| Topic: |
Science > Physics |
| User: |
"Jay R. Yablon" |
| Date: |
16 Oct 2005 01:05:12 PM |
| Object: |
Posted another Preview Paper |
Hello again:
I just today posted a second preview paper at
http://home.nycap.rr.com/jry/FermionMass.htm.
This paper actually previews perhaps four or five distinct papers that I am
considering for future publication. These include the relationship between
duality symmetry and the geometrodynamic vacuum, how to deal with large
running couplings outside the range of perturbation theory, electroweak /
strong and leptoquark unification, and superconductivity. They are all
related to my two papers at http://arxiv.org/abs/hep-ph/0509223 and
http://arxiv.org/abs/hep-ph/0508257 dealing with duality symmetry and
electric and magnetic charges.
The website contains an introduction which can help you navigate through.
Comments and suggestions are appreciated.
Jay.
_____________________________
Jay R. Yablon
Email:
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
16 Oct 2005 08:01:51 PM |
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Hi Jay, studied your site.
3rd rank asymmetrical tensors are fun...
Probably the most famous 3rd rank asymmetric tensor is
Maxwell's
0 = F_ab,c + F_bc,a + F_ca,b
and describes an EM-wave, which is a typical boson.
("," is a partial and ";" is a covariant later).
1) We can define a 2nd rank tensor,
A_a B_b = - A_b B_a
and specify
(A_a B_b);c =0 , (a symmetry and conservation)
and then find,
A_a;c B_b = - B_b;c A_a
is that asymmetric in indices "a,b,c" ?
and symmetric if A and B are exchanged?
2) Set
(A_a A_b);c =0
and find
A_a;c A_b = - A_b;c A_a
and see an asymmetry in indices "a" and "b",
but does it follow "c" and "a" or "b" are
asymmetric?
In problem (1) the vectors A and B are distinct,
so one is applying "relational" GR.
In problem (2) vector "A" can exist on a continuum,
as it is not relative to another vector so it can
apply to a point.
I wonder, do the mathematicians have all this stuff
figured out in catagories or something. I'm thinkin'
if your moving tensors into particle physics you
may as well have lots of ammo, IOW's know your
descriptive options.
The problems above are parallel to my interest,
so perhaps we could work out a scheme if one is
not available for applications to GR and particle
physics.
Best Regards
Ken S. Tucker
Jay R. Yablon wrote:
Hello again:
I just today posted a second preview paper at
http://home.nycap.rr.com/jry/FermionMass.htm.
This paper actually previews perhaps four or five distinct papers that I am
considering for future publication. These include the relationship between
duality symmetry and the geometrodynamic vacuum, how to deal with large
running couplings outside the range of perturbation theory, electroweak /
strong and leptoquark unification, and superconductivity. They are all
related to my two papers at http://arxiv.org/abs/hep-ph/0509223 and
http://arxiv.org/abs/hep-ph/0508257 dealing with duality symmetry and
electric and magnetic charges.
The website contains an introduction which can help you navigate through.
Comments and suggestions are appreciated.
Jay.
_____________________________
Jay R. Yablon
Email:
.
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| User: "Robert Low" |
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| Title: Re: Posted another Preview Paper |
17 Oct 2005 05:32:19 AM |
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Ken S. Tucker wrote:
1) We can define a 2nd rank tensor,
A_a B_b = - A_b B_a
Before you get too carried away with what you
can do with this idea, consider the following.
Choose some basis, and consider the components
A_a, B_b, a,b=1..n (presumably n=4, but it
doesn't make any difference).
A_0 B_0 = -A_0 B_0, so at least one of A_0 and B_0
must be 0. Suppose that A_0 is not 0. Then B_0
must be 0.
Now, A_0 B_1 = -A_1 B_0 = 0, so B_1
must also be 0.
Next, A_0 B_2 = -A_2 B_0 = 0, so B_2
must also be 0.
And so on, so B_b must be the 0 vector, and so
the tensor A_a B_b is in fact the 0 tensor.
(Of course, if we suppose that B_0 is not 0,
we deduce that A_a must be the 0 vector, which
leads to the same conclusion.)
Note: there is no tensor analysis involved in
this argument whatever, so fiddling about
with the metric or the definition of covariant
derivative won't help. It's just algebra.
You can't make a non-trivial skew tensor this
way.
As a general suggestion, whenever you're tempted
to consider a particular construction, try
to do an example, just to see if it works.
It can save a lot of wasted effort.
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
17 Oct 2005 02:57:27 PM |
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Thank you for your kind reply Mr. Low.
Robert Low wrote:
Ken S. Tucker wrote:
1) We can define a 2nd rank tensor,
A_a B_b = - A_b B_a (0)
Before you get too carried away with what you
can do with this idea, consider the following.
I studied your post carefully, let me rebut this way...
A_a B_b = 1/2 (A_a B_b - A_b B_a) (1)
+ 1/2 (A_a B_b + A_b B_a) (2)
Obviously I conditioned (2) to be zero, meaning
my original tensor (0) is equivalent to (1), but
abbreviated, but (1) is clearly asymmetrical.
Nonsymmetrical tensors are tricky. Mr. Low
attempted a solution using absolute values,
which is not a good general prodecure.
Consider the following physics application,
Let F^ab be the EM field tensor and permit
the following time and space relations,
A_0 = B_0 and A_1 = -B_1
then,
F^ab A_a B_b =/= 0 ,
= F^01 A_0 B_1 + F^10 A_1 B_0
= F^10 A_0 A_1 + F^10 A_1 A_0
= 2*F^10 A_1 A_0 ,
set F^10 = E = q/r^2, r = A_1, ct = A_0 , r=ct
then
= 2q ,
which is a nonzero invariant, the fundamental charge.
