Florin wrote:Joy, why not reply at my blog? I do not delete or censor comments there.
-FM
FrediFizzx wrote:To answer the question of the title; they are probably both. Also pretty amazing that they still don't know what a physics postulate is after all this debate.
A = +sign(g(a,e,s)) # Alice's measurement results A(a, e, s) = +/-1 # Here g(u,v,s) is a metric on S^3, reducing to the usual g on R^3
B = -sign(g(b,e,s)) # Bob's measurement results B(b, e, s) = -/+1 # A(a, L) = +/-1 and B(b, L) = +/-1 are exactly as defined by Bell
Cuu = length((A*B)[A > 0 & B > 0]) # Coincidence count of (+,+) events
Cdd = length((A*B)[A < 0 & B < 0]) # Coincidence count of (-,-) events
Cud = length((A*B)[A > 0 & B < 0]) # Coincidence count of (+,-) events
Cdu = length((A*B)[A < 0 & B > 0]) # Coincidence count of (-,+) events
corrs[i,j] = (Cuu + Cdd - Cud - Cdu) / (Cuu + Cdd + Cud + Cdu) # = -a.b
# There are no "0 outcomes" within S^3: Cou = Cod = Cuo = Cdo = Coo = 0
CoB = length(A[g(a,e,s) & A == 0]) # Number of A = 0 events within S^3 (regardless of B events) = 0
CAo = length(B[g(b,e,s) & B == 0]) # Number of B = 0 events within S^3 (regardless of A events) = 0
Rick Lockyer wrote:But this is NOT what Joy is doing. A and B are PRODUCTS, which complicates things a bit. Let a and b be right handed algebraic elements, and a’ and b’ be left handed algebraic element representations using the same coefficients as a and b. We then have four possible combinations of handedness products: C = a*b, D = a*b’, E = a’*b and F = a’*b’. Once again, to make any kind of correspondence between C, D, E and F we must first map each expression to a SINGULAR basis set. Trust me when I say it does not matter if we map e to f or f to e, the cross product cannot be removed by any combination of chiral forms. If you do not trust me, well do the damn math for yourself!
function getRandomLambda()
{
if( rand()>0.5) {return 1;} else {return -1;}
}
function getRandomUnitVector() //uniform random unit vector:
//http://mathworld.wolfram.com/SpherePointPicking.html
{
v=randGaussStd()*e1+randGaussStd()*e2+randGaussStd()*e3;
return normalize(v);
}
batch test()
{
set_window_title("Test of Joy Christian's arXiv:1103.1879 paper");
N=20000; //number of iterations (trials)
I=e1^e2^e3;
s=0;
a=getRandomUnitVector();
b=getRandomUnitVector();
minus_cos_a_b=-1*(a.b);
for(nn=0;nn<N;nn=nn+1) //perform the experiment N times
{
lambda=getRandomLambda(); //lambda is a fair coin,
//resulting in +1 or -1
mu=lambda * I; //calculate the lambda dependent mu
C=-I.a; //C = {-a_j B_j}
D=I.b; //D = {b_k B_k}
E=mu.a; //E = {a_k B_k(L)}
F=mu.b; //F = {b_j B_j(L)}
A=C E; //eq. (1) of arXiv:1103.1879, A(a, L) = {-a_j B_j}{a_k B_k(L)}
B=F D; //eq. (2) of arXiv:1103.1879, B(b, L) = {b_j B_j(L)}{b_k B_k}
q=0;
if(lambda==1) {q=((-C) A B (-D));} else {q=((-D) B A (-C));} //eq. (6)
s=s+q; //summation of all terms.
}
mean_mu_a_mu_b=s/N;
print(mean_mu_a_mu_b); //print the result
print(minus_cos_a_b);
prompt();
}
//Typical result is:
//mean_mu_a_mu_b = 0.87 + 0.00*e2^e3 + 0.00*e3^e1 + 0.00*e1^e2
//minus_cos_a_b = 0.87
//The scalar parts match and others vanish! Proving the result is -a.b.
