Monday, May 27, 2013

Charge, magnetic moment, anapole moment and mass

As observed in the previous posts, it is very easy to solve the EPR paradox. Just two non-superimposable images (chirality) are required. This occurs because the flat electron intersection is in two places at the same time and the internal magnetic field got inverted after performing its mirror image. These two separated circles have to be at present time. This is why they are perpendicular to the direction of movement. If one is ahead of the other, that one is in the future, the other in the past and no wave is produced. A consequence of this is that the magnetic moment of the flat electron will have just two orientations, not any! Would this be enough for the mystery of the "SPIN"!?

The problem I have now is how  to put a magnetic moment on a toroidal current? I found that a toroidal current has a charge moment called an anapole moment. Just at the center of the torus, where I need it. The units of this moment are coulombs per square meter. I think that the "flat electron magnetic moment" is the "anapole moment" consuming an intersection time, some sort of intermittent moment.

If this moment is not in the plane, it will not consume the intersection time and it will be an anapole moment (Concentric circles in Figure 1). But, if this moment is in the plane, it will consume the intersection time and it will be a magnetic moment (two separated circles in Figure 1).

Figure 1  The anapole moment T occurs at the intersection of two concentric circles (aiming into the future in this case). When it is in the plane, it is a magnetic moment.

The intersection time will be the torus arc length divided by the speed of light c. This multiplied by the toroidal current i_{theta} is the electron charge e. By this way, the toroidal current is obtained,

Therefore, the major radius of the torus r_{phi} is half the Compton wavelength and the flat electron cross section would have the extension of the Compton wavelength. Theta is equal to pi and is distributed in the four circles observed in the post "Matter wave", i.e. 45ยบ circle section areas.

The energy contained in the toroid is:

Thus, the mass energy of the electron is the energy contained in the toroid's internal magnetic field.

Saturday, May 4, 2013

Correlation Measurements in Flatland (EPR paradox)

I will follow the instructions very similar to what you can see here:(
We imagine a radioactive substance that emits a pair of flat electrons in each decay. These two electrons go in opposite directions, and are emitted nearly simultaneously. So we can have a sample emitting these pairs of flat electrons. Figure 1 shows such a sample and flat electron filters measuring the spin of each member of the pair:

Figure 1 Entangled flat electrons traveling through filters in opposite directions.

Continuing, for the radioactive substance we will be considering in Figure 1, one-half of the flat electrons incident on the up hand filter emerge and one-half do not. Similarly, one-half of the flat electrons incident on the down hand filter emerge and one-half do not.

But if we look at the correlation between these flat electrons in Figure 2, we find that if the up hand electron does pass through the filter, then its down hand companion does not pass its filter. The handedness of the coupled flat electrons at the center determined this correlation.

Figure 2 Example of a 0% correlation.

We say that each radioactive decay has a total spin of zero: if one electron is spin right its companion is spin left. Now, the case where the two filters have opposite emerged spin.

Figure 3 Example of a 100 % correlation

This time if a particular left hand flat electron passes its filter down Figure 3, then its companion right hand flat electron always passes its filter. It is obvious that the handedness of the flat electrons produced this correlation! 

Finally, if the two filters defined the same emerged flat electron are one perpendicular to the other. 

Figure 4. Correlation experiment with different magnetic field orientation.

One-half of the down hand flat electrons emerge from their filter. One-half of the up hand flat electrons emerge from their filter. If a particular down hand electron passes its filter, one-half of the time its companion up hand flat electron will emerge from its filter, one-half of the time it will not. Because this particular orientation produced the donut intersection and a 50:50 percent probability to have either structure again (see Figure 4).

Thus, Einstein was right! "Imagine that your friend had a pair of gloves and two boxes. He put one glove in each box, and then separated the boxes. Now imagine that you didn’t know which glove was in which box, and you were asked to open one of the boxes. Simple logic would tell you that there was a 50-percent chance of getting the right-handed glove and a 50-percent chance of getting the left-handed glove. And say you opened the box and saw the right-handed glove. You would automatically know which glove was in the other box, but there wouldn’t be any occult connection between the two gloves. Rather, each glove always had its handedness before the observation. "

The orientations of the filters in Figure 1, 2 and 3 were arbitrarily but consistently set to get the Space-Land experimental results.

Playing with Spin Filters

I will follow the instructions very similar to what you can see here:( For simplicity, I would not collimate the particle after it has been separated. Thus, the filtered electron will continue in a straight line down the filter. The flat electrons movement in a magnetic field just obeys the spin separation and do not move as negative particles in a magnetic field.

Figure 1 Spin-right flat electron filter

Figure 1 shows a Stern-Gerlach apparatus, in which a block of lead stops the "spin left" flat electrons. One-half of the incident beam, the "spin left" electrons will be stopped inside the filter, while all the "spin-right" flat electrons will emerge in the same direction before they entered the magnetic field. Thus, this is a "filter" that selects "spin-right" flat electrons.

Figure 2. A second "spin-right" electron filter will not affect the selected spin.

On Figure. 2, We now put a second filter after the first with the same orientation. The second filter has no effect. Half of the electrons from the electron gun emerge from the first box, and all of those electrons pass through the second filter. So, once "right" is defined by the first filter, it is the same as the "right" defined by the second.

Figure 3. A Spin right filter follow by a spin left filter will block all the electrons.

On Figure 3, Now we put the second filter after the first and a block to the right relative to the first. As always, half of the beam of electrons from the electron gun emerge from the first filter, and none of those electrons emerge from the second filter. So, evidently once the first filter defines "right" that definition is the second filter's definition of "left".

Fig. 4 Flatland version of the Space-Land Stern-Gerlach experiment.

Here is another orientation for the second filter, this time it is oriented at 90° relative to the first one. To repeat once again, half of the beam of electrons from the electron gun emerge from the first filter. It turns out that one-half of those electrons pass through the second filter. So if we have two definitions of "right" from two filters at right angles to each other, one half of the electrons will satisfy both definitions. This is because the intersection of the flat electron will change to the intersection of the donut through it equator (see Particle-wave duality post). This will occur with a 50:50 percent probability. Thus, from the 16 flat electrons that went into this second filter, 8 are retained and the other 8 passed through. These 8 flat electrons that passed through have  again 50:50 percent probability to produce the original orientations again. As a result a left filter that should have blocked all the remained electrons can block just half of them. 

This is in perfect agreement with Quantum Mechanic predictions.