by **Joy Christian** » Thu Sep 03, 2015 6:18 pm

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

and

to be 90 degrees and define

for this case. Now multiply (using the geometric product) the first equation, on both sides, by

, from the left. Since we have set

and

to be orthogonal to each other, and since all bivectors such as

square to

, the first equation (in this special case of orthogonal

and

) reduces to

Now if we follow the same procedure for the second equation -- this time using the bivector

, it also reduces to

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

.

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

for

any directional vector

. 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,

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

.

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.

[quote="Joy Christian"]

:o :o :o

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:

[quote="Richard Gill"]

[size=150][color=#FF0000]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 ... [/color][/size][/quote]

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

[quote="Joy Christian"]* * *

[img]http://libertesphilosophica.info/blog/lpmain/wp-content/uploads/sites/10/2015/07/Moldy3.png[/img]

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 [url=http://arxiv.org/abs/1203.2529]these[/url] two [url=http://arxiv.org/abs/1501.03393]papers[/url]) forms a right-handed system; and the second equation (again, as written by me in [url=http://arxiv.org/abs/1203.2529]these[/url] two [url=http://arxiv.org/abs/1501.03393]papers[/url]) 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 [tex]${\bf a}$[/tex] and [tex]${\bf b}[/tex] to be 90 degrees and define [tex]${\bf c} = {\bf a}\,\times\,{\bf b}$[/tex] for this case. Now multiply (using the geometric product) the first equation, on both sides, by [tex]${(+{\it I}\,\cdot\,{\bf c})}$[/tex], from the left. Since we have set [tex]${\bf a}$[/tex] and [tex]${\bf b}$[/tex] to be orthogonal to each other, and since all bivectors such as [tex]${(+{\it I}\,\cdot\,{\bf c})}$[/tex] square to [tex]${-1}$[/tex], the first equation (in this special case of orthogonal [tex]${\bf a}$[/tex] and [tex]${\bf b}$[/tex]) reduces to

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

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

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

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 [tex]${+1\,}$[/tex].

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

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

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

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

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

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.[/quote][/quote]