## Lockyer's math error

Foundations of physics and/or philosophy of physics, and in particular, posts on unresolved or controversial issues

### Lockyer's math error

Lockyer has recently been ranting about this over on a blog at FQXi. Since he has not been able to discover nor figure out his math error below for several months now, perhaps it is time to help him out with it some more. We already told him several times that he has a math error. It actually wasn't all that hard to find his error.

Rick Lockyer wrote:Joy is on record stating the cross product averages out over “Nature’s fair coin choice of orientation”. This is categorically incorrect. The math involved for S^3 is the quaternion algebra isomorphic to the bivector subalgebra of GA Joy uses. This is not disputable. The orientation choices for S^3 are a direct byproduct of the two chiral choices for the definition of quaternion algebra.

The two chiral choices for the definition of quaternion algebra come about in the products of non-scalar basis elements. Define the scalar quaternion basis element as e_0 == 1, and an ordered triplet of the non-scalar basis elements e_1, e_2, and e_3 basis elements as

(e_1 , e_2, e_3 ) == (e_2 , e_3, e_1 ) == (e_3 , e_1, e_2 )

These equivalent forms imply the following quaternion basis element product rules cyclic left to right for positive signed outcomes and cyclic right to left for negative signed outcomes:

e_1 * e_2 = e_3
e_2 * e_3 = e_1
e_3 * e_1 = e_2
e_1* e_3 = -e_2
e_2 * e_1 = -e_3
e_3 * e_2 = -e_1

If we assign x:1, y:2 and z:3 significance to the indexes, we have just defined a right handed chiral system, absolutely relevant to S^3 orientation. The opposite orientation would be described by a different basis set which I will call f_x, with the following ordered triplet with identical xyz index significance:

(f_1, f_3, f_2) == (f_3, f_2, f_1) == (f_2, f_1, f_3)

The rule of algebraic element addition for any algebra over the field of real numbers as we have here is the following for real coefficients A_j and B_j:

A = A_j e_j sum over j for the dimension of the algebra in question
B = B_j e_j sum over j for the dimension of the algebra in question

A + B = (A_j + B_j) e_j sum over j for the dimension of the algebra in question

The extremely important thing to take away from this non-disputable rule of algebraic element addition, which is very relevant to our discussion, is that coefficients attached to IDENTICAL basis elements add in the final result. Therefore, if we have A represented in a right handed quaternion system, and B represented in a left handed quaternion system, we have

A + B =
A_0 e_0 + A_1 e_1 + A_2 e_2 + A_3 e_3 +
B_0 f_0 + B_1 f_1 + B_2 f_2 + B_3 f_3

If we wish to simplify this expression, we must find a way to map either e_j to f_j or f_j to e_j such that the coefficients for A and B are now attached to a consistent set of basis elements such that we may combine their coefficients. There is such a map, it is:

e_0 -> f_0
e_k -> -f_k for k = 1 to 3

With this map or the converse, we have

A + B = (A_0 + B_0) e_0 == (A_0 + B_0) f_0 since e_0 == f_0

So we have indeed dispatched the non-scalar portion. 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! I am going to map f to e for C, D, E, and F.

C =
+ (a_0 b_0) e_0*e_0 + (a_1 b_1) e_1*e_1 + (a_2 b_2) e_2*e_2 + (a_3 b_3) e_3*e_3
+ (a_0 b_1) e_0*e_1 + (a_1 b_0) e_1*e_0 + (a_2 b_3) e_2*e_3 + ( a_3 b_2) e_3*e_2
+ (a_0 b_2) e_0*e_2 + (a_2 b_0) e_2*e_0 + (a_3 b_1) e_3*e_1 + (a_1 b_3) e_1*e_3
+ (a_0 b_3) e_0*e_3 + (a_3 b_0) e_3*e_0 + (a_1 b_3) e_1*e_3 + (a_3 b_1) e_3*e_1
=
+ (a_0 b_0 – a_1 b_1 – a_2 b_2 – a_3 b_3) e_0
+ (a_0 b_1 + a_1 b_0 + a_2 b_3 – a_3 b_2) e_1
+ (a_0 b_2 + a_2 b_0 + a_3 b_1 – a_1 b_3) e_2
+ (a_0 b_3 + a_3 b_0 + a_1 b_2 – a_2 b_1) e_3

