Is action at a distance possible as envisaged by the EPR Paradox.

In summary: QM?In summary, John Bell was not a big fan of QM. He thought it was premature, and that the theory didn't yet meet the standard of predictability set by Einstein.
  • #1,156
JesseM said:
So, seems to be exploiting some variant of the detector efficiency loophole.

I can only tell about the version DrC made for Excel, which you can https://www.physicsforums.com/showpost.php?p=2724402&postcount=389", and I assume it’s their older "Model 1" algorithm.

The one and only 'thing' that can be responsible for the achieved result, is in what they call the "time window", and a pseudo-random number in r0 that is altered depending on the current angle:

6oztpt.png


I guess that only Mr. BS would call this a real LR model, explaining a real underlying theory in nature...
 
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  • #1,157
JesseM said:
It would be possible (in principle) to perform an experiment 100% faithful to all his assumptions about the observable conditions of the experiment, yes.
I disagree that this is possible, as I explained in post #1076:
https://www.physicsforums.com/showpost.php?p=2804344&postcount=1076

And since theorists have come up with modified Bell inequalities that deal with imperfect detector efficiency ...
Completely irrelevant. The simulation model is not dealing with detection efficiency loophole. It deals with the coincidence time window, or if you prefer "coincidence time loophole".

Not if the De Raedt model fails to simultaneously exploit the locality loophole (or it does but requires a very contrived and complicated algorithm).
Huh? What are you talking about?

I'm trusting DrChinese's analysis, unless you can show where it's wrong--do you claim there is some section of the paper that demonstrates that every simulated photon emitted by the source is actually detected? If so, perhaps you could quote that section?
Again this doesn't make sense because De Raedt were modelling an actual experiment, so expecting them make their model so it deliberately does not correspond to what is actually done and observed in the real experiment is queer. If in a real experiment only 5% of photons are detected, and a model is presented for the experiment in which 100% of photons are detected, that will be grounds for invalidating the model. This is simple logic and how science is normally done.
We construct an event-based computer simulation model of the Einstein-Podolsky-Rosen-Bohm
experiments with photons. The algorithm is a one-to-one copy of the data gathering and analysis
procedures used in real laboratory experiments. We consider two types of experiments, those with a source emitting photons with opposite but otherwise unpredictable polarization and those with a source emitting photons with fixed polarization. In the simulation, the choice of the direction of polarization measurement for each detection event is arbitrary. We use three different procedures to identify pairs of photons and compute the frequency of coincidences by analyzing experimental data and simulation data. The model strictly satisfies Einstein’s criteria of local causality, does not rely on any concept of quantum theory and reproduces the results of quantum theory for both types of experiments. We give a rigorous proof that the probabilistic description of the simulation model yields the quantum theoretical expressions for the single- and two-particle expectation values.
Feel free to trust DrC's analysis. To me the simple fact that he misrepresents their explicitly stated claims is disqualifying.

You can swindle "Bell is respected with it ... Which is to say that their model does not claim to match QM" all you want. The authors' claims are pretty clear. Read the paper yourself and see if you still agree with DrC that their model does not claim to match QM.
Summarizing: We have demonstrated that a simulation model that strictly satisfies Ein-
stein’s criteria of locality can reproduce, event-by-event, the quantum theoretical results for
EPRB experiments with photons, without using any concept from quantum theory. We
have given a rigorous proof that this model reproduces the single-particle expectations and
the two-particle correlation of two S = 1/2 particles in the singlet state and product state.
 
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  • #1,158
JesseM said:
It would be possible (in principle) to perform an experiment 100% faithful to all his assumptions about the observable conditions of the experiment, yes. Of course the experiment need not match the theoretical assumptions about hidden variables, since the whole point would be to compare a real experiment which matches these observable experimental conditions to what LR hidden-variables theories would predict about an experiment which matches these observable experimental conditions.
billschnieder said:
I disagree that this is possible, as I explained in post #1076:
https://www.physicsforums.com/showpost.php?p=2804344&postcount=1076
Your argument there is based on a failure to distinguish "assumptions about the observable conditions of the experiment" from "theoretical assumptions about hidden variables"--this sort of confusion is common in your arguments, which is exactly why I phrased my comment in the way I did.

In a case like the Leggett-Garg inequality, the "observable conditions of the experiment" include the idea that on each trial we measure the system at two out of three possible times. Then an inequality is derived based on the theoretical assumption that the system has a well-defined state at all three times (and that the state isn't influenced by your measurements), even the time that we don't actually measure. So if you want to test the theoretical assumption, you do an experiment that matches the specified "observable conditions", and if you find the inequality is violated, that falsifies the theoretical assumption that each measured system had a well-defined state at all three times which wasn't influenced by the measurements.

You earlier showed that you understood the idea of my scratch lotto card example--in that example, each experimenter had a card with three possible boxes (for a total of six boxes), but the experiment only involved each scratching one box (a total of two boxes revealed). The theoretical assumption being tested was that there was an unchanging "hidden fruit" behind all six boxes, and the conclusion was that if experiments always found the same fruit on trials where both experimenters chose the same box, then on trials where both experimenters chose different boxes they should find the same fruit at least 1/3 of the time. If they perform this experiment many times (with perfect 'efficiency' so no cards have to be thrown out) and find that they only get the same fruit 1/4 of the time when they choose different boxes, isn't this a valid falsification of the hypothesis that there was an unchanging hidden fruit behind all six boxes? You wouldn't say the experiment failed to show anything because they assumed six variables with well-defined values but only sampled two, would you?
billschnieder said:
Completely irrelevant. The simulation model is not dealing with detection efficiency loophole. It deals with the coincidence time window, or if you prefer "coincidence time loophole".
Does the computer model assume every emitted photon is detected? If not, why are you so sure they aren't exploiting this loophole? Anyway, you may be right that they exploit the coincidence-time loophole, a slightly different experimental loophole I hadn't been thinking of (discussed here). I think you could view it as a type of detector efficiency loophole in any case, since it means that the detectors aren't correctly identifying all entangled pairs as pairs.
JesseM said:
Not if the De Raedt model fails to simultaneously exploit the locality loophole (or it does but requires a very contrived and complicated algorithm).
billschnieder said:
Huh? What are you talking about?
I was responding to your statement 'since non-localists rely on the same "loopholed experiments" to proclaim the demise of locality, a locally causal explanation of those same experiments however "loopholed" they are, is an effective counter argument.' A model is not an "effective counter argument" if it can't actually explain all the different types of experimental results seen so far, including the experiments where the detector efficiency loophole was closed but the locality loophole was not.
JesseM said:
I'm trusting DrChinese's analysis, unless you can show where it's wrong--do you claim there is some section of the paper that demonstrates that every simulated photon emitted by the source is actually detected? If so, perhaps you could quote that section?
billschnieder said:
Again this doesn't make sense because De Raedt were modelling an actual experiment, so expecting them make their model so it deliberately does not correspond to what is actually done and observed in the real experiment is queer.
I'm not "expecting them" to do anything different than what they did. I was responding to your comment "No. Read the paper!" in response to my comment "seems to be exploiting some variant of the detector efficiency loophole." If you were denying that they exploited the detector efficiency loophole, wouldn't that mean you were claiming that their model assumed conditions of perfectly efficient detection?
billschnieder said:
Feel free to trust DrC's analysis. To me the simple fact that he misrepresents their claims is disqualifying.
But he doesn't misrepresent their claims, his claim is perfectly correct if you understand what he means by "Bell is respected" (namely, that their model would obey Bell inequalities in an experimental setup that actually matched the observable experimental conditions assumed by Bell). If you're just saying that the claim "Bell is respected" would be wrong if we had a different interpretation of the meaning of that phrase, then you're just quibbling over DrChinese's choice of language, not saying his discussion is wrong in any more substantive sense.
 
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  • #1,159
billschnieder said:
It is clear to me from their paper that
a) They have provided an "LR model" of the experiments under consideration. (Section V)
b) For the two types of experiments they considered, they showed that their model agrees with the QM prediction and violates Bell for some values of d. (Section IV)

The authors are not ambiguous about what they claim to have demonstrated. At least to me, it is clear that they have a LR model which violates Bell's inequalities but agrees with QM.

If you do not think they have presented an LR model which agrees with QM and disagrees with Bell for the two types of experiments they considered, you will have to clarify what you mean by
1) LR Model
2) obtain |S| > 2 in theory

I suspect we agree on what constitutes an LR model and that indeed they have an LR model. We should also agree that the data provided by their LR model yields |S| > 2 when subjected to the analysis in question. Where we might disagree is on whether or not they have an LR model that yields |S| > 2 in theory.

That is to say, their LR model can produce unambiguous pairs of events, so there is no need in theory to use the experimental procedure for finding correlated events -- they can easily program their data so that every pair of events is correlated (whereas only a tiny fraction of those under experimental analysis are correlated). Under this type of analysis their LR model will not produce |S| > 2.

So, again, they have not found an LR model that violates Bell's inequality. What they have found is a loop hole in these two particular experiments whereby an LR model can appear to violate Bell's inequality because of the analysis that must be used in a real experiment when you don't know which pairs of events are correlated. Again, this limitation is NOT applicable to their LR model because they CAN know which events are correlated in their model and, using this knowledge, they can easily show that their LR model doesn't violate Bell's inequality.
 
  • #1,160
JesseM said:
Your argument there is based on a failure to distinguish "assumptions about the observable conditions of the experiment" from "theoretical assumptions about hidden variables"--this sort of confusion is common in your arguments, which is exactly why I phrased my comment in the way I did.
My argument is broken down into points as follows:
1) Bell's inequalities can be derived from triples of dichotomous variables without any physical assumption
2) In Bell-test experiments only pairs of values are ever collected at a time (a dataset of pairs)
3) A dataset of pairs can be made to violate inequalities derived from a dataset of triples for purely mathematical reasons
4) I have provided mathematical proof of (1), (2) is an accepted fact. I have provided proof of (3) via simulation
5) Therefore, the violation of Bell's inequalities derived from triples, by experiments such as Bell-test experiments which only collect pairs, is not surprising, it is expected for purely mathematical reasons, having nothing to do with realism or locality.
6) Therefore, Bell's inequality can never be violated by a dataset of triples, even if the physical assumption of of spooky action at a distance is mandated!
If you are now seriously considering responding to it, please clearly point out which of the above points is wrong and why it is wrong. I could not discern a clear response against any of the above points in anything you have written.

Does the computer model assume every emitted photon is detected? If not, why are you so sure they aren't exploiting this loophole?
Had you read the paper, you will have understood this. They have a single parameter dwhich corresponds to the coincidence time window, and they clearly show that with values of d similar to what is used in real experiments, Bell is violated and the simulation agrees with QM but if no coincidence time window is introduced ie d = 0, which is NOT what is done in real experiments, the simulation respects Bell and disagrees with QM, even though not all photons are detected. So yeah, I am pretty sure that their simulation has nothing to do with detection efficiency, and you will be too if only you read the article.

A model is not an "effective counter argument" if it can't actually explain all the different types of experimental results seen so far, including the experiments where the detector efficiency loophole was closed but the locality loophole was not.
Again, despite your wishes, this model has nothing to do with detector efficiency.

his claim is perfectly correct if you understand what he means by "Bell is respected" (namely, that their model would obey Bell inequalities in an experimental setup that actually matched the observable experimental conditions assumed by Bell).
Exactly zero such experimental setups have been realized to date. They present a model of a real experimental setup and THEY CLAIM that their model agrees with QM as evidenced by their own words which I have quoted to you. You can disagree with their claims but you certainly can not say they claim the opposite to what they actually claim, by introducing some other setup, which has never been realized and which they never set out to model. Dr C is free to state that in his opinion, their model agrees with Bell and disagrees with QM. But to say their model does not claim to match QM is false. The former states an opinion about their model, the latter purports to represent their claims but does not. This is obvious, and no amount of quibbling can change this. So feel free to continue the quibbling but count me out of it.
 
  • #1,161
RUTA said:
I suspect we agree on what constitutes an LR model and that indeed they have an LR model. We should also agree that the data provided by their LR model yields |S| > 2 when subjected to the analysis in question. Where we might disagree is on whether or not they have an LR model that yields |S| > 2 in theory.

That is to say, their LR model can produce unambiguous pairs of events, so there is no need in theory to use the experimental procedure for finding correlated events -- they can easily program their data so that every pair of events is correlated (whereas only a tiny fraction of those under experimental analysis are correlated). Under this type of analysis their LR model will not produce |S| > 2.

So, again, they have not found an LR model that violates Bell's inequality.
Would you say the Weihs et al experiment violated Bell's inequality and agreed with QM, or would you say the Weihs et al experiment appeared to violate Bell's inequality because it exploited the coincidence time loophole? If you have no problem with this interpretation of the Weihs et all experiment, which they are modelling, I see not reason to expect anything different about their model.

What they have found is a loop hole in these two particular experiments whereby an LR model can appear to violate Bell's inequality because of the analysis that must be used in a real experiment when you don't know which pairs of events are correlated. Again, this limitation is NOT applicable to their LR model because they CAN know which events are correlated in their model and, using this knowledge, they can easily show that their LR model doesn't violate Bell's inequality.
Sure, you can say that. But they are modelling the experiment, and their model of the experiment violates Bell and agrees with QM. I think you would agree that d=0 does not correspond to the experiments they were modelling.
 
  • #1,162
billschnieder said:
Would you say the Weihs et al experiment violated Bell's inequality and agreed with QM, or would you say the Weihs et al experiment appeared to violate Bell's inequality because it exploited the coincidence time loophole? If you have no problem with this interpretation of the Weihs et all experiment, which they are modelling, I see not reason to expect anything different about their model.


Sure, you can say that. But they are modelling the experiment, and their model of the experiment violates Bell and agrees with QM. I think you would agree that d=0 does not correspond to the experiments they were modelling.

Nature is producing the experimental data and, unlike the LR model, we don't know how She's doing that. Thus, we say simply that Weil's experiment violated Bell's inequality and agreed with QM, just like we can (and I did) say the LR model violated Bell's inequality and agreed with QM (although, I had to add the qualifier -- "using the analysis of this experiment"). The reason for the qualifier is that we KNOW if the LR model is programmed to produce correlated pairs unambiguously, which it can do, then it will NOT violate Bell's inequality and NOT agree with QM.

I'm not trying to play semantic games, there is a distinction between the following two claims:

1. I have an LR model that produces data which when analyzed per experiment X violates Bell's inequality and agrees with QM.

2. I have an LR model that violates Bell's inequality and agrees with QM.

You have to choose your words carefully so as not to conflate these two claims.
 
  • #1,163
RUTA said:
Nature is producing the experimental data and, unlike the LR model, we don't know how She's doing that. Thus, we say simply that Weil's experiment violated Bell's inequality and agreed with QM, just like we can (and I did) say the LR model violated Bell's inequality and agreed with QM (although, I had to add the qualifier -- "using the analysis of this experiment"). The reason for the qualifier is that we KNOW if the LR model is programmed to produce correlated pairs unambiguously, which it can do, then it will NOT violate Bell's inequality and NOT agree with QM.

I'm not trying to play semantic games, there is a distinction between the following two claims:

1. I have an LR model that produces data which when analyzed per experiment X violates Bell's inequality and agrees with QM.

2. I have an LR model that violates Bell's inequality and agrees with QM.

You have to choose your words carefully so as not to conflate these two claims.

Dear RUTA,

Sorry, but (for me) this is NOT your clearest piece of writing (which I value). Could you take more words, please, to make your points more clearly and expansively?

They seem to be important; though I am sure to be in disagreement, even when they have been clarified.

PS: Could you explain, and elaborate fairly fully, why you capitalized KNOW here?

