Hi Jay & all.

On Jul 31, 6:28 pm, "Jay R. Yablon" wrote:
....
I am starting to wonder if the "geometry" of quantum physics is that of
Riemann surfaces, in which case Riemann will be seen to have laid the
geometric foundation for classical and quantum physics alike.

Ken, you love geometry and GR. Start thinking about how spacetime would
look it were treated as a Riemann surface. Need to add complex planes
and more dimensions, of course. I believe that quantum physics is
telling us that nature's geometry on the quantum scale employs Riemann
surfaces in some fashion.

That's agreeable.

What then, might be meant by the "classical limit" of such Riemann
surfaces? New territory!
Jay.

Everyone has their own way to construct dimensions.
My favorite procedure is succesive integration, e.g.

$ 0 dX = q
$ q dX = qX , (constant of integation =0, going forward)
$ qX dX = qX^2/2
....
Set q=1 and sum all those integrals and get exp(X),
http://en.wikipedia.org/wiki/Exponen...mal_definition

Since the integrals above each generate an extra
dimension, X, X^2, X^3... if X is regarded as a length
then, exp(X) has a infinite dimensionality.

Charles pointed out that the "X" in exp(X) is a scalar,
but I noticed exp(X) could be graphed, and that graph
used lengths like Y = exp(X). In relativity, "Y" would
need to be an invariant, but it could also be graphed.

Recall dY/dX = exp(X).

Looks to me like we're trying to define that invariance
even when subjected to a SR effect like the Lorentz-
Fitzgerald Contraction, ((I previously posted a bit about
that)), and I think it follows that is invariant in GR too.

Turning to mathematicians, one may take a 2D curved
surface on a globe, and describe that surface in flat 3D.

Flat space means R_abcd=0, to define a constant metric,
but I employ another standard for "finite geometry".
I reason, X^2 = X_u X_u = X^u X^u because the covariant
and contravariant components are equal, if equal in one
FoR,,, therefore, if Y=exp(X) in one *orthogonal* FoR, X
will be invariant in all FoR's.

The reason I switched to a "finite geometry" from R_abcd
is because the continuum has given way to finite lengths.

I'm wandering a bit, I figure a path integral and Signal
distance "S" are equivalent (?), so I'm researching this,

S^2 = g_uv X^u X^v = (X_u X_v) X^u X^v + (A_u B_v) X^u X^v

where g_uv = (X_u X_v) + (A_u B_v), and g_uv is defined

over a finite length, see,

http://physics.trak4.com/GR_Charge_Couple.pdf

Blasphermy, I'm suggesting "g_uv" is a relation over
finite length, instead of at a point on a continuum.
Best Regards
Ken S. Tucker
PS: Moderators, Save your Souls, runaway, don't
post that :-), LOL, enjoy!

Ken S. Tucker wrote:
Hi Jay & all.

On Jul 31, 6:28 pm, "Jay R. Yablon" wrote:
...
I am starting to wonder if the "geometry" of quantum physics is that of
Riemann surfaces, in which case Riemann will be seen to have laid the
geometric foundation for classical and quantum physics alike.

Ken, you love geometry and GR. Start thinking about how spacetime would
look it were treated as a Riemann surface. Need to add complex planes
and more dimensions, of course. I believe that quantum physics is
telling us that nature's geometry on the quantum scale employs Riemann
surfaces in some fashion.

That's agreeable.

What then, might be meant by the "classical limit" of such Riemann
surfaces? New territory!
Jay.

Everyone has their own way to construct dimensions.
My favorite procedure is succesive integration, e.g.

$ 0 dX = q
$ q dX = qX , (constant of integation =0, going forward)
$ qX dX = qX^2/2
...
Set q=1 and sum all those integrals and get exp(X),
http://en.wikipedia.org/wiki/Exponen...mal_definition

Since the integrals above each generate an extra
dimension, X, X^2, X^3... if X is regarded as a length
then, exp(X) has a infinite dimensionality.

Charles pointed out that the "X" in exp(X) is a scalar,
but I noticed exp(X) could be graphed, and that graph
used lengths like Y = exp(X). In relativity, "Y" would
need to be an invariant, but it could also be graphed.

Recall dY/dX = exp(X).

