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Solution for Spin Connection

I intend to work on this today, fixing the sign change error of the last note.

To: EMyrone@aol.com
Sent: 27/06/2017 18:36:37 GMT Daylight Time
Subj: Re: Solution for Spin Connection

This sounds interesting, then we have to compute bold B from the standing wave to see how this situation can be realized.

Horst

Am 27.06.2017 um 14:13 schrieb EMyrone:

I have solved Eqs. (16) to (23) by hand to find a standing wave solution for the Q vector. The spin connection in this case is the simple omega bold = kappa and omega sub zero = omega, so the spin connection is the four wave vector. The general solution can be found by computer and this does not look to be a difficult problem. This is well worth doing because the method gives the spin connection four vector and the kappa vector in a perfectly general way.

Checking Note 380(4)

Right, agreed, this will be fixed in the final paper as usual. Our system of checking, graphics and computer algebra is also first class, and the entire theory is based on irrefutable Cartan geometry. Our papers have been refereed and checked hundreds of thousands of times by a vast and permanent readership coming from the best universities, institutes, corporations, military facilities and similar in the world. Our scientometrics are uniquely accurate and detailed.

To: EMyrone@aol.com
Sent: 27/06/2017 18:28:33 GMT Daylight Time
Subj: Re: Discussion of 380(4)

When inserting eqs. (10,11) into (9), I obtain the equation

curl (- omega sub 0 Q) – partial / partial t (omega x Q) = 0

which gives a sign change when moving the second term to the RHS.

Horst

Am 27.06.2017 um 10:16 schrieb EMyrone:

I think that Eq. (13) is OK, because it is derived from

curl (- omega sub 0 Q) = – partial / partial t (omega x Q)

To: EMyrone
Sent: 26/06/2017 15:44:41 GMT Daylight Time
Subj: Re Re: Discussion of 380(4)

Shouldn’t there be a minus sign in eq.(13)?

Horst

Am 26.06.2017 um 16:20 schrieb Horst Eckardt:

It seems that in eqs.(16-18) of the note the dime derivatives at the LHS are missing.

Horst

Am 26.06.2017 um 09:56 schrieb EMyrone:

Agreed entirely. Eqs. (16) to (23) make up the set of simultaneous spatial equations in three dimensions, and can be solved for the Q three vector and spin connection four vector in general, for any situation in gravitation and electrodynamics, and combination therefore. In electrodynamics they can be used to solve for the A three-vector and the spin connection four-vector in general. I gradually worked towards this new general solution in the four notes. This is equivalent to a general solution of the EEC2 field equations in three dimensions. So we can now address any problem in physics, chemistry and engineering. Guesswork is no longer needed for the spin connection. In this set of equations there are only space variables, and I think that the computer can deal with the problem of solving a set of seven, fairly simple, simultaneous partial differential equations in seven unknowns, the three scalar components of Q or A and the four components of the spin connection. The general solution can be simplified, and any coordinates can be used, not only the Cartesain. However the Cartesain is the clearest as you inferred some time ago. The Eckardt / Lindstrom methods can be added to this new method. The solutions will almost certainly produce some very interesting spatial graphics. Having found omega sub X and onega sub Y from this genreal method, one can go back to teh orbital equations to see qhat kind of orbits emeerge, and one can addres the Biefeld Brown effect in a self consistent way. Finally, Q or A is also time dependent: Q = Q (t, X, Y, Z); A = A(t, X, Y, Z). An example is plane waves. The scalar potential is also t and space dependent in general.

To: EMyrone
Sent: 25/06/2017 15:55:21 GMT Daylight Time
Subj: Re: 380(4): Compleet Solution in Three dimensions.

It is not se easy to find a set of equations which can give well defined field solutions. For example there must be time and space derivatives for each variable to give unique solutions. The Lagrangian seems mainly to be used to obtain time trajectories of orbits but not for general field solutions. In so far I am not sure if it makes sense to use the LHS of eq.(3) for determining distributed fields Phi and bold omega.
Eqs.(16-23) is a more general scheme which seems to fulfill the above condition. For a solution, the boundary conditions are essential, because there are no terms of inhomogeneity like a charge density.

Horst

Am 25.06.2017 um 10:25 schrieb EMyrone:

This note derives a completely general set of seven simultaneous differential equations, (16) – (18), (19) – (21) and (23) for seven unknowns, the three Cartesian components of the Q three-vector and the four components of the spin connection four-vector. These can all be expressed as functions of space and time. This is an exactly determined problem in three dimensions. The method uses the two homogeneous field equations of ECE2 gravitation, Eq. (22) and the Faraday law of induction Eq. (9), and the antisymmetry condition (19) to (21). In two dimensions X and Y, there is only one antisymmetry condition (27) and the Faraday law reduces to Eq. (28). Using the Coulomb law of ECE2 gravitation gives Eq. (36). So in the planar limit thee are three equations in five unknowns. The Newtonian limit of Eqs. (30) and (31) is used to give five equations in five unknowns. In the next note the Ampere Maxwell Law of ECE2 gravitation will be introduced into the planar analysis, to seek a general solution without having to assume the Newtonian approximation.

