, a clear and important result. It would be very interesting to graph the components of A and the spin connection vector omega, and the components of B. This new ECE2 level magnetic flux density must be compared with experimental data. The vector potential is an example solution of the ECE2 Ampere law, curl B = mu0 J.

Antisymmetry is proven to be conserved in general in magnetostatics provided that there exists a novel vector potential alpha defined by the spin connection term as in Eq. (7). This defines the vacuum or aether or spacetime current density (20), which can be expressed as (21). The standard model of magnetostatics is refuted entirely because it does not contain a spin connection and is not generally covariant. The standard model violates conservation of antisymmetry, the fundamental antisymmetry of the electromagnetic field tensor (an antisymmetric two-form of differential geometry). This was first shown in UFT131 ff. The standard model is Lorentz covariant only, and is a nineteenth century un-unified field theory, now refuted in many ways by many workers. ECE2 is generally covariant because Cartan geometry is generally covariant. In other words ECE2 is a generally covariant unified field theory (ECE and ECE2 theories). The existence of the spin connection has been proven experimentally and to high precision in UFT311, UFT321, UFT364, UFT382 and UFT383, and in many other UFT papers covering a wide range of physics and chemisty. Similarly the standard gravitational and electroweak theories are entirely obsolete (e.g. UFT225 shows up fundamental errors in the electroweak theory) and there now exist two major Schools of Thought in physics (ECE and ECE2 and the obsolete standard model)

a386thpapernotes4.pdf

OK many thanks for all the work, this is very interesting, I will think about this problem and develop some ideas in the next note. There is another equation available, del dot (omega x A) = 0. In fact this equation must always be solved simultaneously with the antisymmetry equations. It comes from del B = 0. It is always possible to look for a simpler vector potential, anyone will do because this work aims to develop a general methodology and there is no need to use a vector potential that turns out to be too complicated.

To: EMyrone@aol.com
Sent: 24/08/2017 18:44:35 GMT Daylight Time
Subj: Re: 386(3): Violation of Conservation of Antisymmetry by the Standard Model

I tested a lot with this potential. The good message is: The partial derivatives are handable. The bad message: solutions of the antisymm. eqs. for omega are difficile. The first and second eq. lead to the same result for omega_z, see o19 and o20. Therefore only one equation is left for omega_x and omega_y, they are underdetermined. I testes several arbitrary values for omega_y. In the protocol I used omega_y=0. (see o23). In all cases the same results for omega x A come out for the X and Y components of B, but the Z component depends on the choice of omega_y. The difference

B – B1
with
B=curl A and
B1 = omega x A

shows that the X and Y component are equal. Obviously the problem is to find a suitable omega_y so that also the Z components are equal too. Then A would be antisymmetric. Otherwise we have to change A.

Horst

Am 24.08.2017 um 13:23 schrieb EMyrone:

This note evaluates the magnetic dipole potential (7) in Cartesian coordinates at a point far from a current loop. The spin connection vector can be found from Eq (7) and Eqs. (9) to (11). Finally the ECE2 magnetic flux density B is found from Eq. (13). It is seen immediately from Eqs. (14) to (16) that the standard model violates antisymmetry, a disaster for standard (Maxwell Heaviside) electrodynamics first pointed out in UFT131 ff. Conservation of antisymmetry requires the existence of a spin connection and the existence of torsion, a clear and important result. It would be very interesting to graph the components of A and the spin connection vector omega, and the components of B. This new ECE2 level magnetic flux density must be compared with experimental data. The vector potential is an example solution of the ECE2 Ampere law, curl B = mu0 J.