//Thus Dr. Christian's arXiv:1103.1879 paper is a classical local
//realistic counter-example that in fact contradicts Bell's theorem.
from __future__ import division
import numpy
from clifford import *
layout, blades = Cl(3,0)
e0, e1, e2 = [blades['e%i'%k] for k in range(3)]
I = (e0^e1^e2)
def randVec3d(lo=0, hi=2*numpy.pi):
theta = numpy.random.uniform(lo, hi)
z = numpy.random.uniform(-1, 1)
sn = numpy.sqrt(1-z**2)
y = sn*numpy.sin(theta)
x = sn*numpy.cos(theta)
return (x^e0) + (y^e1) + (z^e2)
N = 20000
s = 0
a = randVec3d()
b = randVec3d()
for i in range(N):
L = numpy.random.choice([-1., 1.])
mu = L&I
C = (-I)*a
D = I*b
E = mu*a
F = mu*b
A = C&E
B = F&D
q = ((-C)&A&B&(-D)) if L == 1 else ((-D)&B&A&(-C))
s = s+q
print s^(1./N)
print -(a*b)
# Typical Run
#
# 0.44353 - (0.00776^e01) - (0.0068^e02) + (0.00131^e12) # Model
# 0.44353 # -a.b
minkwe wrote:Fred,
I was just going to ask the same thing! The code you posted debunks such arguments.
minkwe wrote:Fred,
I was just going to ask the same thing! The code you posted debunks such arguments.
Joy Christian wrote:Thanks Fred, thanks Michel.
Before someone produces yet another silly objection, let me add here that the non-scalar (or bivector) term in the codes posted by Fred actually vanishes EXACTLY in the analytical Eq. (19) displayed by Albert Jan from my 2007 paper. The numerical simulation of Eq.(19), however, necessarily requires replacing the Riemann integral in Eq. (19) by a discrete sum over a large N. And therefore a small --- but statistically insignificant --- residue of the non-scalar term remains for any finite N.
Rick Lockyer wrote:Fred,
if(lambda==1) {q=((-C) A B (-D));} else {q=((-D) B A (-C));} //eq. (6)
This is NOT Joy's equation 6.
Rick Lockyer wrote:Orientation choice, as I have been telling Joy for years now, is a non-starter.
FrediFizzx wrote:Rick Lockyer wrote:Orientation choice, as I have been telling Joy for years now, is a non-starter.
I think I found your problem. There is no "orientation choice" in Joy's model. And that can well be seen if you understand the GAViewer code and program. GAViewer works in a fixed right handed basis only. There is no "choice". There is only a random 50-50 chance (not "choice") in Nature that a particle pair will be created with a left or right handed orientation. What happens when you look at a left handed system from a right hand only perspective? The order is reversed. Very simple.
FrediFizzx wrote:Rick Lockyer said: "Orientation choice, as I have been telling Joy for years now, is a non-starter."
I think I found your problem. There is no "orientation choice" in Joy's model. And that can well be seen if you understand the GAViewer code and program. GAViewer works in a fixed right handed basis only. There is no "choice". There is only a random 50-50 chance (not "choice") in Nature that a particle pair will be created with a left or right handed orientation. What happens when you look at a left handed system from a right hand only perspective? The order is reversed. Very simple.
Ben6993 wrote:My wife received last week a metal bracelet in the form of a Moibus strip birthday present.
Ben6993 wrote:It seems to me that the problem is having a convention of +I for physical space (which is S^3 in your model). (Though I can see that is necessary in the limited dimensions seemingly allowed in GAViewer, and in the sub-space of GA you chose for your model.) Is it not possible to work instead in higher dimensions, or in an all-enclosing hypervolume, as you may be doing in different circumstances in other papers. In such an all-enclosing hypervolume, which itself could have a sign convention for its torsion [eg for an all enclosing, say, S^7, or whatever], wouldn't the trivectors be free to take whatever signs they are naturally created with, and would there then not be an artificial need to swap any signs because of any trivector convention? Or would the same problem still arise?
Ben6993 wrote:..I do not see how, in physics, the S^3 can have a torsion without there being at extra dimensions for it to twist in. But in maths maybe it is OK?
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