Discussion of Eq. (2) of Note 356(5)

OK many thanks again, I think that all is OK because the velocity is equivalent to a vector potential so we do not expect a Culomb potential.

To: EMyrone@aol.com
Sent: 29/08/2016 17:11:18 GMT Daylight Time
Subj: Re: Eq. (2) of Note 356(5)

The eq. (2) can be solved. It is the same eq. as I sent over for note 356(4). However, this equation only follows for

v_r = v_r(r)
v_theta = 0
v_phi = 0

If more dependencies are allowed and all components are different from zero, there are more complicated equations as can be seen from vector equation o10 in the protocol. If all components of v have only an r dependence, the result is o12, not much simpler. Only with zero components v_theta and v_phi the equ. (2) of the note follows, see o15. There is an analytic solution o19/o20, but it is not a pure Coulomb potential, as already discussed with note 356(4). We have to interpret this result in any way, or the operator (v*grad) has to be calculated in any other way. However it seems not to be used in VAPS execpt for cartesian coordinates.


Am 29.08.2016 um 09:45 schrieb EMyrone:

Can this equation be solved with a partial differential equation package? It looks like an interesting non linear differential equation to which there may be an analytical solution. In order to eliminate the complexities of the spherical polar coordinate system I can set up the problem in the Cartesian system. In any case we have already proven the method of UFT356 and there is a great deal of interest in the latest UFT papers as you can see from this morning’s report. It would be interesting to model boundary conditions on an actual circuit such as that of UFT311, which is the ultimate aim of the research.


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