## Coming Soon!

Foundations of physics and/or philosophy of physics, and in particular, posts on unresolved or controversial issues

### Re: Coming Soon!

Gordon Watson wrote:Fred, as you increase the number of trials, N, so you will approach the infinite case in the limit. Gordon

No sh*t Sherlock! Perfect example of a nonsense post. I don't recall any other day in my life where I was more pissed off than today!!!! And sicker than hell to boot!!!
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FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Coming Soon!

Now, maybe one of you guys can come up with a real example instead of junk nonsense. Remember, we are doing a freakin' simulation!!!

I'm claiming in the context of a simulation where $x=\text{RandomReal}[\{0,2 \pi \}]$ is dependent on the number of trials, that if < f > = < g >, then f = g!!!!!!!!!!!!!!!!!!
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FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Coming Soon!

If I have done everything correctly this entire forum has been set to Read Only Access. It will be this way for a while. However, I think private messaging still works but I will probably ignore PM's to me unless they are about something really good.
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FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Coming Soon!

Ok, back to the original question. I think I need to do some slight fixin' on it.

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 $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$, etc. are actually averages.
____________________________________________________________________________________________________________________________________________
Ok, what we actually have here for the simulation is $< (++) >\; =\; < \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2}) >\; = \frac{1}{4}$.
So, the question is actually this. Does,

$(++) \; =\; \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$??

Now the only way to equate those two would be by their totals or length of the lists. So, let's see. For a million trials, there are 250,000 events that are (++). and there are 250,000 events that are $\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ which makes sense since basically there would be a 1/4 each of the sin^2 and cos^2 probabilities. So, they match quantity-wise. But what we are actually trying to find out is this,

$P(++) \; =\; \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$. Is that true for our simulation? Now, we know that the probability of getting (++) for the simulation is 1/4 so

$P(++) \; =\; \frac{1}{4}$

Now, we also know the average of getting,

$< \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2}) >\;=\;\frac{1}{4}$

So,

$P(\frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2}))\;=\;\frac{1}{4}$

Then by virtue of what QM says about it, I think we can say that,

$P(++) \; =\; \frac{1}{2}\sin^2 (\frac{\eta_{ab}}{2})$ for both the simulation and the analytical formulas. But maybe more proof is needed?

You can send me a PM if any questions or comments.
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FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Coming Soon!

OK Folks. I've created a new more simple forum at,

http://sciphysicsfoundations.com/

I still have some work to do on it but I think it is working well enough to get started with it. You will have to create new user accounts, etc. and wait for me to approve the new accounts.

The reason for this is that this forum has gotten to be a bit old software-wise with no further updates possible. And I wanted to move off of Godaddy hosting to a new server. However, I could not figure out how to get the database from this forum to the new server and to work with the new phpBB forum software. So, we are starting over more simple. This forum will stay at read only forever so you can see the posts. Unless I figure out how to get this database over to the new server. But way too sick to tackle that now or even in the near future.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

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