Regards
Ken S. Tucker
Choose some basis, and consider the components
A_a, B_b, a,b=1..n (presumably n=4, but it
doesn't make any difference).
A_0 B_0 = -A_0 B_0, so at least one of A_0 and B_0
must be 0. Suppose that A_0 is not 0. Then B_0
must be 0.
Now, A_0 B_1 = -A_1 B_0 = 0, so B_1
must also be 0.
Next, A_0 B_2 = -A_2 B_0 = 0, so B_2
must also be 0.
And so on, so B_b must be the 0 vector, and so
the tensor A_a B_b is in fact the 0 tensor.
(Of course, if we suppose that B_0 is not 0,
we deduce that A_a must be the 0 vector, which
leads to the same conclusion.)
Note: there is no tensor analysis involved in
this argument whatever, so fiddling about
with the metric or the definition of covariant
derivative won't help. It's just algebra.
You can't make a non-trivial skew tensor this
way.
As a general suggestion, whenever you're tempted
to consider a particular construction, try
to do an example, just to see if it works.
It can save a lot of wasted effort.
.
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| User: "Robert Low" |
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| Title: Re: Posted another Preview Paper |
17 Oct 2005 03:18:53 PM |
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Ken S. Tucker wrote:
Robert Low wrote:
Ken S. Tucker wrote:
1) We can define a 2nd rank tensor,
A_a B_b = - A_b B_a (0)
Before you get too carried away with what you
can do with this idea, consider the following.
I studied your post carefully,
Study it again. It is no more than high school
algebra.
let me rebut this way...
A_a B_b = 1/2 (A_a B_b - A_b B_a) (1)
+ 1/2 (A_a B_b + A_b B_a) (2)
Yes, this is true. But it doesn't help.
Obviously I conditioned (2) to be zero, meaning
my original tensor (0) is equivalent to (1), but
abbreviated, but (1) is clearly asymmetrical.
Fine. So write out explicitly a pair of vectors
A_a and B_b that has this property. (The elements
of the empty set *all* have lots of interesting
properties, but it doesn't do to spend too much
of one's life studying them.)
Nonsymmetrical tensors are tricky.
Not particularly.
Mr. Low
attempted a solution using absolute values,
which is not a good general prodecure.
Mr Low gave you a simple algebraic argument
that there are no two vectors A_a and B_b
such that A_a B_b is skew and non-zero.
But I do make mistakes, and it's conceivable
that I made a mistake in this case. When
you provide two non-zero vectors A_a and B_b such
that A_a B_b = - A_b B_a, I will immediately
concede that I made a mistake.
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
17 Oct 2005 04:17:53 PM |
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Hi Robert, to less formal...
Robert Low wrote:
Ken S. Tucker wrote:
Robert Low wrote:
Ken S. Tucker wrote:
1) We can define a 2nd rank tensor,
A_a B_b = - A_b B_a (0)
Before you get too carried away with what you
can do with this idea, consider the following.
I studied your post carefully,
Study it again. It is no more than high school
algebra.
Rob, please be careful, above you said,
"Then B_0 must be 0."
Now you CANNOT equate a component (tensor B_0)
to an invariant "0" in mid-calculation. What
you did is specialized the CS then proceeded
to general conclusions. OTOH I maintained
General Covariance to arrive, for example, at
"2q".
let me rebut this way...
A_a B_b = 1/2 (A_a B_b - A_b B_a) (1)
+ 1/2 (A_a B_b + A_b B_a) (2)
Yes, this is true. But it doesn't help.
Obviously I conditioned (2) to be zero, meaning
my original tensor (0) is equivalent to (1), but
abbreviated, but (1) is clearly asymmetrical.
Fine. So write out explicitly a pair of vectors
A_a and B_b that has this property. (The elements
of the empty set *all* have lots of interesting
properties, but it doesn't do to spend too much
of one's life studying them.)
Nonsymmetrical tensors are tricky.
Not particularly.
Mr. Low
attempted a solution using absolute values,
which is not a good general prodecure.
Mr Low gave you a simple algebraic argument
that there are no two vectors A_a and B_b
such that A_a B_b is skew and non-zero.
But I do make mistakes, and it's conceivable
that I made a mistake in this case. When
you provide two non-zero vectors A_a and B_b such
that A_a B_b = - A_b B_a, I will immediately
concede that I made a mistake.
Yes, you maid a mistake, you snipped my physics
example, I didn't snip you...anyway
Mr. Low is right, I'd prefer to be Mr. Wrong.
LOL
Ken
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| User: "Robert Low" |
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| Title: Re: Posted another Preview Paper |
17 Oct 2005 04:47:00 PM |
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Ken S. Tucker wrote:
Hi Robert, to less formal...
Robert Low wrote:
Ken S. Tucker wrote:
Robert Low wrote:
Ken S. Tucker wrote:
1) We can define a 2nd rank tensor,
A_a B_b = - A_b B_a (0)
Before you get too carried away with what you
can do with this idea, consider the following.
I studied your post carefully,
Study it again. It is no more than high school
algebra.
Rob, please be careful, above you said,
"Then B_0 must be 0."
Yes, I said: pick a basis, and consider
the components of A_a and B_b in that
basis. Since A_0 B_0 = -A_0 B_0, the
producet of the two components must be
zero. Assuming that A_0 is not 0, B_0
must be 0.