386(3).pdf

The equivalent of 132,708 printed pages was downloaded (483.852 megabytes) from 2,666 downloaded memory files and 547 distinct visits each averaging 3.8 memory pages and 7 minutes, printed pages to hits ratio of 49.78, top referrals total 2,289,327, main spiders Baidu, Google, MSN and Yahoo. Collected ECE2 3298, Top ten 2107, Collected Evans Morris 759, Autobiography volumes one and two 684, Barddoniaeth (Collected Poetry) 618, Collected Scientometrics 518, F3(Sp) 403, CV 329, Collected Eckardt / Lindstrom 202, Principles of ECE 189, Evans Equations 112, Collected Proofs 98,Engineering Model 85, UFT88 80, PLENR 65, PECE2 65, MJE 53, UFT311 46, PECE 44, UFT321 40, ADD 38, Llais 37, SCI 36, UFT313 35, UFT314 52, UFT315 61, UFT316 28, UFT317 52, UFT318 25, UFT319 52, UFT320 43, UFT322 52, UFT323 33, UFT324 53, UFT325 43, UFT326 29, UFT327 22, UFT328 44, UFT329 37, UFT330 26, UFT331 43, UFT332 43, UFT333 29, UFT334 34, UFT335 47, UFT336 32, UFT337 19, UFT338 31, UFT339 26, UFT340, UFT341 46, UFT342 30, UFT343 50, UFT344 42, UFT345 41, UFT346 44, UFT347 62, UFT348 62, UFT349 34, UFT351 47, UFT352 65, UFT353 55, UFT354 76, UFT355 60, UFT356 51, UFT357 52, UFT358 58, UFT359 52, UFT360 49, UFT361 44, UFT362 48, UFT363 72, UFT364 59, UFT365 55, UFT366 65, UFT367 48, UFT368 65, UFT369 51, UFT370 58, UFT371 55, UFT372 52, UFT373 58, UFT374 64, UFT375 39, UFT376 36, UFT377 46, UFT378 57, UFT379 44, UFT380 55, UFT381 40, UFT382 71, UFT383 63, UFT384 43, UFT385 13 to date in August 2017. Institute for Theoretical Physics University of Goettingen ECE Article; City University of Hong Kong general; University of Sriwijaya Indonesia general; Welsh Networking Rebuttal Documents; University of Leeds UFT85. Intense interest all sectors, updated usage file attached for August 2017.

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It will be interesting to see whether omega x A = 0 and del dot (omega x A) = 0. So in the next note I will refine the method.

This note evaluates the magnetic dipole potential (7) in Cartesian coordinates at a point far from a current loop. The spin connection vector can be found from Eq (7) and Eqs. (9) to (11). Finally the ECE2 magnetic flux density B is found from Eq. (13). It is seen immediately from Eqs. (14) to (16) that the standard model violates antisymmetry, a disaster for standard (Maxwell Heaviside) electrodynamics first pointed out in UFT131 ff. Conservation of antisymmetry requires the existence of a spin connection and the existence of torsion, a clear and important result. It would be very interesting to graph the components of A and the spin connection vector omega, and the components of B. This new ECE2 level magnetic flux density must be compared with experimental data. The vector potential is an example solution of the ECE2 Ampere law, curl B = mu0 J.

a386thpapernotes3.pdf

OK many thanks! The easiest thing to do is to chose a simple potential and magnetic field strength. The method is generally applicable.

To: EMyrone@aol.com
Sent: 23/08/2017 14:19:44 GMT Daylight Time
Subj: Re: Solved Eq. (24) of Note 386(2)

I think that the example (5.39/5.40) of Jackson is exactly what I have used.
The far field approximation (5.41) can be used, this is a dipole field again. The corresponding potential is:

Horst

Am 23.08.2017 um 13:44 schrieb EMyrone:

This can be solved by hand, to give omega sub X, omega sub Y and omega sub Z given A and B from Jackson’s chapter five for example. Any relatively simple A and B can be chosen from that chapter. Then the solution conserves antisymmetry, which is the main purpose of this work. I will give a preliminary hand solution tomorrow which can be checked by computer algebra. For example Eqs. (5.39) and (5.40) of Jackson can be used and the spin connections evaluated. I will write this up tomorrow.

