Joy Christian's colourful exploding balls experiment

[Joy Christian]

Indeed: "Select the states." Not "Reject the data."

The initial or complete state of the system in the model is the pair (e, theta), NOT just the vector e by itself.

In the experiment, Nature selects the states. There will be no post-selection by the experimenter. No detection loophole. N runs -> N states.

There is no post-selection in the model either, nor is there post-selection in the simulation. The model mimics Nature exactly, and the simulation mimics the model.

[Heinera]

So you say that E(0, 135), E(0, 45), E(90, 45), and E(90, 135) will have the quantum correlations in Joy's experiment. Richard (and I as well) says E(0, 135) will be 0.5, the three others -0.5. Where is the "debunked CHSH logic" in that?

I say for four correlations each measured on a disjoins sets of particles, each a fair sample of the population E(0, 135), E(0, 45), E(90, 45), and E(90, 135) will each agree with QM. I say for correlations all determined from the same set of particles E(0, 135), E(0, 45), E(90, 45), and E(90, 135) will not all agree with QM. You and Richard say it doesn't matter whether we use a single set of multiple disjoint sets. Your argument is debunked here: viewtopic.php?f=6&t=44

Let us assume you are correct. Let's assume that you can replicate the QM correlations in Joy's experiment if each correlation is computed on separate sets. Let's do a simple numerical experiment: I have picked two angles a and b. Send me two lists of vectors, Bob's and Alice's (length to be chosen by you). With these lists, I will compute E(a,b). If it agrees with the QM correlation I will shut up forever.

PS: In order to mimic Joy's experiment as closely as possible, I won't reveal the values of a and b at this stage. In Joy's experiment, the vectors are generated without any knowledge of the a and b values to be used later. In fact, there are no detector angle settings in his experiment; just a battery of high speed cameras that are filming the whole thing. After I have received the lists from you, I will reveal my choice of a and b.

Just one set of data, just one correlation to be computed. Can you do that? Because if you can't, I will conclude that there is something you don't understand about Joy's experiment.

[gill1109]

Indeed: "Select the states." Not "Reject the data."

The initial or complete state of the system in the model is the pair (e, theta), NOT just the vector e by itself.

In the experiment, Nature selects the states. There will be no post-selection by the experimenter. No detection loophole. N runs -> N states.

There is no post-selection in the model either, nor is there post-selection in the simulation. The model mimics Nature exactly, and the simulation mimics the model.

(1.) A model has been published in some papers and a book.

(2.) Some simulations have been done and published.

(3.) There will be an experiment.

Note the future tense, and different subject, of (3.).

I am now only interested in the experiment and the bet about its outcome.

I need to know what the experimental data looks like and how it will be processed in order to determine the winner of the bet, before I can unreservedly state that as far as I am concerned "the bet is on".

Let me remind everyone what I understand by "the experimental data".

The experiment involves a lot of exploding balls (N pairs, or four sets of possibly different numbers of pairs) and video cameras *and* image analysis, all of which is done (according to Joy's experimental paper) without any knowledge of which settings are going to be used, which correlations are going to be calculated, and how they are going to be calculated. At the end of the day we have some files of spatial directions. It's not quite clear to me whether there are going to be just two files or four sets of two files. I hope that Joy will make up his mind (a little while ago, he had made up his mind, namely: just two files, one for Alice, one for Bob. That's also what he wrote in his experimental paper).

The experimental data = some computer files of directions.

Joy still has time to move the goal-posts if he wants to, but it will be annoying if the goal-posts keep getting moved, again and again. The point of fixing this protocol is to prevent any goal-post-creep.

Every time the goal-posts are moved by one party, the other can honourably step down. Then there will be no bet any more. I fear that if there is no bet there will be no experiment, since the bet raises the chance of media interest and hence funding for the experiment.

I already wrote code for determining the outcome of the bet from just two files, as per Joy's experimental paper, as far as I understand it. The code takes the experimental data as given, and determines a conclusion: Joy has won the bet or Richard has won the bet.

Joy already agreed to it, but he could ask to move the goal-posts. He might like to take advice from trusted computer experts.

If everyone is happy with my R code we can stick with my R code but it seems to me that the adjudicators maybe don't "speak R". But that doesn't really matter: they just have to run the program and announce the conclusion. If Joy and I have already agreed on the code then they don't even have to look at it.

What I do think is important is that my critics don't all "speak R". I still think it is essential to have Mathematica and Python versions, maybe also Java, which have been thoroughly tested and such that all three or four (R, Python, Mathematica, Java) give an identical decision if they are fed identical data files (up to possible numerical precision issues in case the result in some sense is a "tie".)

