Joy Christian's colourful exploding balls experiment

[gill1109]

Wireframe plot of both theoretical correlation surfaces, superimposed. I have added the points to the surfaces corresponding to the four pairs of settings in the bet concerning the outcome of the experiment. They lie above a square in the alpha-beta plane, judiciously chosen so that one of the correlations is very high, the other three are very low. If you run the code yourself, the plot you see in the rgl graphics window is rotatable.

At some point I would like to have the graphic displayed in an interactive web page, but I still don't have my R Shiny server installed.



To do see this yourself, first of all save the following as "save.txt". It defines the viewing point of the 3d graphic.

Code: Select all

"V1" "V2" "V3" "V4"
"1" 0.89551317691803 -0.445017099380493 -0.00404098257422447 0
"2" 0.00934700295329094 0.00972943194210529 0.999908864498138 0
"3" -0.444936990737915 -0.89546924829483 0.0128725711256266 0
"4" 0 0 0 1


Then run the following code

Code: Select all

library(rgl)
alpha <- seq(from = 0, to = 180, length = 20)
beta <- alpha
surfQ <- expand.grid(x = alpha, y = beta)
surfQ$z <- with(surfQ, { - cos((x - y)*pi/180)})

corQ <- matrix(surfQ$z, 20 ,20)

surfC <- surfQ
surfC$z <- with(surfC, { - 1 + 2 * abs((x - y)/180)})

corC <- matrix(surfC$z, 20 , 20)

alphaE <- c(0, 90)
betaE <- c(45, 135)
ptsQ <- expand.grid(x = alphaE, y = betaE)
ptsQ$z <- with(ptsQ, { - cos((x - y)*pi/180)})

ptsC <- ptsQ
ptsC$z <- with(ptsC, { - 1 + 2 * abs((x - y)/180)})


persp3d(alpha, beta, corQ, front = "line", back = "line", col = "blue", lit = FALSE, zlab = "")
surface3d(alpha, beta, corC, front = "line", back = "line", col = "green", lit = FALSE)
spheres3d(ptsC$x, ptsC$y, ptsQ$z, col = "blue", radius = 2)
spheres3d(ptsC$x, ptsC$y, ptsC$z, col = "green", radius = 2)
lines3d(rep(ptsC$x[1], 2), rep(ptsC$y[1], 2), c(-1, 1))
lines3d(rep(ptsC$x[2], 2), rep(ptsC$y[2], 2), c(-1, 1))
lines3d(rep(ptsC$x[3], 2), rep(ptsC$y[3], 2), c(-1, 1))
lines3d(rep(ptsC$x[4], 2), rep(ptsC$y[4], 2), c(-1, 1))

save <- read.table("save.txt")  ## This line and the next fixes the view-point
par3d(userMatrix = save)         ##                    in the graphic posted here.

## Rotate the picture how you like
## When you like what you see,
## the following lines save the view-point and make a pdf file of the plot

save2 <- par3d("userMatrix")
write.table(save2, "save2.txt")
rgl.postscript("wireframe2.pdf", fmt = "pdf")


[Heinera]

On the FQXi blog Joy writes:

I have not changed my mind about the experiment even once since 2008.

My experimental paper exists since 2008 and it describes the experiment unambiguously, as anyone can see for themselves: http://arxiv.org/abs/0806.3078.

I am interested in observing the actual correlation, E(a, b) = -a.b, for just one pair of observation direction, (a, b).

I don't need any computer code for that. All I need is actual data: N u_k's on both sides.

Earlier in this thread I suggested:

An alternative version of the bet could be that Richard only computes one correlation E(a,b) on the whole set, but he is free to pick any values of a and b he wants. He decides on values of a and b after he has received the data files from the experiment (the experiment doesn´t know a thing about detector settings anyway).

If the computed correlation differs from the QM correlation by more than +/- 0.2, we agree that QM correlations are not reproduced and Richard wins.

Even Michel seems to think that QM correlations could be achieved this way.

Now, how is this suggestion in conflict with your claim that "I am interested in observing the actual correlation, E(a, b) = -a.b, for just one pair of observation direction, (a, b)"?

(Caveat emptor: Richard would win this bet. Just by looking at the data from the experiment, there is a simple strategy to find some (a, b) so that the computed correlation will differ from the QM correlation by at least +/- 0.2. Hint: CHSH inequality.)

[gill1109]

On the FQXi blog Joy writes:

I have not changed my mind about the experiment even once since 2008.

My experimental paper exists since 2008 and it describes the experiment unambiguously, as anyone can see for themselves: http://arxiv.org/abs/0806.3078.

I am interested in observing the actual correlation, E(a, b) = -a.b, for just one pair of observation direction, (a, b).