To: EMyrone
Sent: 23/08/2017 12:35:13 GMT Daylight Time
Subj: Re: Computation of 386(2)

Thanks, I woud indeed prefer a simpler example so that the partial derivatives do not become unhandable.
Horst

Am 23.08.2017 um 11:48 schrieb EMyrone:

Perfect! Eqs (12) to (14) are the results needed. These equations can be used in the antisymmetry laws to obtain the spin connections. This is the main purpose. If the equations get too complicated for Maxima we can look for a simpler approximation from Jackson’s chapter five. The dipole approximation far from the current loop has already been used. I will now develop Note 386(2) further.

To: EMyrone
Sent: 23/08/2017 09:47:04 GMT Daylight Time
Subj: Re: Discussion on Computation of 386(2)

I did some transformations. According to an earlier work, The transformation matrix of unit vectors from spherical coordinates to cartesian coordinates is given by a matrix S (o1 in the protocol) which applied to a vector A_sph written in spherical components gives

A_cart = S * A_sph (1*)

for the vector in cartesian components (o3). If A only has a phi-component, this simplifies to o5. The spherical coordinates appearing in A_cart have still to be transformed to cartesians via o6-o8. For the phi-component of the vector potential then o10 results, and the product (1*) leads to the expression o11 which is the final result, having only an X and Y component as expected. The partial derivatives, however, are becoming extremely complicated, see o12-o14 (exceed the sheet size).

Horst

Am 23.08.2017 um 10:02 schrieb EMyrone:

It may be easier to transform Eq. (8.352) in to Cartesian coordinates, or use other expressions for the magnetic potential. I will have a look at this problem shortly and develop the simple solution suggested in Note 386(2).

To: EMyrone
Sent: 22/08/2017 15:26:50 GMT Daylight Time
Subj: Re: Computation of 386(2)

I found in
http://mathworld.wolfram.com/SphericalCoordinates.html
expressions on expressing cartesian partial derivatives by spherical ones (Eqs. 98-100). It is however not clear to me if these are to be applied to the cartesian or spherical components of A. Maybe that the field components of A have to be transformed to spherical ones first (how?).

Horst

Am 22.08.2017 um 12:33 schrieb EMyrone:

This is excellent progress. It would be very useful to transform the antisymmetry equations (14) to (16) to spherical polar coordinates using Maxima to eliminate human error. Alternatively, Eq. (8.352) can be transformed into Cartesian coordinates and used in Eqs. (14) to (16) to find the spin connection components. Then B1 can be calculated, and B – B1 calculated. The procedure in Eqs. (21) to (24) of Note 386(2) can also be used. I will develop Eq. (4) of this note in Note 386(3). This would mean that the textbook A and B could be used to find the spin connection.

EMyrone
Sent: 22/08/2017 09:53:32 GMT Daylight Time
Subj: Re: 386(2): Simple Conservation of Antisymmetry in Magnetostatics

I made several tests. For the static magnetic field (eqs. 18-20) I verified that omega fulfills the antisymmetry conditions and

B = curl(A) – omega x A = 2 B(0) bold k

as requested. The direct solutions of the antisymm. eqs. cannot be used since the Z component of A is zero and the components appear in the denominator, but inserting (18) into the antisymm. eqs. gives

so this is the right solution.

Then I investigated the standard potential alpha given by Eq. (8.352) of PECE2. Its curl gives r and theta components.
Proceeding as in note 386(1) here gives only partial success:

curl_s(A)-cross(omega,A)=curl_s(alpha) (*1)

gives 3 equations (%o36-%o38) which have to be solved together with the antisymm. eqs. (14-17) rewritten to spherical coordinates. Solving (*1) for all omega’s is not possible but restricting to the first two equations and setting omega_phi=0 gives a solution, see %o40. This procedure is, however, quite unhandy and not exact.

To perform the procedure you proposed in the discussion of note 386(1):
1) Start with A from Eq. (8.352).
2) Calculate omega from Eqs. (5) to (7).
3) Calculate B1 = omega x A.
4) Use the experimentally observed B, and calculate B – B1.
5) If this is not zero, find A such that B1 = omega x A = 0.

we need the antisymm. eqs. in spherical coordinates. I guess that these are formally identical to those of cartesian coordinates because antisymmetry is generally covariant, or am I overlooking something?

