Joy Christian's colourful exploding balls experiment

[minkwe]

If you understand this so well, I am sure you can come up with a numerical demonstration. Generate two lists of vectors (10 000 elements each) that you think can achieve this. I will then run this through Richards program (the version that computes the correlations on four random disjoint subsets). If the correlations E(0, 45) , E(0, 135), E(90, 45) and E(90, 135) all come out with the QM values, I will shut up forever.

Richard's program is a joke and a waste of time. I'm explaining to you why it is a joke and you are asking me to use it to prove my point? Seriously!? I've already produced 3 numerical demonstrations which illustrate the problem. I've dumbed it down it to a very simple numerical illustration using coins but you still do not want to see it.

[gill1109]

Give us the two lists of vectors, then, Michel. If the correlations E(0, 45) , E(0, 135), E(90, 45) and E(90, 135) all come out with the QM values, I will shut up forever, too. Goodnight all!

[Heinera]

If you understand this so well, I am sure you can come up with a numerical demonstration. Generate two lists of vectors (10 000 elements each) that you think can achieve this. I will then run this through Richards program (the version that computes the correlations on four random disjoint subsets). If the correlations E(0, 45) , E(0, 135), E(90, 45) and E(90, 135) all come out with the QM values, I will shut up forever.

Richard's program is a joke and a waste of time. I'm explaining to you why it is a joke and you are asking me to use it to prove my point? Seriously!? I've already produced 3 numerical demonstrations which illustrate the problem. I've dumbed it down it to a very simple numerical illustration using coins but you still do not want to see it.

Sigh. So I'll just divide the lists into four random disjoint subsets myself, and then compute the four correlations separately, myself. Satisfied? Just give me the two lists.

[minkwe]

Gasp. How does it make a difference which order I calculate the four correlations?

Again Richard, you can calculate the correlations anyway you like, but don't be deceived that they mean the same thing no matter how you calculate them.

Richard. I toss a coin 10,000 times on a glass table with a cameras above and below the table, each recording the results of the coin. The experiment produces two files, one from each camera. You have a program which reads the file and calculates an expectation value. You think it doesn't make a difference which file you are running through your program?

Now, I toss the exact same coin 10,000 times again, same experiment as above, we get two more files. You think it doesn't make a difference which of the two files you are running through your program?

Now I chose just the files corresponding to the results from ABOVE the table, you still think they will give the same results as those corresponding to the result from UNDER the table?

[minkwe]

Give us the two lists of vectors, then, Michel. If the correlations E(0, 45) , E(0, 135), E(90, 45) and E(90, 135) all come out with the QM values, I will shut up forever, too. Goodnight all!

Using my analogy, you are asking me to give you two outcomes from the same coin which violates <A> + <B> = 0 and agrees with QM prediction for two separate coins E(A) + E(B) = 0.5
If you understood my argument, you will be asking for 4 lists of pairs of vectors which agree with QM. Or better yet, you'd have shut up already.

[minkwe]

I win, with certainty under Variant 1; with large probability under Variant 2.

Richard, if as you say the average values from Variant 1 are the same as those from Variant 2, why is there a difference in your likelihood of winning the bet between the two. Those are your own words, so please explain why according to you will certainly win it under Variant 1 but can lose it under Variant 2.

It seems you know that there is a difference. Explain the source of the difference.

[gill1109]

What Joy's experiment is about: which of these two pictures is correct?

The surfaces are theoretical correlation functions rho(a, b).

The points are theoretical correlations - four of them, according to two different theories.

Joy's experiment will measure four points, generating four observed correlations E(a, b), not shown in the images.

Will they be close to the blue or to the red points? The answer to that question will determine the outcome of the bet.




The images are made by the R script http://rpubs.com/gill1109/Wireframe
I'm investigating even better ways to visualise this. Help and advice is welcome. Maybe it's easier to do with Mathematica...

[Joy Christian]

What Joy's experiment is about: which of these two pictures is correct?

The images are made by the R script http://rpubs.com/gill1109/Wireframe
I'm investigating even better ways to visualise this. Help and advice is welcome. Maybe it's easier to do with Mathematica...

These images are very nice. You may want to use a third colour for the four points to make them visually better distinguishable from the grids.

Also, just for the record, my model makes exactly the same prediction for the four points as quantum mechanics does: http://rpubs.com/jjc/13965.

[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.

[gill1109]

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.

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

4N runs -> 4N states if Joy wishes to use separate runs for the four correlations. N runs for each correlation.

[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.