Looks to me like we're trying to define that invariance
even when subjected to a SR effect like the Lorentz-
Fitzgerald Contraction, ((I previously posted a bit about
that)), and I think it follows that is invariant in GR too.

Turning to mathematicians, one may take a 2D curved
surface on a globe, and describe that surface in flat 3D.

Flat space means R_abcd=0, to define a constant metric,
but I employ another standard for "finite geometry".
I reason, X^2 = X_u X_u = X^u X^u because the covariant
and contravariant components are equal, if equal in one
FoR,,, therefore, if Y=exp(X) in one *orthogonal* FoR, X
will be invariant in all FoR's.

The reason I switched to a "finite geometry" from R_abcd
is because the continuum has given way to finite lengths.

I'm wandering a bit, I figure a path integral and Signal
distance "S" are equivalent (?), so I'm researching this,

S^2 = g_uv X^u X^v = (X_u X_v) X^u X^v + (A_u B_v) X^u X^v

where g_uv = (X_u X_v) + (A_u B_v), and g_uv is defined

over a finite length, see,

http://physics.trak4.com/GR_Charge_Couple.pdf

Blasphermy, I'm suggesting "g_uv" is a relation over
finite length, instead of at a point on a continuum.
Best Regards
Ken S. Tucker
PS: Moderators, Save your Souls, runaway, don't
post that :-), LOL, enjoy!

i hate threads starting post replying to other posts

On Aug 2, 12:00 pm, conquistador wrote:
Ken S. Tucker wrote:
Hi Jay & all.

On Jul 31, 6:28 pm, "Jay R. Yablon" wrote:
...
I am starting to wonder if the "geometry" of quantum physics is that of
Riemann surfaces, in which case Riemann will be seen to have laid the
geometric foundation for classical and quantum physics alike.

Ken, you love geometry and GR. Start thinking about how spacetime would
look it were treated as a Riemann surface. Need to add complex planes
and more dimensions, of course. I believe that quantum physics is
telling us that nature's geometry on the quantum scale employs Riemann
surfaces in some fashion.

That's agreeable.

What then, might be meant by the "classical limit" of such Riemann
surfaces? New territory!
Jay.

Everyone has their own way to construct dimensions.
My favorite procedure is succesive integration, e.g.

$ 0 dX = q
$ q dX = qX , (constant of integation =0, going forward)
$ qX dX = qX^2/2
...
Set q=1 and sum all those integrals and get exp(X),
http://en.wikipedia.org/wiki/Exponen...mal_definition

Since the integrals above each generate an extra
dimension, X, X^2, X^3... if X is regarded as a length
then, exp(X) has a infinite dimensionality.

Charles pointed out that the "X" in exp(X) is a scalar,
but I noticed exp(X) could be graphed, and that graph
used lengths like Y = exp(X). In relativity, "Y" would
need to be an invariant, but it could also be graphed.

Recall dY/dX = exp(X).

Looks to me like we're trying to define that invariance
even when subjected to a SR effect like the Lorentz-
Fitzgerald Contraction, ((I previously posted a bit about
that)), and I think it follows that is invariant in GR too.

Turning to mathematicians, one may take a 2D curved
surface on a globe, and describe that surface in flat 3D.

Flat space means R_abcd=0, to define a constant metric,
but I employ another standard for "finite geometry".
I reason, X^2 = X_u X_u = X^u X^u because the covariant
and contravariant components are equal, if equal in one
FoR,,, therefore, if Y=exp(X) in one *orthogonal* FoR, X
will be invariant in all FoR's.

The reason I switched to a "finite geometry" from R_abcd
is because the continuum has given way to finite lengths.

I'm wandering a bit, I figure a path integral and Signal
distance "S" are equivalent (?), so I'm researching this,

S^2 = g_uv X^u X^v = (X_u X_v) X^u X^v + (A_u B_v) X^u X^v

where g_uv = (X_u X_v) + (A_u B_v), and g_uv is defined

over a finite length, see,

http://physics.trak4.com/GR_Charge_Couple.pdf

Blasphermy, I'm suggesting "g_uv" is a relation over
finite length, instead of at a point on a continuum.
Best Regards
Ken S. Tucker
PS: Moderators, Save your Souls, runaway, don't
post that :-), LOL, enjoy!

i hate threads starting post replying to other posts

Ok, If you're serious, go to SPF.
Ken