If you assume B_0 is not 0, then A_0
must be.
Either way, A_a B_b ends up being
identically 0.
If a tensor is identically 0 in any
basis, it is 0 in every basis.
Therefore, if A_a B_b = -A_b B_a,
it must be the 0 tensor.
Unless, of course, you can exhibit a pair
of vectors with the property that
A_a B_b = - A_b B_a, without it being
0.
Now you CANNOT equate a component (tensor B_0)
to an invariant "0" in mid-calculation.
I dind't do that. I argued that the component
must be 0 *in whatever basis you picked*.
The argument is that it then follows that
*every* component of A_a B_b must be 0,
and that therefore the tensor itself is 0.
What
you did is specialized the CS then proceeded
to general conclusions.
Because a tensor that vanishes in any basis must
vanish in all bases.
OTOH I maintained
General Covariance to arrive, for example, at
"2q".
You might have arrived there, but you didn't
start with a pair of vectors A_a and B_b such
that A_a B_b is non-zero and skew.
Yes, you maid a mistake, you snipped my physics
example,
It was irrelevant to the claim.
I didn't snip you...
No, but you haven't responded to the question:
can you find a pair of vectors with the property
you claim?
Mr. Low is right, I'd prefer to be Mr. Wrong.
Well, there's something I suspect we can agree
on.
.
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
18 Oct 2005 01:30:43 AM |
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Mr. Low, previously I proved from Eq.(0) we are
working with the following tensor,
2*A_a B_b = A_a B_b - A_b B_a (1)
is that clear? If so then we move to argue
on the RHS of (1).
Ref,
A_a B_b = - A_b B_a (0)
more below,
Before you get too carried away with what you
can do with this idea, consider the following.
I studied your post carefully,
Study it again. It is no more than high school
algebra.
Rob, please be careful, above you said,
"Then B_0 must be 0."
Yes, I said: pick a basis, and consider
the components of A_a and B_b in that
basis. Since A_0 B_0 = -A_0 B_0, the
product of the two components must be
zero. Assuming that A_0 is not 0, B_0
must be 0.
Ok, let's go to dyads again, using two VECTORS
"A" and "B" with my definition from Eq.(0) in
dyadic form,
AB = A.B + AxB = -BA
with the dot "." being the scalar product
and the "x" being the vector product, and
see Spiegel's book, bottom of pg 73, where
"dyads are generalized in tensor analysis".
What Rob is doing is concluding from A.B=0
that A or B must be zero and that's wrong,
he's making a mistake there, but to his
credit a very good one!
In tensor notation I show the AxB part of
the dyadic appears as my Eq.(1) above.
What Mr. Low has taught me is that we should
take great care in going from Vector analysis
into Tensor analysis, and we shouldn't skip
over dyadics.
Regards
Ken S. Tucker
If you assume B_0 is not 0, then A_0
must be.
Either way, A_a B_b ends up being
identically 0.
If a tensor is identically 0 in any
basis, it is 0 in every basis.
Therefore, if A_a B_b = -A_b B_a,
it must be the 0 tensor.
Unless, of course, you can exhibit a pair
of vectors with the property that
A_a B_b = - A_b B_a, without it being
0.
Now you CANNOT equate a component (tensor B_0)
to an invariant "0" in mid-calculation.
I dind't do that. I argued that the component
must be 0 *in whatever basis you picked*.
The argument is that it then follows that
*every* component of A_a B_b must be 0,
and that therefore the tensor itself is 0.
What
you did is specialized the CS then proceeded
to general conclusions.
Because a tensor that vanishes in any basis must
vanish in all bases.
OTOH I maintained
General Covariance to arrive, for example, at
"2q".
You might have arrived there, but you didn't
start with a pair of vectors A_a and B_b such
that A_a B_b is non-zero and skew.
Yes, you maid a mistake, you snipped my physics
example,
It was irrelevant to the claim.
I didn't snip you...
No, but you haven't responded to the question:
can you find a pair of vectors with the property
you claim?
Mr. Low is right, I'd prefer to be Mr. Wrong.
Well, there's something I suspect we can agree
on.
.
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| User: "John Park" |
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| Title: Re: Posted another Preview Paper |
18 Oct 2005 05:21:48 PM |
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"Ken S. Tucker" (dynamics@vianet.on.ca) writes:
[...]
Ok, let's go to dyads again, using two VECTORS
"A" and "B" with my definition from Eq.(0) in
dyadic form,
AB = A.B + AxB = -BA
with the dot "." being the scalar product
and the "x" being the vector product, and
see Spiegel's book, bottom of pg 73, where
"dyads are generalized in tensor analysis".
[...]
Could you elaborate on the above equation? To a naive nonGRist, it looks as
though you have a dyadic on the LHS, and on the RHS the sum of a scalar
and a (pseudo)vector. In other words, it looks as though you are adding
apples and screwdrivers and hoping to get elephants. As a naive nonGRist, I
am also puzzled that you apparently ignore the fact that the cross product is
not defined for 4-vectors.
--John Park
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
18 Oct 2005 06:23:23 PM |
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John Park wrote:
"Ken S. Tucker" (dynamics@vianet.on.ca) writes:
[...]
Ok, let's go to dyads again, using two VECTORS
"A" and "B" with my definition from Eq.(0) in
dyadic form,
AB = A.B + AxB = -BA
with the dot "." being the scalar product
and the "x" being the vector product, and
see Spiegel's book, bottom of pg 73, where
"dyads are generalized in tensor analysis".
[...]
Could you elaborate on the above equation?
Yes.
