Superdeterminism

[FrediFizzx]

Then you should not be surprised what A(a,h) = 1/sign(a.h) means physically. But wait, what scalar point does Bell's equation (4) give when those two vectors ("a", "h") are orthogonal?

If a happens to be orthogonal to h (the measure of that happening, by the way, is nearly zero), then sign(h.a) is assumed to be equal to the sign of the first nonzero component from the set {a_x, a_y, a_z}. Bell does not specify this in his 1964 paper, but he does so in a later paper, and so do GHSZ in their paper.

Nearly zero? Exactly zero, I think. For this reason, it does not matter how you define Bell's (4) when "a" and "h" are orthogonal. You could define it to always to equal +1 in that case. Or introduce a further hidden variable which is an independent fair coin toss [which actually never gets used!]. As Bell wrote "as the probability of this is zero we will not make special prescriptions for it". He doesn't bother to write down a special prescription because *any special prescription would do". He doesn't say that we should *discard outcomes from his mathematical model*.

There is simply no point in giving further details.

1/sign(a.h) is indeterminate when "a" is orthogonal to "h" because you have 1/0.
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[Joy Christian]

1/sign(a.h) is indeterminate when "a" is orthogonal to "h" because you have 1/0.

Indeed. So Michel's function A(a, h) = 1/sign(a.h) is not as well defined as he thinks it is. It blows up to infinity for a.h = 0.

By contrast, Bell's prescription A(a, h) = sign(a.h), with the addendum I wrote down below, is well defined and provides a definite result +/-1 for all values of a and h, without exception:

If a happens to be orthogonal to h (the measure of that happening, by the way, is nearly zero), then sign(h.a) is assumed to be equal to the sign of the first nonzero component from the set {a_x, a_y, a_z}. Bell does not specify this in his 1964 paper, but he does so in a later paper, and so do GHSZ in their paper.

But all this is beside the point. The main issue is that the setting-dependent distribution function p(a, b | h) does not make much physical sense.
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[minkwe]

Indeed. So Michel's function A(a, h) = 1/sign(a.h) is not as well defined as he thinks it is. It blows up to infinity for a.h = 0.

Yeah, if you go outside the domain of the function, what do you expect? Every function is not "well-behaved" outside its domain. Garbage in, garbage out.

[minkwe]

You keep making an assumption of bad faith on my part.

My assumption is reasonable based on the history concerning this topic:
http://www.sciphysicsforums.com/spfbb1/ ... ple#p14472

[Joy Christian]

Indeed. So Michel's function A(a, h) = 1/sign(a.h) is not as well defined as he thinks it is. It blows up to infinity for a.h = 0.

Yeah, if you go outside the domain of the function, what do you expect? Every function is not "well-behaved" outside its domain. Garbage in, garbage out.

Restricting the domain of A(a, h) in that way prevents A(a, h) from being well defined for all h in H, which is the space of all hidden variables for the physical system. If h (or lambda) is a spin vector as in Bell's example of a local model, then restricting the domain of A(a, h) restricts h from taking all values on the unit S^2. But physically an exploding bomb does scatter in principle in all possible directions h in R^3. No direction h is excluded.

But again, none of this really addresses the main objection raised. Namely, what justifies physically a setting-dependent probability distribution of hidden variables (or microstates) that are being integrated over. I don't believe mathematics is an issue here. The relevant mathematics is trivial.
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[minkwe]

The main issue is that the setting-dependent distribution function p(a, b | h) does not make much physical sense.

I don't know what physical role you think p(h|a, b) plays that you just can't accept? Whether you like it or not, p(h|a, b) ALWAYS plays a physical role in the EPRB experiments. The only difference is that you prefer to assume that p(h|a, b) = p(h). Therefore your problem is not with the physical role played by p(h|a, b). Rather you have chosen to make an additional assumption that p(h|a, b) = p(h).

What is the physical justification for that assumption? This is the crux of the matter.

In a typical EPRB experiment, you have a distribution produced by a source, and from that, you select a subset that produced the outcomes you observed when the settings were (a,b). p(h|a, b) =/= p(h) (also known as "superdeterminism") simply means that this subset is not the same as the source distribution. It makes perfect physical sense.

If you perform a different experiment with settings (c,d) you are selecting a different subset p(h|c, d) and this subset is yet different from p(h|a, b). If you want to believe that all those subsets are the same as the source distribution, you have to justify it physically.

p(h|a, b) does not mean that the source distribution causally depends on the settings (a, b). It only means the distribution of "h" which corresponds to the outcomes actually measured at settings (a,b).

Superdeterminism simply says the distributions are different. It doesn't posit why they are different. I tried above to give one possible reason why they could be different. But perhaps I should have asked instead why you believe they must be the same.

[Joy Christian]

What is the physical justification for that assumption? This is the crux of the matter.

