SciPhysicsForums.com

by **FrediFizzx** » Mon Oct 25, 2021 5:33 pm

Yeah, Yeah, Yeah! Another freakin' update. Now, with theoretical support from the Product Calculation for -a.b. Pretty freakin' awesome!!! 1jillion gazillion trials! :lol:

Cloud File.

https://www.wolframcloud.com/obj/fredif ... rum-pc2.nb

Direct Files.

download/newCS-33-S3quat-3D-forum-pc2.pdf
download/newCS-33-S3quat-3D-forum-pc2.nb

Enjoy the simulation that is a Bell and Gill theory killer!!!! It just keeps getting better and better!! :mrgreen:

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by **Gordon Watson** » Mon Oct 25, 2021 6:39 pm

FrediFizzx wrote:Yeah, Yeah, Yeah! Another freakin' update. Now, with theoretical support from the Product Calculation for -a.b. Pretty freakin' awesome!!! 1jillion gazillion trials!

Cloud File.

https://www.wolframcloud.com/obj/fredif ... rum-pc2.nb

Direct Files.

download/newCS-33-S3quat-3D-forum-pc2.pdf
download/newCS-33-S3quat-3D-forum-pc2.nb

Enjoy the simulation that is a Bell and Gill theory killer!!!! It just keeps getting better and better!!
.

Fred, given your simulation of N trials over N random particle-pairs, how many data-points appear in your plot? Is it always N? Thanks; Gordon
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by **FrediFizzx** » Mon Oct 25, 2021 6:48 pm

Gordon Watson wrote:Fred, given your simulation of N trials over N random particle-pairs, how many data-points appear in your plot? Is it always N? Thanks; Gordon

Sometimes. But we usually lose a little bit in the analysis process. The way to tell is to look at Total Events = 995602 after the plot. That is actually tweakable with beta and xi to get it closer to the number of trials. Thank you very much for a no-nonsense question.
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by **Joy Christian** » Mon Oct 25, 2021 8:37 pm

FrediFizzx wrote:
Austin Fearnley wrote: ... [incomprehensible nonsense snipped] ... Also, are your graphs showing that Bell's Inequalities are broken? Every once in a while you state that nothing can break the Bell Inequalities. So what is the conclusion?

The graphs DO NOT show that the inequalities are broken. The current graphs show that both Bell's junk physics theory and Gill's junk theory are broken. Shot dead to a million pieces that never can be put back together.

Just to be clear, Austin, inequalities are never "broken." Mathematical inequalities cannot be "broken." But they can be exceeded, by switching to a different inequality with a higher bound. That is of course cheating. But cheating is what experimentalists do and then falsely claim that they have "violated" Bell inequalities. Make no mistake. All Bell-believers are con-artists.

I hope you recover from the virus soon. We currently have about 40,000 new cases daily in England. The Tories, however, are determined not to bring back lockdown or social restrictions.
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by **FrediFizzx** » Tue Oct 26, 2021 11:14 am

The updated paper explaining all of this is still COMING SOON. It's a doozy and a real eye opener.
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by **FrediFizzx** » Fri Oct 29, 2021 11:55 am

Worthless Bell fanatics still can't take down the simulation!!! Guess they are taking cover somewhere for being totally wrong all these years. :mrgreen:

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by **FrediFizzx** » Sat Oct 30, 2021 11:44 am

I deleted this thinking it wasn't proven then I realized it is proven.

Proof that the simulation obtains -a.b. There is an analytical proof also but way longer.
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by **FrediFizzx** » Sun Oct 31, 2021 5:24 am

This expression seems a bit odd to me.

In order to get the probabilities for each of the four outcome pairs say in a large simulation, they first have to be averaged over many trials per (a-b) angle. It seems to me that in a proper simulation each of the four probabilities are going to converge to 1/4 for very large number of trials. At least that is what I am finding with our latest simulation.

Ave ++ = 0.248903
Ave -- = 0.248803
Ave +- = 0.246508
Ave -+ = 0.255786

That was for 10,000 trials. For 5 million trials,

Ave ++ = 0.249787
Ave -- = 0.249991
Ave +- = 0.250293
Ave -+ = 0.249929

Much closer to 1/4 each. So, for analytical purposes, it doesn't seem unreasonable to assign 1/4 to each of the four outcome pair probabilities.
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by **FrediFizzx** » Sun Oct 31, 2021 7:11 am

FrediFizzx wrote:This expression seems a bit odd to me.

