Hi Richard
I am a guest here and cannot start a new thread. Maybe you or Fred should start a new thread on 'Bell and retrocausality'? As I wrote before, I know that I do not qualify for your prize money as my method does not defeat Bell's Inequalities but simply bypasses them via retrocausality using your conspiracy loophole. But my method does reproduce the -cos theta correlation.
Richard wrote:
Don’t try to fathom anything.
I am trying to close down my physics work as I am feeling too old now and I have the feeling that what you are suggesting will involve me in a lot of computer programming work. I am prepared to do the work if it is meaningful. I don't know if it is meaningful unless I fathom why you are asking. I do not want to input a lot of effort if at the end you again simply say that you do not like the method. Shut-up-and-calculate is for working according to QM. Shut-up-and-calculate is not for following the rules of Richard Gill [no offense intended though :)]
I have already in the appendix of my June 2020 vixra paper
https://vixra.org/abs/2101.0179 given my computer code for using my model to give Malus Law results in a particle-at-a-time simulation. In the main body of that paper I describe my classical model of the electron and the photon and show how that model can be used to give - cos theta for the Bell correlation as long as the antiparticles are travelling backwards in time.
I completely accept that I cannot get -0.707 for a forwards in time Bell correlation for the detector settings of 0 deg for Alice and 45 deg for Bob. I have never bothered with simulating a CHSH design set up as I believe it is a waste of time in a simulation. I have normally used a single setting for each detector of Alice and Bob.
However it is easy for me to describe in words what you are asking, despite me believing it is completely an over-the-top exercise. If I cannot get a result for a simulation using 0 and 45 deg then why further complicate matters.
Anyway:
Richard wrote:
Measure electron #1 in up (+z) direction. Outcome is either + or - in +z direction.
Then re-measure it in x direction. Outcome is either + or - in x direction.
Use this outcome to choose between Alice's 0 or pi/2 setting for measuring electron #2.
The first step is the most complicated as the incoming electron#1 has to be assumed to be unpolarised as I have no history for it of previous measurements.
Generate a random number between 0 and 1 and multiply it by 2pi to give 'angle1'.
That gives a simulated polarisation angle in the xz plane. 'up' corresponds to 'angle1' = 0 ie along the z axis.
Given a host of such electrons with polarisation angle = angle1, the average number of electrons measured in the up direction is (using Malus's Law) proportional to cos^2 theta where theta = angle1-zero deg, ie theta equals angle1.
Of course, since electron#1 is in general unpolarised its average value of angle1 will be a random value and so 50 per cent of all electrons will be polarised 'up'.
But to get a measurement for an individual electron#1, knowing the polarisation is 'angle1' when measured, is not enough to know what the measurement outcome will be. We need a second level of random input of the hidden variable value.
QM measurements tell us that when an individual electron is measured as 'up' then if it is measured again and again in the 'up' direction it will still be recorded as 'up'. This means that the hidden variables of an 'up' electron [which are responsible for the measurement outcome] are all pointing within 90 degrees of 'up'. My model has these h.v. angles as dynamic but fitting within a statistical envelope calculated using Malus's Law.
Next measure electron#1, with polarisation angle = angle1, in the up direction. That is along the z axis in the xz plane. All the h.v.s in electron#1 must be pointing in the direction angle1 plus or minus 90deg. (so that repeated measurements in the angle1 polarisation setting would not change any measurement results in repeated measurements along angle1.)
Further, these h.v.s have to conform to a distribution set by Malus's Law. Basiclly more h.v.s need to point at angle1 than not. The distribution is NOT random (which is why random on a sphere simulation methods do not work).
Next a little calculus. Malus's Law for electrons gives a cumulative result. Ranging from 'all pass' at theta = zero deg (difference in settings of first and second filters) to 'half shall pass' at theta = 90 deg. These are cos^2 zero/2 deg and cos^2 90/2 deg resp. The divisor of '2' is used to switch from the normal law for photons to an adapted law for electrons.
But for the statistical envelope of h.v.s about angle1 we need a density function rather than the cumulative density function given by Malus's Law. To get the density function we just differentiate.