The physical justification for using p(h) as the probability distribution is quite simple. It is the distribution of microstates or hidden variables that are being integrated over at a later time. In experiments of any kind, many parameters are used for the purposes of doing the experiment. Among these parameters are the fixed settings a and b. They are not part of the statistical ensemble of the physical states h. They do not produce results, h does. They are fixed parameters. It makes no sense to include them in the probability distribution of microstates at any stage. The only variable being integrated over is h.
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[minkwe]

Restricting the domain of A(a, h) in that way prevents A(a, h) from being well defined for all h in H, which is the space of all hidden variables for the physical system.

So you chose to assume that A(a,h) must be defined for all values of "h". Why? What physical justification do you have for this assumption?

If h (or lambda) is a spin vector as in Bell's example of a local model, then restricting the domain of A(a, h) restricts h from talking all values on the unit S^2. But physically an exploding bomb does scatter in principle in all possible directions h in R^3. No direction h is excluded.

No direction is excluded but there are directions for which it is impossible to measure certain physical properties. I could easily envision an impulse ratio which is the inverse of momentum transfer to my detector. This property will be undefined for certain directions. If you set up an experiment to measure this property, of the bomb fragments and actually record the results, it would make perfect physical sense why not all directions would correspond to outcomes. And if the only way you knew of the presence of bomb fragments was by a reading from my device, you would never know that. Yet the source distribution of h will be different than whatever you measure and the distribution of h corresponding to what you actually measured would depend on what you actually measured. This makes perfect physical sense.

[minkwe]

What is the physical justification for that assumption? This is the crux of the matter.

The physical justification for using p(h) as the probability distribution is quite simple. It is the distribution of microstates or hidden variables that are being integrated over at a later time. In experiments of any kind, many parameters are used for the purposes of doing the experiment. Among these parameters are the fixed settings a and b. They are not part of the statistical ensemble of the physical states h. They do not produce results, h does. They are fixed parameters. It makes no sense to include them in the probability distribution of microstates at any stage. The only variable being integrated over is h.
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My answer to your Bomb fragments challenge demonstrates that what you say here cannot be correct. The outcome is generated by a combination of setting (direction of measurement) and hidden variable (momentum of bomb fragment). Besides, what integration are you talking about?

Are you talking about the theoretical analysis to derive the inequality (Bell's equation 2)? If this is the one, then you are making a circular argument. The fact that you assume a fixed distribution in the theoretical analysis is not by itself a physical justification for doing so.

Are you talking about the empirical integration done after the experiment to calculate E(a,b) from experimental outcomes? If so, then p(h) does not appear in this integration. Only p(h|a,b) ever does because in this case, you are integrating ONLY over actual experimental outcomes and p(h|a,b) is "the distribution of "h" which corresponds to the outcomes actually measured at settings (a,b)".

Therefore you haven't really provided any physical justification for the assumption.

[Joy Christian]

what integration are you talking about?

I am talking about , with the following physical justification:

The physical justification for using p(h) as the probability distribution is quite simple. It is the distribution of microstates or hidden variables that are being integrated over at a later time. In experiments of any kind, many parameters are used for the purposes of doing the experiment. Among these parameters are the fixed settings a and b. They are not part of the statistical ensemble of the physical states h. They do not produce results, h does. They are fixed parameters. It makes no sense to include them in the probability distribution of microstates at any stage. The only variable being integrated over is h.

I am not concerned about any inequalities.
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[minkwe]

I am talking about , with the following physical justification:

So what does H and p(h) represent in the bomb explosion example where "h" is the momentum of the fragment and A(a,h) is the measurement function representing my machine?

Are you suggesting that the H, and p(h) in your equation above will be exactly the same as the p(h) in the expression below?



For your bomb fragments and my measuring machine, It can't be. This is the whole point. It makes perfect sense to write E(a,b) as.


This is the only expression that is ever measured in any experiment. Where p(h|a,b) just means "the distribution of h corresponding to the outcomes obtained when the settings are (a,b)". To go from to , you are making an assumption that p(h|ab) = p(h).

You haven't provided any physical justification for that assumption. Note that even if your assumption were true, the expression



Is still correct. And it is always correct whether your assumption is right or wrong. But if your assumption is wrong then,


will be wrong, So your expression is the stronger position and demands a physical justification. What is it?

[Joy Christian]

Michel, I am not making any assumptions. You are making the assumption that the probability distribution p(h) depends on the settings a and b. You have to justify that assumption. You think you have but you haven't. You and superdeterminism guys are claiming that h influences the choices of a and b, but you haven't understood that that is what you are claiming. You have to prove that h influences the freely chosen random settings a and b. Sabine is obliged to do the same. Only then you or anyone else can claim that the probability distribution p depends on a and b, in addition to depending on h.
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[FrediFizzx]

will be wrong, So your expression is the stronger position and demands a physical justification. What is it?

When p(h) is created, "c" and "d" don't exist yet. For me, that is a pretty good physical justification.
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[minkwe]

will be wrong, So your expression is the stronger position and demands a physical justification. What is it?