The reason for the qualifier is that we KNOW if the LR model is programmed to produce correlated pairs unambiguously, which it can do, then it will NOT violate Bell's inequality and NOT agree with QM.

What do we KNOW here, and how do we KNOW IT, please?

PPS: Please clear your PF mail box.

Thank you very much,

JenniT
 
  • #1,164
RUTA said:
I'm not trying to play semantic games, there is a distinction between the following two claims:

1. I have an LR model that produces data which when analyzed per experiment X violates Bell's inequality and agrees with QM.

2. I have an LR model that violates Bell's inequality and agrees with QM.

You have to choose your words carefully so as not to conflate these two claims.
I see your point. My point though is the following: Claim (1) and (2) are not different within the context of the paper we are discussing so I do not see why you insist on making a distinction. The authors did not claim to be deriving "an LR model" in a general sense. Their focus in that paper is to present "an LR model of the experiment" and in that context you can not separate their model from the constraints imposed by the experimental situation being modeled. It seems from your phrasing of (1) that you prefer to to do that. But do you apply the same standard to QM? QM only gives predictions for clearly stated experimental setups. I don't think you will expect QM to generate a dataset which you will then analyze according to experiment X.
 
  • #1,165
RUTA said:
I'm not trying to play semantic games

Well, I’m afraid you’ll have to... Semantic games are Mr. BS favorite engagement. The more completely meaningless words he produces, the happier he gets. Just watch and learn...
 
  • #1,166
billschnieder said:
My argument is broken down into points as follows:
1) Bell's inequalities can be derived from triples of dichotomous variables without any physical assumption
2) In Bell-test experiments only pairs of values are ever collected at a time (a dataset of pairs)
3) A dataset of pairs can be made to violate inequalities derived from a dataset of triples for purely mathematical reasons
4) I have provided mathematical proof of (1), (2) is an accepted fact. I have provided proof of (3) via simulation
5) Therefore, the violation of Bell's inequalities derived from triples, by experiments such as Bell-test experiments which only collect pairs, is not surprising, it is expected for purely mathematical reasons, having nothing to do with realism or locality.
6) Therefore, Bell's inequality can never be violated by a dataset of triples, even if the physical assumption of of spooky action at a distance is mandated!
If you are now seriously considering responding to it, please clearly point out which of the above points is wrong and why it is wrong. I could not discern a clear response against any of the above points in anything you have written.
3) is ambiguous. It's true a dataset of pairs can be made to violate inequalities from a set of triples under certain sampling conditions, but not under the conditions assumed by Bell, where the choice of which two values to measure is random on each trial, and there is no correlation between the probability of a given triple and the choice of which pair to measure on a given trial.
billschnieder said:
Had you read the paper, you will have understood this. They have a single parameter dwhich corresponds to the coincidence time window, and they clearly show that with values of d similar to what is used in real experiments, Bell is violated and the simulation agrees with QM but if no coincidence time window is introduced ie d = 0, which is NOT what is done in real experiments, the simulation respects Bell and disagrees with QM, even though not all photons are detected. So yeah, I am pretty sure that their simulation has nothing to do with detection efficiency, and you will be too if only you read the article.
Suppose the simulation was altered so that 100% of all photons were detected by the simulated detectors--would the model continue to violate Bell inequalities? If not, I would say that by definition it is exploiting the detector efficiency loophole, even if it is also exploiting the coincidence time loophole.
JesseM said:
his claim is perfectly correct if you understand what he means by "Bell is respected" (namely, that their model would obey Bell inequalities in an experimental setup that actually matched the observable experimental conditions assumed by Bell).
billschnieder said:
Exactly zero such experimental setups have been realized to date.
Do you think this is relevant to judging whether DrChinese's claim is correct or not? Bell's original proof did not concern any experiment, it was about comparing the theoretical predictions of local realism to the theoretical predictions of QM, and noting that their predictions must differ in certain theoretically-possible experimental setups. So, it's worth pointing out (as DrChinese did) that the model in the paper does not disprove Bell's theoretical claims, since it would obey the Bell inequalities in the theoretically-possible experimental setup Bell was discussing.
billschnieder said:
They present a model of a real experimental setup and THEY CLAIM that their model agrees with QM as evidenced by their own words which I have quoted to you.
They only claim that it agrees with QM for the specific experiments they analyze. I doubt they claim it would agree with QM in all the Aspect-type experiments that have been done to date, let alone in any experiment which is theoretically possible in QM
billschnieder said:
Dr C is free to state that in his opinion, their model agrees with Bell and disagrees with QM. But to say their model does not claim to match QM is false.
When he says it "disagrees with QM", I think he means that it would not make the same predictions as QM in all theoretically-possible experiments. Assuming that this is what he meant, then if your criticism is meant to be something more than a semantic quibble about the words he used to express this idea, do you think anything the authors said contradicts the claim that there model would not make the same predictions as QM in all theoretically-possible experiments?
 
  • #1,167
billschnieder said:
I see your point. My point though is the following: Claim (1) and (2) are not different within the context of the paper we are discussing so I do not see why you insist on making a distinction. The authors did not claim to be deriving "an LR model" in a general sense. Their focus in that paper is to present "an LR model of the experiment" and in that context you can not separate their model from the constraints imposed by the experimental situation being modeled. It seems from your phrasing of (1) that you prefer to to do that. But do you apply the same standard to QM? QM only gives predictions for clearly stated experimental setups. I don't think you will expect QM to generate a dataset which you will then analyze according to experiment X.

I'm not a good writer. In fact, I'm not a good communicator in general. Sorry, I'll try again. Keep in mind that I'm conveying my opinion about their work. If you know that my opinion is wrong, maybe you could explain that to me.

I like your phrase, "an LR model of the experiment." That's right on the money. Now, the experiment is also in agreement with QM. Does that mean the LR model and QM are equivalent? No. What's the difference between QM and the LR model? If you wrote a computer program to simulate a perfect set of QM data, i.e., no guess work as to which pairs of events are correlated, then it would give |S| > 2. If you do the same with the LR model, it will not give |S| > 2.

Therefore, what the LR model shows is that you cannot use this experiment to claim, "We have proof that QM's prediction of |S| > 2 is right," because a computer simulator (their LR model) which doesn't give |S| > 2 does create data which give |S| > 2 when analyzed via this experiment.

How's that?
 
  • #1,168
RUTA, JesseM and DrC,

I see that Mr. BS is back on track with the "Bell's Inequalities Triples Scam - BITS". Can you please verify if I’m a moron and Alain Aspect is a liar – or if it’s billschnieder who possesses both these noble attributes.

This is a slide from Alain Aspect himself, showing the measurements of the first famous EPR-Bell experiment which violated Bell's Inequality in 1982:

r6xwxz.jpg


I see "Measured value" and the violation of "Bell's limits", but I don’t see Alain Aspect measuring "entangled triples"...?:bugeye:?

So, what do you think...?:rolleyes:?
 
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  • #1,169
Deepak Kapur said:
Is action at a distance possible as envisaged by the EPR Paradox?

Dear Deepak,

Answering your OP in my terms:

Is action at a distance possible?

No; no way!

PS: Though it is immaterial to my general response above, I'd be happy to learn what this phrase might mean: ... as envisaged by the EPR Paradox?

With best regards,

JenniT
 
  • #1,170
JesseM said:
3) is ambiguous. It's true a dataset of pairs can be made to violate inequalities from a set of triples under certain sampling conditions, but not under the conditions assumed by Bell, where the choice of which two values to measure is random on each trial, and there is no correlation between the probability of a given triple and the choice of which pair to measure on a given trial.
I will let you respond to this one by yourself:
Bell's original proof did not concern any experiment, it was about comparing the theoretical predictions of local realism to the theoretical predictions of QM
An inequality such as X <= Y, means that "X" MUST always be less than or equal to "Y". If it is shown that in some cases X is greater than Y, the inequality is violated. It is the same as saying X can be greater than Y. There is nothing ambiguous there. It just means X is not necessarily less than Y as the inequality states.

I also notice that you actually agree with my point (3), except you claim that there are certain sampling assumptions in Bell's work which I have not taken into account. As you admitted in your later statement, Bell was never concerned about any actual experimental measurements or trials, so your earlier statement suggesting that Bell assumed data to be sampled in pairs and measured randomly on each trial is flatly wrong. There are no such claimed sampling assumptions in Bell's work which my point(3) supposedly violates. If you disagree point it out with a quote from Bell's work. Even if there was such an assumption, my point (3) still does not violate such a requirement.

I had hoped you would have a more substantive critique of those points.
 
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  • #1,171
billschnieder said:
I will let you respond to this one by yourself:
Bell's original proof did not concern any experiment, it was about comparing the theoretical predictions of local realism to the theoretical predictions of QM
I meant any real experiment that had actually been done. Obviously Bell's proof did concern a theoretical experiment, otherwise conditions like there being a spacelike separation between the measurements don't make any sense.
billschnieder said:
An inequality such as X <= Y, means that "X" MUST always be less than or equal to "Y".
If you specify some conditions which are necessary for the inequality to hold, then obviously you're only saying that X will be less than or equal to Y under those specific conditions.
billschnieder said:
I also notice that you actually agree with my point (3), except you claim that there are certain sampling assumptions in Bell's work which I have not taken into account. As you admitted in your later statement, Bell was never concerned about any actual experimental measurements or trials
Again, he was certainly concerned with some conditions that the experiments he was considering in theory should satisfy. Surely you aren't failing to understand something so extremely basic about Bell's proof? Do you think Bell would claim that the inequality would still be guaranteed to hold if there was no spacelike separation between measurements, for example?
billschnieder said:
so your earlier statement suggesting that Bell assumed data to be sampled in pairs and measured randomly on each trial is flatly wrong. There are no such claimed sampling assumptions in Bell's work which my point(3) supposedly violates. If you disagree point it out with a quote from Bell's work.
Look for example at p. 9 of http://cdsweb.cern.ch/record/142461/files/198009299.pdfpapers , where Bell uses the sock analogy:
Suppose, however, that the socks come in pairs. And suppose that we know by experience that there is little variation between the members of a pair, in that if one member passes a given test then the other also passes that same test if it is performed. Then from d'Espagnat's inequality we can infer the following:

(the number of pairs in which one could pass at 0 degrees and the other not at 45 degrees)

plus

(the number of pairs in which one could pass at 45 degrees and the other not at 90 degrees)

is not less than

(the number of pairs in which one could pass at 0 degrees and the other not at 90 degrees)

This is not yet empirically testable, for although the two tests in each bracket are now on different socks, the different brackets involv a different tests on the same sock. But we now add the random sampling hypothesis: if the sample of pairs is sufficiently large and if we choose at random a big enough subsample to suffer a given pair of tests, then the pass/fail fractions of the subsample can be extended to the whole sample with high probability.
billschnieder said:
Even if there was such an assumption, my point (3) still does not violate such a requirement.
How about the assumption that there is no correlation between the choice of two measurements on a particular trial and the probability of different possible combinations of three hidden variables? If you don't violate that assumption, I don't see how you can get a violation of the inequalities.
 
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  • #1,172
JenniT said:
Is action at a distance possible?

No; no way!

Finally a real convincing scientific proof, this is what we have all been waiting for! Thanks!
 
  • #1,173
I’m tired.

I’m tired of all this anti-intellectual propaganda. Hillbillies screaming out their stubborn ignorance – No way! Wrong! Read the paper!

How stupid can it get? And the worst thing of all – it’s contagious.

(This does not apply to DrC, RUTA, JesseM and my_wan, thank god.)

I just have to do this... for the "casual reader"... finally some balanced scientific intellectual reasoning, where "maybe" is not an invective... as an answer to OP, finally after +1100 posts:
http://arxiv.org/abs/quant-ph/0609163"

Quantum mechanics: Myths and facts
Hrvoje Nikolic
Theoretical Physics Division, Rudjer Boskovic Institute
Journal reference: Found.Phys.37:1563-1611,2007
...

6.3 (Non)locality without hidden variables?

Concerning the issue of locality, the most difficult question is whether QM itself, without hidden variables, is local or not. The fact is that there is no consensus among experts on that issue. It is known that quantum effects, such as the Einstein-Podolsky-Rosen-Bell effect or the Hardy effect, cannot be used to transmit information. This is because the choice of the state to which the system will collapse is random (as we have seen, this randomness may be either fundamental or effective), so one cannot choose to transmit the message one wants. In this sense, QM is local. On the other hand, the correlation among different subsystems is nonlocal, in the sense that one subsystem is correlated with another subsystem, such that this correlation cannot be explained in a local manner in terms of preexisting properties before the measurement. Thus, there are good reasons for the claim that QM is not local.

Owing to the nonlocal correlations discussed above, some physicists claim that it is a fact that QM is not local. Nevertheless, many experts do not agree with this claim, so it cannot be regarded as a fact. Of course, at the conceptual level, it is very difficult to conceive how nonlocal correlations can be explained without nonlocality. Nevertheless, hard-orthodox quantum physicists are trying to do that (see, e.g., [24, 25, 26, 27]). In order to save the locality principle, they, in one way or another, deny the existence of objective reality. Without objective reality, there is nothing to be objectively nonlocal. What remains is the wave function that satisfies a local Schrödinger equation and does not represent reality, but only the information on reality, while reality itself does not exist in an objective sense. Many physicists (including myself) have problems with thinking about information on reality without objective reality itself, but it does not prove that such thinking is incorrect.

To conclude, the fact is that, so far, there has been no final proof with which most experts would agree that QM is either local or nonlocal. (For the most recent attempt to establish such a consensus see [40].) There is only agreement that if hidden variables (that is, objective physical properties existing even when they are not measured) exist, then they must be nonlocal. Some experts consider this a proof that they do not exist, whereas other experts consider this a proof that QM is nonlocal. They consider these as proofs because they are reluctant to give up either of the principle of locality or of the existence of objective reality. Nevertheless, more open-minded (some will say – too open-minded) people admit that neither of these two “crazy” possibilities (nonlocality and absence of objective reality) should be a priori excluded.


P.S. action at a distance = nonlocality
 
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  • #1,174
JesseM said:
If you specify some conditions which are necessary for the inequality to hold, then obviously you're only saying that X will be less than or equal to Y under those specific conditions.
Point (1) which you did not respond or object to deduces that the only condition necessary is existence of triples of two-valued variables.
Again, he was certainly concerned with some conditions that the experiments he was considering in theory should satisfy.
No EPRB experiment in theory can produce triples of two-valued variables. You do not deny this, which is point (2).
Do you think Bell would claim that the inequality would still be guaranteed to hold if there was no spacelike separation between measurements, for example?
Bell does not have to claim it for it to be true. This is claim (6), and if you are objecting to it say so and let's have a little mathematical exercise to verify it. Are you contesting claim (6)?
How about the assumption that there is no correlation between the choice of two measurements on a particular trial and the probability of different possible combinations of three hidden variables? If you don't violate that assumption, I don't see how you can get a violation of the inequalities
Again I point you to claim (1) which establishes the minimum requirement for deriving Bell-type inequalities, and claim (2) which clearly states that the requirement is not met in any bell-test experiment. Are you now contesting claim (1) and/or (2)? If you are, simply say so.
 
  • #1,175
DevilsAvocado said:
I’m tired.

I’m tired of all this anti-intellectual propaganda. Hillbillies screaming out their stubborn ignorance – No way! Wrong! Read the paper!

How stupid can it get? And the worst thing of all – it’s contagious.

(This does not apply to DrC, RUTA, JesseM and my_wan, thank god.)