To a naive nonGRist
This is just simple math, no GR conditions.
, it looks as
though you have a dyadic on the LHS, and on the RHS the sum of a scalar
and a (pseudo)vector. In other words, it looks as though you are adding
apples and screwdrivers and hoping to get elephants. As a naive nonGRist, I
am also puzzled that you apparently ignore the fact that the cross product is
not defined for 4-vectors.
Cross product in 4D needs a glance at the permutation
tensor, sometimes called the Levi-Cevita, and the "dual
tensors" as Pauli calls them.
Mr. Parker, we'll need a ref for when cross products
aren't defined in 4D, who told you that?
Ken
--John Park
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| User: "John Park" |
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| Title: Re: Posted another Preview Paper |
18 Oct 2005 10:39:20 PM |
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"Ken S. Tucker" (dynamics@vianet.on.ca) writes:
John Park wrote:
"Ken S. Tucker" (dynamics@vianet.on.ca) writes:
[...]
Ok, let's go to dyads again, using two VECTORS
"A" and "B" with my definition from Eq.(0) in
dyadic form,
AB = A.B + AxB = -BA
with the dot "." being the scalar product
and the "x" being the vector product, and
see Spiegel's book, bottom of pg 73, where
"dyads are generalized in tensor analysis".
[...]
Could you elaborate on the above equation?
Yes.
To a naive nonGRist
This is just simple math, no GR conditions.
, it looks as
though you have a dyadic on the LHS, and on the RHS the sum of a scalar
and a (pseudo)vector. In other words, it looks as though you are adding
apples and screwdrivers and hoping to get elephants. As a naive nonGRist, I
am also puzzled that you apparently ignore the fact that the cross product is
not defined for 4-vectors.
Cross product in 4D needs a glance at the permutation
tensor, sometimes called the Levi-Cevita, and the "dual
tensors" as Pauli calls them.
Mr. Parker, we'll need a ref for when cross products
aren't defined in 4D, who told you that?
I should have said there's no 4-vector corresponding to the cross product of
two 4-vectors. (Feynman Lectures, Vol II; section 31-5, p. 31-8)
--John Park>
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 11:51:00 AM |
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John Park wrote:
"Ken S. Tucker" (dynamics@vianet.on.ca) writes:
John Park wrote:
"Ken S. Tucker" (dynamics@vianet.on.ca) writes:
[...]
Ok, let's go to dyads again, using two VECTORS
"A" and "B" with my definition from Eq.(0) in
dyadic form,
AB = A.B + AxB = -BA
with the dot "." being the scalar product
and the "x" being the vector product, and
see Spiegel's book, bottom of pg 73, where
"dyads are generalized in tensor analysis".
[...]
Could you elaborate on the above equation?
Yes.
To a naive nonGRist
This is just simple math, no GR conditions.
, it looks as
though you have a dyadic on the LHS, and on the RHS the sum of a scalar
and a (pseudo)vector. In other words, it looks as though you are adding
apples and screwdrivers and hoping to get elephants. As a naive nonGRist, I
am also puzzled that you apparently ignore the fact that the cross product is
not defined for 4-vectors.
Cross product in 4D needs a glance at the permutation
tensor, sometimes called the Levi-Cevita, and the "dual
tensors" as Pauli calls them.
Mr. Parker, we'll need a ref for when cross products
aren't defined in 4D, who told you that?
I should have said there's no 4-vector corresponding to the cross product of
two 4-vectors. (Feynman Lectures, Vol II; section 31-5, p. 31-8)
--John Park>
What is usually done is to use Levi-Civita's tensor
also known as the permutation tensor "e". Then in 4D,
V^a = e^abcd X_b Y_c Z_d
where V^a is the volume of the cube XYZ and "a"
is the direction. It's the analog of a vector
product in 3D that has a vector norm to the surface
with magnitude proportional to the area.
Ken
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| User: "Jay R. Yablon" |
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| Title: SO? |
18 Oct 2005 11:16:38 PM |
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I'm glad you all are having such fun with antisymmetric tensors.
So, let me bring you all back to what Ken rightly calls
"Probably the most famous 3rd rank asymmetric tensor is
Maxwell's
0 = F_ab,c + F_bc,a + F_ca,b Eq(0)"
At http://home.nycap.rr.com/jry/FermionMass.htm, I am showing how this
tensor might be related to baryons, because in non-abelian gauge theory,
this tensor becomes non-zero and seems to contain three fermion currents,
and when you transfer the math over to Feynman diagrams, "baryon" seems to
jump right out.
I am still trying to midwife a new foundation for baryonic physics from
Maxwell's magnetic equation in non-Abelian gauge theories, and think I have
made a decent start. Any help?
Jay.
_____________________________
Jay R. Yablon
Email:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1129677803.256504.135550@o13g2000cwo.googlegroups.com...
John Park wrote:
"Ken S. Tucker" (dynamics@vianet.on.ca) writes:
[...]
Ok, let's go to dyads again, using two VECTORS
"A" and "B" with my definition from Eq.(0) in
dyadic form,
AB = A.B + AxB = -BA
with the dot "." being the scalar product
and the "x" being the vector product, and
see Spiegel's book, bottom of pg 73, where
"dyads are generalized in tensor analysis".
[...]
Could you elaborate on the above equation?
Yes.
To a naive nonGRist
This is just simple math, no GR conditions.
, it looks as
though you have a dyadic on the LHS, and on the RHS the sum of a scalar
and a (pseudo)vector. In other words, it looks as though you are adding
apples and screwdrivers and hoping to get elephants. As a naive nonGRist,
I
am also puzzled that you apparently ignore the fact that the cross
product is
not defined for 4-vectors.