In order to get the probabilities for each of the four outcome pairs say in a large simulation, they first have to be averaged over many trials per (a-b) angle. It seems to me that in a proper simulation each of the four probabilities are going to converge to 1/4 for very large number of trials. At least that is what I am finding with our latest simulation.

Ave ++ = 0.248903
Ave -- = 0.248803
Ave +- = 0.246508
Ave -+ = 0.255786

That was for 10,000 trials. For 5 million trials,

Ave ++ = 0.249787
Ave -- = 0.249991
Ave +- = 0.250293
Ave -+ = 0.249929

Much closer to 1/4 each. So, for analytical purposes, it doesn't seem unreasonable to assign 1/4 to each of the four outcome pair probabilities.

Ok, now for the next part of this.

QM assigns for those 4 outcome probabilities,

Again, in a simulation with many trials, we have to average $\frac{1}{2}\cos^2 (\frac{\eta_{ab}}{2})$ and $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ over all the (a-b) angles. Lo and behold, when we do that we obtain,

$<\frac{1}{2}\cos^2 (\frac{\eta_{ab}}{2})> = <\frac{1}{4}(1+\cos(\eta_{ab}))> = \frac{1}{4}$ ,
$<\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})> = <\frac{1}{4}(1-\cos(\eta_{ab}))> = \frac{1}{4}$ ,

Because $<\cos(\eta_{ab})> = 0$ .

So, it seems to me that all of the parts of the original E(a, b) expression are all equal to 1/4. Analytically-wise.
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by **FrediFizzx** » Sun Oct 31, 2021 9:09 am

FrediFizzx wrote:
FrediFizzx wrote:This expression seems a bit odd to me.

In order to get the probabilities for each of the four outcome pairs say in a large simulation, they first have to be averaged over many trials per (a-b) angle. It seems to me that in a proper simulation each of the four probabilities are going to converge to 1/4 for very large number of trials. At least that is what I am finding with our latest simulation.

Ave ++ = 0.248903
Ave -- = 0.248803
Ave +- = 0.246508
Ave -+ = 0.255786

That was for 10,000 trials. For 5 million trials,

Ave ++ = 0.249787
Ave -- = 0.249991
Ave +- = 0.250293
Ave -+ = 0.249929

Much closer to 1/4 each. So, for analytical purposes, it doesn't seem unreasonable to assign 1/4 to each of the four outcome pair probabilities.

Ok, now for the next part of this.

QM assigns for those 4 outcome probabilities,

Again, in a simulation with many trials, we have to average $\frac{1}{2}\cos^2 (\frac{\eta_{ab}}{2})$ and $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ over all the (a-b) angles. Lo and behold, when we do that we obtain,

$<\frac{1}{2}\cos^2 (\frac{\eta_{ab}}{2})> = <\frac{1}{4}(1+\cos(\eta_{ab}))> = \frac{1}{4}$ ,
$<\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})> = <\frac{1}{4}(1-\cos(\eta_{ab}))> = \frac{1}{4}$ ,

Because $<\cos(\eta_{ab})> = 0$ .

So, it seems to me that all of the parts of the original E(a, b) expression are all equal to 1/4. Analytically-wise.

Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ = 1/4, etc., does that prove that $P(++) = \frac{1}{4} = \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ , etc. for our analytical situation? Keeping mind that P(++) and $\frac{1}{2}\sin^2$ , etc. are actually averages.
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by **Joy Christian** » Sun Oct 31, 2021 10:32 am

FrediFizzx wrote:Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ = 1/4, etc., does that prove that $P(++) = \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ , etc. for our analytical situation?

No, it doesn't. I can think of dozens of functions for which $< f(\eta_{ab}) > = \frac{1}{4}$ would be true.

Consider, for example, $f(\eta_{ab}) := \frac{1}{4} - \frac{\eta_{ab}}{2\pi}$ .

