New simulation shows Einstein was correct about hidden variables

[Theoretical]

 

Not sure why I'm wasting my time doing this but here are the variables that are used to control the polarizer increment. 2pi/3, which in units of degrees comes to 120.

 

long double total_polarizer_incs = 3.0L;

long double polarizer_angle2_inc = TWO_PI / total_polarizer_incs;

 

as a simulation is not evidence but a prediction against which we can compare an actual experiment, your conclusion is invalid.

That's true a sim or paper written math is no replacement for an actual experiment. It is sufficiently valid for me because I know how simple the code is and that it uses well established equations.

 

 

Please close the thread lol! It's a waste of time, but perhaps someday someone will use the code to come to the same conclusion I came to: No spooky action at a distance. Einstein was correct as far as I'm concerned, in terms of spooky action. As for what modern physicists consider hidden variables, that's a matter of opinion, and I'm certain Einstein would have agreed. I was never interested in other hidden variable theories.

[david345]

The problem with your simulation is it should NOT include the cos(angle)^2 equation. As a simulation it should get it's results using only addition, multiplication, and division.

Step 1: simulation fires out several electrons.

Step 2: add how many electrons are detected at each angle.

Step 3: divide (#electrons at each angle)\(total electrons)we will call this result x

Step 4: (2\3)x+(1\3)1=?

Step 5: compare ? to 1\2

You cheated by replacing steps 1,2, and 3 with the cos(angle)^2 equation. If the spin was determined then x would be 1\3 Because you cheated x became .25 I do not care if Malus law is classical or not. You used it incorrectly. If your code was a legitimate simulation it would not include that equation. It is nowhere in 1-5 the only place it would be used is to get the .5 number which would then be compared against the result of step 4. Because you already had that .5 number you can COMPLETELY remove cos(angle)^2 from the code and only use the number .5 in step 5. Come back when you have removed this from your code.

Edited January 7, 2015 by david345

[Theoretical]

The 1\2 number came from a video about electrons. The correct equation for electrons is .5(sin(angle))^2. http://en.m.wikipedia.org/wiki/Sakurai%27s_Bell_inequality. Cos is for when they match sin is for when they are opposite. The problem with your simulation is it should NOT include the cos(angle)^2 equation. As a simulation it should get it's results using only addition, multiplication, and division.

Step 1: simulation fires out several electrons.

Step 2: add how many electrons are detected at each angle.

Step 3: divide (#electrons at each angle)\(total electrons)we will call this result x

Step 4: (2\3)x+(1\3)1=?

Step 5: compare ? to 1\2

You cheated by replacing steps 1,2, and 3 with the cos(angle)^2 equation. If the spin was determined then x would be 1\3 Because you cheated x became .25 I do not care if Malus law is classical or not. You used it incorrectly. If your code was a legitimate simulation it would not include that equation. It is nowhere in 1-5 the only place it would be used is to get the .5 number which would then be compared against the result of step 4. Because you already had that .5 number you can COMPLETELY remove cos(angle)^2 from the code and only use the number .5 in step 5. Come back when you have removed this from your code.

No. The cos(angle)^2 determines the light intensity through the polarizer when you know the difference in angle between the polarizer and photons polarity. It's the only equation to use.

[david345]

The experiment is intended to determine the polarization. Intensity is irrelevant. The cos equation gives you the expected probability. You cheated by replacing the measured probability with the expected probability. This is why your Sim is not a Sim. Your Sim calculates the expected probability and then you say look my Sim gave the expected probability. Your Sim needs to perform steps 1,2, and 3. Until then your Sim is just a cheater.

Edited January 7, 2015 by david345

[Theoretical]

The experiment is intended to determine the polarization. Intensity is irrelevant. The cos equation gives you the expected probability. You cheated by replacing the measured probability with the expected probability. This is why your Sim is not a Sim.

It's I* cos(angle)^2 where I is intensity and it is used when you KNOW the polarizations. This does not work if you don't know the polarizations. This proves the sim already knew the polarizations.

Cheating? Come on. This isn't preschool. I used the correct equation.

[david345]

If the polarization is 35 then you say 35. Did that use cos(angle)^2? No. You have been caught cheating and refuse to admit it. If it is 35 when it leaves the source then it should be 35 when it gets to the detector. That last sentence is simple and clear. Where is the cos equation in that process. 13 pages and not one person has agreed with you.