I don't need any computer code for that. All I need is actual data: N u_k's on both sides.

Earlier in this thread I suggested:

An alternative version of the bet could be that Richard only computes one correlation E(a,b) on the whole set, but he is free to pick any values of a and b he wants. He decides on values of a and b after he has received the data files from the experiment (the experiment doesn´t know a thing about detector settings anyway).

If the computed correlation differs from the QM correlation by more than +/- 0.2, we agree that QM correlations are not reproduced and Richard wins.

Even Michel seems to think that QM correlations could be achieved this way.

Now, how is this suggestion in conflict with your claim that "I am interested in observing the actual correlation, E(a, b) = -a.b, for just one pair of observation direction, (a, b)"?

(Caveat emptor: Richard would win this bet. Just by looking at the data from the experiment, there is a simple strategy to find some (a, b) so that the computed correlation will differ from the QM correlation by at least +/- 0.2. Hint: CHSH inequality.)

Brilliant!

We do the experiment. Joy hands me over the two files of directions u_k and -u_k.

I will then produce one (a, b) such that the computed correlation differs from the QM correlation by at least +/- 0.1.

[gill1109]

OK Joy. You win. Yet again.

You give me two files, one containing N directions of angular momentum u_k, k = 1, ... N, the other containing N directions - u _k.

I will then pick two measurement directions a and b, and show you that your correlation is more than 0.2 away from the quantum predicted correlation.

Is that a deal?

Obviously, I won't tell you the measurement directions in advance.
But the adjudicators will be able to verify that I'm right. So we just need one agreed computer script which takes as input two files of angular momentum directions from you, and two directions of measurement from me.

[Joy Christian]

Looks like people are more interested in winning bets and making snide remarks than actually wanting to understand the true nature of the physical reality.

If you only use one value of a and one value of b you prove absolutely nothing. There is nothing remarkable about a correlation of +/- 0.7. The remarkable thing is the pattern of four correlations, three - 0.7, and one + 0.7. In fact the whole mystery revolves around the joint distribution of four binary variables - Alice setting, Alice outcome, Bob setting, Bob outcome.

This is utter nonsense. All one needs is one single pair of correlation, E(a, b), to test my hypothesis.

QM predicts E(0, 135) = + 0.5 in my proposed experiment.

Bell predicts E(0, 135) = + 0.5 in my proposed experiment.

I predict E(0, 135) = + 0.7 in my proposed experiment.

One can try to obfuscate these very simple predictions with all sorts of snide remarks and statistics, but in the end one does not need any code to see that 0.5 != 0.7.

[gill1109]

OK Joy. You win.

You give me two files, one containing N directions of angular momentum u_k, k = 1, ... N, the other containing N directions - u_k.

I will then pick two measurement directions a and b, and show you that your correlation is more than 0.2 away from the quantum predicted correlation.

Is that a deal?

Note:

Joy predicts E(0, 45) = - 0.7, E(0, 135) = + 0.7, E(90, 45) = -0.7, E(90, 135) = - 0.70 in his proposed experiment.

I predict that at least one of those four predictions are off by at least 0.2.

[Joy Christian]

OK Joy. You win.

You give me two files, one containing N directions of angular momentum u_k, k = 1, ... N, the other containing N directions - u _k.

I will then pick two measurement directions a and b, and show you that your correlation is more than 0.2 away from the quantum predicted correlation.

Is that a deal?

No.

The deal is what I have stated: E(0, 135) = + 0.5 versus E(0, 135) = + 0.7.

I will give you N u_k's for both sides to check that 0.5 != 0.7. You can check that for the four points you have been plotting, but one point would be enough for me.

[gill1109]

The deal we made before was a CHSH style experiment with four correlations.

You give me the N u_k's.

I'll show you that one of your four predictions is wrong.

[Joy Christian]

The deal we made before was a CHSH style experiment with four correlations.

You give me the N u_k's.

I'll show you that one of your four predictions is wrong.

Done.

[gill1109]

The deal we made before was a CHSH style experiment with four correlations.

You give me the N u_k's.

I'll show you that one of your four predictions is wrong.

Done.

OK, that's splendid. Good night!

[Joy Christian]

One point where your prediction is wrong, and I have won the bet. Or no deal, because you have moved the goal-posts yet again. I'll announce that at Vaxjo, too.

Then you will be announcing a lie. It will not be the first time for you.

Look, I was responding to the snide remarks by you and what's his name. My goal post has been set in my experimental paper since 2008: E(a, b) = -a.b.

[gill1109]

One point where your prediction is wrong, and I have won the bet. Or no deal, because you have moved the goal-posts yet again. I'll announce that at Vaxjo, too.