I just have to do this... for the "casual reader"... finally some balanced scientific intellectual reasoning, where "maybe" is not an invective... as an answer to OP, finally after +1100 posts:

P.S. action at a distance = nonlocality

Nikolic's opinion is what I sense in the community as well. Something "big" has to happen to bring consensus. Shortly after introducing Relational Blockworld at Bub's conference "New Directions in the Foundations of Physics" and at Price's conference "Time-Symmetric QM," I was naively enthusiastic. Aharonov quickly bust my bubble :-) Not to be mean, of course, he just wanted me to understand what I was up against. He had introduced his two vector formalism years ago and even used it to devise new experiments. He said the experiments were all the physics community at large cared about, and they weren't all that interested in them b/c they didn't disprove QM or nonlocality, etc. That's why I think it's got to be something "big," e.g., a new theory of physics, to break the log jam.
 
  • #1,176
billschnieder said:
Point (1) which you did not respond or object to deduces that the only condition necessary is existence of triples of two-valued variables.
That's true if you are considering an inequality where all values of the triples are known, and the pairs that appear in the inequality deal with every single triple that satisfies it. For example, this page states an inequality that's guaranteed to hold if you know the value of three variables A,B,C for some collection of objects:
The result of the proof will be that for any collection of objects with three different parameters, A, B and C:

The number of objects which have parameter A but not parameter B plus the number of objects which have parameter B but not parameter C is greater than or equal to the number of objects which have parameter A but not parameter C.

We can write this more compactly as:

Number(A, not B) + Number(B, not C) greater than or equal to Number(A, not C)
If we have a collection of objects and we know whether each object has A/not A, B/not B, and C/not C, then Number(A, not B) could include every object in the collection which has (A, not B), and likewise for the other pairs. In this case, the inequality is guaranteed to hold with no additional assumptions. On the other hand, if for each triplet we only sample two out of three properties, so Number(A, not B) would only be the number of triplets where the two properties we checked were A and B (so we didn't check C) and the result was (A, not B), and likewise for the other pairs, in this case it's no longer guaranteed that the inequality will hold, you need additional assumptions (particularly the assumption that the sampling was done in such a way that there should be no correlation between which two variables were sampled and the probability of the triplet having different possible combinations of all three variables).

So, if you are claiming in (1) that you could derive the inequality even in conditions where we only sample two of the three values for each triplet, then I would certainly object to that, although I would say that the inequality can be justified with additional assumptions like the one I mentioned about the choice of sample having no correlation to the full values of the triplet.
billschnieder said:
No EPRB experiment in theory can produce triples of two-valued variables. You do not deny this, which is point (2).
Right, I don't deny that. The assumption of triples is a theoretical assumption of hidden variables theories, but one cannot measure all three.
JesseM said:
Do you think Bell would claim that the inequality would still be guaranteed to hold if there was no spacelike separation between measurements, for example?
billschnieder said:
Bell does not have to claim it for it to be true. This is claim (6), and if you are objecting to it say so and let's have a little mathematical exercise to verify it. Are you contesting claim (6)?
Yes, I'm contesting it in the case where the different pairs are only based on the subset of objects where those two variables were sampled, though I wouldn't contest it in the case where the pairs in the inequality like Number(A, not B) represent all objects that satisfy that pair (i.e. all objects in the collection that have property A but not B, as opposed to just the ones where we measured A and B and found A, not B).

And what Bell claimed is relevant to your own statement "As you admitted in your later statement, Bell was never concerned about any actual experimental measurements or trials". Again, do you deny that Bell's proof assumes certain experimental conditions for the theoretical experiment under consideration, like the condition of a spacelike separation between measurements?
 
  • #1,177
This last post of yours is a masterpiece of obfuscation. So let us untangle it shall we. Here again for reference are the claims you are responding to:
billschnieder said:
1) Bell's inequalities can be derived from triples of dichotomous variables without any physical assumption
2) In Bell-test experiments only pairs of values are ever collected at a time (a dataset of pairs)
3) A dataset of pairs can be made to violate inequalities derived from a dataset of triples for purely mathematical reasons
4) I have provided mathematical proof of (1), (2) is an accepted fact. I have provided proof of (3) via simulation
5) Therefore, the violation of Bell's inequalities derived from triples, by experiments such as Bell-test experiments which only collect pairs, is not surprising, it is expected for purely mathematical reasons, having nothing to do with realism or locality.
6) Therefore, Bell's inequality can never be violated by a dataset of triples, even if the physical assumption of of spooky action at a distance is mandated!

Now let us go through one by one and see how you have responded so far

(1) Bell's inequalities can be derived from triples of dichotomous variables without any physical assumption
JesseM said:
That's true if you are considering an inequality where all values of the triples are known, and the pairs that appear in the inequality deal with every single triple that satisfies it.
And it is false when exactly? If all values in a triple are not known then you do not have a triple. The claim states clearly that Bell's inequality is an arithmetic relationship between triples of numbers each of which can take values of (+1 or -1). The claim is essentially that it is impossible to find triples of numbers obeying this requirement which will violate the inequality, irrespective of physical or statistical considerations. If you agree with it, simply saying so will do rather than go through a long winding rabbit trail that has nothing to do with the claim itself. If you disagree, use whatever method you like to provide me a triple of numbers each with values of (+1 or -1) which violates the inequality. Note, if you can not produce such a list of triples then not only are you admitting claim (1), you will also be admitting claim (6).

(2) In Bell-test experiments only pairs of values are ever collected at a time (a dataset of pairs)
JesseM said:
Right, I don't deny that. The assumption of triples is a theoretical assumption of hidden variables theories, but one cannot measure all three.
So you agree with (2), but I noticed how you sneaked in the underlined statement as if to suggest that claim (1) is only true for hidden variable theories. Claim (2) is about experiments not theories. Claim (1) is not concerned with any physical theory. It is a universally valid arithmetic relationship between any three variables with values (+1 or -1). This is why I say you are a master at obfuscation. So I strike the underlined text.

(3) A dataset of pairs can be made to violate inequalities derived from a dataset of triples for purely mathematical reasons
JesseM said:
So, if you are claiming in (1) that you could derive the inequality even in conditions where we only sample two of the three values for each triplet, then I would certainly object to that
Huh? An ingenious way of agreeing with claim (3) while appearing to object. You phrased your agreement with claim (3) as an objection to claim (1). Claim (3) states essentially that you can not derive the inequalities as an arithmetic relationship between only pairs of values. Remember, claim (1) is dealing with triples not pairs. Claim (3) is dealing with pairs not triples. Of course you say that claim (1) will not be valid for pairs, which is exactly what claim (3) says! I take it from the above you agree with claim (3).

(4) I have provided mathematical proof of (1), (2) is an accepted fact. I have provided proof of (3) via simulation
JesseM said:
...

(5) Therefore, the violation of Bell's inequalities derived from triples, by experiments such as Bell-test experiments which only collect pairs, is not surprising, it is expected for purely mathematical reasons, having nothing to do with realism or locality.
JesseM said:
...

(6) Therefore, Bell's inequality can never be violated by a dataset of triples, even if the physical assumption of spooky action at a distance is mandated!
JesseM said:
Yes, I'm contesting it in the case where the different pairs are only based on the subset of objects where those two variables were sampled, though I wouldn't contest it in the case where the pairs in ...
Another masterpiece at obfuscation. Claim (6) is dealing with a dataset of triples not pairs, did you see the word pairs anywhere there? So by contesting a dataset of pairs, you are responding to something other than claim (6). Do you claim that dataset of triples from a situation in which there was non-local communication between the devices generating the data will violate Bell's inequality. If you do, say so rather than put up the strawman of pairs in order to appear to be contesting the claim when you are not.

JesseM said:
Again, do you deny that Bell's proof assumes certain experimental conditions for the theoretical experiment under consideration, like the condition of a spacelike separation between measurements?
I will draw your attention back to claim (1). That is why claim (1) is so important because it demonstrates conclusively that you do not need any locality, probability, factorization etc assumptions to derive Bell's inequalities. All those assumptions are peripheral to the fact that the resulting inequalities are universally valid arithmetic relationships between triples of numbers with values (+1 or -1). As I mentioned to you earlier in this thread, these relations have been known for 100+ years. As soon as Bell assumed that properties existed simultaneously at three angles (a,b,c), he was guaranteed to obtain those relationships. It is completely peripheral, the reason for the three angles. He could have as well assumed that non-local communication was involved between space-like and it wouldn't have changed the resulting inequalities. Clearly, this means the inequalities are not relationships that must exist only for local hidden variable scenarios. They are valid for all possible scenarios involving three variables with values (+1 or -1). If you disagree, provide the triples of values which violate it using any assumption of your choosing, you can even assume FTL for all I care, just give me the triples and we can calculate.
 
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  • #1,178
RUTA said:
Nikolic's opinion is what I sense in the community as well. Something "big" has to happen to bring consensus. Shortly after introducing Relational Blockworld at Bub's conference "New Directions in the Foundations of Physics" and at Price's conference "Time-Symmetric QM," I was naively enthusiastic. Aharonov quickly bust my bubble :-) Not to be mean, of course, he just wanted me to understand what I was up against. He had introduced his two vector formalism years ago and even used it to devise new experiments. He said the experiments were all the physics community at large cared about, and they weren't all that interested in them b/c they didn't disprove QM or nonlocality, etc. That's why I think it's got to be something "big," e.g., a new theory of physics, to break the log jam.


Thanks RUTA,

Your interesting thoughts certainly help breaking the log jam in this thread. :smile:

Yes, it’s pretty obvious that something "BIG" has to happen to bring consensus and progress. This has always been the case in the history of science, and now the signs are flashing in the sky again. One major "lighthouse" (not on Shutter Island :wink:) is EPR-Bell, another is Quantum Gravity (QG).

I’m amazed of the "denial process" practiced in this thread right now. What are they afraid of? That KGB or CIA will read their minds "at a distance"?? What also amazes me is the fact that the denialers find the solution for everything around 1800 - 1850??

Anyhow, your story about http://en.wikipedia.org/wiki/Yakir_Aharonov" , and David Bohm worked closely with Albert Einstein!

This is the closest to "God" I will ever get! :smile:


P.S. What possible 'candidates' do you see for a "new theory of physics"?
 
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  • #1,179
billschnieder said:
(1) Bell's inequalities can be derived from triples of dichotomous variables without any physical assumption

And it is false when exactly? If all values in a triple are not known then you do not have a triple.
Here we again see the type of basic confusion I referred to in post #1158, between "assumptions about the observable conditions of the experiment" and "the theoretical assumptions about hidden variables". It may well be that our experimental conditions are such that we only sample a pair of properties for each object, but we can still consider the theoretical consequence of the assumption that each object actually had well-defined values for all three properties (which could be known by a hypothetical omniscient observer, even if we don't sample all three ourselves).

You accuse me of "obfuscation" but it's not clear you understand what I'm talking about here, so let me give a simple example. Suppose we have a collection of 9 objects, each of which can be tested on three properties A, B, C, and for a test of A we'd get either result A+ or A-, for a test of B we'd get B+ or B-, and for a test of C we'd get either C+ or C-. For each of the 9 objects, here is a list comparing the actual full set of three values known by an omniscient observer with the two properties we chose to sample for each object:

1. Actual values: A+,B+,C-. We sampled A and B, got (A+,B+).
2. Actual values: A-,B-,C-. We sampled B and C, got (B-,C-)
3. Actual values: A+,B-,C-. We sampled A and C, got (A+,C-).
4. Actual values: A-,B-,C+. We sampled A and B, got (A-,B-).
5. Actual values: A-,B+,C-. We sampled B and C, got (B+,C-).
6. Actual values: A+,B+,C-. We sampled A and C, got (A+,C-).
7. Actual values: A+,B-,C+. We sampled A and B, got (A+,B-).
8. Actual values: A-,B-,C+. We sampled B and C, got (B-,C+).
9. Actual values: A+,B-,C-. We sampled A and C, got (A+,C-).

Now consider the following inequality:

(Total number of objects with properties A+ and B-) + (Total number of objects with properties B+ and C-) greater than or equal to (Total number of objects with properties A+ and C-)

By dividing each term by the total number of objects (9 in this case), we get an almost-identical inequality, which I will call inequality #1:

(Fraction of all nine objects with properties A+ and B-) + (Fraction of all nine objects with properties B+ and C-) greater than or equal to (Fraction of all nine objects with properties A+ and C-)

Assuming that all objects have well-defined unchanging values for all three properties, we can prove as a purely mathematical matter that this inequality must hold (the proof is trivial--every triplet with A+ and C- must either be of type A+B+C- or type A+B-C-, and if the former it will also contribute to the number with B+ and C-, if the latter it will also contribute to the number with A+ and B-). And indeed you can see that in the example it does hold:

--Objects 3, 7 and 9 had A+ and B-, so (Fraction of all nine objects with properties A+ and B-) = 3/9

--Objects 1, 5, and 6 had B+ and C-, so (Fraction of all nine objects with properties B+ and C-) = 3/9

--Objects 1, 3, 6 and 9 had A+ and C-, so (Fraction of all nine objects with properties A+ and C-) = 4/9

And, it is indeed true that 3/9 + 3/9 is greater than or equal to 4/9.

On the other hand, consider the following different inequality #2 which concerns only properties that were actually sampled:

(Fraction of A,B samples which gave result A+, B-) + (Fraction of B,C samples which gave result B+, C-) greater than or equal to (Fraction of A,C samples which gave result A+,C-)

You can see that this inequality is violated in this example, because:

--Objects 1, 4 and 7 were sampled for A,B and of these only 7 gave result A+,B- so (Fraction of A,B samples which gave result A+, B-) = 1/3

--Objects 2, 5 and 8 were sampled for B,C and of these only 5 gave result B+,C- so (Fraction of B,C samples which gave result B+, C-) = 1/3

--Objects 3, 6 and 9 were sampled for A,C and 3, 6 and 9 all gave result A+,C- so (Fraction of A,C samples which gave result A+, C-) = 3/3

Since it's not true that 1/3 + 1/3 is greater than or equal to 3/3, the inequality is violated.

The Bell inequalities derived by Bell and other physicists are all inequalities where the terms deal with the variables that are sampled in some theoretical experiment (like the second inequality I wrote above), not with the full set of three values that are theorized to exist in local realism (like the first inequality above). But if you add some additional assumptions beyond local realism (assumptions which can themselves be justified by the details of how the experiment is conducted plus the assumption of local realism), like the assumption that the probability of different hidden triplets is not correlated with the two you actually choose to sample, then you can show that in the limit as your sample size approaches infinity, the probability that this inequality will be respected approaches 1:

(Fraction of A,B samples which gave result A+, B-) + (Fraction of B,C samples which gave result B+, C-) greater than or equal to (Fraction of A,C samples which gave result A+,C-)
billschnieder said:
The claim states clearly that Bell's inequality is an arithmetic relationship between triples of numbers each of which can take values of (+1 or -1). The claim is essentially that it is impossible to find triples of numbers obeying this requirement which will violate the inequality, irrespective of physical or statistical considerations. If you agree with it, simply saying so will do rather than go through a long winding rabbit trail that has nothing to do with the claim itself.
But your claim above seems to have two parts:

1a) If we have a set of triplets, we can derive a purely arithmetic inequality where each term represents the fraction of all triplets that have a certain pair of properties, and this inequality is guaranteed to hold mathematically.
1b) Bell's own inequality is an inequality of this type (implied by the part where you say 'Bell's inequality is an arithmetic relationship...")