D =
+ (a_0 b_0) e_0*f_0 + (a_1 b_1) e_1*f_1 + (a_2 b_2) e_2*f_2 + (a_3 b_3) e_3*f_3
+ (a_0 b_1) e_0*f_1 + (a_1 b_0) e_1*f_0 + (a_2 b_3) e_2*f_3 + ( a_3 b_2) e_3*f_2
+ (a_0 b_2) e_0*f_2 + (a_2 b_0) e_2*f_0 + (a_3 b_1) e_3*f_1 + (a_1 b_3) e_1*f_3
+ (a_0 b_3) e_0*f_3 + (a_3 b_0) e_3*f_0 + (a_1 b_3) e_1*f_3 + (a_3 b_1) e_3*f_1
=
+ (a_0 b_0 + a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (– a_0 b_1 + a_1 b_0 – a_2 b_3 + a_3 b_2) e_1
+ (– a_0 b_2 + a_2 b_0 – a_3 b_1 + a_1 b_3) e_2
+ (– a_0 b_3 + a_3 b_0 – a_1 b_2 + a_2 b_1) e_3

E =
+ (a_0 b_0) f_0*e_0 + (a_1 b_1) f_1*e_1 + (a_2 b_2) f_2*e_2 + (a_3 b_3) f_3*e_3
+ (a_0 b_1) f_0*e_1 + (a_1 b_0) f_1*e_0 + (a_2 b_3) f_2*e_3 + ( a_3 b_2) f_3*e_2
+ (a_0 b_2) f_0*e_2 + (a_2 b_0) f_2*e_0 + (a_3 b_1) f_3*e_1 + (a_1 b_3) f_1*e_3
+ (a_0 b_3) f_0*e_3 + (a_3 b_0) f_3*e_0 + (a_1 b_3) f_1*e_3 + (a_3 b_1) f_3*e_1
=
+ (a_0 b_0 + a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (a_0 b_1 – a_1 b_0 – a_2 b_3 + a_3 b_2) e_1
+ (a_0 b_2 – a_2 b_0 – a_3 b_1 + a_1 b_3) e_2
+ (a_0 b_3 – a_3 b_0 – a_1 b_2 + a_2 b_1) e_3

F =
+ (a_0 b_0) f_0*f_0 + (a_1 b_1) f_1*f_1 + (a_2 b_2) f_2*f_2 + (a_3 b_3) f_3*f_3
+ (a_0 b_1) f_0*f_1 + (a_1 b_0) f_1*f_0 + (a_2 b_3) f_2*f_3 + ( a_3 b_2) f_3*f_2
+ (a_0 b_2) f_0*f_2 + (a_2 b_0) f_2*f_0 + (a_3 b_1) f_3*f_1 + (a_1 b_3) f_1*f_3
+ (a_0 b_3) f_0*f_3 + (a_3 b_0) f_3*f_0 + (a_1 b_3) f_1*f_3 + (a_3 b_1) f_3*f_1
=
+ (a_0 b_0 – a_1 b_1 – a_2 b_2 – a_3 b_3) e_0
+ (– a_0 b_1 – a_1 b_0 + a_2 b_3 – a_3 b_2) e_1
+ (– a_0 b_2 – a_2 b_0 + a_3 b_1 – a_1 b_3) e_2
+ (– a_0 b_3 – a_3 b_0 + a_1 b_2 – a_2 b_1) e_3

Joy starts with pure bivector values, meaning we have A_0 = B_0 = 0. This does simplify things a bit. Substituting we have

C =
– (a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (a_2 b_3 – a_3 b_2) e_1
+ (a_3 b_1 – a_1 b_3) e_2
+ (a_1 b_2 – a_2 b_1) e_3

D =
+ (a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
– (a_2 b_3 – a_3 b_2) e_1
– (a_3 b_1 – a_1 b_3) e_2
– (a_1 b_2 – a_2 b_1) e_3

E =
+ (a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
– (a_2 b_3 – a_3 b_2) e_1
– (a_3 b_1 – a_1 b_3) e_2
– (a_1 b_2 – a_2 b_1) e_3

F =
– (a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (a_2 b_3 – a_3 b_2) e_1
+ (a_3 b_1 – a_1 b_3) e_2
+ (a_1 b_2 – a_2 b_1) e_3

So with pure bivectors (quaternions) we have C = –D = –E = F. Since these are the ONLY chiral combinations available, there is absolutely no way to clear the non-scalar component of a sum of similar products of pure bivectors through any possible modification of orientation for individual products.

