Archive for August, 2017

Solved Eq. (24) of Note 386(2)

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.

Final version of PECE2 book

Agreed with Doug, the typesetting is highly accurate and it is a fine production. UFT354 is currently the most read paper of the ECE2 series, on the effect of torsion on the relation between the metric and Christoffel symbol, and proving that the torsion tensor must be totally antisymmetric. This was the essence of Doug’s contribution. It is very important. Torsion completely refutes Einsteinian general relativity.

In a message dated 22/08/2017 19:18:25 GMT Daylight Time, writes:

I see no errors in what I submitted. The book looks great.
Doug

On Sun, Aug 20, 2017 at 1:43 PM, Horst Eckardt <mail> wrote:

Dear authors,

I finished editing of the book. Please check that you

chapters are completely valid and let me know if something needs to be changed.

Best regards,
Horst

Computation of 386(2)

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@aol.com
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.

transform-A.pdf

Discussion on Computation of 386(2)

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@aol.com
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.

Daily Report 21/7/17

The equivalent of 152,598 printed pages was downloaded (556.371 gigabytes) from 2,333 downloaded memory files (hits) and 550 distinct visits each averaging 3.7 memory pages and 7 minutes, printed pages to hits ratio of 65.41, top referrals total of 2,288,617, main spiders Baidu, Google, MSN and Yahoo. Collected ECE2 3045, Top ten 1969, Collected Evans Morris 693, Autobiography volumes one and two 671, Barddoniaeth (Collected Poetry) 612, CV 328, F3(Sp) 304, Principles of ECE 180, Collected Eckardt / Lindstrom 180, Evans Equations 107, Collected Proofs 96, Engineeringf Model 80, UFT88 71 PLENR 64, PECE2 61, MJE 52, UFT311 45, PECE 42, UFT321 39, Llais 36, SCI 34, ADD 34, UFT313 33, UFT314 46, UFT315 58, UFT316 28, UFT317 49, UFT318 22, UFT319 50, UFT320 40, UFT322 40, UFT323 31, UFT324 47, UFT325 42, UFT326 28, UFT327 21, UFT328 44, UFT329 36, UFT330 25, UFT331 40, UFT332 40, UFT333 29, UFT334 33, UFT335 46, UFT336 29, UFT337 18, UFT338 30, UFT339 25, UFT340 34, UFT341 42, UFT342 29, UFT343 45, UFT344 41, UFT345 38, UFT346 43, UFT347 57, UFT348 56, UFT349 31, UFT351 43, UFT352 56, UFT353 51, UFT354 73, UFT355 57, UFT356 48, UFT357 45, UTF358 53, UFT359 48, UFT360 41, UFT361 42, UFT362 43, UFT363 69, UFT364 55, UFT365 50, UFT366 61, UFT367 46, UFT368 63, UFT369 48, UFT370 57, UFT371 52, UFT372 49, UFT373 54, UFT374 57, UFT375 37, UFT376 34, UFT377 43, UFT378 47, UFT379 41, UFT380 52, UFT381 36, UFT382 65, UFT383 59, UFT384 42, UFT385 6 to date in August 2017. Public Ministry of the State of Rio de Janeiro Brazil My Page; City University of Hong Kong general; Information Technology City University of Hong Kong UFT8; University of Guadalajara Mexico F3(Sp); Mathematics University of Oxford Essay 15. Intense interest all sectors, updated usage file attached for August 2017.

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Computation of 386(2)

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.

386(2).pdf

Daily Report Sunday 20/7/17

The equivalent of 123,769 printed pages was downloaded (451.260 megabytes) from 2,442 downloaded memory files (hits) and 539 distinct visits each averaging 2.8 printed pages and 6 minutes, printed pages to hits ratio of 50.68, top referrals total 2,288,461, main spiders Baidu, Google, MSN and Yahoo. Collected ECE2 2927, Top ten 1905, Autobiography volumes one and two 687, Collected Evans Morris 660(est), Barddoniaeth (Collected Poetry) 604, Collected scientometrics 448, CV 326, F3(Sp) 254, Evans Equations 135, Engineering Model 80, UFT88 64, PECE2 60, PLENR 59, MJE 52, CEFE 46, UFT311 44, PECE 40, UFT321 37, Llais 35, SCI 33, ADD 33, 83Ref 30, UFT313 33, UFT314 35, UFT315 52, UFT316 25, UFT317 46, UFT318 22, UFT319 48, UFT320 35, UFT322 36, UFT323 30, UFT324 46, UFT325 40, UFT326 28, UFT327 21, UFT328 40, UFT329 34, UFT330 25, UFT331 38, UFT332 38, UFT333 28, UFT334 33, UFT335 44, UFT336 28, UFT337 18, UFT338 28, UFT339 25, UFT340 33, UFT341 42, UFT342 27, UFT343 43, UFT344 39, UFT345 35, UFT346 39, UFT347 55, UFT348 55, UFT349 31, UFT351 43, UFT352 55, UFT353 49, UFT354 72, UFT355 56, UFT356 47, UFT357 43, UFT358 49, UFT359 45, UFT360 39, UFT361 42, UFT362 38, UFT363 65, UFT364 52, UFT365 46, UFT366 60, UFT367 42, UFT368 62, UFT369 46, UFT370 53, UFT371 52, UFT372 46, UFT373 54, UFT374 56, UFT375 31, UTF376 30, UFT377 41, UFT378 43, UFT379 39, UFT380 49, UFT381 30, UFT382 59, UFT383 57, UFT384 41 to date in August 2017. Australian National University UFT213; Federico Santa Maria Technical University at Vina el Mar, Chile F3(Sp), State University of New York at Stony Brook UFT41; City University of Hong Kong general; Kiev Englynion. Intense interest all sectors, updated usage file attached for August 2017.

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