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 http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