So we took Lockyer's advice and did the math for ourself. For the following analysis, only C and F above really matter as we would never have mixed left-right handed pairs since the particle pairs themselves form the left or right handed system. What Lockyer's error is that when he calculated F (left handed) the first time, he double translated it. IOW, he did the mapping twice. And you might notice that the LHS of the equations for C and F have typos in them ((a_3 b_1) f_3*f_1 and (a_1 b_3) f_1*f_3 appear twice each) but it is corrected in the RHS. I have underlined them above in the quote. Anyways... back to the double mapping. As can be seen by the multiplication tables here,

http://captaincomputersensor.net/quaternions.html ,

the mapping Lockyer specified above is already contained in the difference between the two tables. So when he was calculating F, he did the left handed multiplication and then applied the mapping again by applying the minus signs to the appropriate elements. Of course that will take you right back to the beginning of what you were doing and no translation occured at all. We can see what he did by taking the element (a_2 b_3) f_2*f_3 then looking up f_2*f_3 in the left handed table equals -e_1 so the result should just be -a_2 b_3 for the e_1 line but you can see that he had it as +a_2 b_3 in the RHS in the e_1 line so he multiplied again by -1.

When only mapped once and dispensing with the zero components we get for F,

F =
– (a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (-a_2 b_3 + a_3 b_2) e_1
+ (-a_3 b_1 + a_1 b_3) e_2
+ (-a_1 b_2 + a_2 b_1) e_3

And we can see that C is no longer equal to F as it should be since common sense logic dictates that if you start with the same formulas for left and right that upon translation of one of them, they should no longer be equal. Now... from what we learned about this from using GAViewer, we can check this simply by reversing the order of the multipication. And again dispensing with the zero components we have,

F = (b1 e1 + b2 e2 + b3 e3)(a1 e1 + a2 e2 + a3 e3)
=
+a1b1 e1e1 + a2b2 e2e2 + a3b3 e3e3
+a2b3 e3e2 + a3b2 e2e3
+a3b1 e1e3 + a1b3 e3e1
+a1b2 e2e1 + a2b1 e1e2
=
-(a1b1 + a2b2 + a3b3)e0
+(-a2b3 + a3b2)e1
+(-a3b1 + a1b3)e2
+(-a1b2 + a2b1)e3

Check complete and they match. Now when we add C to F and divide by two to get the average we obtain,

(C + F)/2 = -(a1b1 + a2b2 + a3b3) + 0 = -a.b

As expected. QED. So we have proven that Joy's model is correct a little different way.
FrediFizzx
Independent Physics Researcher

Posts: 2687
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Location: N. California, USA

### Re: Lockyer's math error

***

Thanks, Fred, for exposing the sophomoric math error of a yet another incompetent fool who has been relentlessly harassing us online for the past several years.

Let me also point to a very simple and transparent derivation of the EPR-B correlation, which I have already posted elsewhere: viewtopic.php?f=6&t=222#p5846

***
Joy Christian
Research Physicist

Posts: 2719
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Lockyer's math error

Joy Christian wrote:***

Thanks, Fred, for exposing the sophomoric math error of a yet another incompetent fool who has been relentlessly harassing us online for the past several years.

Let me also point to a very simple and transparent derivation of the EPR-B correlation, which I have already posted elsewhere: viewtopic.php?f=6&t=222#p5846

***

You're welcome. Now if Lockyer continues to spew his nonsense anywhere, we can just point to the post above.
FrediFizzx
Independent Physics Researcher

Posts: 2687
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Lockyer's math error

As posted on FQXi:

There is a block copy editing error only on the last sum line for C,D,E and F, did not change some 3's to 2's. The simplifications that follow immediately have no errors, so sorry but there is no math error.