Horst

Am 21.08.2017 um 13:27 schrieb EMyrone:

This note gives a simple solution (21) – (24). In general, Eqs. (11) and (14) to (16) must be solved simultaneously by computer. So antisymmetry is conserved in general in magnetostatics, Q. E. D. Eq. (24) is the magnetic equivalent of E = – omega sub 0 A. Given any potential vector, the spin connection vector must always be worked out by antisymmetry.

The equivalent of 176,191 printed pages was downloaded (642.392 megabytes) from 2,446 downloaded memory files (hits) and 550 distinct visits each averaging 3.4 printed pages and 8 minutes, printed pages to hits ratio of 72.03, top referals total 2,288,843, main spiders Baidu, Google, MSN and Yahoo. Collected ECE2 3265, Top ten 2057, Collected Evans / Morris 726(est), Autobiography volumes one and two 678, Barddoniaeth (Collected poetry) 616, Collected Scientometrics 503, F3(Sp) 369, CV 328, Collected Eckardt / Lindstrom 195, Principles of ECE 185, Evans Equations 109, Collected Proofs 96, Engineering Model 83, UFT88 77, PECE2 64, MJE 52, CEFE 50, UFT311 46, UFT321 40, Llais 36, ADD 35, UFT313 35, UFT314 49, UFT315 60, UFT316 28, UFT317 50, UFT318 25, UFT319 50, UFT320 40, UFT322 42, UFT323 31, UFT324 53, UFT325 43, UFT326 29, UFT327 22, UFT328 44, UFT329 39, UFT330 26, UFT331 41, UFT332 41, UFT333 29, UFT334 34, UFT335 47, UFT336 30, UFT337 19, UFT338 30, UFT339 26, UFT340 39, UFT341 45, UFT342 30, UFT343 47, UFT344 42, UFT345 41, UFT346 44, UFT347 61, UFT348 61, UFT349 34, UFT351 45, UFT352 63, UFT353 53, UFT354 74, UFT355 57, UFT356 50, UFT357 51, UFT358 55, UFT359 50, UFT360 47, UFT361 44, UFT362 45, UFT363 72, UFT364 56, UFT365 52, UFT366 64, UFT367 46, UFT368 65, UFT369 50, UFT370 58, UFT371 55, UFT372 51, UFT373 58, UFT374 61, UFT375 37, UFT376 34, UFT377 46, UFT378 51, UFT379 43, UFT380 55, UFT381 38, UFT382 66, UFT383 62, UFT384 43, UFT385 11 to date in August 2017. Emmanuel College Melbourne general; University of Guadalajara Mexico F3(Sp); Wayback Machine spidering (www.archive.org). Intense interest all sectors, updated usage file attached for August 2017.

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Additionally, a 401 Unauthorized error was encountered while trying to use an ErrorDocument to handle the request.

This can be solved by hand, to give omega sub X, omega sub Y and omega sub Z given A and B from Jackson’s chapter five for example. Any relatively simple A and B can be chosen from that chapter. Then the solution conserves antisymmetry, which is the main purpose of this work. I will give a preliminary hand solution tomorrow which can be checked by computer algebra. For example Eqs. (5.39) and (5.40) of Jackson can be used and the spin connections evaluated. I will write this up tomorrow.

To: EMyrone@aol.com
Sent: 23/08/2017 12:35:13 GMT Daylight Time
Subj: Re: Computation of 386(2)

Thanks, I woud indeed prefer a simpler example so that the partial derivatives do not become unhandable.
Horst

Am 23.08.2017 um 11:48 schrieb EMyrone:

Perfect! Eqs (12) to (14) are the results needed. These equations can be used in the antisymmetry laws to obtain the spin connections. This is the main purpose. If the equations get too complicated for Maxima we can look for a simpler approximation from Jackson’s chapter five. The dipole approximation far from the current loop has already been used. I will now develop Note 386(2) further.