Daily Report 26/6/17

The equivalent of 71,685 printed pages was downloaded (261.363 megabytes) from 1,831 memory files downloaded (hits) and 375 distinct visits each averaging 4.1 printed pages and 7 minutes, printed pages to hits ratio of 39.15, top referrals total of 2,254,288, main spiders Google, MSN and Yahoo. Top ten 2290, Collected ECE2 2160, Collected Evans Morris 858, Collected scientometrics 580, Barddoniaeth 468, Autobiography volumes one and two 459, Evans Equations 361, Principles of ECE 228, F3(Sp) 225, CEFE 169, UFT88 93, PECE 81, Engineering Model 66, ECE2 63, CV 61, 83Ref 43, MJE 42, Llais 37, UFT311 37, UFT321 21, UFT313 20, UFT314 40, UFT315 39, UFT316 29, UFT317 29, UFT318 21, UFT319 37, UFT320 23, UFT322 22, UFT323 30, UFT324 30, UFT325 39, UFT326 16, UFT327 17, UFT328 44, UFT329 32, UFT330 17, UFT331 42, UFT332 25, UFT333 13, UFT334 26, UFT335 31, UFT336 32, UFT337 16, UFT338 21, UFT339 13, UFT340 28, UFT341 33, UFT342 29, UFT343 35, UFT344 44, UFT345 42, UFT346 36, UFT347 45, UFT348 35, UFT349 42, UTF351 47, UFT352 74, UFT353 39, UFT354 45, UFT355 38, UFT356 51, UFT357 31, UFT358 63, UFT359 41, UFT360 31, UFT361 26, UFT362 39, UFT363 46, UFT364 52, UFT365 25, UFT366 63, UFT367 26, UFT368 46, UFT369 25, UFT370 23, UFT371 23, UFT372 40, UFT373 24, UFT374 33, UFT375 22, UFT376 15, UFT377 31, UFT378 36, UFT379 16 to date in June 2017. Tor Project Univeristy of Waterloo Canada general; Wolfram Company general; Johns Hopkins University Medicine UFT213; Association for the Union of Student Residences University of South Paris (Paris-Psud) UFT42. Intense interest all sectors, updated usage file attached for June 2017.

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Solution for Spin Connection

I have solved Eqs. (16) to (23) by hand to find a standing wave solution for the Q vector. The spin connection in this case is the simple omega bold = kappa and omega sub zero = omega, so the spin connection is the four wave vector. The general solution can be found by computer and this does not look to be a difficult problem. This is well worth doing because the method gives the spin connection four vector and the kappa vector in a perfectly general way.

Another suggestion for solving the antigravity problem

I agree that the spin connection vector omega bold is a universal geometrical property that is the same for electromagnetism and gravitation. So is omega sub 0. The rigorous answer is to solve Eqs. (16) to (23) of Note 280(4) for bold Q, bold omega, and omega sub 0, then find phi for gravitation from Eq. (10) of that note. Finally use Eq. (14) of Note 380(2) to find the electric charge density needed to cahnge the gravitational phi to any desired value. Alternatively find bold omega from solving Eqs. (16) to (23) simultaneously, then use a model charge density in Eq. (14) to find phi. These are only two out of many possibilities. The set of Eqs. (16) to (23) can be simplified by intelligent approximation. I will think about this next. The give the gravitational vector potential Q (t, X, Y, Z) in a completely general way.

To: EMyrone@aol.com
Sent: 26/06/2017 20:00:53 GMT Daylight Time
Subj: Re: Another suggestion for solving the antigravity problem

A simple solution could be looking as follows:
The gravitational acceleration is

g = – nabla Phi + bold omega * Phi

with spin connection omega and gravitational potential Phi. Since there is only one space geometry, there is only one and the same omega for gravitation and electromagnetism. If it is possible to enhance bold omega significantly by electromagnetism, this should have an impact on g. So one of the equations (the above one) is nearly trivial. The question is how to construct an additional bold omega by electromagnetism. My idea was by a rotating magnetic field. But how to compute this? We need a quantitative theory.