C =
+ (a_0 b_0) e_0*e_0 + (a_1 b_1) e_1*e_1 + (a_2 b_2) e_2*e_2 + (a_3 b_3) e_3*e_3
+ (a_0 b_1) e_0*e_1 + (a_1 b_0) e_1*e_0 + (a_2 b_3) e_2*e_3 + (a_3 b_2) e_3*e_2
+ (a_0 b_2) e_0*e_2 + (a_2 b_0) e_2*e_0 + (a_3 b_1) e_3*e_1 + (a_1 b_3) e_1*e_3
+ (a_0 b_3) e_0*e_3 + (a_3 b_0) e_3*e_0 + (a_1 b_2) e_1*e_2 + (a_2 b_1) e_2*e_1
=
+ (a_0 b_0 – a_1 b_1 – a_2 b_2 – a_3 b_3) e_0
+ (a_0 b_1 + a_1 b_0 + a_2 b_3 – a_3 b_2) e_1
+ (a_0 b_2 + a_2 b_0 + a_3 b_1 – a_1 b_3) e_2
+ (a_0 b_3 + a_3 b_0 + a_1 b_2 – a_2 b_1) e_3

D =
+ (a_0 b_0) e_0*f_0 + (a_1 b_1) e_1*f_1 + (a_2 b_2) e_2*f_2 + (a_3 b_3) e_3*f_3
+ (a_0 b_1) e_0*f_1 + (a_1 b_0) e_1*f_0 + (a_2 b_3) e_2*f_3 + (a_3 b_2) e_3*f_2
+ (a_0 b_2) e_0*f_2 + (a_2 b_0) e_2*f_0 + (a_3 b_1) e_3*f_1 + (a_1 b_3) e_1*f_3
+ (a_0 b_3) e_0*f_3 + (a_3 b_0) e_3*f_0 + (a_1 b_2) e_1*f_2 + (a_2 b_1) e_2*f_1
=
+ (a_0 b_0 + a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (– a_0 b_1 + a_1 b_0 – a_2 b_3 + a_3 b_2) e_1
+ (– a_0 b_2 + a_2 b_0 – a_3 b_1 + a_1 b_3) e_2
+ (– a_0 b_3 + a_3 b_0 – a_1 b_2 + a_2 b_1) e_3

E =
+ (a_0 b_0) f_0*e_0 + (a_1 b_1) f_1*e_1 + (a_2 b_2) f_2*e_2 + (a_3 b_3) f_3*e_3
+ (a_0 b_1) f_0*e_1 + (a_1 b_0) f_1*e_0 + (a_2 b_3) f_2*e_3 + (a_3 b_2) f_3*e_2
+ (a_0 b_2) f_0*e_2 + (a_2 b_0) f_2*e_0 + (a_3 b_1) f_3*e_1 + (a_1 b_3) f_1*e_3
+ (a_0 b_3) f_0*e_3 + (a_3 b_0) f_3*e_0 + (a_1 b_2) f_1*e_2 + (a_2 b_1) f_2*e_1
=
+ (a_0 b_0 + a_1 b_1 + a_2 b_2 + a_3 b_3) e_0
+ (a_0 b_1 – a_1 b_0 – a_2 b_3 + a_3 b_2) e_1
+ (a_0 b_2 – a_2 b_0 – a_3 b_1 + a_1 b_3) e_2
+ (a_0 b_3 – a_3 b_0 – a_1 b_2 + a_2 b_1) e_3

F =
+ (a_0 b_0) f_0*f_0 + (a_1 b_1) f_1*f_1 + (a_2 b_2) f_2*f_2 + (a_3 b_3) f_3*f_3
+ (a_0 b_1) f_0*f_1 + (a_1 b_0) f_1*f_0 + (a_2 b_3) f_2*f_3 + (a_3 b_2) f_3*f_2
+ (a_0 b_2) f_0*f_2 + (a_2 b_0) f_2*f_0 + (a_3 b_1) f_3*f_1 + (a_1 b_3) f_1*f_3
+ (a_0 b_3) f_0*f_3 + (a_3 b_0) f_3*f_0 + (a_1 b_2) f_1*f_2 + (a_2 b_1) f_2*f_1
=
+ (a_0 b_0 – a_1 b_1 – a_2 b_2 – a_3 b_3) e_0
+ (– a_0 b_1 – a_1 b_0 + a_2 b_3 – a_3 b_2) e_1
+ (– a_0 b_2 – a_2 b_0 + a_3 b_1 – a_1 b_3) e_2
+ (– a_0 b_3 – a_3 b_0 + a_1 b_2 – a_2 b_1) e_3