This function gives $< \frac{1}{4} - \frac{\eta_{ab}}{2\pi} > = \frac{1}{4}$ .

Therefore, following your logic, $P(++) = \frac{1}{4} - \frac{\eta_{ab}}{2\pi}$ .
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by **FrediFizzx** » Sun Oct 31, 2021 11:20 am

Joy Christian wrote:
FrediFizzx wrote:Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ = 1/4, etc., does that prove that $P(++) = \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ , etc. for our analytical situation?

No, it doesn't. I can think of dozens of functions for which $< f(\eta_{ab}) > = \frac{1}{4}$ would be true.

Consider, for example, $f(\eta_{ab}) := \frac{1}{4} - \frac{\eta_{ab}}{2\pi}$ .

This function gives $< \frac{1}{4} - \frac{\eta_{ab}}{2\pi} > = \frac{1}{4}$ .

Therefore, following your logic, $P(++) = \frac{1}{4} - \frac{\eta_{ab}}{2\pi}$ .
.

Well, that is only true if $\eta_{ab} = 0$ . So, we still have $P(++) = \frac{1}{4}$ . Try again.
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by **Joy Christian** » Sun Oct 31, 2021 11:38 am

FrediFizzx wrote:
Joy Christian wrote:
FrediFizzx wrote:Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ = 1/4, etc., does that prove that $P(++) = \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ , etc. for our analytical situation?

No, it doesn't. I can think of dozens of functions for which $< f(\eta_{ab}) > = \frac{1}{4}$ would be true.

Consider, for example, $f(\eta_{ab}) := \frac{1}{4} - \frac{\eta_{ab}}{2\pi}$ .

This function gives $< \frac{1}{4} - \frac{\eta_{ab}}{2\pi} > = \frac{1}{4}$ .

Therefore, following your logic, $P(++) = \frac{1}{4} - \frac{\eta_{ab}}{2\pi}$ .
.

Well, that is only true if $\eta_{ab} = 0$ . So, we still have $P(++) = \frac{1}{4}$ . Try again.

Your logic does not work. I gave an example of why your logic does not work.
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by **FrediFizzx** » Sun Oct 31, 2021 11:44 am

Joy Christian wrote:Your logic does not work. I gave an example of why your logic does not work.

Well, actually I'm still waiting for the example that shows it doesn't work. Your last example didn't do anything.
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by **Joy Christian** » Sun Oct 31, 2021 12:17 pm

Joy Christian wrote:
FrediFizzx wrote:Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ = 1/4, etc., does that prove that $P(++) = \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ , etc. for our analytical situation?

No, it doesn't. I can think of dozens of functions for which $< f(\eta_{ab}) > = \frac{1}{4}$ would be true.

Consider, for example, $f(\eta_{ab}) := \frac{1}{4} - \frac{\eta_{ab}}{2\pi}$ .

This function gives $< \frac{1}{4} - \frac{\eta_{ab}}{2\pi} > = \frac{1}{4}$ .

Therefore, following your logic, $P(++) = \frac{1}{4} - \frac{\eta_{ab}}{2\pi}$ .

Here is another function which does the same thing:
$f(\eta_{ab}) := \frac{1}{4} + \sin(\eta_{ab}) + \cos(\eta_{ab})$ ,
implying that
$P(++) = \frac{1}{4} + \sin(\eta_{ab}) + \cos(\eta_{ab})$ .
In fact, there are many functions that would do the same thing.
.

by **FrediFizzx** » Sun Oct 31, 2021 12:54 pm

Joy Christian wrote:Here is another function which does the same thing:
$f(\eta_{ab}) := \frac{1}{4} + \sin(\eta_{ab}) + \cos(\eta_{ab})$ ,
implying that
$P(++) = \frac{1}{4} + \sin(\eta_{ab}) + \cos(\eta_{ab})$ .
In fact, there are many functions that would do the same thing.

Hmm... that doesn't look right substitution-wise.
$f(\eta_{ab})- \sin(\eta_{ab}) - \cos(\eta_{ ab}) = \frac{1}{4}$ ,
So,
$P(++) = f(\eta_{ab})- \sin(\eta_{ab}) - \cos(\eta_{ ab})$ ,
$P(++) = \frac{1}{4} + \sin(\eta_{ab}) + \cos(\eta_{ab}) - \sin(\eta_{ab}) - \cos(\eta_{ ab})$ ,
$P(++) = \frac{1}{4}$ .