Edited January 7, 2015 by david345

[Theoretical]

If the polarization is 35 then you say 35. Did that use cos(angle)^2? No. You have been caught cheating and refuse to admit it.

Lol you're talking to a child. Look, the photon polarization is known. So the sim needs to know if the photon goes through the polarizer or is reflected. The cos(angle)^2 works on non entangled photons by the way. I have no idea where you're getting at with cheating because it's a sim, and the sim needs to know where the NON-entangled photons go. If I don't use the cos(angle)^2 equation then the sim will have to simulate countless atoms that make up the polarizer. That's absurd. Any sincere person reading this thread knows I use the correct equation.

 

Again, this sim is about simulating NON-entangled photons. There's no cheating. I used the correct equation that any professor would have their students use in order to find what percentage of photons go through the polarizer.

 

The sim shows that when this experiment is done with non-entangled photons that it gets the same results.

[david345]

Lol you're talking to a child.

Then maybe you should stop acting like one. 13 pages and you have been proven wrong time and time again. 13 pages and not one person has agreed with you. 13 pages and you have failed to produce one single demonstration of your code. You have produced nothing but excuses and lies. Edited January 7, 2015 by david345

[Theoretical]

If it is 35 when it leaves the source then it should be 35 when it gets to the detector.

Of course it is. I never said its not. Look at the code.

Where is the cos equation in that process.

It's at the point when the photons are at the polarizer and need to know if the photon goes through or reflected.

Then maybe you should stop acting like one. 13 pages and you have been proven wrong time and time again. 13 pages and not one person has agreed with you. 13 pages and you have failed to produce one single demonstration of your code. You have produced nothing but excuses and lies.

Nobody's addressed my code until now. But it's questionable if you're addressing the code.

[david345]

It's at the point when the photons are at the polarizer and need to know if the photon goes through or reflected.

There is 3 possibilities. It goes through O, 120, or 240 If there is three equal possibilities then each should have a 33% chance of happening. Your cos(angle)^2 equation gives a 25% chance. This is why it has no business being in your simulation. Three equal possibilities each having a 25% chance of happening? This is what you are doing by adding this to your code. Edited January 7, 2015 by david345

[Theoretical]

There is 3 possibilities. It goes through O, 120, or 240 If there is three equal possibilities then each should have a 33% chance of happening. Your cos(angle)^2 equation gives a 25% chance. This is why it has no business being in your simulation. Three equal possibilities each having a 25% chance of happening? This is what you are doing by adding this to your code.

No. Using a photon at 0 deg polarization: When polarizer is at 0deg there's 100% chance of going through. At 120deg its 25%. At 240deg its 25%.

 

Do you get it yet? (100% + 25% + 25%) / 3 = 50%

[david345]

No. Using a photon at 0 deg polarization: When polarizer is at 0deg there's 100% chance of going through. At 120deg its 25%. At 240deg its 25%.

Do you get it yet? (100% + 25% + 25%) / 3 = 50%

There is 9 possible configurations of the two detectors a=0 b=120 c=240.

 

. ABC

A sds

B dsd

C sds

 

S is where they match d is where they are different.

there are 5 s and 4 d. S happens 5 of the 9 possibilities. I don't see why this is so hard to see.This is why you can not use the cos(angle)^2 equation when they have a definite polarization when they leave the detector. This is basic physics.

Edited January 7, 2015 by david345

[Theoretical]

You would get a F in physics because the correct equation to use for light through a polarizer regardless if they're non entangled is cos(angle)^2

 

cos(0)^2*100% = 100%

 

cos(120)^2*100% = 25%

 

cos(240)^2*100% = 25%

 

That is why you're getting the wrong results because you don't know the basic physics of light and polarizer. God knows what equation you're using but nobody who knows what they're doing would use.

 

And of course the above is degrees. MFC math library uses radians which is why the code uses variables such as TWO_PI.

 

 

So you're saying that 33% of photons that are at 0 degrees polarization go through a polarizer that's at 120 degrees. Everyone here who knows how to calculate this knows you're wrong. No big deal. People make mistakes. Looks like my sim gets the correct answer.