No change: C=F, and when added, the non-scalar term does not go to zero. This typo was pretty damn obvious, don't you think? Let me spell out the simplifications for you:

f_1 * f_2 = -f_3 = +e_3 = e_1 * e_2
f_2 * f_3 = -f_1 = +e_1 = e_2 * e_3
f_3 * f_1 = -f_2 = +e_2 = e_3 * e_1
f_2 * f_1 = +f_3 = -e_3 - e_2 * e_1
f_3 * f_2 = +f_1 = -e_1 = e_3 * e_2
f_1 * f_3 = +f_2 = -e_2 = e_1 * e_3

You can see from the e and f indexes how the signs on the bivector results will be the same for right and left handed representations when cast in the e basis. But then this is more simply represented by

(I.a)(I.b) = (-I.a)(-I.b)

Sorry for the typo, but it was not a supportive argument for you anyway. The first 3 sums resulting in e_0, e_1 and e_2 were correct as they stood. This is extremely easy and straight forward, not my problem the three of you can't do simple algebra.
Rick Lockyer

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Location: Nipomo

### Re: Lockyer's math error

***

Well, well. Just as I thought. Lockyer is not only incompetent, but also just as disingenuous and duplicitous as his masters Gill and Moldoveanu.

I think the correct response to Lockyer's stupidity was the one provided by Michel Fodje some time ago: viewtopic.php?f=6&t=183&start=80#p5080

***
Joy Christian
Research Physicist

Posts: 2719
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Lockyer's math error

Joy Christian wrote:***

Well, well. Just as I thought. Lockyer is not only incompetent, but also just as disingenuous and duplicitous as his masters Gill and Moldoveanu.

I think the correct response to Lockyer's stupidity was the one provided by Michel Fodje some time ago: viewtopic.php?f=6&t=183&start=80#p5080

***

Yep, it surely has gone from incompetence to disingenuous after his double mapping error was explicitly pointed out to him and he repeats the double mapping error in his response above. No point in trying to help Lockyer out anymore. It is plainly a hopeless task.
FrediFizzx
Independent Physics Researcher

Posts: 2687
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Lockyer's math error

***
I have posted the following comments at FQXi where Lockyer is still spewing his nonsense:

Fred, in addition to what you say, Rick Lockyer is also dishonest. His math errors have been pointed out to him literally hundreds of times, here at FQXi and elsewhere, but he dishonestly disregards every opportunity to recognize them. It is also worth noting that he has no qualifications in physics to understand something as basic as what is meant by a "hidden variable", or what is the actual physics behind the EPR-Bohm type experiments. All of these things contribute to his ignorance and inability to recognize his math errors.

By contrast, my 3-sphere calculations have been explicitly verified by Lucien Hardy, a FQXi member and one of the foremost experts in the foundations of quantum mechanics. Moreover, one of my papers on the arXiv where I show this calculations has been published in the International Journal of Theoretical Physics, whose founding editor as well as several of its distinguished editorial board members are also members of FQXi.

To be sure, one does not have to depend on any such "argument from authority." All one has to do is to look at just a few lines of my elementary calculations in my latest paper (for example) to verify for themselves that my mathematical and physical arguments are trivially correct.

***
Joy Christian
Research Physicist

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Location: Oxford, United Kingdom

### Re: Lockyer's math error

Joy Christian wrote:***
I have posted the following comments at FQXi where Lockyer is still spewing his nonsense:

Fred, in addition to what you say, Rick Lockyer is also dishonest. His math errors have been pointed out to him literally hundreds of times, here at FQXi and elsewhere, but he dishonestly disregards every opportunity to recognize them. It is also worth noting that he has no qualifications in physics to understand something as basic as what is meant by a "hidden variable", or what is the actual physics behind the EPR-Bohm type experiments. All of these things contribute to his ignorance and inability to recognize his math errors.

By contrast, my 3-sphere calculations have been explicitly verified by Lucien Hardy, a FQXi member and one of the foremost experts in the foundations of quantum mechanics. Moreover, one of my papers on the arXiv where I show this calculations has been published in the International Journal of Theoretical Physics, whose founding editor as well as several of its distinguished editorial board members are also members of FQXi.

