Schmelzer's and Gill's mathematical nonsense

[guest]

What Gill is questioning here is the following elementary equality:

The expression on the left hand side equals lambda^k.

The expression on the right hand side equals lambda^k times sign(s_1 . a).

Left and right hand side are equal if and only if sign(s_1 . a) = +1, i.e., if s_1 is in the hemisphere centered on a.

[FrediFizzx]

What Gill is questioning here is the following elementary equality:

The expression on the left hand side equals lambda^k.

The expression on the right hand side equals lambda^k times sign(s_1 . a).

Left and right hand side are equal if and only if sign(s_1 . a) = +1, i.e., if s_1 is in the hemisphere centered on a.

No problem there since s_1 --> a in the limit upon detection. It all still depends on the experimenter's setting. You have to realize that there is a s1 initial and a s1 final. And s1_i is not necessarily equal to s1_f.
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[Joy Christian]

What Gill is questioning here is the following elementary equality:

The expression on the left hand side equals lambda^k.

The expression on the right hand side equals lambda^k times sign(s_1 . a).

Left and right hand side are equal if and only if sign(s_1 . a) = +1, i.e., if s_1 is in the hemisphere centered on a.

That is correct. But we are comparing two entirely different models --- THE correct model on the LHS, and Bell's silly model on the RHS. You have to read what I have written in my post above: "(at least in the present context)." More importantly, you have to read what I have discussed in the paper: https://arxiv.org/abs/1405.2355.

Alternatively, one can see that the two models are effectively the same, and can even be made strictly equal mathematically, by simply confining to only acute angles between s_1 and a, which, as Fred points out, is perfectly valid since we are only interested in heuristic comparison of two very different models in the s_1 --> a limit.

Finally, there are two more ways to see the strict equality. One is what Jay has done in his analysis, and another is how I have identified s_1^k with lambda^k x s_1.

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[FrediFizzx]

Hi Joy,

I think you could make more sense mathematically to a mathematician if you did something like this.



That is what is really going on anyways.

[Joy Christian]

Hi Joy,

I think you could make more sense mathematically to a mathematician if you did something like this.



That is what is really going on anyways.

Perhaps you are right. But anyone who really wants to understand what is going on does not have to be a mathematician to do so.

Any physicist should be able to understand that the measurement function A(a, lambda) encodes a detection process, not a free evolution of the spin.

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[FrediFizzx]

Hi Joy,

I think you could make more sense mathematically to a mathematician if you did something like this.



That is what is really going on anyways.

Perhaps you are right. But anyone who really wants to understand what is going on does not have to be a mathematician to do so.

Any physicist should be able to understand that the measurement function A(a, lambda) encodes a detection process, not a free evolution of the spin.

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I not sure you actually need the limiting process then. Can't you just state that s1_f = a and s2_f = b?
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[Joy Christian]

I am not sure you actually need the limiting process then. Can't you just state that s1_f = a and s2_f = b?

How would you specify the conservation of spin in that case? Initially L^2(s, lambada) = -1 must hold, and the functions A and B must be mathematically well defined.

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[FrediFizzx]

I am not sure you actually need the limiting process then. Can't you just state that s1_f = a and s2_f = b?

How would you specify the conservation of spin in that case? Initially L^2(s, lambada) = -1 must hold, and the functions A and B must be mathematically well defined.

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If you specify s1_i --> s1_f = a and s2_i --> s2_f = b for the limiting process on the A and B function definitions, then you can drop the limit process on most all of the rest of the calculations.

I am just trying to see if there is a way to make it more clear for those that just look at the math and then have stupid objections. Of course there is no problem for people that actually understand the physics.
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[Joy Christian]

I am not sure you actually need the limiting process then. Can't you just state that s1_f = a and s2_f = b?

How would you specify the conservation of spin in that case? Initially L^2(s, lambada) = -1 must hold, and the functions A and B must be mathematically well defined.

***

If you specify s1_i --> s1_f = a and s2_i --> s2_f = b for the limiting process on the A and B function definitions, then you can drop the limit process on most all of the rest of the calculations.

I am just trying to see if there is a way to make it more clear for those that just look at the math and then have stupid objections. Of course there is no problem for people that actually understand the physics.
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Specifying s1_i --> s1_f = a and s2_i --> s2_f = b is harmless. It is purely cosmetic, because s_1 and s_2 are not hidden variables to being with. Only lambda is.

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[FrediFizzx]

Specifying s1_i --> s1_f = a and s2_i --> s2_f = b is harmless. It is purely cosmetic, because s_1 and s_2 are not hidden variables to being with. Only lambda is.

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Good. Then why not do it? You can get rid of the A and B limit process on eqs. (69) thru (71). Actually you could go directly from (68) to (72) but you might want to retain some of the steps for clarity.