Then the experiment and the bet have the broadest possible concensus behind them, and nobody who is part of the concensus can complain about any rigging by anyone, after it is all over and done with. That's why I'd like my own strongest critics, and Joy's own strongest supporters, to take the lead in preparing some translations.

We also already have perl and Excel versions, if anyone likes to compare and/or study them. They look OK to me.

[Joy Christian]

Joy still has time to move the goal-posts if he wants to, but it will be annoying if the goal-posts keep getting moved, again and again. The point of fixing this protocol is to prevent any goal-post-creep.

I have not changed any goal-posts in my life. My experimental paper exists since 2008. It describes the experiment in sufficient detail, together with how it should be performed. I am happy with the refinements made to the procedure as long as they do not modify what I have proposed. Here is a reminder of what I have proposed:

**********************************************************************************

For the record, let me repeat that equation (16) of my attached
experimental paper describes exactly how the expectation values
E(a, b), E(a', b), E(a, b'), and E(a', b') are to be computed in my
proposed experiment. Four separate sums are to be calculated as
follows

E(a, b) = 1/N Sum_j A_j B_j ,

E(a, b') = 1/N Sum_j A_j B'_j ,

E(a', b) = 1/N Sum_j A'_j B_j ,

and

E(a', b') = 1/N Sum_j A'_j B'_j .

It is a matter of indifference whether N here is chosen to be the same
or different for each of the four alternatives.

The experimental procedure described in my paper is unambiguous.

**********************************************************************************

Further details of my proposed experiment can be found on my blog.

[minkwe]

Let us assume you are correct. Let's assume that you can replicate the QM correlations in Joy's experiment if each correlation is computed on separate sets.
...
Just one set of data, just one correlation to be computed. Can you do that? Because if you can't, I will conclude that there is something you don't understand about Joy's experiment.

Let us assume that you are correct that it is possible for humans to go to the moon. Prove it by going to the moon. Can you do that? Because if you can't I will conclude that it is impossible to go to the moon.

The point of the experiment is to prove that it is possible to reproduce the QM correlations local realistically, you and Richard say it is impossible. Joy and I say it is possible. Joy has given you a model which does it. That's not enough for you. I've written 2 simulations which do that. Those are not enough for you either. Isn't that the reason the experiment is being done, so that nature can pick a winner?

I'm not trying to convince you in this thread what the experimental results will produce, nature will do that. I'm trying to convince you that the results of the experiment do depend on the way the data is collected and analyzed. You and Richard say it doesn't matter. I say it does. But surprisingly, even though Richard believes it does not matter, he does believe that doing it on separate sets of particles reduces his chances of winning the bet. He doesn't tell us why something which does not matter for the experiment changes his chances of winning.

[Heinera]

Let us assume you are correct. Let's assume that you can replicate the QM correlations in Joy's experiment if each correlation is computed on separate sets.
...
Just one set of data, just one correlation to be computed. Can you do that? Because if you can't, I will conclude that there is something you don't understand about Joy's experiment.

Let us assume that you are correct that it is possible for humans to go to the moon. Prove it by going to the moon. Can you do that? Because if you can't I will conclude that it is impossible to go to the moon.

But why would this innocent experiment be so difficult for you that you compare it to going to the moon? Because, since you have no idea what values of a and b I have chosen, you would have to ensure that the dataset produced QM correlations for any values of a and b. But wait, that is impossible, right?

[Heinera]

Zen:

Your last post messed up tags when quoting earlier posts. Please edit, and then I'll delete this post.

[Joy Christian]

Code: Select all

good <- abs(ca) > f & abs(cb) > f  ## Select the 'states'

Indeed: "Select the states." Not "Reject the data."

The initial or complete state of the system in the model is the pair (e, theta), NOT just the vector e by itself.

Mathematically, when you select the states with a rule that depends on the detectors settings, you make the probability distribution of (e,theta) depend on (guess what?) the detectors settings! Hence, this is not a LHV model. It's difficult to understand why you can't see this. I'm sorry, that's the way probability theory works. I didn't invent it. Blame Borel, Kolmogorov, Levy, etc. Also, I believe that Richard does see it but, curiously, he pretends that the elephant is not in the room.

What is trivial for me to understand is why you don't see the fact that neither the initial state (e, theta), nor the probability distribution of the initial state (e, theta) depends on the detector settings. The reason is quite simple. I have explained it over and over again. Until you get out of the flatland of R^3, you will never be able to understand that neither the initial state (e, theta), nor the probability distribution of the initial state (e, theta) depends on the detector settings. I did not invent the topology [ SU(2) ] of the physical space [ S^3 ] we happen to live in. Blame Nature for being so clever. Borel, Kolmogorov, Levy, etc were not as clever as Nature.