To be sure, one does not have to depend on any such "argument from authority." All one has to do is to look at just a few lines of my elementary calculations in my latest paper (for example) to verify for themselves that my mathematical and physical arguments are trivially correct.

***

Yep, you know it has to be dishonesty when one can easily see that the mapping is already done in the tables here,

http://captaincomputersensor.net/quaternions.html

Very easy to see since the tables are side by side.
FrediFizzx
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Posts: 2687
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Lockyer's math error

***

I am reproducing the following post from elsewhere in this forum because it is pertinent to the discussion of Lockyer's sophomoric mistake being discussed here:

Joy Christian wrote:

Although he is banned from this forum (just as he is banned from Paul Snively's blog), Richard Gill has "kindly" sent me the following admission in a private email:

Richard Gill wrote:
I am delighted to admit that I had overlooked the intended interpretation of your two equations: the interpretation which makes them entirely consistent with one another. In retrospect, very obvious ...

This was sent to me in a private email after he saw my following earlier post in this thread:

Joy Christian wrote:* * *

Well, Lockyer seems to have finally recognized his error. I hope he has the courage to admit his error in public (I expect no such decency from Gill and Moldoveanu).

Let me spell out the key point explicitly here so there remains no doubt in anyone's mind. The mathematical demonstration is exceedingly simple and easy to follow.

What we want to show is that the first equation above (as written by me in these two papers) forms a right-handed system; and the second equation (again, as written by me in these two papers) forms a left-handed system. Gill, Moldoveanu, and Lockyer falsely claim that I have made a sign error on the RHS of the second equation.

To check the handedness of these equations, let us set the angle between ${\bf a}$ and ${\bf b}$ to be 90 degrees and define ${\bf c} = {\bf a}\,\times\,{\bf b}$ for this case. Now multiply (using the geometric product) the first equation, on both sides, by ${(+{\it I}\,\cdot\,{\bf c})}$, from the left. Since we have set ${\bf a}$ and ${\bf b}$ to be orthogonal to each other, and since all bivectors such as ${(+{\it I}\,\cdot\,{\bf c})}$ square to ${-1}$, the first equation (in this special case of orthogonal ${\bf a}$ and ${\bf b}$) reduces to

${(+{\it I}\;\cdot\;{\bf a})\,(+{\it I}\,\cdot\,{\bf b})\,(+{\it I}\,\cdot\,{\bf c})\,=+1.}$

Now if we follow the same procedure for the second equation -- this time using the bivector ${(-{\it I}\,\cdot\,{\bf c})}$, it also reduces to

${(- {\it I}\;\cdot\;{\bf a})\,(-{\it I}\,\cdot\,{\bf b})\,(-{\it I}\,\cdot\,{\bf c})\,=+1.}$

So far so good. Both equations, as long as they remain "unaware" of each other, can be taken to represent a right-handed system, because their RHS equals to ${+1\,}$.

But now suppose we wish to compare the two systems, as done in the successive trials of EPR-B type experiments ( is the spin "up", or "down" in a given trial? ). Then we must find a functional mapping between the two equations. But that is a trivial task in the present set up, since we see that the respective bivectors are related as

${(-{\it I}\;\cdot\;{\bf n})\,=\,- \,(+{\it I}\,\cdot\,{\bf n}),}$

for any directional vector ${\bf n}$. If we now substitute the above mapping in the second of the two equations we have derived, we obtain at once, for this second system,

${(+{\it I}\,\cdot\,{\bf a})\,(+{\it I}\,\cdot\,{\bf b})\,(+{\it I}\,\cdot\,{\bf c})\,=-1.}$

Thus we see at once that the second equation represents a left-handed system with respect to the first system, because now the RHS of this equation equals to ${-1\,}$.

In conclusion, a sign mistake has indeed been made for the past several years, but it is made by Gill, Moldoveanu, and Lockyer. They should go back to their schools.
Joy Christian
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### Re: Lockyer's math error

Christian, I can't believe you keep trotting out the most obvious statement you have made a sign error.

(-I.a)(-I.b) = (-1)(-1)(I.a)(I.b) = (I.a)(I.b)

BUT.... the right hand sides you show for (I.a)(I.b) and (-I.a)(-I.b) differ by a sign.