Your example is still not working for me. Try again.
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by **Joy Christian** » Sun Oct 31, 2021 1:34 pm

FrediFizzx wrote:
Joy Christian wrote:Here is another function which does the same thing:
$f(\eta_{ab}) := \frac{1}{4} + \sin(\eta_{ab}) + \cos(\eta_{ab})$ ,
implying that
$P(++) = \frac{1}{4} + \sin(\eta_{ab}) + \cos(\eta_{ab})$ .
In fact, there are many functions that would do the same thing.

Hmm... that doesn't look right substitution-wise.
$f(\eta_{ab})- \sin(\eta_{ab}) - \cos(\eta_{ ab}) = \frac{1}{4}$ ,
So,
$P(++) = f(\eta_{ab})- \sin(\eta_{ab}) - \cos(\eta_{ ab})$ ,
$P(++) = \frac{1}{4} + \sin(\eta_{ab}) + \cos(\eta_{ab}) - \sin(\eta_{ab}) - \cos(\eta_{ ab})$ ,
$P(++) = \frac{1}{4}$ .

Your example is still not working for me. Try again.

I give up, because I don't understand what setting $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})=\frac{1}{4}$ means within your logic. To me, it means $\eta_{ab}=\frac{\pi}{2}$ , always, throughout all experimental trials.
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by **FrediFizzx** » Sun Oct 31, 2021 2:00 pm

Joy Christian wrote:I give up, because I don't understand what setting $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})=\frac{1}{4}$ means within your logic. To me, it means $\eta_{ab}=\frac{\pi}{2}$ , always, throughout all experimental trials.

I guess you missed this from one of the earlier postings.

Keeping mind that P(++) and $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ , etc. are actually averages.

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by **FrediFizzx** » Sun Oct 31, 2021 2:16 pm

@Joy You are thinking of this.

Actually more solutions than just pi/2. But that is NOT what we are doing.
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by **FrediFizzx** » Sun Oct 31, 2021 3:23 pm

FrediFizzx wrote:
FrediFizzx wrote:
FrediFizzx wrote:This expression seems a bit odd to me.

In order to get the probabilities for each of the four outcome pairs say in a large simulation, they first have to be averaged over many trials per (a-b) angle. It seems to me that in a proper simulation each of the four probabilities are going to converge to 1/4 for very large number of trials. At least that is what I am finding with our latest simulation.

Ave ++ = 0.248903
Ave -- = 0.248803
Ave +- = 0.246508
Ave -+ = 0.255786

That was for 10,000 trials. For 5 million trials,

Ave ++ = 0.249787
Ave -- = 0.249991
Ave +- = 0.250293
Ave -+ = 0.249929

Much closer to 1/4 each. So, for analytical purposes, it doesn't seem unreasonable to assign 1/4 to each of the four outcome pair probabilities.

Ok, now for the next part of this.

QM assigns for those 4 outcome probabilities,

Again, in a simulation with many trials, we have to average $\frac{1}{2}\cos^2 (\frac{\eta_{ab}}{2})$ and $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ over all the (a-b) angles. Lo and behold, when we do that we obtain,

$<\frac{1}{2}\cos^2 (\frac{\eta_{ab}}{2})> = <\frac{1}{4}(1+\cos(\eta_{ab}))> = \frac{1}{4}$ ,
$<\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})> = <\frac{1}{4}(1-\cos(\eta_{ab}))> = \frac{1}{4}$ ,

Because $<\cos(\eta_{ab})> = 0$ .

So, it seems to me that all of the parts of the original E(a, b) expression are all equal to 1/4. Analytically-wise.

Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ = 1/4, etc., does that prove that $P(++) = \frac{1}{4} = \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ , etc. for our analytical situation? Keeping mind that P(++) and $\frac{1}{2}\sin^2$ , etc. are actually averages.

So, I guess this question is still open. Joy seems to be trying to say no but not doing a very good job of it.
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