[Joy Christian]

What is trivial for me to understand is why you don't see the fact that neither the initial state (e, theta), nor the probability distribution of the initial state (e, theta) depends on the detector settings. The reason is quite simple. I have explained it over and over again. Until you get out of the flatland of R^3, you will never be able to understand that neither the initial state (e, theta), nor the probability distribution of the initial state (e, theta) depends on the detector settings. I did not invent the topology [ SU(2) ] of the physical space [ S^3 ] we happen to live in. Blame Nature for being so clever. Borel, Kolmogorov, Levy, etc were not as clever as Nature.

Great, Joy! Keep deluding yourself! But it happens that this "flatlander" knows one thing or two about set theory.

I challenge you once and for all: using your own definition of the set \Lambda contained in your own complete.pdf document, write down just one pair (e_0,\theta_0) belonging to \Lambda such that \theta_0 > 0.

Answer to your challenge is coming up shortly (a la Captain Kirk).

[gill1109]

Just wanted to say, a whole lot of recent posts here are "off-topic". This is the place to talk about Joy's experiment. Not about Joy's model or about simulations of Joy's model.

Some pages back Michel astutely remarked that I seemed to realise that there was some difference between a Variant 1 experiment and a Variant 2 experiment. It's perhaps also off-topic here, but when he's done the R computer experiment which I set in another thread, and in particular done both experiments, and repeat both experiments say 1000 times, and produced two histograms of the 1000 values of - E(0, 45) + E(0, 135) - E(90, 45) - E(90, 135) (one for variant 1 and one for variant 2) then he might understand what difference I have in mind. It's important to bear in mind that the criterion for Joy to win the bet is not whether or not - E(0, 45) + E(0, 135) - E(90, 45) - E(90, 135) exceeds 2, but whether or not it exceeds 2.4.

The thread in question is http://www.sciphysicsforums.com/spfbb1/viewtopic.php?f=6&t=40 and it's called A silly computer experiment ... or, the heart of the matter?. The R code there does a kind of test run for Joy's experiment with N = 10 000. So in variant 1, data from a single set of 10 000 pairs of particles is used to calculate four correlations; in variant 2, four sets of 10 000 pairs are used to calculate four correlations, each on a separate set.

The exercise is not just to run each code snippet once or only a few times, but to write a "for loop" which runs each one 1000 times, stores the results one by one, and plots a histogram of the finally obtained 1000 values of - E(0, 45) + E(0, 135) - E(90, 45) - E(90, 135).

You could call this a simulation of a simulation. The two variants of Joy's experiment, each with N = 10 000 pairs of particles per correlation, are simulated, not once, but 1000 times. Of course I do not actually simulate Joy's model (I don't know how to - I agree with earlier posts here, stating that it is impossible to do so). I simulated the usual model generating the usual triangle wave correlation function.

Sorry all this is "off topic" too, but since you guys have been off-topic for one or two pages, I thought I might allow myself to be it, too.

[Joy Christian]

What is trivial for me to understand is why you don't see the fact that neither the initial state (e, theta), nor the probability distribution of the initial state (e, theta) depends on the detector settings. The reason is quite simple. I have explained it over and over again. Until you get out of the flatland of R^3, you will never be able to understand that neither the initial state (e, theta), nor the probability distribution of the initial state (e, theta) depends on the detector settings. I did not invent the topology [ SU(2) ] of the physical space [ S^3 ] we happen to live in. Blame Nature for being so clever. Borel, Kolmogorov, Levy, etc were not as clever as Nature.

Great, Joy! Keep deluding yourself! But it happens that this "flatlander" knows one thing or two about set theory.

I challenge you once and for all: using your own definition of the set \Lambda contained in your own complete.pdf document, write down just one pair (e_0,\theta_0) belonging to \Lambda such that \theta_0 > 0.

Here is the answer to your challenge, Captain Kirk style: http://libertesphilosophica.info/blog/w ... mplete.pdf (please note the changes, especially in Lambda).

[Joy Christian]

This is simply ridiculous!

With your new definition, the only elements left in are those pairs for which .

Again, we have .

Find a way to dispense with the universal quantifier in your definition of , or you're totally screwed.

Nice try, Capitain Kirk!

Best wishes from Flatland!

This may be true in the Flatland---i.e., R^3---but it is certainly not true in S^3. In S^3 there always exists some elements of for any . There are infinitely many S^3 in S^3 (one at each point of S^3).

[Joy Christian]

Really, Capitain? So, it must be really easy for you to mathematically prove that for any given pair there is no such that . Please, write down this simple proof to enlighten us Flatlanders.