Two possibilities only:

1. Both RHS's are wrong
2. There is a sign error

This is completely obvious to all but the delusional like you, Diether, Ray and Fodje.

Any of you fools got an explanation for this?? Good luck redefining the rules for multiplication by a real number.

Fred, this is JC's resume. If I was him I would want you to take it down. You should put up a new thread: "Joy Christian's Math Error and Our Humble Apologies" where you four idiots admit what surely 99% of the drive by group already knows, and apologize to all that you have been completely unprofessional with.
Rick Lockyer

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Location: Nipomo

### Re: Lockyer's math error

Rick Lockyer wrote:Christian, I can't believe you keep trotting out the most obvious statement you have made a sign error.

(-I.a)(-I.b) = (-1)(-1)(I.a)(I.b) = (I.a)(I.b)

BUT.... the right hand sides you show for (I.a)(I.b) and (-I.a)(-I.b) differ by a sign.

Two possibilities only:

1. Both RHS's are wrong
2. There is a sign error

This is completely obvious to all but the delusional like you, Diether, Ray and Fodje.

Any of you fools got an explanation for this?? Good luck redefining the rules for multiplication by a real number.

Fred, this is JC's resume. If I was him I would want you to take it down. You should put up a new thread: "Joy Christian's Math Error and Our Humble Apologies" where you four idiots admit what surely 99% of the drive by group already knows, and apologize to all that you have been completely unprofessional with.

I already told you to forget about (-I.a)(-I.b) if it bothers you but you can't let go to due to mis-understanding the physics. It is not all that important to Joy's model. What is important is that the left hand system translates to the right hand perspective as,

(I.b)(I.a) = -a.b - (-I).(a x b) = -a.b + a^b
FrediFizzx
Independent Physics Researcher

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### Re: Lockyer's math error

***

So now Lockyer has once again resorted to lying and cheating since his sophomoric math error has been exposed for all to see.

He is simply making things up which cannot be found in any of my papers. Either he is really thick, or completely clueless about what is presented in my papers.

He is equating heads with tails of a fair coin,

and then crying wolf.

His purpose clearly is to hoodwink the scientific community by this deception.

I, for one, am not fooled by his paper-wolf.

Anyone serious enough about physics can check out my argument presented at the link above, or in some of my papers listed below:

(1)

(2)

(3)

These are just a handful of my papers where I have explicitly debunked the misguided arguments like the nonsense Lockyer has conjured up above.

Let me once again quote Michel Fodje's response to Lockyer's deceitful nonsense:

There is no sign error. If you do not see what has been explained to you yet then it is you who is incompetent. If you do see it but can't admit your error, then you are being disingenuous. Your arguments on this thread are not about Joy's model, but rather a figment of your imagination.

***
Joy Christian
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Location: Oxford, United Kingdom

### Re: Lockyer's math error

Sure Diether, forget about the most obvious proof there is a sign error in Christian's work.

"... due to mis-understanding the physics"

There is no valid physics here, just bad math. No physical model is valid if the math behind it is bad. Orientation change in S^3 is not a valid model, and required results only come about after math errors are committed.

"It is not all that important to Joy's model"

Then why did HE post it?? It is the at the foundation of his model

The right handed perspective is the e system above. The left handed perspective is the f system above. They do not directly add, and when you translate one result to the system of the other, you get the same thing.

You started this sorry thread, yet have not demonstrated a single instance of a math error by me. It is really making you look stupid.
Rick Lockyer

Posts: 46
Joined: Tue May 27, 2014 6:22 am
Location: Nipomo

### Re: Lockyer's math error

***
Lockyer seems to have quite a knack of making totally bogus claims. Perhaps because he has no background in physics and he is prone to making elementary algebraic mistakes as Fred has exposed above. For example on the FQXi blog Lockyer made a totally silly claim that "the chiral change map is an automorphism." But it is well known since 1956 that this is false. It is truly mindboggling where his breathtaking arrogance stems from, considering that he can't even do simple algebra correctly.

