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.

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.

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).


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
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?).


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.

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?


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.

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