But first, I want the distribution to run from 0 to 180 deg not from -90 to +90 deg, see EXTRACT below, so I need to differentiate
1 - cos^2 (theta/2), rather than cos^2 (theta/2). This gives 0.5*sin theta for the statistical envelope where theta runs from 0 to 180 deg.
This function yields 0 at theta = 0 and 180 deg and yields 0.5 at theta = 90. Total density = 1 over the whole curve or envelope. This shows that no h.v.s point at 90 deg away from the polarisation angle of an electron while most point along the polarisation angle.
Next pick an individual h.v. at random in this curve at the instant of measurement. My electron structure is dynamic but if I pick the instant of measurement than it is a static vector.
[ABC] Next, I choose a random number between 0 and 1 to help determine a h.v.
at the instant of measurement.
Need to switch back to the cumulative density functon for this step. (It was useful to obtain the density function to see how the h.v.s are distributed about the polarisation angle, but sorry its back to the cumulative density function.) Let the random number represent the cumulative number of h.v.s in the statistical envelope starting at theta=0. [Now see Extract#2 below for more details, if that helps.]
If we integrate the electron spin vector density 0.5 * sin x between x = 0 and x = psi, we get 0.5(-cos psi - -cos(0)).
= -0.5 * cos psi +0.5.
Set this integral to be p, the random cumulative probability for the generated electron.
So p = -0.5 * cos psi +0.5
and hence cos psi = 1-2p.
and psi is the angle whose cosine is 1-2p. Which is arccos(1-2p).
So now we have an electron with a known spin vector at angle psi. If this angle is greater [this should read 'less'] than the angle between the electron polarisation angle and the measuring angle (0 degrees for the first measurement along the z axis) then the particle passes the filter as +1.
This is where the two polarisation angles are separated by between 0 and 180 deg. If that angle is greater than 180 deg then we need to use -up as the electron polarisation angle and start again.
So, after all that, we have finished stage 1 of preparing one detector setting.
Next, we re-measure in direction +x axis. As this is a 90 deg difference in polarisation angles we know that 50 per cent of electrons will pass the filter, on average. But what about our particular electron#1?
We need to go back to instructions at [ABC] above and work it through. To summarise, we end up with another angle for the h.v. as phi' =
arccos(1-2p') where phi' and p' are new values. If phi' is less than 90 deg then measurement = +1.
Measurement +1 refers to Alice getting setting 0 deg rather than pi/2.
Note that I have not used retrocausality yet as all this calculation was necessary just to program a particle-at-a-time method for Malus's Law used in a forwards-in-time method.
EXTRACT#1 from appendix of my june 2020 vixra paper:
'
' To find the distribution of individual spin vectors (at angles of phi) is is necessary to differentiate not
' cos squared (phi/2)
' but 1 - cos squared (phi/2),
' and 1 - cos squared (phi/2) = 0.5 - 0.5 * cos phi
' which differentiates to 0.5 sin phi.
' This is the curve which is plotted for electrons in Figure A of the report: "Malus's Law and Bell's Theorem with local hidden variables".
EXTRACT#2 from appendix of my june 2020 vixra paper:
' First, by generating a random number between 0 and 1 to represent a random cumulative probability of a particle
' lying on the spin vector distribution curve. Next, we need to find out what the spin vector angle (psi) is, for the
' generated particle, corresponding to that random cumulative probability (p).
'
' If we integrate the electron spin vector density 0.5 * sin x between x = 0 and x = psi, we get 0.5(-cos psi - -cos(0)).
' = -0.5 * cos psi +0.5.
'
' Set this integral to be p, the random cumulative probability for the generated electron.
'
' So p = -0.5 * cos psi +0.5
' and hence cos psi = 1-2p.
'
' and psi is the angle whose cosine is 1-2p.
'
' That is commonly called arccos(1-2p)
'
' But in MS visual basic the arccos function does not exist and so the function used is a convoluted, but standard,
' combination of arctan functions.
'
' So now we have a particle with a known spin vector angle psi. If this angle is greater than theta then the particle
' passes the filter, where theta is the angle between the incoming polarisation vector and the filter spin vector axis.