***
Joy Christian
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### Re: Lockyer's math error

Joy Christian wrote:***
Lockyer seems to have quite a knack of making totally bogus claims. Perhaps because he has no background in physics and he is prone to making elementary algebraic mistakes as Fred has exposed above. For example on the FQXi blog Lockyer made a totally silly claim that "the chiral change map is an automorphism." But it is well known since 1956 that this is false. It is truly mindboggling where his breathtaking arrogance stems from, considering that he can't even do simple algebra correctly.

***

Yeah, Lockyer would like us to believe that a left handed screw is the same as a right handed screw. His justification would be that there are no left and right handed screw drivers. LOL! It is pathetic and sad that he can't understand some basic physics that even a 6 year old could grasp.
FrediFizzx
Independent Physics Researcher

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### Re: Lockyer's math error

***

Let me ask this very simple physics question: Is the following equation correct?

$J_i J_j = -\,\delta_{ij} + \epsilon_{ijk} \,J_k \,,$

where $J_i$ are the components of angular momenta. Note the $+$ sign before the second term on the RHS.

Please vote Yes or No to this question.

I am asking this question because Lockyer and his "friends" think that the above equation is incorrect, whereas I have used it as one of the two alternatively possible equations for angular momenta in my papers; for example in this one (cf. page 2). All the dispute of the past many years is about the $+$ sign before the second term.

***
Joy Christian
Research Physicist

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Location: Oxford, United Kingdom

### Re: Lockyer's math error

Joy Christian wrote:***

Let me ask this very simple physics question: Is the following equation correct?

$J_i J_j = -\,\delta_{ij} + \epsilon_{ijk} \,J_k \,,$

where $J_i$ are the components of angular momenta. Note the $+$ sign before the second term on the RHS.

Please vote Yes or No to this question.

I am asking this question because Lockyer and his "friends" think that the above equation is incorrect, whereas I have used it as one of the two alternatively possible equations for angular momenta in my papers; for example in this one (cf. page 2). All the dispute of the past many years is about the $+$ sign before the second term.

***

One of the two terms of the RHS is always zero: the second if i = j, the first otherwise.
The first term is -1 if i = j, i.e., the square of each component is -1, so each component of J is imaginary.
The sum of the squares of the components, i.e., the square of J, is -3.
In physics the the square of an angular momentum is a non-negative real. In quantum mechanics the values are constrained to a discrete set but not to a single value.
So the equation is not correct.
Mikko

Posts: 161
Joined: Mon Feb 17, 2014 2:53 am

### Re: Lockyer's math error

Mikko wrote:One of the two terms of the RHS is always zero: the second if i = j, the first otherwise.
The first term is -1 if i = j, i.e., the square of each component is -1, so each component of J is imaginary.
The sum of the squares of the components, i.e., the square of J, is -3.
In physics the the square of an angular momentum is a non-negative real. In quantum mechanics the values are constrained to a discrete set but not to a single value.
So the equation is not correct.

Interesting.

So do you think that the following equation is also incorrect?

$J_i J_j = -\,\delta_{ij} - \epsilon_{ijk} \,J_k \,.$

PS: I did make a mistake in my initial post. I should have said "generators of angular momenta" rather than "components of angular momenta."

***
Joy Christian
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Location: Oxford, United Kingdom

### Re: Lockyer's math error

***

So we have two equations in classical mechanics:

$J_i J_j = -\,\delta_{ij} + \epsilon_{ijk} \,J_k \,,$

and

$J_i J_j = -\,\delta_{ij} - \epsilon_{ijk} \,J_k \,,$

where $J_i$ are the generators of angular momenta.

Are these two equations correct? Are both of them correct? Are both of them wrong? Is only one of them correct? If that is the case, then which one?

***
Joy Christian
Research Physicist

Posts: 2719
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Lockyer's math error

Joy Christian wrote:***

So we have two equations in classical mechanics:

$J_i J_j = -\,\delta_{ij} + \epsilon_{ijk} \,J_k \,,$

and

$J_i J_j = -\,\delta_{ij} - \epsilon_{ijk} \,J_k \,,$

where $J_i$ are the generators of angular momenta.

Are these two equations correct? Are both of them correct? Are both of them wrong? Is only one of them correct? If that is the case, then which one?

***

Perhaps this paper is relevant? See equation (5.18)

http://campus.mst.edu/physics/courses/4 ... apter7.pdf
FrediFizzx
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