I would agree with 1a but not with 1b, for the reasons explained above. Perhaps you did not actually mean for your claim 1) to include the subclaim 1b), in which case please clarify.
billschnieder said:
If you disagree, use whatever method you like to provide me a triple of numbers each with values of (+1 or -1) which violates the inequality.
I don't deny that for this type of arithmetic inequality it's impossible to find a set of triples which violate it, I just deny that Bell's own inequality is this type of inequality.
billschnieder said:
(3) A dataset of pairs can be made to violate inequalities derived from a dataset of triples for purely mathematical reasons
JesseM said:
So, if you are claiming in (1) that you could derive the inequality even in conditions where we only sample two of the three values for each triplet, then I would certainly object to that
Huh? An ingenious way of agreeing with claim (3) while appearing to object.
No, it just depends on what kind of "inequalities" you are talking about when you say "a dataset of pairs can be made to violate inequalities derived from a dataset of triples." If you're talking about inequalities where each term deals with the fraction of all triplets which have a certain pair of properties, like this one:

(Fraction of all objects with properties A+ and B-) + (Fraction of all objects with properties B+ and C-) greater than or equal to (Fraction of all objects with properties A+ and C-)

...in that case (3) is incorrect, it's impossible for pairs of this form to violate the inequalities if each pair is the correct fraction of triples that have the stated properties.

On the other hand, if you're talking about an inequality of this form:

(Fraction of A,B samples which gave result A+, B-) + (Fraction of B,C samples which gave result B+, C-) greater than or equal to (Fraction of A,C samples which gave result A+,C-)

...then in that case I agree with (3), such pairs can violate the inequalities. However, if we impose some additional conditions like that appear in derivations of Bell inequalities, we can show that the probability this inequality will be violated approaches zero in the limit as the number of samples approaches infinity.
billschnieder said:
(4) I have provided mathematical proof of (1), (2) is an accepted fact. I have provided proof of (3) via simulation
The simulation is only relevant to Bell's proof if your pairs are based on sampling a pair of values from a triple, where each term of the inequality is of a form like (Fraction of A,B samples which gave result A+, B-), and where you have included the appropriate conditions of the proof like the condition that the probability of different possible triplets is independent of which two variables are sampled.
billschneider said:
(5) Therefore, the violation of Bell's inequalities derived from triples, by experiments such as Bell-test experiments which only collect pairs, is not surprising
It should be very surprising for any advocate of local realism, since as I said if you include the (reasonable) conditions of Bell's proof like the no-conspiracy assumption (which can themselves be justified using the conditions of the experiment and the assumption of local realism), you find that the probability of the inequality being violated approaches zero as your sample gets very large.
billschnieder said:
(6) Therefore, Bell's inequality can never be violated by a dataset of triples, even if the physical assumption of spooky action at a distance is mandated!
True if we assume the values for all three values of a triple are unchanging, but if the choice of which property to sample first can alter the values of the other three properties (which, for spacelike-separated measurements would require spooky action at a distance), then the inequality can certainly be violated. Would you like a numerical example of this?
billschnieder said:
All those assumptions are peripheral to the fact that the resulting inequalities are universally valid arithmetic relationships between triples of numbers with values (+1 or -1). As I mentioned to you earlier in this thread, these relations have been known for 100+ years. As soon as Bell assumed that properties existed simultaneously at three angles (a,b,c), he was guaranteed to obtain those relationships.
Not if you allow for the possibility that the properties can change over time, or that there could be a correlation between the values of the three properties and the choice of which two to sample. That's why deriving the Bell inequalities isn't as simple as deriving the simple arithmetic inequality, you need to invoke additional physical assumptions about the experimental setup.
 
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  • #1,180
JesseM said:
Here we again see the type of basic confusion I referred to in post #1158, between "assumptions about the observable conditions of the experiment" and "the theoretical assumptions about hidden variables". It may well be that our experimental conditions are such that we only sample a pair of properties for each object, but we can still consider the theoretical consequence of the assumption that each object actually had well-defined values for all three properties (which could be known by a hypothetical omniscient observer, even if we don't sample all three ourselves).

JesseM, I’m afraid this is a much more severe dysfunction than "basic confusion". DrC has explained this rock-solid and crystal-clear to everyone who wishes to understand, several times, but it just doesn’t work for Mr. BS. I think there is no hope...

DrChinese said:
For those following this discussion, billschnieder is basically addressing this question:

For a stream of particles, we'll call them Alice, does Alice have well defined polarization values for angle settings a=0, b=120 and c=240 degrees which match the QM expectation value of .25?

EPR says YES, Alice does, unless you unreasonably require this to be proven by simultaneous prediction of those values. Their explanation is that since a, b OR c can be predicted (but not all simultaneously), that should be adequate as proof. You can predict a, b or c of course by first observing Alice's twin partner, Bob.

Now the reason I choose the angle settings I did for a, b and c is that allows me to construct a very simple dataset to test for Alice, and then compare to the QM expectation value. Assuming 4 photons in Alice's stream (to get us started):

a b c
+ + -
+ - +
- + -
+ - -

So the above would be possible realistic values per EPR. I made these up, in an attempt to provide values that would be as close as possible to the Quantum Mechanical predictions. Obviously, 4 is way too few to be rigorous but yet is sufficient to discuss. So let's consider what Alice's twin Bob would look like:

a b c
+ + -
+ - +
- + -
+ - -

I.e. the same as Alice. So suppose we compare all the permutations of observing Alice and Bob at different angle pairs ab, bc, and ac. What would we see? And the answer is that even for as few as 4 items, it is not possible to get closer than a 33% average correlation rate.

ab=.25
bc=.25
ac=.50
------
average=.333333

QM would predict 25%, i.e. cos^2(theta=120 degrees). Obviously, you would have a standard deviation which is too large to be meaningful for this few items. But in a larger sample the actual experimentally observed value is in fact 25%. Using the EPR logic, something is clearly wrong.

Bill's argument has something to do with 3 observers and either 2 or 3 attributes (I am not really sure which). But the actual question is whether 1 particle has 3 well defined values. If 1 does, then 2 do too. And that is what is being considered in Bell tests. Whenever you question this point, simple consider that EPR is saying:

Alice stream =
a b c
+ + -
+ - +
- + -
+ - -

Bob stream =
a b c
+ + -
+ - +
- + -
+ - -

On the other hand, QM goes no farther than this:

Alice stream =
a b
+ +
+ -
- +
+ -

Bob stream =
a b
+ +
+ -
- +
+ -

Which matches QM just fine.
 
  • #1,181
DevilsAvocado said:
P.S. What possible 'candidates' do you see for a "new theory of physics"?

I read the threads in "Beyond the Standard Model" looking for candidates, but haven't seen any to my liking. It seems the unification community isn't addressing foundational issues and the foundations community isn't addressing unification. I think Smolin was right when he said the foundational problems of quantum mechanics probably constitute “the most serious problem facing modern science" and this problem “is unlikely to be solved in isolation; instead, the solution will probably emerge as we make progress on the greater effort to unify physics.” [Smolin, L., The Trouble with Physics, Houghton Mifflin, Boston, 2006.] Unfortunately, I think most unification researchers take this to mean the foundational problems of QM will be solved in a unification theory, but they won't have any bearing on the construct of the theory. Even Smolin didn't know how LQG would bear on EPR-Bell when I asked him at the Wheeler Symposium in 2002. He wasn't dismissive, however, saying that was something he planned to look into. I guess I got my answer in the aforementioned book :smile:

As I said in an earlier post, we are currently working on a new approach to classical gravity (essentially nonseparable Regge calculus) based on what started as an interpretation of QM (Relational Blockworld). But, if you've ever worked with Regge calculus, you can appreciate that I'm hoping for something better :smile:
 
  • #1,182
JesseM;2823991 It may well be that our experimental conditions are such that we only [i said:
sample[/i] a pair of properties for each object, but we can still consider the theoretical consequence of the assumption that each object actually had well-defined values for all three properties (which could be known by a hypothetical omniscient observer, even if we don't sample all three ourselves).
You are the one confused. A dataset of triples means just that. If you know only pairs you have a dataset of pairs even if you assume that there was a third value for each pair which you do not know. Simply because you can not directly calculate the LHS of Bell's inequality without the a third value and if you are calculating the LHS using a separate experiment for each term, you can not claim to have a dataset of triples. If you assume that the three values exist and you would like to consider them together theoretically as triples (like Bell did), then you have a dataset of triples not pairs. So for all your explanations, you have not provided anything of contrary to my claim. Claim (1) is dealing with datasets of triples not pairs despite your wishes. Again if you agree with it just say you do. If you disagree say so as well.

The Bell inequalities derived by Bell and other physicists are all inequalities where the terms deal with the variables that are sampled in some theoretical experiment

In the theoretical experiment, the values exist as triples, the terms in the inequality are obtained from the triples and in this case the inequality can never be violated even if FTL is involved. You aren't saying anything interesting here that is supposed to counter what I said.

Go back to Bell's original work from equation (14) onwards where he introduces the third angle and follow the derivation from there. You will realize that everything prior to that point is peripheral. So contrary to your claims, Bell's inequality is in fact an arithmetic relationship between triples. I have provided mathematical proof by deriving the exact same equation without any other assumption than the presence of triples. Bell also assumes the presence of triples and he obtains the same mathematical relationship. Is this a coincidence? Note also that you fail to specify the extra assumption without which Bell's inequality can not be derived, provided you already have triples like Bell assumed.

Now if you are ready to admit claim (1) but want to argue that under certain conditions a dataset of pairs will also satisfy the inequality, but not under every condition, then that is understandable but non earth-shattering because my claim (1) already lays out the conditions under which those terms involving pairs will obey the inequality -- ie, when they are extracted from a dataset of triples! And my claim (3) already states that not all datasets of pairs will obey the inequality. So you will be admitting both claims (1) and (3) here. In this case then,the discussion will be to examine whether those conditions are met in the Bell test experiments in which only pairs are measured (claim (2)). I show below that the conditions are not met.

But your claim above seems to have two parts:

1a) If we have a set of triplets, we can derive a purely arithmetic inequality where each term represents the fraction of all triplets that have a certain pair of properties, and this inequality is guaranteed to hole mathematically.
1b) Bell's own inequality is an inequality of this type (implied by the part where you say 'Bell's inequality is an arithmetic relationship...")

I would agree with 1a but not with 1b, for the reasons explained above. Perhaps you did not actually mean for your claim 1) to include the subclaim 1b), in which case please clarify.

I don't deny that for this type of arithmetic inequality it's impossible to find a set of triples which violate it, I just deny that Bell's own inequality is this type of inequality.
So you accept claim (1) but deny that the inequality so derived is Bell's inequality. I suggest you stick to Bell's inequality rather than your toy version):
I have proven the arithmetic relationship
|ab+ac|-bc <= 1
from which it immediately follows that if you have a list of triples of numbers of any length where each number can take values (+1 or -1), the following is also true
|<ab> + <ac>| - <bc> <= 1

Bell's variables (a,b,c) constitute a triple of variables each of which can take values (+1 or -1), each symbol in the inequality has exactly the same meaning, the same as the input to my derivation. Is it your claim that using variables with the same meaning, and properties like Bell's and obtaining the inequalities like Bell without any other assumption is accidental? Yes or No?

If you disagree, provide a dataset of triples which obeys Bell's inequalities but violates the one I derived or vice versa. It should be easy for you to do. Since you claim there are other assumptions in Bell's inequality that makes my claim (1) inapplicable to Bell's inequalities, all you need do is use one of those assumptions to provide a dataset or even a single data point where both disagree. I will consider your failure to do that as an admission of my claim (1). Then we can go on to discuss the areas where you have genuine disagreement.

No, it just depends on what kind of "inequalities" you are talking about when you say "a dataset of pairs can be made to violate inequalities derived from a dataset of triples." If you're talking about inequalities where each term deals with the fraction of all triplets which have a certain pair of properties, like this one:

(Fraction of all objects with properties A+ and B-) + (Fraction of all objects with properties B+ and C-) greater than or equal to (Fraction of all objects with properties A+ and C-)

...in that case (3) is incorrect, it's impossible for pairs of this form to violate the inequalities if each pair is the correct fraction of triples that have the stated properties.
Huh? I'm talking about Bell's inequality not your toy version. Your version bears no resemblance to what Bell actually did, or to any actual Bell-test experimental situation. You will have to rephrase your objection in Bell's form or the form of what is actually done in Bell test experiments so that we can examine it if it makes any sense. The current version is just obfuscation and is far removed from what we are discussing.

...then in that case I agree with (3), such pairs can violate the inequalities. However, if we impose some additional conditions like that appear in derivations of Bell inequalities, we can show that the probability this inequality will be violated approaches zero in the limit as the number of samples approaches infinity.
As I have demonstrated and I hope you now agree, so long as you have a list of triples in hand with values restricted to (+1 or -1), whether theoretical or measured, no matter how the list was generated, Bell's inequality is never violated (claim 6).

Now let us examine how the experiments are actually performed. For reference, we have the inequality
|<ab> + <ac>| - <bc> <= 1
In actual experiments each term from the above is actually from a different experiment. In one run (say run 1), the experimenters measure <ab> (call it <a1b1>), from the next run, they measure <a2c2> and from the third run, they measure <b3c3>. Now you certainly will not deny that this is how such experiments are typically performed. If you disagree, say so.

Your argument, as much as one can be discerned, is that in the limit as N becomes large, the fact that three different experiments are used does not matter. But that is short-sighted and misses the point of my argument.

Note that in the inequality
|<ab> + <ac>| - <bc> <= 1

a,b,c within angled brackets represent lists of numbers +1, -1. This equation can be factored as:

|<a(b+c)>| - <bc> <= 1

The above implies that if <bc> = -1, then <(b+c)> must be zero, otherwise the inequality will be violated. Obviously it is easy to see that whenever bc = -1, (b+c) is zero for. Now, in the experimental situation describe above, the LHS being calculated is

|<a1b1> + <a2c2>| - <b3c3>

in order to be able to factor this like we did above and ensure that the inequality is true for any finite list of numbers measurable in an experiment, a1 within the angled brackets must be the same as a2, similarly b1 must be the same as b3 and c2 must be the same as c3. This means, not only must a1 have the exact same number of +1's and -1's as a2, the exact sequence must also match (same for b1,b3 and c2,c3).
You may naively think that it is possible to sort the numbers so that the sequence matches but note the following:

We take our first pair from the first run (a1,b1), we then sort our second pair (a2,c2) such that the a2 and a1 exactly match. Note we now have (a1,b1,a2,c2) four columns of data, but two columns are identical so we can drop a2 altogether and we now have our dataset of triples (a1,b1,c2). This dataset NEVER violates Bell's inequalities because we can calculate

|<a1b1> + <a1c2>| - <b1c2>
|<a(b1+c2)>| - <b1c2>
But this is not what is done in Bell test experiments, so let us try to continue our sorting to incorporate the third dataset. We immediately face a difficulty! We have a list of triples (a1, b1, c2), we need to sort (b3,c3) so that not only does b3 line up with b1, but also c3 lines up with c2! This is a practically impossible task. It is obvious from the above that the third term is not independent of the first two but is directly derived from it.

It should be very surprising for any advocate of local realism, since as I said if you include the (reasonable) conditions of Bell's proof like the no-conspiracy assumption (which can themselves be justified using the conditions of the experiment and the assumption of local realism), you find that the probability of the inequality being violated approaches zero as your sample gets very large.
Be my guest, provide a dataset of triples which violates |ab+ac|-bc <= 1. Feel free to include conspiracy, non-locally communication and any other kind of assumption you like in generating the dataset. You will never violate the inequality. If you can not do that, you must admit that the inequality has no bearing on presence or absence of locality, conspiracy, or any other physical assumption you might think of.