To: EMyrone
Sent: 23/08/2017 09:47:04 GMT Daylight Time
Subj: Re: Discussion on Computation of 386(2)

I did some transformations. According to an earlier work, The transformation matrix of unit vectors from spherical coordinates to cartesian coordinates is given by a matrix S (o1 in the protocol) which applied to a vector A_sph written in spherical components gives

A_cart = S * A_sph (1*)

for the vector in cartesian components (o3). If A only has a phi-component, this simplifies to o5. The spherical coordinates appearing in A_cart have still to be transformed to cartesians via o6-o8. For the phi-component of the vector potential then o10 results, and the product (1*) leads to the expression o11 which is the final result, having only an X and Y component as expected. The partial derivatives, however, are becoming extremely complicated, see o12-o14 (exceed the sheet size).

Horst

Am 23.08.2017 um 10:02 schrieb EMyrone:

It may be easier to transform Eq. (8.352) in to Cartesian coordinates, or use other expressions for the magnetic potential. I will have a look at this problem shortly and develop the simple solution suggested in Note 386(2).

To: EMyrone
Sent: 22/08/2017 15:26:50 GMT Daylight Time
Subj: Re: Computation of 386(2)

I found in
http://mathworld.wolfram.com/SphericalCoordinates.html
expressions on expressing cartesian partial derivatives by spherical ones (Eqs. 98-100). It is however not clear to me if these are to be applied to the cartesian or spherical components of A. Maybe that the field components of A have to be transformed to spherical ones first (how?).

Horst

Am 22.08.2017 um 12:33 schrieb EMyrone:

This is excellent progress. It would be very useful to transform the antisymmetry equations (14) to (16) to spherical polar coordinates using Maxima to eliminate human error. Alternatively, Eq. (8.352) can be transformed into Cartesian coordinates and used in Eqs. (14) to (16) to find the spin connection components. Then B1 can be calculated, and B – B1 calculated. The procedure in Eqs. (21) to (24) of Note 386(2) can also be used. I will develop Eq. (4) of this note in Note 386(3). This would mean that the textbook A and B could be used to find the spin connection.

EMyrone
Sent: 22/08/2017 09:53:32 GMT Daylight Time
Subj: Re: 386(2): Simple Conservation of Antisymmetry in Magnetostatics

I made several tests. For the static magnetic field (eqs. 18-20) I verified that omega fulfills the antisymmetry conditions and

B = curl(A) – omega x A = 2 B(0) bold k

as requested. The direct solutions of the antisymm. eqs. cannot be used since the Z component of A is zero and the components appear in the denominator, but inserting (18) into the antisymm. eqs. gives

so this is the right solution.

Then I investigated the standard potential alpha given by Eq. (8.352) of PECE2. Its curl gives r and theta components.
Proceeding as in note 386(1) here gives only partial success:

curl_s(A)-cross(omega,A)=curl_s(alpha) (*1)

gives 3 equations (%o36-%o38) which have to be solved together with the antisymm. eqs. (14-17) rewritten to spherical coordinates. Solving (*1) for all omega’s is not possible but restricting to the first two equations and setting omega_phi=0 gives a solution, see %o40. This procedure is, however, quite unhandy and not exact.

To perform the procedure you proposed in the discussion of note 386(1):
1) Start with A from Eq. (8.352).
2) Calculate omega from Eqs. (5) to (7).
3) Calculate B1 = omega x A.
4) Use the experimentally observed B, and calculate B – B1.
5) If this is not zero, find A such that B1 = omega x A = 0.

we need the antisymm. eqs. in spherical coordinates. I guess that these are formally identical to those of cartesian coordinates because antisymmetry is generally covariant, or am I overlooking something?

Horst

Am 21.08.2017 um 13:27 schrieb EMyrone:

This note gives a simple solution (21) – (24). In general, Eqs. (11) and (14) to (16) must be solved simultaneously by computer. So antisymmetry is conserved in general in magnetostatics, Q. E. D. Eq. (24) is the magnetic equivalent of E = – omega sub 0 A. Given any potential vector, the spin connection vector must always be worked out by antisymmetry.