At 12:17 pm 29/12/2005 +1100, you wrote: >In reply to Jones Beene's message of Wed, 28 Dec 2005 12:19:31 >-0800: >Hi, >[snip] >>"During the quantum transition energy flows from state one to >>another. These states are associated with elastic discontinuities. >>The transitional quantum state is described by its velocity as >>measured with respect to an elastic discontinuity. The velocity of >>the quantum transition is a property of its frequency and >>displacement. The frequency is the Compton frequency Fc. The >>displacement is equal to the extent of the elastic displacement. >>This extent equals the classical radius of the electron rp. For >>centric systems the quantum transition expresses itself through >>its circumferential velocity. A factor of 2 p was incorporated to >>obtain circumferential velocity of the transitional state. The >>velocity of the quantum transition was derived, below, from this >>understanding. >> >>Velocity = 2 p Fc l meters/second >> >>Velocity = ( 2 p ) [Mc2/h] ( 1.409 x 10-15 ) meters/second > >First, if l is the classical electron radius, and M the mass of >the electron then the velocity works out to 2.188E6 m/s (also >assuming p stands for Pi). > >Now let us suppose that the electron is a toroid, which rotates >about both axes. Let us further suppose that the Compton frequency >of the electron is the higher of the two rotation rates, i.e. >about the minor axis. Suppose further that the minor radius is the >classical electron radius. > >Then the equation above is the surface rotation velocity about the >minor axis. > >My reason for choosing a toroid is that it is the simplest closed >form that has two states (charge?) when rotating about both axes. > >My guess is that the major radius is the Bohr radius in the ground >state hydrogen atom, so that the ratio of the minor radius to the >major radius is the square of the fine structure constant. > > >Regards, > >Robin van Spaandonk
I like it. 8-) Frank Grimer