[FrediFizzx]

Specifying s1_i --> s1_f = a and s2_i --> s2_f = b is harmless. It is purely cosmetic, because s_1 and s_2 are not hidden variables to being with. Only lambda is.

***

Good. Then why not do it? You can get rid of the A and B limit process on eqs. (69) thru (71). Actually you could go directly from (68) to (72) but you might want to retain some of the steps for clarity.

Oh, you have to keep the limit on for eq. (69) because there you are doing {L(s1_i, lamba^k) L(s2_i, lambda^k)}. But actually the way you have it now, there is no limit process anyways for (70) and (71).

[Joy Christian]

Specifying s1_i --> s1_f = a and s2_i --> s2_f = b is harmless. It is purely cosmetic, because s_1 and s_2 are not hidden variables to being with. Only lambda is.

***

Good. Then why not do it? You can get rid of the A and B limit process on eqs. (69) thru (71). Actually you could go directly from (68) to (72) but you might want to retain some of the steps for clarity.

Actually, I take it back. Specifying s1_i --> s1_f = a and s2_i --> s2_f = b gives the wrong impression. It gives the impression that initial spin s1_i is somehow evolving into final spin s2_f, which just happens to be equal to a. But that is wrong. After the initial decay of the neutral pion there is no interaction between the spin and the detector, or between the two spins, until the spin hits the detector. The spins are evolving freely until detection. Only the conservation of the initial spin-0 remains in force during the evolution, so that s1 = s2 is maintained. It is only at the detector that the components of the two spins along the two respective directions a and b are measured by Alice and Bob, just as Bell has it in his own local model of 1964. So at least physically specifying s1_i --> s2_f = a is wrong. What is happening physically is the detection process s1 --> a, which picks out the normalized component of s1 along a, just as I have it in the paper.

PS: Something is still wrong with the site. It takes ages to load. Or is it just my browser that is slow?

[Ben6993]

Hi Fred
Just thinking about the expression si -> sf -> a.
As si and sf (= a) are vectors they should be in bold, especially as you say you were trying to make more sense for the mathematicians.
I can understand the urge to insert Appendix in v5, as the '->' symbol makes me think of a classical '[asymptotically] tending to a limit' when in fact there is no 'tending to' in this case. This is one of the weird things about measuring an electron. Whatever the s value, the electron for Alice either immediately snaps from s to a with the emission of a single photon, or it doesn't emit a photon at all. (For simplicity I am just imagining the case where she either detects an emitted photon or there is no emitted photon.)

So on detection by Alice, an electron's states are instantly transformed from s1 and lambda1 to a and lambda_new. (And as an aside, as an electron changes handedness on detection, lambda_new does not necessarily equal lambda1?) But one can think of detecting an electron as peparing an electron for a new experiment. So a and Lambda_new are appropriate for the newly prepared electron in a new experiment, whereas the old electron, i.e. the one participating in the Bell experiment, had states s1 and lambda1.

However, since lambda1 is an unknown, information has to be built up statistically about the old electrons using the outcomes of Alice on many electrons all in terms of measurement results for which the newly measured electron vectors are a.
So, after measurement, all Alice's vectors are a single vector a. But before measurement there is a full range of vector directions: s1.
And s1 does not need to equal a for an electron to emit a photon (otherwise there would be very few detections indeed).
That makes me think that the '->' symbol is a very confusing one.
Is there a symbol for 's1 snaps to a'?


PS Yes, website loading time is slow at the moment

[FrediFizzx]

Actually, I take it back. Specifying s1_i --> s1_f = a and s2_i --> s2_f = b gives the wrong impression. It gives the impression that initial spin s1_i is somehow evolving into final spin s2_f, which just happens to be equal to a. But that is wrong. After the initial decay of the neutral pion there is no interaction between the spin and the detector, or between the two spins, until the spin hits the detector. The spins are evolving freely until detection. Only the conservation of the initial spin-0 remains in force during the evolution, so that s1 = s2 is maintained. It is only at the detector that the components of the two spins along the two respective directions a and b are measured by Alice and Bob, just as Bell has it in his own local model of 1964. So at least physically specifying s1_i --> s2_f = a is wrong. What is happening physically is the detection process s1 --> a, which picks out the normalized component of s1 along a, just as I have it in the paper.

PS: Something is still wrong with the site. It takes ages to load. Or is it just my browser that is slow?

That is not true. There are always polarizers (in the EPR-Bohm case it would be Stern-Gerlach devices) before the detectors and they direct the particle via its spin to either the up (+1) detector or the down (-1) detector. So s1_i --> s2_f = a is correct. "a" is the action of the polarizer via its angle setting. s1_i does go to s2_f at the polarizer. Think about it; you don't change the angle of the detector, you change the angle of the polarizer.

