Recently I posted to spr, aka the black hole of knowledge,
((Igor must be from ontario))
regarding Yablon's recent paper where he derives the invariant

K = sqrt(g) (E.B).

that I define as a constant and invariant.

Yablon defines,

kappa_v = k_v = K,v = 0 , (K0)

Yablon, sets kappa_v =/=0 occasionally, and Tucker disagrees.

Let the integral "$" of K,v be

$ K,v dx^v = $ dK = K (K1)

I'll use k = kappa, then

$ k_v dx^v = K (K1)

follows.

Set up two geodesics A and B denoted k(A)_v and k(B)_v, then

B
$ k_v dx^v = K(B) - K(A) = n*K = invariants,
A

where "n" is an integer and "K" is an invariant constant.

What is it that you guys think is wrong with the above??

IMHO Yablon has defined quantization in (K0), although
Yablon might disagree.

I better stop here, pending more people to insult!

Ken S. Tucker

Ken S. Tucker wrote:
Recently I posted to spr, aka the black hole of knowledge,
((Igor must be from ontario))
regarding Yablon's recent paper where he derives the invariant

K = sqrt(g) (E.B).

that I define as a constant and invariant.

Yablon defines,

kappa_v = k_v = K,v = 0 , (K0)

Yablon, sets kappa_v =/=0 occasionally, and Tucker disagrees.

Let the integral "$" of K,v be

$ K,v dx^v = $ dK = K (K1)

I'll use k = kappa, then

$ k_v dx^v = K (K1)

follows.

Set up two geodesics A and B denoted k(A)_v and k(B)_v, then

B
$ k_v dx^v = K(B) - K(A) = n*K = invariants,
A

where "n" is an integer and "K" is an invariant constant.

What is it that you guys think is wrong with the above??

IMHO Yablon has defined quantization in (K0), although
Yablon might disagree.

I better stop here, pending more people to insult!

My objection was in what you think the two geodesics represent
in the original context. I didn't press the issue recognizing its
roots were in an ancient debate about falling photons.

Ann Landers says:
Try to be more disambiguous and
folks will misunderestimate your intent to insult.

Sue...

"A Metric Determined by Photon Exchange"
http://arxiv.org/abs/physics/0108012




Ken S. Tucker

Sue... wrote:
Ken S. Tucker wrote:
Recently I posted to spr, aka the black hole of knowledge,
((Igor must be from ontario))
regarding Yablon's recent paper where he derives the invariant

K = sqrt(g) (E.B).

that I define as a constant and invariant.

Yablon defines,

kappa_v = k_v = K,v = 0 , (K0)

Yablon, sets kappa_v =/=0 occasionally, and Tucker disagrees.

Let the integral "$" of K,v be

$ K,v dx^v = $ dK = K (K1)

I'll use k = kappa, then

$ k_v dx^v = K (K1)

follows.

Set up two geodesics A and B denoted k(A)_v and k(B)_v, then

B
$ k_v dx^v = K(B) - K(A) = n*K = invariants,
A

where "n" is an integer and "K" is an invariant constant.

What is it that you guys think is wrong with the above??

IMHO Yablon has defined quantization in (K0), although
Yablon might disagree.

I better stop here, pending more people to insult!

My objection was in what you think the two geodesics represent
in the original context. I didn't press the issue recognizing its
roots were in an ancient debate about falling photons.

Ok photons is fine.

sin^2 + cos^2 =1

(E.B)^2 + (ExB)^2 = c^2

How's that?

"A Metric Determined by Photon Exchange"
http://arxiv.org/abs/physics/0108012

Radar ranging is a good way to see things.
Ken

Ken S. Tucker wrote:
Sue... wrote:
Ken S. Tucker wrote:
Recently I posted to spr, aka the black hole of knowledge,
((Igor must be from ontario))
regarding Yablon's recent paper where he derives the invariant

K = sqrt(g) (E.B).

that I define as a constant and invariant.

Yablon defines,

kappa_v = k_v = K,v = 0 , (K0)

Yablon, sets kappa_v =/=0 occasionally, and Tucker disagrees.

Let the integral "$" of K,v be

$ K,v dx^v = $ dK = K (K1)

I'll use k = kappa, then

$ k_v dx^v = K (K1)

follows.

Set up two geodesics A and B denoted k(A)_v and k(B)_v, then

B
$ k_v dx^v = K(B) - K(A) = n*K = invariants,
A

where "n" is an integer and "K" is an invariant constant.

What is it that you guys think is wrong with the above??

IMHO Yablon has defined quantization in (K0), although
Yablon might disagree.

I better stop here, pending more people to insult!

My objection was in what you think the two geodesics represent
in the original context. I didn't press the issue recognizing its
roots were in an ancient debate about falling photons.

Ok photons is fine.

sin^2 + cos^2 =1

(E.B)^2 + (ExB)^2 = c^2

How's that?

A fair approximation for homogenous space ?
But your (or any) quantization is a fairly intimate
interaction with matter or some probability of
an intimate interaction. So... if the squares
are like a QED trick to square probability
arrows, there seems a little redemption.

Where B = f(E) does it still make sense?
Because Maxwell's equations are disjointed
and can be expressed as either integrals or
retarded potentials, we can't just hop back and forth
when it is conveinent. They don't have any
safety interlocks to keep us from doing weird
things... like sending photons back in time.

Fair warning:
I am biased anyway because I don't
believe gravity gives a hoot whether quanta is
emitted or absorbed but I can be entertained for
long periods of time watching someone try to
put a square peg in a round hole, so don't let
me discourage you from looking for what might
indeed be a valuable expediant. )

Sue...




"A Metric Determined by Photon Exchange"
http://arxiv.org/abs/physics/0108012

Radar ranging is a good way to see things.
Ken

Ken S. Tucker wrote:
Recently I posted to spr, aka the black hole of knowledge,
((Igor must be from ontario))

Turns out Igor is in ontario, he's a student, no wonder he doesn't
understand GR yet, and has superficial ability in tensors.
++++++

Jay Yablon wrote,
You do not need to check the derivation to (2.5). I did not "find" this
identity. This identity can be found in John Archibald Wheeler's
Geometrodynamics, page 251, note 22. It is a central identity to any
consideration of electric magentic duality.

Igor, wrote,
I see even less relevance of possible electric-magnetic duality to the
question at hand.

Igor crack a book, Pauli, Theory of Relativity pg 34,
Eq.(54b), learn duality.

What does the invariant "J" mean in Eq.(46a) ????

That's what Mr. Yablon is researching.

Yes, as requested, I'll try to post to spr but Igor will censor me
once again, he hates people smarter than him.
Ken S. Tucker

"Ken S. Tucker" wrote in message
oups.com...

Ken S. Tucker wrote:
Recently I posted to spr, aka the black hole of knowledge,
((Igor must be from ontario))

Turns out Igor is in ontario, he's a student, no wonder he doesn't
understand GR yet, and has superficial ability in tensors.


A study of Google Earth PROVES that Ontario is at sea level in the
vicinity of Toronto.
http://earth.google.com/
That's why no Ontarian can understand GR.
Androcles.