I have already proved that in S^3 there always exists some elements of for any . I have already provided the proof on this page: http://libertesphilosophica.info/blog/w ... mplete.pdf.

[minkwe]

But why would this innocent experiment be so difficult for you that you compare it to going to the moon?

Because my ability to reproduce QM correlations local realistically does not prove it is impossible to reproduce them. Just as your ability to go to the moon does not prove it is impossible to do it. That's simple logic. So again I ask the question. If the correlations from a single set are the same as those from multiple sets, why is Richard's likelihood of winning the bet reduced going from a single set to multiple sets?

[minkwe]

Really, Michel? You too? You can't see that the probability distributions of (e,theta) in the simulations of the model depend on the detectors settings? It's a standard Monte Carlo Acception-Rejection method. Please, read the code again.

Zen, are you denying that my simulations reproduce the QM correlations? Or are you saying that they do reproduce the QM correlations but by using Acception-Rejection.

Are you aware of anything in QM or any experimental evidence that forbids acception-rejection?

Since you are probabilist. Do you know of any scenario whatsoever for which the expression P(AB|X) = P(A|X)P(B|X) is true, but at the same time the expression P(AB|X) = P(A|X)P(B|AX) is false?

I asked you a question in the other thread, concerning statistics, error and the CHSH (viewtopic.php?f=6&t=40&start=10). Did you miss it?

[FrediFizzx]

To All,

Zen wished to no longer participate and his posts have been deleted per his request.

[gill1109]

To All,

Zen wished to no longer participate and his posts have been deleted per his request.

Pity, Zen put his finger on the sore spot in the "Christian 2.0" derivation of Joy's model: the universal quantifier. "For all" does not mean "for some". Joy does not even provide a proof of existence. It's easy to see that his set of states is the empty set. Because it is empty, anything one can say about its members is true.

But this is off-topic. We should be talking here about the experiment, not about the theory!

[FrediFizzx]

'Tis a pity that he couldn't get a grip on parallelized 3-sphere geometry and topology just like you. I have to scratch my head on that because it is really not all that hard. But maybe someday... One can only hope for the best.

[gill1109]

'Tis a pity that he couldn't get a grip on parallelized 3-sphere geometry and topology just like you. I have to scratch my head on that because it is really not all that hard. But maybe someday... One can only hope for the best.

Yes, keep on hoping. We could start a new thread on that topic. This thread is for the experiment. We know Joy's predictions for what will come out of it. No need to discuss the theory which lay behind the predictions. Here we discuss practical experimental issues. The experiment has to satisfy two criteria:

(1) if Joy is right, it has to convince his critics that he was right
(2) if Joy is wrong, it has to convince his supporters that he was wrong

So the experiment has to be looked at from both sides. The experimental arrangements, the protocol, the rule for deciding who has won the bet. But the theory (theories) are irrelevant.

[Joy Christian]

To All,

Zen wished to no longer participate and his posts have been deleted per his request.

Pity, Zen put his finger on the sore spot in the "Christian 2.0" derivation of Joy's model: the universal quantifier. "For all" does not mean "for some". Joy does not even provide a proof of existence. It's easy to see that his set of states is the empty set. Because it is empty, anything one can say about its members is true.

This is complete nonsense. Yet another bogus propaganda, just like all previous bogus propagandas and errors manufactured by various people to derail my program.

It is astonishing that for mathematicians like Gill and Zen it is difficult to understand that the metrical and topological structures of S^3---and consequently the properties of vectors and angles between the vectors---are dramatically different from R^3. In any event I have repeatedly stressed that when I say "for all" I mean "for all." So why the bogus claim? I have repeatedly provided the demanded "proof of existence." Here: http://libertesphilosophica.info/blog/w ... mplete.pdf.

Does the set Lambda defined in equations (6) and (10) on this one-page document look like an empty set to anyone? If it does, then you need to first learn how the cosine function is defined in a non-flat Riemannian manifold (just pick up any good book on differential geometry). Secondly, note that vectors in a parallelized 3-sphere defined in equation (1) satisfy the geometric product: ab = a.b + a/\b. Finally, note that Lambda is actually a set of a pair of unit quaternions, (p_o, q_o), which necessarily satisfy the triangle inequality. Consequently, the set Lambda of complete states, (e_o, s_o), defined in equation (10) is simply an expression of the geometrical and topological structures of S^3, captured by the triangle inequality. It remains valid (and is non-empty) for all vectors n (i.e., the detector settings).

Given the completely transparent proof I have linked above and my repeated explanations, it is quite disingenuous of both Gill and Zen to continue the false propaganda about the universal quantifier. It is a cheap trick to discredit. I do not want to hear any more bogus claims about my model. It is time to grow up.