True if we assume the values for all three values of a triple are unchanging, but if the choice of which property to sample first can alter the values of the other three properties (which, for spacelike-separated measurements would require spooky action at a distance), then the inequality can certainly be violated. Would you like a numerical example of this?
I would love to see your dataset of triples which violates the inequality. Make any assumption you like, non-locality, spooky action, consipiracy, time-variation etc while generating the list. All I ask is that you give me a list of triples of numbers each with values (+1, -1) which violates |<ab>+<ac>|-<bc> <= 1 for the whole list, or |ab+ac|-bc <= 1 for each individual triple.

If you can not generate such a dataset, then you can not with a straight face claim that experiments violate genuine Bell inequalities. It simply points to a discrepancy with the way the data from experiments are treated as I have outlined above.
 
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  • #1,183
RUTA said:
I read the threads in "Beyond the Standard Model" looking for candidates

Wow! Why did I miss this forum!?
Thanks! Now I need some "gadget" the will give me time2... :smile:

RUTA said:
It seems the unification community isn't addressing foundational issues and the foundations community isn't addressing unification. I think Smolin was right when he said the foundational problems of quantum mechanics probably constitute “the most serious problem facing modern science" and this problem “is unlikely to be solved in isolation; instead, the solution will probably emerge as we make progress on the greater effort to unify physics.” [Smolin, L., The Trouble with Physics, Houghton Mifflin, Boston, 2006.] Unfortunately, I think most unification researchers take this to mean the foundational problems of QM will be solved in a unification theory, but they won't have any bearing on the construct of the theory.

This seems like a BIG problem for something "BIG" to happen... I just wonder if all this is too much for a "New Einstein" to handle alone...? On the other hand, QM was not a one-man-show... :rolleyes:

RUTA said:
Even Smolin didn't know how LQG would bear on EPR-Bell when I asked him at the Wheeler Symposium in 2002. He wasn't dismissive, however, saying that was something he planned to look into. I guess I got my answer in the aforementioned book

Cool that you’ve discussed EPRB with Smolin! :cool: Loop Quantum Gravity (LQG) looks promising. Can LQG "generate" classical spacetime that matches GR? And its predictions of violation of the constancy of the speed of light could maybe be a possible 'answer' to EPRB...!?

Anyhow, to layman like me, this is all very simple: You just have to figure out what space is, exactly. And then do the same for gravity and matter + some simple formulas for the overall interaction between The Three Musketeers. And your done, bada bing bada boom! o:)

(sorry, bad joke :redface:)

RUTA said:
As I said in an earlier post, we are currently working on a new approach to classical gravity (essentially nonseparable Regge calculus) based on what started as an interpretation of QM (Relational Blockworld). But, if you've ever worked with Regge calculus, you can appreciate that I'm hoping for something better

This is all very interesting, and I think you deserve a medal just for trying. I hope you will be successful!

(Asking me about Regge calculus, is like asking me for a weather report for the Moon... :biggrin:)

Thanks for the info, very interesting!
 
  • #1,184
billschnieder said:
You are the one confused. A dataset of triples means just that.
Bell's proof does not assume a "dataset of triples" if by "dataset" you mean the experimental data that each term in the inequality is assumed to be based on. In some variants of the proof it may assume there is some objective truth about all three members of the triple even if we can't measure them all, but the inequality only deals with measurable pairs.
billschnieder said:
If you know only pairs you have a dataset of pairs even if you assume that there was a third value for each pair which you do not know.
Yes, exactly! And the Bell inequality is a prediction about the statistics on such a dataset of pairs, given some assumptions about how they were gathered and the laws of physics that determine their values.
billschnieder said:
Simply because you can not directly calculate the LHS of Bell's inequality without the a third value and if you are calculating the LHS using a separate experiment for each term, you can not claim to have a dataset of triples. If you assume that the three values exist and you would like to consider them together theoretically as triples (like Bell did), then you have a dataset of triples not pairs.
Not sure what you mean by "consider them together theoretically as triples", if this is an important part of your argument you'll have to explain in more detail. We may be assuming theoretically the pairs are sampled from triples, but a term in the inequality like (Fraction of A,B samples which gave result A+, B-) still deals with a collection of measured pairs (specifically the pairs where we measured A,B). Do you disagree?
billschnieder said:
So for all your explanations, you have not provided anything of contrary to my claim. Claim (1) is dealing with datasets of triples not pairs despite your wishes. Again if you agree with it just say you do. If you disagree say so as well.
Based on your comment above I no longer know what you mean when you say "datasets of triples", you seem to be using that phrase in a rather odd way that you have never defined. According to my commonsense use of the term, you only have a "dataset of triples" if you actually measured three properties of each "object" (the 'thing' being a pair of entangled particles in most cases, or a short time interval where we can measure a SQUID ring at any of three times in the case of the Leggett-Garg inequality), if you only measured two for each "object" then by definition you have a dataset of pairs. So, for example, the inequality I mentioned earlier:

(Fraction of A,B samples which gave result A+, B-) + (Fraction of B,C samples which gave result B+, C-) greater than or equal to (Fraction of A,C samples which gave result A+,C-)

...would be an inequality dealing with a dataset of pairs, even though I explicitly assumed each pair was drawn from a well-defined triple. If you disagree, please give a careful definition of what you mean by the phrases "dataset of pairs" and "dataset of triples".
billschnieder said:
In the theoretical experiment, the values exist as triples, the terms in the inequality are obtained from the triples and in this case the inequality can never be violated even if FTL is involved.
I don't know what you mean by "the terms in the inequality are obtained from the triples" either. In the theoretical experiment I described, only two properties were measured from each object, and a term in the inequality like (Fraction of A,B samples which gave result A+, B-) dealt only with the subset of triples where A and B were sampled, not with the full set of triples. I showed explicitly how the inequality was violated in this case. Do you disagree that the terms in Bell's inequality also deal only with subsets of all the entangled particle pairs that were measured, so that for example a term like P(experimenter A found spin-up at 0 degrees, experimenter B found spin-down at 45 degrees) would deal only with the particle pairs that were actually measured with detector settings of 0 for A and 45 for B? So if we can't make the theoretical assumption that the full values of the triple are uncorrelated with the choice of detector settings, it would be possible for Bell's inequality to be violated too even in a local realist universe? (for example, if the detectors are actually set in the past light cone of the source emitting the particles, it's conceivable the source would "know" which settings the particles will encounter on each trial and tailor the triple of values to that, in just the right way to violate the inequality)
billschnieder said:
Go back to Bell's original work from equation (14) onwards where he introduces the third angle and follow the derivation from there. You will realize that everything prior to that point is peripheral.
Bell's "original work" is highly condensed, written for an audience of physicists who can be expected to understand his implicit assumptions, it's silly to ignore all his later writings where he stated the assumptions more explicitly, like the paper I quoted in post #1171 where he referred to the need to assume a "random sampling hypothesis" (also, since we are dealing with scientific ideas rather than religious scriptures, it is perfectly legitimate to consider the derivation of Bell inequalities by authors other than Bell). And even in that original paper, if you examine his equations you see he does assume that the probabilities of different hidden variables λ (which in his original proof completely determine the triplet of predtermined values for each detector setting) are not in any way correlated to the choice of detector settings a and b, even if he doesn't state this explicitly (for example, in equation (21) note that he uses the same probability distribution [tex]\rho(\lambda)[/tex] in calculating P(a,b) and P(a,c)). As I mentioned way back in post #861 on page 54 of this thread, that's what later authors have called the "no-conspiracy assumption".
billschnieder said:
So contrary to your claims, Bell's inequality is in fact an arithmetic relationship between triples.
So what do you think he was talking about in that quote from post #1171 where he talked about the need for a "random sampling hypothesis"?
billschnieder said:
I have provided mathematical proof by deriving the exact same equation without any other assumption than the presence of triples.
In physics, you can't assume one equation is the "exact same" as the other just because they are written in the same abstract form. For example, the two inequalities in my example:

(Fraction of all nine objects with properties A+ and B-) + (Fraction of all nine objects with properties B+ and C-) greater than or equal to (Fraction of all nine objects with properties A+ and C-)

and

(Fraction of A,B samples which gave result A+, B-) + (Fraction of B,C samples which gave result B+, C-) greater than or equal to (Fraction of A,C samples which gave result A+,C-)

Can both be written as:

F(A+,B-) + F(B+,C-) >= F(A+,C-)

...but their meaning is very different! The first is guaranteed to hold due to basic arithmetical considerations, but the second is not, and I explicitly showed it was violated in my example. If you make some additional assumptions like the no-conspiracy assumption and the assumption of a very large number of trials, you can show that the second should hold as well, but it's clearly a different inequality than the first since it requires different assumptions to derive.
billschnieder said:
Bell also assumes the presence of triples and he obtains the same mathematical relationship. Is this a coincidence? Note also that you fail to specify the extra assumption without which Bell's inequality can not be derived, provided you already have triples like Bell assumed.
One of the extra assumptions is the no-conspiracy assumption, which is what I was referring to in the previous post when I said "But if you add some additional assumptions beyond local realism (assumptions which can themselves be justified by the details of how the experiment is conducted plus the assumption of local realism), like the assumption that the probability of different hidden triplets is not correlated with the two you actually choose to sample"...
billschnieder said:
Now if you are ready to admit claim (1)
Nope, not if (1) includes the idea that Bell's inequality has the same meaning as the arithmetical inequality just because it can be written in the same form.
billschnieder said:
but want to argue that under certain conditions a dataset of pairs will also satisfy the inequality, but not under every condition, then that is understandable but non earth-shattering because my claim (1) already lays out the conditions under which those terms involving pairs will obey the inequality -- ie, when they are extracted from a dataset of triples!
Would you say the pairs in my example were "extracted from a dataset of triples"? The pairs were all extracted from a list of triples which would be known by an omniscient observer, though they weren't known by the experimenter so I wouldn't call them a "dataset". But if you would say that the pairs were "extracted from a datset of triples" then this is an explicit counterexample to your claim above, since the resulting pairs violated the inequality.
billschnieder said:
So you accept claim (1) but deny that the inequality so derived is Bell's inequality.
I thought it seemed to be part of claim (1) that the "inequality so derived is Bell's inequality", since part of claim (1) was "Bell's inequality is an arithmetic relationship between triples of numbers". I did ask for clarification on this point when I said "Perhaps you did not actually mean for your claim 1) to include the subclaim 1b), in which case please clarify." When I ask for clarification I'm not being rhetorical, I can't really discuss my opinion on a statement of yours if I'm unclear on what you're actually saying.
billschnieder said:
I suggest you stick to Bell's inequality rather than your toy version):
How is mine a "toy version"? A, B and C could stand for polarizer angles, with A+ and A- meaning spin-up or spin-down at some angle, for example. In that case

(Fraction of A,B samples which gave result A+, B-) + (Fraction of B,C samples which gave result B+, C-) greater than or equal to (Fraction of A,C samples which gave result A+,C-)

...is one of various Bell inequalities (it's exactly the one he was discussing in the quote from post #1171).
billschnieder said:
I have proven the arithmetic relationship
|ab+ac|-bc <= 1
from which it immediately follows that if you have a list of triples of numbers of any length where each number can take values (+1 or -1), the following is also true
|<ab> + <ac>| - <bc> <= 1
But it's only guaranteed to be true arithmetically if you actually know ab, ac and bc for every member of the list. If for each member of the list you can only sample two, so <ab> only refers to the average result on the subset of the list where you actually sampled a and b, then it's no longer guaranteed to be true. Do you disagree?
billschnieder said:
Bell's variables (a,b,c) constitute a triple of variables each of which can take values (+1 or -1), each symbol in the inequality has exactly the same meaning, the same as the input to my derivation. Is it your claim that using variables with the same meaning, and properties like Bell's and obtaining the inequalities like Bell without any other assumption is accidental? Yes or No?
No, it's not accidental, as any proof of a Bell-type inequality like this:

(Fraction of A,B samples which gave result A+, B-) + (Fraction of B,C samples which gave result B+, C-) greater than or equal to (Fraction of A,C samples which gave result A+,C-)

...would make use of the a more basic arithmetical inequality like this:

(Fraction of all triples with properties A+ and B-) + (Fraction of all triples with properties B+ and C-) greater than or equal to (Fraction of all triples with properties A+ and C-)

However, although proving the bottom inequality would be a necessary condition for deriving the top one, it would not be sufficient to derive the top one, to do so you need some additional assumptions like the no-conspiracy assumption.
billschnieder said:
If you disagree, provide a dataset of triples which obeys Bell's inequalities but violates the one I derived or vice versa.
Of course, no collection of unchanging triples could violate a basic arithmetical inequality like the one you derived. The problem is the opposite: if we are sampling pairs of variables from a collection of triples, it is possible to see a violation of a Bell inequality (which deals with observed statistics in the pairs we sampled) even when the collection of triples does not violate the more basic arithmetical inequality. My example was of exactly this type.
billschnieder said:
It should be easy for you to do. Since you claim there are other assumptions in Bell's inequality that makes my claim (1) inapplicable to Bell's inequalities, all you need do is use one of those assumptions to provide a dataset or even a single data point where both disagree.
I already showed that for the arithmetical inequality I was dealing with, are you saying you don't believe I can do something analogous for your inequality? In other words, are you saying I can't come up with a set of triples, along with a choice for each triple of which two variables are sampled by the experimenter, that violates this relation?

|(average value of a*b for all triples in which experimenter measured a and b) + (average value of a*c for all triples in which experimenter measured a and c)| - (average value of b*c for all triples in which experimenter measures b and c) <= 1

On the other hand, if you agree that it is possible for the above relation to be violated, despite the fact that this arithmetic relation never can be:

|(average value of a*b for all triples) + (average value of a*c for all triples)| - (average value of b*c for all triples) <= 1

...then that's exactly the point I am making, that Bell's inequality is of the first type rather than the second, so proving the second alone isn't sufficient to prove Bell's inequality.

billschnieder said:
Huh? I'm talking about Bell's inequality not your toy version. Your version bears no resemblance to what Bell actually did
There are plenty of Bell inequalities, and the one I was talking about was in fact an inequality discussed by Bell. Again see http://cdsweb.cern.ch/record/142461/files/198009299.pdfpapers which I referred to in post #1171, where in equation (9) on p. 10 of the pdf he writes:
(The probability of being able to pass at 0 degrees and not able at 45 degrees)
plus
(The probability of being able to pass at 45 degrees and not able at 90 degrees)
is not less than
(The probability of being able to pass at 0 degrees and not able at 90 degrees)
Aside from the fact that I talked about fractions with a pair of properties and here he is talking about the probability of having a pair of properties, this is exactly the same as the inequality I wrote.

Anyway, like I said, I'd be happy to come up with a similar example tailored to your inequality if you really need it.
billschnieder said:
You will have to rephrase your objection in Bell's form or the form of what is actually done in Bell test experiments so that we can examine it if it makes any sense. The current version is just obfuscation and is far removed from what we are discussing.
You love to jump to the conclusion that I am "obfuscating", don't you? The thought doesn't even cross your mind that there might be some gap in your knowledge and understanding of all things Bell-related (like not knowing that my inequality is one of the Bell inequalities, as explained above), so my examples and arguments might have some relevance that just hasn't occurred to you? You really are ridiculously uncharitable when it comes to interpreting other people's arguments, you always jump to the conclusion that they are saying something foolish rather than asking questions to see if you might be missing something.
billschnieder said:
As I have demonstrated and I hope you now agree, so long as you have a list of triples in hand with values restricted to (+1 or -1), whether theoretical or measured, no matter how the list was generated, Bell's inequality is never violated (claim 6).
Disagree, since once again the Bell inequalities deal with the probabilities of measuring a given pair of values, not with the probability that all triples have a given pair of values even if those weren't the two you measured.
 