When p(h) is created, "c" and "d" don't exist yet. For me, that is a pretty good physical justification.
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p(h|ab) is the distribution that corresponds to the outcomes measured at settings (a,b). p(h|ab) is not the source distribution. Why do you insist on using the source distribution to integrate over outcomes instead of the distribution that actually corresponds to the outcomes? If you plan to compare this result with any experiment, it is obvious which one you must use.

[FrediFizzx]

will be wrong, So your expression is the stronger position and demands a physical justification. What is it?

When p(h) is created, "c" and "d" don't exist yet. For me, that is a pretty good physical justification.
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p(h|ab) is the distribution that corresponds to the outcomes measured at settings (a,b). p(h|ab) is not the source distribution. Why do you insist on using the source distribution to integrate over outcomes instead of the distribution that actually corresponds to the outcomes? If you plan to compare this result with any experiment, it is obvious which one you must use.

Quite frankly that equation above is equal to zero and easy to prove. Joy's expression with the limit replacement functions is a much better expression to use.
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[minkwe]

You are making the assumption that the probability distribution p(h) depends on the settings a and b. You have to justify that assumption.

Superdeterminism means

This absolutely does not mean depends on the settings (a, b). It simply means the distribution of corresponding to the outcomes at settings (a, b), is not the same as the source distribution. Note the difference between this and what you think the claim is.

You and superdeterminism guys are claiming that h influences the choices of a and b

Absolutely not, this is an unfortunate misunderstanding of superdeterminism originating from Bell's unfortunate use of the term "super-determinism" to describe . It does not mean the hidden variables influence what choices experimenters can make or their freedom. All it means is that the distribution of hidden variables corresponding to the outcomes actually obtained is not the same as the source distribution.

You have to prove that h influences the freely chosen random settings a and b. Sabine is obliged to do the same. Only then you or anyone else can claim that the probability distribution p depends on a and b, in addition to depending on h.

I don't have to prove that because that is absolutely not the claim I have made here. And Sabine explicitly explains why the freedom of the experimentalist makes no difference.

[Joy Christian]

You are making the assumption that the probability distribution p(h) depends on the settings a and b. You have to justify that assumption.

Superdeterminism means

This absolutely does not mean depends on the settings (a, b). It simply means the distribution of corresponding to the outcomes at settings (a, b), is not the same as the source distribution. Note the difference between this and what you think the claim is.

You and superdeterminism guys are claiming that h influences the choices of a and b

Absolutely not, this is an unfortunate misunderstanding of superdeterminism originating from Bell's unfortunate use of the term "super-determinism" to describe . It does not mean the hidden variables influence what choices experimenters can make or their freedom. All it means is that the distribution of hidden variables corresponding to the outcomes actually obtained is not the same as the source distribution.

You have to prove that h influences the freely chosen random settings a and b. Sabine is obliged to do the same. Only then you or anyone else can claim that the probability distribution p depends on a and b, in addition to depending on h.

I don't have to prove that because that is absolutely not the claim I have made here.

We have stopped communicating. You have been ignoring the points I have been making, so I will do the same. We are not going to agree about this.
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[FrediFizzx]

will be wrong, So your expression is the stronger position and demands a physical justification. What is it?

When p(h) is created, "c" and "d" don't exist yet. For me, that is a pretty good physical justification.
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p(h|ab) is the distribution that corresponds to the outcomes measured at settings (a,b). p(h|ab) is not the source distribution. Why do you insist on using the source distribution to integrate over outcomes instead of the distribution that actually corresponds to the outcomes? If you plan to compare this result with any experiment, it is obvious which one you must use.

Here is the proof that that equation is a really poor one to use for this. A and B are equal to +/-1. Make the substitution and you have,



The vectors "c" and "d" have dropped out right away. If we are going to talk about this, let's use a better expression where "a" and "b" don't drop out.
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[minkwe]

We have stopped communicating. You have been ignoring the points I have been making, so I will do the same. We are not going to agree about this.

Not true. I'm making a good-faith effort to address your points. This is not about agreement. It is possible to state an argument accurately even if you disagree with it. I don't think you state my argument accurately. Not even remotely. That's why when you say

You and superdeterminism guys are claiming that h influences the choices of a and b

I can only conclude that you haven't understood what I'm saying.

[minkwe]

When p(h) is created, "c" and "d" don't exist yet. For me, that is a pretty good physical justification.
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p(h|ab) is the distribution that corresponds to the outcomes measured at settings (a,b). p(h|ab) is not the source distribution. Why do you insist on using the source distribution to integrate over outcomes instead of the distribution that actually corresponds to the outcomes? If you plan to compare this result with any experiment, it is obvious which one you must use.

Here is the proof that that equation is a really poor one to use for this. A and B are equal to +/-1. Make the substitution and you have,



The vectors "c" and "d" have dropped out right away. If we are going to talk about this, let's use a better expression where "a" and "b" don't drop out.
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Fred, it doesn't matter. You can write it as:



if you like.

is the distribution of h which corresponds to the outcomes actually measured at settings (a,b). This is what is measured experimentally when we measure E(a,b).