Cross product in 4D needs a glance at the permutation
tensor, sometimes called the Levi-Cevita, and the "dual
tensors" as Pauli calls them.
Mr. Parker, we'll need a ref for when cross products
aren't defined in 4D, who told you that?
Ken
--John Park
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| User: "Ken S. Tucker" |
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| Title: Re: SO? |
19 Oct 2005 10:51:29 AM |
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Hi Jay
Jay R. Yablon wrote:
I'm glad you all are having such fun with antisymmetric tensors.
We're training to wreck other peoples theories :-).
I've a 19" LCD monitor, but your Eq.(1) in bold is
blurred.
So, let me bring you all back to what Ken rightly calls
"Probably the most famous 3rd rank asymmetric tensor is
Maxwell's
0 = F_ab,c + F_bc,a + F_ca,b Eq(0)"
At http://home.nycap.rr.com/jry/FermionMass.htm, I am showing how this
tensor might be related to baryons, because in non-abelian gauge theory,
this tensor becomes non-zero and seems to contain three fermion currents,
and when you transfer the math over to Feynman diagrams, "baryon" seems to
jump right out.
I am still trying to midwife a new foundation for baryonic physics from
Maxwell's magnetic equation in non-Abelian gauge theories, and think I have
made a decent start. Any help?
Jay.
Sure, can you describe a proton?
(that's the big one).
And later does that procedure lead to neutrons
and hyperons at least in sight.
I'd focus to start on applying it to a proton,
because it's stable, and you avoid the decay
complications.
Have you read AE's Nonsymetrical Field Theory?
Regards
Ken
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| User: "Jay R. Yablon" |
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| Title: Re: SO? |
19 Oct 2005 12:06:30 PM |
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Have you read AE's Nonsymetrical Field Theory?
You bet, Ken. Very, very important work, much underutilized and
under-recognized so far. Especially AE's "surprising" finding that the
strength of the vacuum equation R_uv=0 is equal in strength to Maxwell's
equations for (source free) electrodynamics, and I'd say, even for
electrodynamics with sources including the third rank antisymmetric tensors
for which I am presently trying to develop a relationship with baryons.
Look at section 2 in
http://home.nycap.rr.com/jry/Papers/Monopoles%20Preview%20Paper.pdf, which
shows what I am doing with AE's work on non-symmetric field theory in the
present context of magnetic monopoles and duality symmetry and symmetry
breaking.
See also Chapter 6 of http://home.nycap.rr.com/jry/Papers/Reinich.pdf, which
I wrote in 1984, where I use AE's field strength calculation to argue that
R_uv = 0 appears to be the geometrodynamic expression of Maxwell's
equations, even with sources, all in one equation. I think that AE wanted
to say this, but prudently stopped just short because he couldn't yet fully
prove it.
These two together comprise the subject matter of one of my planned future
papers.
If this stands up to all you folks who are "training to wreck other peoples
theories :-)," then what we have here is a formal unification of classical
gravitation and classical electrodynamics which should serve as the starting
point for all attempts at wider unification. Curious what you all have to
say about this.
Jay.
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| User: "Ken S. Tucker" |
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| Title: Re: SO? |
19 Oct 2005 01:28:36 PM |
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Jay R. Yablon wrote:
Have you read AE's Nonsymmetrical Field Theory?
You bet, Ken. Very, very important work, much underutilized and
under-recognized so far. Especially AE's "surprising" finding that the
strength of the vacuum equation R_uv=0 is equal in strength to Maxwell's
equations for (source free) electrodynamics, and I'd say, even for
electrodynamics with sources including the third rank antisymmetric tensors
for which I am presently trying to develop a relationship with baryons.
Ok, I suspect we may be going to numeralogy, but in view
QT that might be reasonable. For example, the tensor
F_uv;w has 48 non-zero terms, (IIRC), and then dividing
by 4D renders AE's z1=12 in that AE article.
2nd, I find you do NOT need nonsymmetrical Christoffel's
because the fundametric guv is nonsymmetric, which is a
hell of a simplification, because the Maxwell Eq's can
be subsumed into the "symmetrical Christoffel".
Look at section 2 in
http://home.nycap.rr.com/jry/Papers/Monopoles%20Preview%20Paper.pdf, which
shows what I am doing with AE's work on non-symmetric field theory in the
present context of magnetic monopoles and duality symmetry and symmetry
breaking.
Ok, I agree with your criticism of (2.20) and agree with (2.21)
but I'm unsure about (2.25) could you check that and confirm.
See also Chapter 6 of http://home.nycap.rr.com/jry/Papers/Reinich.pdf, which
I wrote in 1984, where I use AE's field strength calculation to argue that
R_uv = 0 appears to be the geometrodynamic expression of Maxwell's
equations, even with sources, all in one equation. I think that AE wanted
to say this, but prudently stopped just short because he couldn't yet fully
prove it.
IMHO, it is a mistake to antisymmetrize the Christoffel.
John Moffat (see his work on ArXivs) pursues that in his
NGT, (Nonsymmetrical Gravitational Theory). (John's an nice
guy, we had dinner together, you can email him, he's likely
the world's leading expert on NGT).
These two together comprise the subject matter of one of my planned future
papers.
If this stands up to all you folks who are "training to wreck other peoples
theories :-)," then what we have here is a formal unification of classical
gravitation and classical electrodynamics which should serve as the starting
point for all attempts at wider unification. Curious what you all have to
say about this.