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  • #1,185
(continued)

billschnieder said:
Now let us examine how the experiments are actually performed. For reference, we have the inequality
|<ab> + <ac>| - <bc> <= 1
In actual experiments each term from the above is actually from a different experiment. In one run (say run 1), the experimenters measure <ab> (call it <a1b1>), from the next run, they measure <a2c2> and from the third run, they measure <b3c3>. Now you certainly will not deny that this is how such experiments are typically performed. If you disagree, say so.
Why would I disagree, when this is exactly the point I keep trying to underscore? When <ab> refers to the average of a*b only over those trials where a and b was measured, not to the average of a*b over all triples regardless of which two were measured, then the inequality is no longer guaranteed to hold by arithmetic alone, you need additional assumptions.
billschnieder said:
Your argument, as much as one can be discerned, is that in the limit as N becomes large, the fact that three different experiments are used does not matter.
Here you seem to be talking about conditions under which an inequality like this:

|(average value of a*b for all triples in which experimenter measured a and b) + (average value of a*c for all triples in which experimenter measured a and c)| - (average value of b*c for all triples in which experimenter measures b and c) <= 1

...can be derived. This is an entirely separate issue from the other point I was arguing, which was just the idea that the above inequality is not guaranteed to hold in spite of the fact that its arithmetical analogue is guaranteed:

|(average value of a*b for all triples) + (average value of a*c for all triples)| - (average value of b*c for all triples) <= 1

Anyway, if you agree that these types of inequalities are conceptually separate, that Bell's inequality was of the top type, and that a proof of the bottom one doesn't constitute a proof of the top (even if we assume that there were well-defined triples on each run despite the fact that we only sampled too), then at least we'd be getting somewhere. In that case we could move onto the separate question, what assumptions are needed to justify the top type of inequality, which matches the type of inequalities derived by Bell? In this case the assumption of a large number of trials is part of it, but a more important part is the idea that the probability of having a given hidden triple on each trial is not correlated to the choice of which two values were actually measured on that trial.
billschnieder said:
But that is short-sighted and misses the point of my argument.

Note that in the inequality
|<ab> + <ac>| - <bc> <= 1

a,b,c within angled brackets represent lists of numbers +1, -1. This equation can be factored as:

|<a(b+c)>| - <bc> <= 1
It doesn't make sense to me to factor <ab> + <ac> as <a(b+c)>. Suppose that a1b1 means on run #1 you measured properties a and b and got results a1 and b1, likewise a2c2 would mean on run #2 you measured properties a and c and got results a2 and c2, etc. Let's imagine a very short experiment where we measured a and b on the first three runs, and measured a and c on the next three runs, followed by b and c on the last three. Then <ab> + <ac> would be equivalent to:

(a1b1 + a2b2 + a3b3)/3 + (a4c4 + a5c5 + a6c6)/3

Or equivalently

(a1b1 + a2b2 + a3b3 + a4c4 + a5c5 + a6c6)/3

But with the averages written out in this explicit form, I don't see how it makes sense to reduce this to <a(b+c)>. If you think it does make sense, can you show what that factorization would look like written out in the same sort of explicit form?
billschnieder said:
The above implies that if <bc> = -1, then <(b+c)> must be zero, otherwise the inequality will be violated.
That doesn't seem to be true either, not if by <(b+c)> you mean the following:

(b1 + b2 + b3 + c4 + c5 + c6)/3

In that case suppose we have the following data:

a1=-1, b1=+1, a2=+1, b2=+1, a3=+1, b3=-1, a4=+1, c4=+1, a5=+1, c5=-1, a6=+1, c6=+1, b7=+1, c7=-1, b8=-1, c8=+1, b9=-1, c9=+1

In this case, <bc> = (b7c7 + b8c8 + b9c9)/3 = ((+1*-1) + (-1*+1) + (-1*+1))/3 = -1

And |<ab> + <ac>| = |(a1b1 + a2b2 + a3b3 + a4c4 + a5c5 + a6c6)/3| =
|((-1*+1) + (+1*+1) + (+1*-1) + (+1*+1) + (+1*-1) + (+1*+1))/3| = 0

So, the inequality is satisfied, since |<ab> + <ac>| - <bc> 0 - (-1) = +1

But if <(b + c)> = (b1 + b2 + b3 + c4 + c5 + c6)/3, this is equal to:

((+1) + (+1) + (-1) + (+1) + (-1) + (+1))/3 = 2/3, not zero.

billschnieder said:
Be my guest, provide a dataset of triples which violates |ab+ac|-bc <= 1. Feel free to include conspiracy, non-locally communication and any other kind of assumption you like in generating the dataset.
Again, are you just asking me to show triplets which violate this inequality?

|(average value of a*b for all triples in which experimenter measured a and b) + (average value of a*c for all triples in which experimenter measured a and c)| - (average value of b*c for all triples in which experimenter measures b and c) <= 1
 
  • #1,186
JesseM said:
Bell's proof does not assume a "dataset of triples" if by "dataset" you mean the experimental data that each term in the inequality is assumed to be based on. In some variants of the proof it may assume there is some objective truth about all three members of the triple even if we can't measure them all, but the inequality only deals with measurable pairs.
You are just quibbling here. If you did not understand what I meant why were you objecting? Bell's inequality is derived by assuming the existence of triples (a,b,c) and the inequality imposes constraints on how the pairs (a,b), (a,c) and (b,c) from these triples should behave. You start out with the triples and the terms involving pairs are extracted from the triples. If you have arbitrary datasets of pairs (i,j), (k,l),(m,n) you can not calculate anything comparable to Bell's inequality UNLESS you rearrange them such that they form pairs derived from a triples just like Bell did. This is not rocket science.

And the Bell inequality is a prediction about the statistics on such a dataset of pairs, given some assumptions about how they were gathered and the laws of physics that determine their values.
Wrong. I have proven that you do not need any laws of physics to determine the same constraints between (a,b,c). Others have done it as well. Your only response to that is the claim that my derived inequality is different from Bell's. But that is hot air and you know it. I have exactly the same inequality like Bell, ALL the terms in my inequality mean exactly the same as that of Bell. Yet for some mysterious reason, you claim the two are different just because they were derived differently. That is outrageous. You have not, and can not point out any valid difference between the two inequalities because there is none. You continue to obfuscate by bringing in tangential issues. Why don't you deal with the exact equation I presented? Why do you find the need to bring in your own equation with terms defined as you like, other than to obfuscate the issue?

Not sure what you mean by "consider them together theoretically as triples", if this is an important part of your argument you'll have to explain in more detail.
Oh, I am sure you understand very well what I mean. It is exactly what Bell meant when he said on page 406 of his original article that:
Bell said:
It follows that c is another unit vector
P(a,b) - P(a,c) = ...
Contrary to what you seem to be claiming here, according to Bell (a,b,c) are existing together. So when he goes on to derive his inequality half a page later to be

1 + P(b,c) >= |P(a,b) -P(a,c)|

he has nothing in mind other than that those terms originate from the triple (a,b,c) which exist together. You should know this.

We may be assuming theoretically the pairs are sampled from triples, but a term in the inequality like (Fraction of A,B samples which gave result A+, B-) still deals with a collection of measured pairs (specifically the pairs where we measured A,B). Do you disagree?
Please enough of the "fraction of A,B samples ... blah blah". Bell isn't dealing with any fractions and neither am I, so why obfuscate with this nonsense. Deal with the inequality I presented or Bell's which is stated above.

According to my commonsense use of the term, you only have a "dataset of triples" if you actually measured three properties of each "object"
If you are suggesting that dataset only means something that is actually measured, then why do you insist that Bell was modelling a dataset of pairs. Bell did not actually measure anything in order to obtain his inequalities did he? He considered triples of properties ad his inequalities are defined over these triples. That is all I need. No need to obfuscate by dragging us into discussions about SQUID etc.

I don't know what you mean by "the terms in the inequality are obtained from the triples" either.
Read page 406 of Bell's original paper. He assumes that there exist 3 vectors (a,b,c), also known as a TRIPLE then he derives the inequalities

1 + P(b,c) >= | P(a,b) - P(a,c)|

Each term in the above contains only two vectors from the same TRIPLE. In other words, the symbols (a,b,c) MUST mean exactly the same thing in each term!


So if we can't make the theoretical assumption that the full values of the triple are uncorrelated with the choice of detector settings, it would be possible for Bell's inequality to be violated too even in a local realist universe? (for example, if the detectors are actually set in the past light cone of the source emitting the particles, it's conceivable the source would "know" which settings the particles will encounter on each trial and tailor the triple of values to that, in just the right way to violate the inequality)
Rubbish. Provide me a dataset of triples which violates the above inequality, for which the terms (a,b,c) mean exactly the same thing in each term. Use whatever assumptions of conspiracy or "source knowing settings" that you like. All I want is for you back up your claim that it is possible for the inequality to be violated without extra assumptions in addition to the existence of triples (a,b,c). I have been asking you this for the last 3-4 posts and you haven't provided one, yet you keep claiming that without your extra assumptions the inequality can be violated.

Bell's "original work" is highly condensed, written for an audience of physicists who can be expected to understand his implicit assumptions, it's silly to ignore all his later writings where he stated the assumptions more explicitly, like the paper I quoted in post #1171
...
In physics, you can't assume one equation is the "exact same" as the other just because they are written in the same abstract form.
What is silly is to suggest that two inequalities which are exactly identical, down to the meanings of the terms are in fact different because they were derived differently. That is silly in any field of science including physics.

For example, the two inequalities in my example:

(Fraction of all nine objects with properties A+ and B-) + (Fraction of all nine objects with properties B+ and C-) greater than or equal to (Fraction of all nine objects with properties A+ and C-)

and

(Fraction of A,B samples which gave result A+, B-) + (Fraction of B,C samples which gave result B+, C-) greater than or equal to (Fraction of A,C samples which gave result A+,C-)

Can both be written as:

F(A+,B-) + F(B+,C-) >= F(A+,C-)
More obfuscation masterpiece. The symbols don't mean the same thing for both cases. The terms in my inequality mean exactly the same as those in Bell's.

But it's only guaranteed to be true arithmetically if you actually know ab, ac and bc for every member of the list. If for each member of the list you can only sample two, so <ab> only refers to the average result on the subset of the list where you actually sampled a and b, then it's no longer guaranteed to be true. Do you disagree?

Huh? Isn't this what I have been telling you and you've been objecting? Isn't that exactly what my claim (1) and claim (3) are talking about? Why would you claim to object to something you actually agree with unless you like quibbling.

However, although proving the bottom inequality would be a necessary condition for deriving the top one, it would not be sufficient to derive the top one, to do so you need some additional assumptions like the no-conspiracy assumption.
All I ask is that you give conspiracy-infested dataset which violates Bell's inequality. Remember, (a,b,c) must mean the same thing in each term of the inequality, unless by conspiracy you mean failure to make sure (a,b,c) mean the same in each term, such as by using a different "a's" for different terms.

Of course, no collection of unchanging triples could violate a basic arithmetical inequality like the one you derived.
Is it your claim that they can violate Bell's inequality? And by unchanging do you mean the meanings (a,b,c) is changing from one term to the next? If that is what you mean, then that is strange because according to Bell, (a,b,c) must mean the same thing for all terms, so changing the value from term to term is not being faithful to Bell.

I already showed that for the arithmetical inequality I was dealing with, are you saying you don't believe I can do something analogous for your inequality? In other words, are you saying I can't come up with a set of triples, along with a choice for each triple of which two variables are sampled by the experimenter, that violates this relation?
You can not come up with a dataset of triples for which Bell's inequality will be violated no matter how the triples are generated, no matter the physical conditions you apply. Provide the dataset of triples and we can talk. This is my claim (1) and claim (6).
On the other hand, it is possible to come up with a dataset of pairs for which Bell's inequality will be violated. This is my claim (3). And the reason why this dataset of pairs will violate the inequality is entirely mathematical and has to do with the fact that the symbols will not mean exactly the same thing in all terms of the inequality like Bell assumed.

...then that's exactly the point I am making, that Bell's inequality is of the first type rather than the second, so proving the second alone isn't sufficient to prove Bell's inequality.
So far you seem to be agreeing with all of my claims and the so called objections are mere appearances.

Anyway, like I said, I'd be happy to come up with a similar example tailored to your inequality if you really need it.
All I ask is that you provide the dataset of triples which violates Bell's inequality. You can impose any physical constrains of your choosing, such as non-locality, conspiracy, and any other feature of your choosing. Just provide the dataset of triples. If you can not, then don't be surprised when I claim that not even FTL can violate Bell's inequalities (claim 6).

You love to jump to the conclusion that I am "obfuscating", don't you?
Because I have noted your special expertise before and I am on the lookout for it. Let's just say I know your tactics, I'm not stupid.

The thought doesn't even cross your mind that there might be some gap in your knowledge and understanding of all things Bell-related
This sounds very much like your autobiography.

(like not knowing that my inequality is one of the Bell inequalities, as explained above), so my examples and arguments might have some relevance that just hasn't occurred to you?
It clearly occurs to me that if somebody brings out every obscure form of Bell's inequalities in a discussion where simply sticking to the form being discussed will be clearer, such is an attempt either at obfuscation

You really are ridiculously uncharitable when it comes to interpreting other people's arguments, you always jump to the conclusion that they are saying something foolish rather than asking questions to see if you might be missing something.
This also sound very much like your autobiography. For example, see the following statement of yours which purports to be responding to a claim of mine but is actually not because I never argued anything of the sort being alledged.

[quoteDisagree, since once again the Bell inequalities deal with the probabilities of measuring a given pair of values, not with the probability that all triples have a given pair of values even if those weren't the two you measured.[/QUOTE]
 
  • #1,187
JesseM said:
Why would I disagree, when this is exactly the point I keep trying to underscore? When <ab> refers to the average of a*b only over those trials where a and b was measured, not to the average of a*b over all triples regardless of which two were measured, then the inequality is no longer guaranteed to hold by arithmetic alone, you need additional assumptions.

If by this you mean the inequality in which (a,b,c) mean something different from term to term needs extra assumptions to be valid, then you are not saying anything relevant to what Bell derived or what I derived. In Bell's case as in mine, (a,b,c) must mean exactly the same thing for each term. I still do not see an actual objection to my claims despite your many words. You build a straw man and then knock it down and when I point out to you that what you are knocking down is not my claim, you express surprise.

Here you seem to be talking about conditions under which an inequality like this:

|(average value of a*b for all triples in which experimenter measured a and b) + (average value of a*c for all triples in which experimenter measured a and c)| - (average value of b*c for all triples in which experimenter measures b and c) <= 1

...can be derived. This is an entirely separate issue from the other point I was arguing, which was just the idea that the above inequality is not guaranteed to hold in spite of the fact that its arithmetical analogue is guaranteed:

|(average value of a*b for all triples) + (average value of a*c for all triples)| - (average value of b*c for all triples) <= 1

You are confused. If I have a dataset of triples such as:
a b c
1: + + -
2: + - +
3: + - -
4: - + -
5: - - +
...

in iteration (1), if the experimenter measured (a,b) they will obtain (++) and if they measured (b,c) they would have obtained (+-). So contrary to your statements above, there is no difference between
"average value of a*b for the all triples" and "average value of a*b for all triples for which the experimenter measured a and b". So the distinction you are trying to impose on the inequalities is not there. That is why you are unable to provide a dataset of triples which satisfies one but not the other. Except of course you do not have a dataset of triples and I have explained to you what you must do in that case to get the dataset of triples from separate datasets of pairs. Otherwise the symbols in:
|<ab> + <ac>| - <bc> < = 1
will not mean the same thing from term to term! The only way you can ever violate the above inequality is if the symbols do not mean the same from term to term. If you disagree provide a dataset of triples for which the symbols in the terms mean the same thing but the inequality is violated.