Horst

Am 26.06.2017 um 10:11 schrieb EMyrone:

Agreed with this, Note 380(4) can be appleid to this anti gravity problem in order to simulate the apparatus and optimize conditions for counter gravitation. Note 380(4) used the homogeneous field equations of ECE2 gravitation:

del cap omega = 0

curl g + partial cap omega / partial t = 0

and the antisymmetry laws from

cap omega = curl Q – omega x Q

Here cap omega is the gravitomagnetic field, g is the gravitational field, Q is the gravitational vector potential, and omega the space part of the spin connection four vector. The inhomogeneous laws of ECE2 gravitation were not used in Note 380(4), and it was shown that the above three equations are sufficient to completely determine Q and the spin connection four vector. Having found them, they can be used in the inhomogeneous laws, the ECE2 gravitational Coulomb law and Ampere Maxwell law. Exactly the same remarks apply to ECE2 electromagnetism, and combinations of electromagnetism and gravitation. This ought to produce efficient counter gravitational designs. We can describe any existing counter gravitational apparatus with these powerful ECE2 equations.

To: EMyrone
Sent: 25/06/2017 16:01:12 GMT Daylight Time
Subj: Another suggestion for solving the antigravity problem

There are rumours out that antigravity can be achieved by rotating
magnetic fields (like in a 3-phase motor). In this case the spin
connection is the vector of the rotation axis if I see this right. So we
have a predefined bold omega and can apply the Faraday and/or
Ampere-Maxwell law to find bold A and bold Q. Perhaps worth a thought. I
am not sure if the coupling from e-m to gravity can be applied in the
same way as before.

Horst

Discussion of 380(4)

I think that Eq. (13) is OK, because it is derived from

curl (- omega sub 0 Q) = – partial / partial t (omega x Q)

To: EMyrone@aol.com
Sent: 26/06/2017 15:44:41 GMT Daylight Time
Subj: Re Re: Discussion of 380(4)

Shouldn’t there be a minus sign in eq.(13)?

Horst

Am 26.06.2017 um 16:20 schrieb Horst Eckardt:

It seems that in eqs.(16-18) of the note the dime derivatives at the LHS are missing.

Horst

Am 26.06.2017 um 09:56 schrieb EMyrone:

Agreed entirely. Eqs. (16) to (23) make up the set of simultaneous spatial equations in three dimensions, and can be solved for the Q three vector and spin connection four vector in general, for any situation in gravitation and electrodynamics, and combination therefore. In electrodynamics they can be used to solve for the A three-vector and the spin connection four-vector in general. I gradually worked towards this new general solution in the four notes. This is equivalent to a general solution of the EEC2 field equations in three dimensions. So we can now address any problem in physics, chemistry and engineering. Guesswork is no longer needed for the spin connection. In this set of equations there are only space variables, and I think that the computer can deal with the problem of solving a set of seven, fairly simple, simultaneous partial differential equations in seven unknowns, the three scalar components of Q or A and the four components of the spin connection. The general solution can be simplified, and any coordinates can be used, not only the Cartesain. However the Cartesain is the clearest as you inferred some time ago. The Eckardt / Lindstrom methods can be added to this new method. The solutions will almost certainly produce some very interesting spatial graphics. Having found omega sub X and onega sub Y from this genreal method, one can go back to teh orbital equations to see qhat kind of orbits emeerge, and one can addres the Biefeld Brown effect in a self consistent way. Finally, Q or A is also time dependent: Q = Q (t, X, Y, Z); A = A(t, X, Y, Z). An example is plane waves. The scalar potential is also t and space dependent in general.

To: EMyrone
Sent: 25/06/2017 15:55:21 GMT Daylight Time
Subj: Re: 380(4): Compleet Solution in Three dimensions.

It is not se easy to find a set of equations which can give well defined field solutions. For example there must be time and space derivatives for each variable to give unique solutions. The Lagrangian seems mainly to be used to obtain time trajectories of orbits but not for general field solutions. In so far I am not sure if it makes sense to use the LHS of eq.(3) for determining distributed fields Phi and bold omega.
Eqs.(16-23) is a more general scheme which seems to fulfill the above condition. For a solution, the boundary conditions are essential, because there are no terms of inhomogeneity like a charge density.

Horst

Am 25.06.2017 um 10:25 schrieb EMyrone:

This note derives a completely general set of seven simultaneous differential equations, (16) – (18), (19) – (21) and (23) for seven unknowns, the three Cartesian components of the Q three-vector and the four components of the spin connection four-vector. These can all be expressed as functions of space and time. This is an exactly determined problem in three dimensions. The method uses the two homogeneous field equations of ECE2 gravitation, Eq. (22) and the Faraday law of induction Eq. (9), and the antisymmetry condition (19) to (21). In two dimensions X and Y, there is only one antisymmetry condition (27) and the Faraday law reduces to Eq. (28). Using the Coulomb law of ECE2 gravitation gives Eq. (36). So in the planar limit thee are three equations in five unknowns. The Newtonian limit of Eqs. (30) and (31) is used to give five equations in five unknowns. In the next note the Ampere Maxwell Law of ECE2 gravitation will be introduced into the planar analysis, to seek a general solution without having to assume the Newtonian approximation.