What about giving the un-wahed masses a Proton.
Jay.
Ken
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| User: "John C. Polasek" |
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| Title: Re: Posted another Preview Paper |
18 Oct 2005 10:22:20 PM |
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On 18 Oct 2005 22:21:48 GMT, (John Park)
wrote:
"Ken S. Tucker" (dynamics@vianet.on.ca) writes:
[...]
Ok, let's go to dyads again, using two VECTORS
"A" and "B" with my definition from Eq.(0) in
dyadic form,
AB = A.B + AxB = -BA
with the dot "." being the scalar product
and the "x" being the vector product, and
see Spiegel's book, bottom of pg 73, where
"dyads are generalized in tensor analysis".
[...]
Could you elaborate on the above equation? To a naive nonGRist, it looks as
though you have a dyadic on the LHS, and on the RHS the sum of a scalar
and a (pseudo)vector. In other words, it looks as though you are adding
apples and screwdrivers and hoping to get elephants. As a naive nonGRist, I
am also puzzled that you apparently ignore the fact that the cross product is
not defined for 4-vectors.
--John Park
Ken, the juxtaposition of two vectors is an outer product which is a
square matrix. Further you show AxB = -BA, where AxB is a vector. Are
you making these things up?
As for cross products in 4space, they require a definition of proper
cyclic order as 123 231 312, or 132 etc. You can't have such an
unambiguous lefthand righthand choice with 4 indices. I think I'm
right on that. What you might have is a jackleg quaternion.
John Polasek
http://www.dualspace.net
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 12:53:04 PM |
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Hi Mr. Polasek
John C. Polasek wrote:
On 18 Oct 2005 22:21:48 GMT, (John Park)
wrote:
"Ken S. Tucker" (dynamics@vianet.on.ca) writes:
[...]
Ok, let's go to dyads again, using two VECTORS
"A" and "B" with my definition from Eq.(0) in
dyadic form,
AB = A.B + AxB = -BA
with the dot "." being the scalar product
and the "x" being the vector product, and
see Spiegel's book, bottom of pg 73, where
"dyads are generalized in tensor analysis".
[...]
Could you elaborate on the above equation? To a naive nonGRist, it looks as
though you have a dyadic on the LHS, and on the RHS the sum of a scalar
and a (pseudo)vector. In other words, it looks as though you are adding
apples and screwdrivers and hoping to get elephants. As a naive nonGRist, I
am also puzzled that you apparently ignore the fact that the cross product is
not defined for 4-vectors.
--John Park
Ken, the juxtaposition of two vectors is an outer product which is a
square matrix. Further you show AxB = -BA, where AxB is a vector. Are
you making these things up?
You can confirm with Fred that Spiegel has a scalar
product between vectors that outputs a vector on pg
74 of his book "Vector Analysis", problem 35.
These are also known as "Dyads" and "Triads".
As for cross products in 4space, they require a definition of proper
cyclic order as 123 231 312, or 132 etc. You can't have such an
unambiguous lefthand righthand choice with 4 indices. I think I'm
right on that. What you might have is a jackleg quaternion.
As I posted to Mr. Low today, and as I see from Jay's
post today, we may need nonsymmetrical metrics, as I
think we do, like g_12 = - g_21. Those become important
in "relational" GR, which I subscribe to, to evolve
classical GR out of the continuum, and more able to
deal with finite lengths that a relation needs.
John Polasek
http://www.dualspace.net
Regards
Ken
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| User: "Robert Low" |
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| Title: Re: Posted another Preview Paper |
18 Oct 2005 03:49:09 AM |
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Ken S. Tucker wrote:
Mr. Low, previously I proved from Eq.(0) we are
working with the following tensor,
2*A_a B_b = A_a B_b - A_b B_a (1)
is that clear?
It's clear that you want it. It's
also clear that it only happens when
at least on of A and B is the 0 vector.
If so then we move to argue
on the RHS of (1).
Ref,
A_a B_b = - A_b B_a (0)
Look, just exhibit a pair of vectors that
have that property.
I repeat: if you think you can find two
vectors which have that property, show
me them.
And now, on to the pointless bit.
If arguing based on the components in a basis
is too complicated for you, I strongly suspect
that what I'm about to write will make as much
sense to you as it would to a daffodil, but
what the heck.
Suppose that A_a B_b = - A_b B_a
and that neither of A_a nor B_b
is the zero vector. Then pick X^a
such that neither A_a X^a nor B_a X^a is zero.
It follows that
B_b (A_aX^a) = - A_b (A_a X^a),
so that A_a and B_b must be proportional,
i.e. A_a = k B_a for some non-zero k.
But now
A_a B_b = k A_a A_b
which is clearly symmetric. But we also
require it to be skew. This can only happen
if it is 0, and hence A_a is zero, which is
a contradiction.
What Rob is doing is concluding from A.B=0
that A or B must be zero and that's wrong,
Except of course that it isn't. I'm concluding
from the fact that the product of two real numbers
(particular components of A and B in some basis)
is 0 that one of the numbers must be 0, and
that's right. Your inability to understand
the relevance does rather baffle me, though.
he's making a mistake there, but to his
credit a very good one!
Ooh, that's not the mistake I'm making. The
mistake I'm making is that you're capable
of following a simple mathematical argument.
What Mr. Low has taught me is that we should
Alas, I fear I have taught you nothing. But
I can always hope that at least the occasional
lurker has been saved some effort.