It doesn't make sense to me to factor <ab> + <ac> as <a(b+c)>. Suppose that a1b1 means on run #1 you measured properties a and b and got results a1 and b1, likewise a2c2 would mean on run #2 you measured properties a and c and got results a2 and c2, etc. Let's imagine a very short experiment where we measured a and b on the first three runs, and measured a and c on the next three runs, followed by b and c on the last three. Then <ab> + <ac> would be equivalent to:

(a1b1 + a2b2 + a3b3)/3 + (a4c4 + a5c5 + a6c6)/3

Or equivalently

(a1b1 + a2b2 + a3b3 + a4c4 + a5c5 + a6c6)/3

But with the averages written out in this explicit form, I don't see how it makes sense to reduce this to <a(b+c)>.
The factorization <ab> + <ac> = <a(b+c)> is not my idea, it is Bell's. Look at page 406 of his original paper, the section leading up to equation (15) and shortly there after. And that is why you need to sort them because without doing that, you can not factorize it!

Also, you are confusing runs with iterations. Note that within angled brackets, terms such as a1,b1, etc are lists of numbers with values (+1,-1) since we are calculating averages. So if you performed three runs of the experiment in which you measured (a,b) on the first, (a,c) on the second and (b,c) on the third the averages from each run will be

<a1*b1> for run 1
<a2*c2> for run 2
<b3*c3> for run 3

The numbers after the letter correspond to the run number, not the iteration.

Again you are agreeing while appearing to disagree with me. My argument is that you can not apply Bell's inequality to dataset of pairs obtained in this way UNLESS you sort them such that a1 becomes equivalent to a2 and b2 to b3 etc. If you do not do that, you do not have terms that can be used in Bell's inequality or the one I derived.

That doesn't seem to be true either, not if by <(b+c)> you mean the following:

(b1 + b2 + b3 + c4 + c5 + c6)/3

In that case suppose we have the following data:

a1=-1, b1=+1, a2=+1, b2=+1, a3=+1, b3=-1, a4=+1, c4=+1, a5=+1, c5=-1, a6=+1, c6=+1, b7=+1, c7=-1, b8=-1, c8=+1, b9=-1, c9=+1

In this case, <bc> = (b7c7 + b8c8 + b9c9)/3 = ((+1*-1) + (-1*+1) + (-1*+1))/3 = -1

And |<ab> + <ac>| = |(a1b1 + a2b2 + a3b3 + a4c4 + a5c5 + a6c6)/3| =
|((-1*+1) + (+1*+1) + (+1*-1) + (+1*+1) + (+1*-1) + (+1*+1))/3| = 0

So, the inequality is satisfied, since |<ab> + <ac>| - <bc> 0 - (-1) = +1

Essentially you have the datasets of pairs as follows:

Run1:
a1b1
-+
++
+-

Run2:
a2c3
++
+-
++

Run3:
b3c3
+-
-+
-+


<a1b1> = -1/3
<a2c2> = 1/3
<b3c3> = -1

Note that the symbols in the inequality do not mean the same thing so we can not factor them the way Bell did. You can only factor them if you have a dataset of triples, or you resort the dataset of pairs so that it becomes a dataset of triples. Using your dataset above, let us focus the last two runs. we can write them down as follows:
a2 c2 b3 c3
+ + + -
+ - - +
+ + - +
Clearly we can resort the last two columns so that the c2 column matches the c3 column to get
a2 c2 b3 c3
+ + - +
+ - + -
+ + - +

And since c2 and c3 are now equivalent, we can drop the c3 column altogether and we now have our dataset of triples which can never violate the inequality. We do not even need the first dataset of pairs.
a c b
+ + -
+ - +
+ + -

<ab> = -1/3
<ac> = 1/3
<bc> = -1

It is now obvious why this contrived example obeyed the inequality even though the symbols were not the same.

But note now that after sorting

<a(b+c)> = <ab> + <ac> = 0

In any case, I wasn't asking you to give me a dataset of pairs which obeys the inequality. I was asking you to give me a dataset of triples which violates it.
 
  • #1,188
JesseM said:
Again, are you just asking me to show triplets which violate this inequality?

Yes, this is really what this genius is expecting.
billschnieder said:
I was asking you to give me a dataset of triples which violates it.


Mr. BS is probably the only man on this planet who sees this magnificent disproval of Bell's Theorem – where a dataset of TRIPLES is required from an entangled PAIR of TWO photons.

Absolutely brilliant.

Mr. BS has not only proven John Bell, Alain Aspect & Anton Zeilinger wrong, but also the whole board of the Wolf Foundation. Professor Alain Aspect, a member of the French Academy of Sciences and French Academy of Technologies, received together with professor Anton Zeilinger http://www.wolffund.org.il/cat.asp?id=25&cat_title=PHYSICS" :
For their fundamental conceptual and experimental contributions to the foundations of quantum physics, specifically an increasingly sophisticated series of tests of Bell’s inequalities or extensions there of using entangled quantum states.


The Wolf Prizes in physics and chemistry are often considered the most prestigious awards in those fields after the Nobel Prize.

It’s absolutely amazing that one unknown man, Mr. BS, posses all this knowledge and superior intelligence.

This groundbreaking discovery must result in the 2010 Nobel Prize in Physics!
 
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  • #1,189
billschnieder said:
You are just quibbling here. If you did not understand what I meant why were you objecting?
Because I thought I understood, but then you objected to my statement:
It may well be that our experimental conditions are such that we only sample a pair of properties for each object, but we can still consider the theoretical consequence of the assumption that each object actually had well-defined values for all three properties (which could be known by a hypothetical omniscient observer, even if we don't sample all three ourselves).
by saying:
You are the one confused. A dataset of triples means just that. If you know only pairs you have a dataset of pairs even if you assume that there was a third value for each pair which you do not know. Simply because you can not directly calculate the LHS of Bell's inequality without the a third value and if you are calculating the LHS using a separate experiment for each term, you can not claim to have a dataset of triples. If you assume that the three values exist and you would like to consider them together theoretically as triples (like Bell did), then you have a dataset of triples not pairs.
So, that response made me realize that I didn't really understand what you were talking about.
billschnieder said:
Bell's inequality is derived by assuming the existence of triples (a,b,c) and the inequality imposes constraints on how the pairs (a,b), (a,c) and (b,c) from these triples should behave. You start out with the triples and the terms involving pairs are extracted from the triples. If you have arbitrary datasets of pairs (i,j), (k,l),(m,n) you can not calculate anything comparable to Bell's inequality UNLESS you rearrange them such that they form pairs derived from a triples just like Bell did. This is not rocket science.
Well, yes, your description of starting with theoretical triples and then extracting measured pairs seems to be identical to what I said in the quote above: "our experimental conditions are such that we only sample a pair of properties for each object, but we can still consider the theoretical consequence of the assumption that each object actually had well-defined values for all three properties". But your response to that was to say I was confused. So, were you disagreeing with my claim that Bell's proof (one version of it anyway) starts with the theoretical assumption of triples (a,b,c) whose values might be known by an omniscient observer (or by us if we're writing down a hypothetical example like the one I gave) but aren't known by the experimenter, and then assumes that "we only sample a pair of properties" for each triple, like (a,c)?
billschnieder said:
Wrong. I have proven that you do not need any laws of physics to determine the same constraints between (a,b,c). Others have done it as well. Your only response to that is the claim that my derived inequality is different from Bell's. But that is hot air and you know it. I have exactly the same inequality like Bell, ALL the terms in my inequality mean exactly the same as that of Bell.
No, they don't. The terms in the purely arithmetical inequality are of this form:

(Fraction of all triples with properties A+ and B-)

While the terms in Bell inequalities are of this form:

(Fraction of A,B samples which gave result A+, B-)

This does make a significant difference, since as I already showed, given a set of triples an inequality with terms of the first form can never be violated, but given a set of triples and a choice of which pair to sample from each triple, a similar inequality with terms of the second form can be violated. Instead of nebulous insults like "that is hot air and you know it", you might actually address what you think is wrong, specifically, with this claim. Do you disagree that even if an inequality with terms of the first form is impossible to violate, an identical-looking inequality with terms of the second form can be violated? Do you disagree that the terms in Bell's inequalities are understood to be of the second form rather than the first form? Without specifics like this I have no idea what you think is wrong with my claims.

edit: if you're going to object to my talking about 'fractions of triples' since this makes sense for other Bell inequalities but not the one you want to discuss, consider the other two forms I discuss below, comparing (average value of a*b for all triples) with (average value of a*b for all triples where experimenter sampled a and b)
JesseM said:
Not sure what you mean by "consider them together theoretically as triples", if this is an important part of your argument you'll have to explain in more detail.
billschnieder said:
Oh, I am sure you understand very well what I mean.
Again with the uncharitable assumptions, basically accusing me of lying. As before, the reason the phrase was confusing was because it was couched as an objection to my own statement which was saying nothing more than that we were assuming theoretically the triples existed but also assuming that only a pair was sampled by the experimenter. You objected to something in my statement, and your argument involved the sentence "If you assume that the three values exist and you would like to consider them together theoretically as triples". If it wasn't for the context I would assume I did understand what this sentence meant--that at a theoretical level we assume the existence of triples, even if we don't assume they're known to the theoretical experimenter--but since this was part of an objection to my statement which said exactly the same thing, that suggested my interpretation of the meaning was wrong, leaving me unsure about what you did mean. Of course it's also possible you just misunderstood the statement of mine you were objecting to, so if you actually understood what I meant you wouldn't object.

Communication isn't always easy, but it's easier if each person asks for clarifications and doesn't instantly leap to uncharitable interpretations (including the interpretation that the other person must be lying if he claims not to understand something) and object strenuously on the basis of those interpretations.
billschnieder said:
It is exactly what Bell meant when he said on page 406 of his original article that:
Bell said:
It follows that c is another unit vector
P(a,b) - P(a,c) = ...
Contrary to what you seem to be claiming here, according to Bell (a,b,c) are existing together.
Sure, what sentence of mine could possibly make you think I was "claiming" otherwise? In the comment of mine you objected to, I said "but we can still consider the theoretical consequence of the assumption that each object actually had well-defined values for all three properties". So, of course I was assuming theoretically that predetermined values for each of the three detector angles (a,b,c) are existing together, even if only two are sampled by the experimenter.
billschnieder said:
So when he goes on to derive his inequality half a page later to be

1 + P(b,c) >= |P(a,b) -P(a,c)|

he has nothing in mind other than that those terms originate from the triple (a,b,c) which exist together.
Of course they "originate from the triple", that's exactly what I said in the comment you objected to.
billschnieder said:
Please enough of the "fraction of A,B samples ... blah blah". Bell isn't dealing with any fractions and neither am I, so why obfuscate with this nonsense.
Bell's inequalities typically deal with probabilities--how do you think we measure values for probabilities empirically in a Bell-type experiment? Obviously a term like P(A+, B-) would be determined empirically by looking at the fraction of A,B samples where the result was A+ and B-.

It seems like you are solely interested in considering Bell's original paper rather than any of his subsequent papers or derivations by other physicists--personally I think this is a bad idea, since as I said we are considering science rather than religious scriptures so there's nothing special about the "original" presentation of a given result, and Bell's later papers along with derivations by other authors give presentations that are more clear in assumptions Bell originally left implicit. But if you insist on considering only Bell's original paper, then it's true that in that paper a term like P(a,b) would not actually refer to a probability, but rather to an expectation value. An expectation value is just a prediction about the average value of (experimenter #1's result on with setting a)*(experimenter #2's result with setting b) in the limit of a large number of trials (to save space I'll abbreviate as the average value of a*b even though strictly speaking a and b were detector angles rather than the results for those angles). Which means in any empirical experiment you'd be considering an average of your data on sampled pairs. So if you don't like my talk about fractions (even though it's completely relevant to other Bell inequalities), you can instead consider the distinction between terms of this type:

(average value of a*b for all triples)

vs. terms of this type:

(average value of a*b for all triples where experimenter sampled a and b)

Then my claim would be that if you have a theoretical list of triples along with a pair sampled by the experimenter from each triple, you can prove by pure arithmetic that an inequality of this form is guaranteed to hold:

1 + (average value of b*c for all triples)
>= |(average value of a*b for all triples) - (average value of a*c for all triples)|

But on the other hand, you can come up with such a list where an inequality of this form is violated:

1 + (average value of b*c for all triples where experimenter sampled b and c)
>= |(average value of a*b for all triples where experimenter sampled a and b) - (average value of a*c for all triples where experimenter sampled a and c)|

Do you disagree? If not, do you disagree that Bell's inequality is meant to be of the second form, since obviously the experimenter doesn't know the average value of b*c for all triples, including ones where he sampled a and b or a and c?

edit: before replying to this, make sure you first read post #1191 since you seem to have some confusion about what I mean by terms like (average value of b*c for all triples) and (average value of b*c for all triples where experimenter sampled b and c)--they are not equivalent, as I explain in that post.
billschnieder said:
If you are suggesting that dataset only means something that is actually measured,
That was just one interpretation of what it might mean--if you wish to specify that a purely theoretical list of triples is a "dataset of triples" that's OK with me as long as we're clear.
billschnieder said:
then why do you insist that Bell was modelling a dataset of pairs. Bell did not actually measure anything in order to obtain his inequalities did he?
Because his model included the idea of a theoretical experimenter, so he was dealing with the pairs that would actually be measured by this theoretical experimenter. "Actually" is relative to the hypothetical world you're considering, like if I said "in how many of Sherlock Holmes' cases did Watson actually make any significant contribution to solving it?"
billschnieder said:
Read page 406 of Bell's original paper. He assumes that there exist 3 vectors (a,b,c), also known as a TRIPLE then he derives the inequalities

1 + P(b,c) >= | P(a,b) - P(a,c)|

Each term in the above contains only two vectors from the same TRIPLE. In other words, the symbols (a,b,c) MUST mean exactly the same thing in each term!
In Bell's original paper the vectors a, b, c are just the detector angles, whereas P(b,c) refers to the expectation value for the products of their results (each result being +1 or -1) on trials where the first experimenter picks detector angle b and the second picks detector angle c. You can see from the form of equation (2) that P(a,b) must be an expectation value, since he takes the product of experimenter A's result (determined by the function A(a,λ)) and experimenter B's result (determined by the function B(b,λ)) and multiplies these by the probability density on that value of λ, then integrating over all possible values of λ. This means that even though the product of the two experimenter's results on a given trial (given by A(a,λ)*B(b,λ) for whatever value of λ occurs on that trial) is guaranteed to be either +1 or -1, P(a,b) can be any real number between +1 and -1, which shows that it must deal with averages over an arbitrarily large set of trials rather than results on individual trials. He also notes immediately after equation (2) that "this should equal the quantum mechanical expectation value ... But it will be shown that this is not possible" (i.e. the expectation value derived from the assumption of local hidden variables cannot possibly equal the quantum-mechanical expectation value".