Then you will be announcing a lie. It will not be the first time for you.

Look, I was responding to the snide remarks by you and what's his name. My goal post has been set in my experimental paper since 2008: E(a, b) = -a.b.

One point where your prediction is wrong, and I have won the bet.
I will announce our just made agreement at Vaxjo, with your permission, of course.
No movement of any goal posts by anyone, no mention of any movement of goal posts.

That's a promise. OK?

If you wish to withdraw from the bet before Vaxjo, I will not mention it there at all, because in that case there won't have been a bet. We were merely negotiating conditions but failed to come to an agreement. Similarly I trust that you will also grant me "thinking time". Like when you buy something online by credit card. You have so many days to revoke your purchase.

[Joy Christian]

One point where your prediction is wrong, and I have won the bet. Or no deal, because you have moved the goal-posts yet again. I'll announce that at Vaxjo, too.

Then you will be announcing a lie. It will not be the first time for you.

Look, I was responding to the snide remarks by you and what's his name. My goal post has been set in my experimental paper since 2008: E(a, b) = -a.b.

One point where your prediction is wrong, and I have won the bet.
I will announce our just made agreement at Vaxjo, with your permission, of course.
No movement of any goal posts by anyone.
If you wish to withdraw from the bet before Vaxjo, I will not mention it there at all.
That's a promise. OK?

Good. I have no problem with that. Just for the record: We are talking about the same four points you have been plotting, correct?

[gill1109]

Good. I have no problem with that. Just for the record: We are talking about the same four points you have been plotting, correct?

Yes, those are the only points I'm interested in. Nice picture, right? And just wait till you see it revolve. I'm working on the movie version, now.



Unfortunately the maximum resolution of images on this forum is not sufficient for this rather complex picture. But if you run the code yourself which I put up at http://rpubs.com/gill1109/Wireframe, you can see it in much higher resolution, and even take it for a spin yourself...

[Joy Christian]

Yes, nice picture. But I prefer the previous plot because the points were easier to see in that one.

Here are my predictions for my proposed experiment: E(0, 45) = - 0.7, E(0, 135) = + 0.7, E(90, 45) = -0.7, E(90, 145) = - 0.7.

[gill1109]

Yes, nice picture. But I prefer the previous plot because the points were easier to see in that one.

Here are my predictions in my proposed experiment: E(0, 45) = - 0.7, E(0, 135) = + 0.7, E(90, 45) = -0.7, E(90, 135) = - 0.7.

And I predict one of those four will be off by at least 0.2

[Joy Christian]

Yes, nice picture. But I prefer the previous plot because the points were easier to see in that one.

Here are my predictions in my proposed experiment: E(0, 45) = - 0.7, E(0, 135) = + 0.7, E(90, 45) = -0.7, E(90, 145) = - 0.7.

And I predict one of those four will be off by at least 0.2

Fine.

[Joy Christian]

Yes, nice picture. But I prefer the previous plot because the points were easier to see in that one.

Here are my predictions in my proposed experiment: E(0, 45) = - 0.7, E(0, 135) = + 0.7, E(90, 45) = -0.7, E(90, 145) = - 0.7.

And I predict one of those four will be off by at least 0.2

Fine.

OK, there is a typo in the last prediction. Correcting: E(0, 45) = - 0.7, E(0, 135) = + 0.7, E(90, 45) = -0.7, E(90, 135) = - 0.7.

[gill1109]

Correcting: E(0, 45) = - 0.7, E(0, 135) = + 0.7, E(90, 45) = -0.7, E(90, 135) = - 0.7.

Yep. I hadn't even noticed. I occasionally make the same slip of the finger. I have corrected a few of them.

[gill1109]

Can we add a little more precision to the numbers? Recall that cos(45 degrees) = sqrt(2) / 2 = 0.7071...

So to be more precise, your four targets are

E(0, 45) = - 0.7071..., E(0, 135) = + 0.7071..., E(90, 45) = - 0.7071..., E(90, 135) = - 0.7071....

I bet that at least one will be missed by an amount 0.2 (= 1/5) or more.

I don't mind how large N will be.

As a matter of interest, will the two files contain exactly equal and opposite directions u_k and - u_k or only approximately equal and opposite directions? In the former case, just one file of N directions is enough. The second file contains the set of exactly opposite directions. Add 180 degrees to the longitude (azimuthal angle) theta, change the sign of the co-latitude (zenith angle, polar angle) phi.

I use mathematician's notation. I read on wikipedia that physicists tend to use theta for the zenith angle, phi for the azimuthal angle. We have to be agreed on what is in the files (azimuth and zenith, or zenith and azimuth).

It's important that everything is completely clear and agreed by the two bettors and the three adjudicators.