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
18 Oct 2005 02:29:50 PM |
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Robert Low wrote:
Ken S. Tucker wrote:
Mr. Low, previously I proved from Eq.(0) we are
working with the following tensor,
2*A_a B_b = A_a B_b - A_b B_a (1)
is that clear?
It's clear that you want it.
Good! Now what we want from you is to post
the vector product AxB in tensor notation.
Can you do that?
Ken
....
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| User: "Robert Low" |
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| Title: Re: Posted another Preview Paper |
18 Oct 2005 02:55:29 PM |
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Ken S. Tucker wrote:
Robert Low wrote:
Ken S. Tucker wrote:
Mr. Low, previously I proved from Eq.(0) we are
working with the following tensor,
2*A_a B_b = A_a B_b - A_b B_a (1)
is that clear?
It's clear that you want it.
Good! Now what we want from you is to post
the vector product AxB in tensor notation.
Can you do that?
No, no, after you.
Give two vectors A and B such that
A_a B_b = - A_b B_a (without it
being the 0 tensor).
I've given you two independent
proofs that it can't be
done. Each time you simply
ignore the argument.
You insist that it can be done,
by 'conditioning' A and B
such that A_a B_b - A_b B_a
be 0. Of course, this is about
as possible as arranging a square
to have three sides by 'conditioning'
one of the sides to be of length
0 without changing the others.
But I'm always open to new ideas,
so just show me an explicit example.
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
18 Oct 2005 03:37:49 PM |
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Robert Low wrote:
Ken S. Tucker wrote:
Robert Low wrote:
Ken S. Tucker wrote:
Mr. Low, previously I proved from Eq.(0) we are
working with the following tensor,
2*A_a B_b = A_a B_b - A_b B_a (1)
is that clear?
It's clear that you want it.
Good! Now what we want from you is to post
the vector product AxB in tensor notation.
Can you do that?
No
Ok then, we know what your problem is, how do
you want to learn tensorized AxB?
I read your posts Rob, but unless we can agree
on AxB then we cannot proceed together.
Post the question to sci.math, there's lots of
smart guys in that group who might help you.
Ken
Give two vectors A and B such that
A_a B_b = - A_b B_a (without it
being the 0 tensor).
I've given you two independent
proofs that it can't be
done. Each time you simply
ignore the argument.
You insist that it can be done,
by 'conditioning' A and B
such that A_a B_b - A_b B_a
be 0. Of course, this is about
as possible as arranging a square
to have three sides by 'conditioning'
one of the sides to be of length
0 without changing the others.
But I'm always open to new ideas,
so just show me an explicit example.
.
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| User: "Robert Low" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 01:11:35 AM |
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Ken S. Tucker wrote:
Robert Low wrote:
Ken S. Tucker wrote:
Robert Low wrote:
Ken S. Tucker wrote:
Mr. Low, previously I proved from Eq.(0) we are
working with the following tensor,
2*A_a B_b = A_a B_b - A_b B_a (1)
is that clear?
It's clear that you want it.
Good! Now what we want from you is to post
the vector product AxB in tensor notation.
Can you do that?
No
Oh, nice bit of dishonest quotation.
And I notice you have *never* even
responded to the point that you cannot
find vectors A and B such that
A_a B_b = -A_b B_a.
Time to update that killfile.
I read your posts Rob,
Looking at the words in them isn't the
same as reading them. But I've invested
all the effort I care to in trying
to help you. You just carry on posting
your tripe.
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 10:30:03 AM |
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Robert Low wrote:
Ken S. Tucker wrote:
Robert Low wrote:
Ken S. Tucker wrote:
Robert Low wrote:
Ken S. Tucker wrote:
Mr. Low, previously I proved from Eq.(0) we are
working with the following tensor,
2*A_a B_b = A_a B_b - A_b B_a (1)
is that clear?
It's clear that you want it.
Good! Now what we want from you is to post
the vector product AxB in tensor notation.
Can you do that?
No
Oh, nice bit of dishonest quotation.
And I notice you have *never* even
responded to the point that you cannot
find vectors A and B such that
A_a B_b = -A_b B_a.
A0 = B0 , A1 = - B1
as explained!
BTW, Low's solution of setting A0*B0 =0 means,
-1 = 0/0 which is certainly not a proof.
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| User: "Robert Low" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 11:51:24 AM |
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Ken S. Tucker wrote:
Robert Low wrote:
And I notice you have *never* even
responded to the point that you cannot
find vectors A and B such that
A_a B_b = -A_b B_a.
A0 = B0 , A1 = - B1
Oh, I see. You're so clueless that
you think that if A=[1,1] and B=[1,-1]
then A^T B is skew-symmetric. Well,
it was worth not kill-filing you
until now just for that little gem.
But don't worry. I won't annoy you
any more by trying to get you to
make contact with any form of mathematical
coherence. Any further replies to your
posts will just be in case any new lurkers
might be led astray.
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 12:13:40 PM |
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Robert Low wrote:
Ken S. Tucker wrote:
Robert Low wrote:
And I notice you have *never* even
responded to the point that you cannot
find vectors A and B such that
A_a B_b = -A_b B_a.
A0 = B0 , A1 = - B1
For brevity set C_ab = A_a B_b, and
C_ab = -C_bc.
Get the "norm" by
C = AB = g^ab C_ab .
IFF g^ab = s^ab (symmetrical) then C=0.
However,
IF g^ab = s^ab + a^ab (nonsymmetrical)
then C=AB= a^ab C_ab =/=0.
Indeed now g^ab C_ab = invariant =/=0,
proving C_ab is not always zero.
Mr. Low's failures were numerous, most
importantly, subbing invariants into
components.