It is not your browser; it is the hosting company that I can't do anything about until Monday.
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[Joy Christian]

Actually, I take it back. Specifying s1_i --> s1_f = a and s2_i --> s2_f = b gives the wrong impression. It gives the impression that initial spin s1_i is somehow evolving into final spin s2_f, which just happens to be equal to a. But that is wrong. After the initial decay of the neutral pion there is no interaction between the spin and the detector, or between the two spins, until the spin hits the detector. The spins are evolving freely until detection. Only the conservation of the initial spin-0 remains in force during the evolution, so that s1 = s2 is maintained. It is only at the detector that the components of the two spins along the two respective directions a and b are measured by Alice and Bob, just as Bell has it in his own local model of 1964. So at least physically specifying s1_i --> s2_f = a is wrong. What is happening physically is the detection process s1 --> a, which picks out the normalized component of s1 along a, just as I have it in the paper.

PS: Something is still wrong with the site. It takes ages to load. Or is it just my browser that is slow?

That is not true. There are always polarizers (in the EPR-Bohm case it would be Stern-Gerlach devices) before the detectors and they direct the particle via its spin to either the up (+1) detector or the down (-1) detector. So s1_i --> s2_f = a is correct. "a" is the action of the polarizer via its angle setting. s1_i does go to s2_f at the polarizer.

It is not your browser; it is the hosting company that I can't do anything about until Monday.
.

Sure, there are polarizers. But all of that is just part of the measurement set up, confined locally to the labs of Alice and Bob, who are space-like separated from one another. There is nothing in-between for the spins to deviate from free evolution, until they hit the polarizers. So physically s1_i --> s2_f = a is not what is happening.

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[FrediFizzx]

Sure, there are polarizers. But all of that is just part of the measurement set up, confined locally to the labs of Alice and Bob, who are space-like separated from one another. There is nothing in-between for the spins to deviate from free evolution, until they hit the polarizers. So physically s1_i --> s2_f = a is not what is happening.

***

Sure it is. "a" is the angle of the polarizer wrt some reference frame. That means when the particle is going thru the polarizer, its spin is aligned to "a" by the magnetic field and it is either "up" (+1) or down (-1). So surely s1_f = a after the polarizer and before the detectors. Well, actually s1_f is "a" or "-a" depending if aligned up or down. And s1_i (initial) is not necessarily equal to s1_f (final) but it could be.

[Joy Christian]

Sure, there are polarizers. But all of that is just part of the measurement set up, confined locally to the labs of Alice and Bob, who are space-like separated from one another. There is nothing in-between for the spins to deviate from free evolution, until they hit the polarizers. So physically s1_i --> s2_f = a is not what is happening.

***

Sure it is. "a" is the angle of the polarizer wrt some reference frame. That means when the particle is going thru the polarizer, its spin is aligned to "a" by the magnetic field and it is either "up" (+1) or down (-1). So surely s1_f = a after the polarizer and before the detectors. Well, actually s1_f is "a" or "-a" depending if aligned up or down. And s1_i (initial) is not necessarily equal to s1_f (final) but it could be.

I suppose one can have two meanings, or two separate definitions of the word "initial", one for the decay event --- which is what is meant by "initial" in the Bell literature --- and one for the detection event as you have it. But why bother with all such complications when I have explained clearly in the paper what is happening physically, with words, equations, and even a very illustrative figure?

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[FrediFizzx]

I suppose one can have two meanings, or two separate definitions of the word "initial", one for the decay event --- which is what is meant by "initial" in the Bell literature --- and one for the detection event as you have it. But why bother with all such complications when I have explained clearly in the paper what is happening physically, with words, equations, and even a very illustrative figure?
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Because even Jay didn't quite understand it from his post on RW and I not sure Ben does yet either. You should do whatever you can to make it more clear to a reader. Most people don't bother with a bunch of detail.

For me s1_i means the creation event when s is created and s1_f means at detection. And quite actually s1_f can be a or -a. I am not sure you are accounting for the extra sign flip possibility from the polarizer. That for sure makes your A and B functions not equal to lambda and -lambda respectively because you could have A = - lambda and B = + lambda or A = + lambda and B = + lambda, etc. If a = b then you will always get anti-correlation though.

[FrediFizzx]

So the limit function should actually be,



Or simply,



If you don't want to bother with initial and final.
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[Joy Christian]

So the limit function should actually be,



Or simply,



If you don't want to bother with initial and final.
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The correct limit is as I have it in the paper. All sign changes are accounted for, and encoded in lambda.

If we have it like then no correlations would be observed. Correlations would average out to zero.

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