I'm not sure if this contradicts what you say in the quote above. It sounded like you were saying that the equation dealt with the results we'd get for a single triple of predetermined results like (+1 on a, -1 on b, -1 on c), but I may be misinterpreting you.
billschnieder said:
Rubbish. Provide me a dataset of triples which violates the above inequality, for which the terms (a,b,c) mean exactly the same thing in each term.
Does "mean exactly the same thing in each term" suggest you want each term to deal with the results for the same single triple? If not, what does it mean? Again, what I am saying is that Bell intended his inequality to say that in a local realist universe where we're doing an experiment of a certain type, in the limit of a large number of trials we should expect the following to hold:

1 + (average value of b*c for all triples where experimenter sampled b and c)
>= |(average value of a*b for all triples where experimenter sampled a and b) - (average value of a*c for all triples where experimenter sampled a and c)|

...and I am saying that it is possible to come up with a list of theoretical triples (or 'dataset of triples' if you don't mean 'dataset' to imply the data known by the theoretical experimenter) where the above is violated (and if you have a mathematical rule specifying a correlation between the detector settings on each trial and the probability of getting different possible triples on each trial, in violation of the no-conspiracy assumption, you can show that the above will be violated even in the limit as the number of trials becomes arbitrary large). Do you disagree with that?
billschnieder said:
All I want is for you back up your claim that it is possible for the inequality to be violated without extra assumptions in addition to the existence of triples (a,b,c). I have been asking you this for the last 3-4 posts and you haven't provided one, yet you keep claiming that without your extra assumptions the inequality can be violated.
I already gave you an example for a different inequality, one which Bell also referred to. Again, I can give an example for the specific inequality you ask about, but I want to make sure in advance that it's OK with you if each term is of the form:

(average value of b*c for all triples where experimenter sampled b and c)

rather than the form:

(average value of b*c for all triples)

If you do think it's impossible even with the top form, then just say so and I'm happy to provide an example which proves you wrong. On the other hand, if you do agree it's possible to come up with a list of triples (along with a list of which pair were sampled for each triple) that violates the inequality with the top form, then we are in agreement about what can and can't be proved impossible with pure arithmetic, and then the question we should debate is which form Bell was implicitly assuming in his own inequality.
JesseM said:
In physics, you can't assume one equation is the "exact same" as the other just because they are written in the same abstract form.
billschnieder said:
What is silly is to suggest that two inequalities which are exactly identical, down to the meanings of the terms are in fact different because they were derived differently.
Well, good thing I wasn't suggesting that then eh? I was suggesting that the two inequalities are not identical because the "meaning of their terms" is not identical. I think I made that pretty clear when I said (immediately after the sentence you just quoted) the following:
For example, the two inequalities in my example:

(Fraction of all nine objects with properties A+ and B-) + (Fraction of all nine objects with properties B+ and C-) greater than or equal to (Fraction of all nine objects with properties A+ and C-)

and

(Fraction of A,B samples which gave result A+, B-) + (Fraction of B,C samples which gave result B+, C-) greater than or equal to (Fraction of A,C samples which gave result A+,C-)

Can both be written as:

F(A+,B-) + F(B+,C-) >= F(A+,C-)

...but their meaning is very different!
Would you disagree the physical meaning of F(A+,B-) depending on whether we interpret it to mean (Fraction of all nine objects with properties A+ and B-) or if we interpret it to mean (Fraction of A,B samples which gave result A+, B-)?

Now, I know you have a habit of refusing to discuss any of the inequalities which appeared in papers other than the very first paper Bell ever wrote about the inequality, so I can rephrase this by comparing the following inequalities:

1 + (average value of b*c for all triples)
>= |(average value of a*b for all triples) - (average value of a*c for all triples)|

vs.

1 + (average value of b*c for all triples where experimenter sampled b and c)
>= |(average value of a*b for all triples where experimenter sampled a and b) - (average value of a*c for all triples where experimenter sampled a and c)|

Would you agree the meaning of these inequalities is different, even though they can both be written in the abstract form 1 + P(b,c) >= |P(a,b) - P(a,c)| ?
 
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  • #1,190
(response to post #1186 continued)
billschnieder said:
The terms in my inequality mean exactly the same as those in Bell's.
In your inequality, does P(b,c) refer to "average value of b*c for all triples where experimenter sampled b and c"? If it does, then it's not hard to find a set of triples that violates your inequality. And if it doesn't, then no, the terms in your inequality don't mean the same thing as those in Bell's.
JesseM said:
But it's only guaranteed to be true arithmetically if you actually know ab, ac and bc for every member of the list. If for each member of the list you can only sample two, so <ab> only refers to the average result on the subset of the list where you actually sampled a and b, then it's no longer guaranteed to be true. Do you disagree?
billschnieder said:
Huh? Isn't this what I have been telling you and you've been objecting? Isn't that exactly what my claim (1) and claim (3) are talking about? Why would you claim to object to something you actually agree with unless you like quibbling.
I don't know what your overall point is supposed to be since your whole argument appears rather confused to me, so my objection wasn't to any overall point you were trying to make but rather to the specific claim you made in (1) (and possibly in (3) too, although I explained in post #1179 that I thought the meaning of (3) was ambiguous and asked for clarification which you didn't provide), considered on their own and not in the context of your whole argument. I already explained that I disagree with (1) because Bell's inequality is not just an arithmetic inequality, the meaning of the terms in his inequality is different--in a purely arithmetic inequality with the same equation each term would have a meaning like (average value of b*c for all triples), but in Bell's inequality each term has a meaning like (average value of b*c for all triples where experimenter sampled b and c). And because of this difference in meaning, proofs of the Bell inequality require a few additional physical assumptions beyond what's required to prove the arithmetic inequality.
billschnieder said:
All I ask is that you give conspiracy-infested dataset which violates Bell's inequality. Remember, (a,b,c) must mean the same thing in each term of the inequality, unless by conspiracy you mean failure to make sure (a,b,c) mean the same in each term, such as by using a different "a's" for different terms.
If "a" refers to a choice of detector angle then I am assuming the same a for each term (and as I noted before, I use the notation a*b to refer not to the product of the two detector angles, but the product of the two experimenter's results when they choose detector angles a and b). On the other hand, if "a" refers to the predetermined value of a given triple for setting a or something along those lines, then I'm not clear on what you mean by 'using different "a's" for different terms". In this inequality, would you say I am using different "a's" for different terms or not?

1 + (average value of b*c for all triples where experimenter sampled b and c)
>= |(average value of a*b for all triples where experimenter sampled a and b) - (average value of a*c for all triples where experimenter sampled a and c)|

If you would say I am using different "a's" here that's fine, but then my point is that this is the meaning of the terms in Bell's inequality. So, in that case it seems what you're asking for is not a "conspiracy-infested dataset which violates Bell's inequality", but rather a data set which violates some inequality where the terms have a different meaning than they do in Bell's, like this one:

1 + (average value of b*c for all triples)
>= |(average value of a*b for all triples) - (average value of a*c for all triples)|

...but I have said all along that you can prove by pure arithmetic that it's impossible to find a collection of unchanging triples that violates this inequality. The point is that this is not Bell's inequality since the terms have the wrong meaning, which is why I objected to (1).
billschnieder said:
And by unchanging do you mean the meanings (a,b,c) is changing from one term to the next?
No, I meant that the triplet hidden variables associated with the three detector settings a,b,c may be changing with time for a single pair of particles, so that if you measure the first particle with detector setting a, that may change the three hidden variables associated with the second particle if you measure it at a slightly later time. If there is not a spacelike separation between the two measurements, then you can have a local hidden variables theory of this form which seems to violate some Bell inequality by exploiting the "locality loophole".
billschnieder said:
On the other hand, it is possible to come up with a dataset of pairs for which Bell's inequality will be violated. This is my claim (3).
Again, I need clarification on this claim before I can tell you if I agree or disagree. Are you assuming that the "dataset of pairs" is derived from a list of triples in some way, or are we just making up pairs in a totally arbitrary way? If the pairs are derived from a list of triples in some way, how exactly?
billschnieder said:
And the reason why this dataset of pairs will violate the inequality is entirely mathematical and has to do with the fact that the symbols will not mean exactly the same thing in all terms of the inequality like Bell assumed.
Also not clear on what "the symbols will not mean exactly the same thing in all terms of the inequality" means. For the inequality 1 + P(b,c) >= |P(a,b) - P(a,c)|, would you say the symbols mean the same thing in all the terms if we interpret the meaning like this?

1 + (average value of b*c for all triples)
>= |(average value of a*b for all triples) - (average value of a*c for all triples)|

How about if we interpret the meaning like this?

1 + (average value of b*c for all triples where experimenter sampled b and c)
>= |(average value of a*b for all triples where experimenter sampled a and b) - (average value of a*c for all triples where experimenter sampled a and c)|
billschnieder said:
Because I have noted your special expertise before and I am on the lookout for it. Let's just say I know your tactics, I'm not stupid.
In other words, you assume that my genuine confusion about your arguments must be a deceptive "tactic"--that's exactly what I mean by "uncharitable interpretation". In fact it's not true, I genuinely am not sure what the hell you are trying to say half the time, but you have such a reactionary/paranoid mindset that I don't imagine I can ever convince you otherwise.
JesseM said:
(like not knowing that my inequality is one of the Bell inequalities, as explained above), so my examples and arguments might have some relevance that just hasn't occurred to you?
billschnieder said:
It clearly occurs to me that if somebody brings out every obscure form of Bell's inequalities in a discussion where simply sticking to the form being discussed will be clearer, such is an attempt either at obfuscation
You didn't make clear at the outset that "the form being discussed" was the one in his original paper, in this recent discussion of ours I was the first one to bring up a specific mathematical inequality, first in post #1171 where I quoted a paper from Bell and then again in post #1176 where I talked about

Number(A, not B) + Number(B, not C) greater than or equal to Number(A, not C)

Then in post #1179 I again referred to that inequality, showing that the purely arithmetic version of the inequality can't be violated by a series of triples, but a Bell-type inequality with the same equation can be. It wasn't until post #1182 that you brought up the inequality |ab+ac|-bc <= 1. It's not really fair that you should have total control over the terms of the discussion in this way, but as seen above I'm fine with discussing this inequality too. Still it's a bit much that you now accuse me of an attempt at obfuscation because I brought up a specific example in what had previously been an overly abstract discussion, and then I didn't immediately drop that example when you brought up a slightly different one.

Also, the inequality I mention is hardly "obscure", if you didn't have a single-minded interest in Bell's original paper only and instead looked at discussions of Bell's inequality by other authors, you'd see that this inequality is mentioned more often in introductory discussions of Bell's proof than the one in the original paper, perhaps because it's so much simpler to see how it's derived (I gave a quick derivation in post #1179 when I said 'the proof is trivial--every triplet with A+ and C- must either be of type A+B+C- or type A+B-C-, and if the former it will also contribute to the number with B+ and C-, if the latter it will also contribute to the number with A+ and B-'). I already gave a link to one website which uses it as a starting point, and wikipedia refers to this inequality as http://en.wikipedia.org/wiki/Sakurai's_Bell_inequality]Sakurai's[/PLAIN] Bell inequality because it appeared in Sakurai's widely-used 1994 textbook on QM (the wikipedia article mentions a number of other well-known papers and books on Bell's proof that have used it).
JesseM said:
You really are ridiculously uncharitable when it comes to interpreting other people's arguments, you always jump to the conclusion that they are saying something foolish rather than asking questions to see if you might be missing something.
billschnieder said:
This also sound very much like your autobiography. For example, see the following statement of yours which purports to be responding to a claim of mine but is actually not because I never argued anything of the sort being alledged.
Disagree, since once again the Bell inequalities deal with the probabilities of measuring a given pair of values, not with the probability that all triples have a given pair of values even if those weren't the two you measured.
This was in response to your statement:
As I have demonstrated and I hope you now agree, so long as you have a list of triples in hand with values restricted to (+1 or -1), whether theoretical or measured, no matter how the list was generated, Bell's inequality is never violated (claim 6).
Are you saying that you would agree with both the following claims A and B?

(A) In Bell's inequality 1 + P(b,c) >= |P(a,b) - P(a,c)|, the correct interpretation of the meaning of the terms would be:

1 + (average value of b*c for all triples where experimenter sampled b and c)
>= |(average value of a*b for all triples where experimenter sampled a and b) - (average value of a*c for all triples where experimenter sampled a and c)|

(B) With that interpretation of the meaning, it is possible to find a list of triples with values restricted to (+1 or -1) that violates the inequality

If you disagree with either of those, then it seems there was no fault in my understanding of (6) or the "As I have demonstrated..." quote, since my comment "Disagree, since once again the Bell inequalities deal with the probabilities of measuring a given pair of values" was meant as an assertion of point (A), and I had already explained earlier in the post about why I believed point (B).

In any case, on the subject of "uncharitable" interpretations, you can probably find occasional examples where I jumped to an erroneous conclusion about what you were saying, but I think you can find many more examples where I said what I thought you might be saying but then asked questions to check if I was misinterpreting. For example, just from the post you were responding to above:
Not sure what you mean by "consider them together theoretically as triples", if this is an important part of your argument you'll have to explain in more detail. We may be assuming theoretically the pairs are sampled from triples, but a term in the inequality like (Fraction of A,B samples which gave result A+, B-) still deals with a collection of measured pairs (specifically the pairs where we measured A,B). Do you disagree?
Based on your comment above I no longer know what you mean when you say "datasets of triples", you seem to be using that phrase in a rather odd way that you have never defined ... If you disagree, please give a careful definition of what you mean by the phrases "dataset of pairs" and "dataset of triples".
I don't know what you mean by "the terms in the inequality are obtained from the triples" either ... Do you disagree that the terms in Bell's inequality also deal only with subsets of all the entangled particle pairs that were measured
billschnieder said:
Now if you are ready to admit claim (1)
Nope, not if (1) includes the idea that Bell's inequality has the same meaning as the arithmetical inequality just because it can be written in the same form. [that 'if' refers to my earlier request for clarification about (1) which you haven't responded to]
Would you say the pairs in my example were "extracted from a dataset of triples"? The pairs were all extracted from a list of triples which would be known by an omniscient observer, though they weren't known by the experimenter so I wouldn't call them a "dataset". But if you would say that the pairs were "extracted from a datset of triples" then this is an explicit counterexample to your claim above, since the resulting pairs violated the inequality.
I thought it seemed to be part of claim (1) that the "inequality so derived is Bell's inequality", since part of claim (1) was "Bell's inequality is an arithmetic relationship between triples of numbers". I did ask for clarification on this point when I said "Perhaps you did not actually mean for your claim 1) to include the subclaim 1b), in which case please clarify." When I ask for clarification I'm not being rhetorical, I can't really discuss my opinion on a statement of yours if I'm unclear on what you're actually saying.
But it's only guaranteed to be true arithmetically if you actually know ab, ac and bc for every member of the list. If for each member of the list you can only sample two, so <ab> only refers to the average result on the subset of the list where you actually sampled a and b, then it's no longer guaranteed to be true. Do you disagree?
I already showed that for the arithmetical inequality I was dealing with, are you saying you don't believe I can do something analogous for your inequality? In other words, are you saying I can't come up with a set of triples, along with a choice for each triple of which two variables are sampled by the experimenter, that violates this relation?
So, in this single post you can see a lot of examples where I don't just leap to uncharitable conclusions about what you're saying, but instead ask questions to try to clarify. I think if we looked over a bunch of your previous posts we'd find very few instances where you thought I might be saying something foolish but asked for clarifications rather than immediately jumping to the conclusion I was. So even though you can probably find some instances of me jumping to uncharitable conclusions, I think it's safe to say that the degree to which you do this to me is a lot greater. And likewise you are now getting into nasty comments about my motives and suggesting that any comments of mine that don't immediately reach the correct conclusion about what you were saying must be a deliberate "tactic" (as if I really did understand you perfectly well but was pretending not to in order to annoy you)--it would more charitable to just assume I was a bit thick! But of course the most charitable and fair assumption is that communication about complex issues like these is sometimes difficult and arguments that may seem clear to you can seem genuinely ambiguous to intelligent readers who aren't privy to all your thought processes.
 
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