Regards
Ken S. Tucker
Oh, I see. You're so clueless that
you think that if A=[1,1] and B=[1,-1]
then A^T B is skew-symmetric. Well,
it was worth not kill-filing you
until now just for that little gem.
But don't worry. I won't annoy you
any more by trying to get you to
make contact with any form of mathematical
coherence. Any further replies to your
posts will just be in case any new lurkers
might be led astray.
.
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| User: "" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 02:23:28 PM |
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In article <dj635h$phg@netnews.net.lucent.com>, Tom Roberts <tjroberts@lucent.com> writes:
Ken S. Tucker wrote:
A_a B_b = -A_b B_a.
A0 = B0 , A1 = - B1
So you actually think AO AO = -AO AO, and yet AO is not zero. Ditto for
A1. <shrug>
I have known for a long time that you don't understand the basics of GR,
differential geometry, and tensor analysis. I had not realized you don't
even understand the basics of arithmetic.
As a rule, cranks are highly consistent in their ignorance, since in
order to remain ignorant of one's ignorance, one needs to be
uniformly, accross the board, ignorant.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 02:33:04 PM |
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Thanks Mati...
mme...@cars3.uchicago.edu wrote:
In article <dj635h$phg@netnews.net.lucent.com>, Tom Roberts <tjroberts@lucent.com> writes:
Ken S. Tucker wrote:
A_a B_b = -A_b B_a.
A0 = B0 , A1 = - B1
So you actually think AO AO = -AO AO, and yet AO is not zero. Ditto for
A1. <shrug>
I have known for a long time that you don't understand the basics of GR,
differential geometry, and tensor analysis. I had not realized you don't
even understand the basics of arithmetic.
As a rule, cranks are highly consistent in their ignorance, since in
order to remain ignorant of one's ignorance, one needs to be
uniformly, accross the board, ignorant.
Mati has that right, Robert's and Low are fuckin
with tensors. Tensors form relations, can't sub
invariants into components!
Way to go Mati!
Ken S. Tucker
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| User: "Tom Roberts" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 01:26:56 PM |
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Ken S. Tucker wrote:
A_a B_b = -A_b B_a.
A0 = B0 , A1 = - B1
So you actually think AO AO = -AO AO, and yet AO is not zero. Ditto for
A1. <shrug>
I have known for a long time that you don't understand the basics of GR,
differential geometry, and tensor analysis. I had not realized you don't
even understand the basics of arithmetic.
Tom Roberts tjroberts@lucent.com
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| User: "Robert Low" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 01:48:29 PM |
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Tom Roberts wrote:
Ken S. Tucker wrote:
A_a B_b = -A_b B_a.
A0 = B0 , A1 = - B1
So you actually think AO AO = -AO AO, and yet AO is not zero. Ditto for
A1. <shrug>
I have known for a long time that you don't understand the basics of GR,
differential geometry, and tensor analysis. I had not realized you don't
even understand the basics of arithmetic.
I certainly found that little exchange educational. And, thinking back
(I'm certainly not looking) at what he's posted previously, I suspect
that this 'skew symmetric A_a B_b' is something he uses all the time.
It's sad to come across somebody with such severe delusions
of competence.
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
19 Oct 2005 01:56:17 PM |
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Tom Roberts wrote:
[nothing]
see my previous posts.
Ken
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| User: "Ken S. Tucker" |
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| Title: Re: Posted another Preview Paper |
16 Oct 2005 07:29:39 PM |
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Jay R. Yablon wrote:
Hello again:
I just today posted a second preview paper at
http://home.nycap.rr.com/jry/FermionMass.htm.
This paper actually previews perhaps four or five distinct papers that I am
considering for future publication. These include the relationship between
duality symmetry and the geometrodynamic vacuum, how to deal with large
running couplings outside the range of perturbation theory, electroweak /
strong and leptoquark unification, and superconductivity. They are all
related to my two papers at http://arxiv.org/abs/hep-ph/0509223 and
http://arxiv.org/abs/hep-ph/0508257 dealing with duality symmetry and
electric and magnetic charges.
The website contains an introduction which can help you navigate through.
Comments and suggestions are appreciated.
Jay.
_____________________________
Jay R. Yablon
Email:
Hi Jay, studied your site.
3rd rank asymmetrical tensors are fun...
Probably the most famous 3rd rank asymmetric tensor is
Maxwell's
0 = F_ab,c + F_bc,a + F_ca,b
and describes an EM-wave, which is a typical boson.
("," is a partial and ";" is a covariant later).
1) We can define a 2nd rank tensor,
A_a B_b = - A_b B_a
and specify
(A_a B_b);c =0 , (a symmetry and conservation)
and then find,
A_a;c B_b = - B_b;c A_a
is that asymmetric in indices "a,b,c" ?
and symmetric if A and B are exchanged?
2) Set
(A_a A_b);c =0
and find
A_a;c A_b = - A_b;c A_a
and see an asymmetry in indices "a" and "b",
but does it follow "c" and "a" or "b" are
asymmetric?
In problem (1) the vectors A and B are distinct,
so one is applying "relational" GR.
In problem (2) vector "A" can exist on a continuum,
as it is not relative to another vector so it can
apply to a point.
I wonder, do the mathematicians have all this stuff
figured out in catagories or something. I'm thinkin'
if your moving tensors into particle physics you
may as well have lots of ammo, IOW's know your
descriptive options.
The problems above are parallel to my interest,
so perhaps we could work out a scheme if one is
not available for applications to GR and particle
physics.
Best Regards
Ken S. Tucker
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