[david345]

You would get a F in physics because the correct equation to use for light through a polarizer

This equation is only correct because the light does not have a definite polarization until it is measured. This is why you continue to calculate probabilities instead of give an example where you track the polarization from the start until the finish. It is because you can't. All you can do is calculate probabilities using the very equation that proves you wrong. Edited January 7, 2015 by david345

[Theoretical]

This equation is only correct because the light does not have a definite polarization until it is measured.

You have it backwards. The equation only works when you know the polarization angle difference between the polarizer and photon, which is should be obvious to you due to the "angle" in cos(angle)^2. And like the sim code shows, both polarities are know before the photon enters the polarizer.

 

Just admit the obvious that you're wrong.

 

Furthermore, I've seen this in the NEC antenna engine. It's very simple. The antenna emits a specific known polarization. An array of elements acts as a polarizer. This exactly how my sim is setup. The NEC engine is in agreement with the cos(angle)^2 equation, which is no surprise.

 

So this part of the sim is not up for debate. It's a fact that the sim is using the correct equation.

 

And who knows what equation you're using to get 33% with 120 degrees. Crazy lol.

[david345]

You have it backwards. The equation only works when you know the polarization angle difference between the polarizer and photon, which is should be obvious to you due to the "angle" in cos(angle)^2. And like the sim code shows, both polarities are know before the photon enters the polarizer.

 

Just admit the obvious that you're wrong.

It is obvious you used the video because it is not a clear example. Once again I will demonstrate clearly why you are wrong.

The detector gives a 0 or 1

 

 

We will assume 4 "unentangled" photons with matching polarizations are sent out in both directions. Their polarization will be determined when they leave the source. The detectors will read a 0 or 1 depending on the axis of polarization. We will set the polarizations to 1010 for the 4 photons.

step 1: Bob and Alice set their detectors at matching angles. They both get 1010.

step 2: The same 1010 signal is sent out. Alice rotates detector 30 degrees. The cos(angle)^2 equation says she should get a 25% error rate. Bob gets 1010 Alice gets 1110. A 25% mismatch.

step 3: The same 1010 signal is sent out. Alice returns detector to original position and now Bob rotates his detector -30%. The cos(angle)^2 equation says Bob should get a 25% error rate. Bob gets 1000 and Alice gets 1010. A 25% mismatch.

Step 4: The same 1010 signal is sent out. Alice rotates her detector 30 degrees. Bob rotates his detector -30 degrees. Alice should get 1110 the same result as step 2. The exact same signal was sent out. Her detector was in the exact same 30 degree position. She should get the exact same result. Bob should also get the exact same 1000 result as step 3. It was the exact same 1010 signal sent out and his detector is in the exact same -30 degree position. Let us compare results. Alice 1110 Bob 1000. This is a 50% mismatch. This is what happens when the polarizations are known when they leave the detector. The cos(angle)^2 equation says there should be a 75% mismatch. The only way you can get this 75% mismatch is if the polarizations are not determined until they are measured and the angle of Bob's detector affected Alice's results and the angle of Alice's detector affected Bob's result. A 75% mismatch can only be explained by spooky action at a distance. Nick Herbert developed this proof and he received a Ph.D. in physics from Stanford University. Good night.

Edited January 7, 2015 by david345

[studiot]

 

Theoretical post 259

Nobody's addressed my code until now.

 

False.

 

I asked a specific question about a specific line of your code in my post 222.

I did not say your line was right or wrong since I am unable to evaluate its correctness without a reply.

 

You did not answer this, despite a couple of reminders.

 

Instead your only subsequent reference to any of my questions has been to beg to close the thread, as I suggested might happen (post 249).

[swansont]

I no longer will respond to posts that make claims such as "that's not what you're doing" without clear evidence. It's a waste of time. I modeled the sim after the video except I used photons as the particle.

 

 

!

Moderator Note

 

That is simply not going to fly. You are blatantly ignoring explanations as to why your sim isn't doing what you claim it's doing, and are simply soapboxing. There's no point in continuing.

 

What I truly don't get is why you think you understand this after watching a video, and yet you have had several knowledgeable people explaining things to you. Why you consider one source of information credible and another not is baffling.

 

 

 

 

 

I remember answering that. The intent of the sim is to get the results WITHOUT using QM or entanglement.

 

Since Bell's theorem and EPR are INHERENTLY quantum mechanical, this is the clearest admission that your sim doesn't do what you think it does.