Using Stoke's theorem, one can show that the closed line integral must be
conserved from one end of a world-tube to the other end in a scalar field.
If the line integral must be conserve from one closed curve to the next,
then does this result in the need of a geodesic surface from one end of the
tube to the other? I'm thinking in terms of classical string theory and
trying to justify the Euler-Lagrange equations. If an extremal surface can
be justified for other reasons, then perhaps the Euler-Lagrange equation can
be derived from that. I am told that the Euler-Lagrange vector is normal to
the surface and that it is zero for an extremal surface which must then be a
geodesic surface. But if the surface must be extremal for other reasons,
then perhaps we can derive the meaning of energy and momentum as some
geometric features on this surface. The surface may need to be extremal, for
example, because the most probable path is one along equi-potential paths
where the gradient in the direction of the normal is zero.

I would appreciate some comments, thank you.

Mike.

Mike wrote:
Using Stoke's theorem, one can show that the closed line integral must be
conserved from one end of a world-tube to the other end in a scalar field.
If the line integral must be conserve from one closed curve to the next,
then does this result in the need of a geodesic surface from one end of the
tube to the other? I'm thinking in terms of classical string theory and
trying to justify the Euler-Lagrange equations. If an extremal surface can
be justified for other reasons, then perhaps the Euler-Lagrange equation can
be derived from that. I am told that the Euler-Lagrange vector is normal to
the surface and that it is zero for an extremal surface which must then be a
geodesic surface. But if the surface must be extremal for other reasons,
then perhaps we can derive the meaning of energy and momentum as some
geometric features on this surface. The surface may need to be extremal, for
example, because the most probable path is one along equi-potential paths
where the gradient in the direction of the normal is zero.

I would appreciate some comments, thank you.

Mike.


Try looking at the simplest possible case of what you're talking about.
Consider a 2-dimensional manifold M equipped with a positive-definite
Riemannian metric g. Construct on M a simple open curve C such that C
has endpoints A,B. In order for C to be a maximal curve two conditions
must be satisfied; if L is the length of C between the endpoints then we
require

1: dL = 0,
2: d(dL) = 0.

If condition 1 holds then C is said to be weakly maximal. If both
conditions hold then C is said to be strongly maximal. Carry out the
usual procedure of taking variations with respect to local coordinates
on M and you get a rather straightforward result. One can simplify the
situation somewhat by considering coordinates x^a(s), 1 = a = 2 such that

x^1(s) = s,
x^2(s) = 0.

Once you've carried out these variations, the smart thing to do is to
describe the curve in terms of its extrinsic curvature. We began by
assuming that C was imbedded in M, so the condition for the curve to be
maximal then reduces to finding curves which satisfy

tr(K) = 0,

where tr(K) = g^(ij)K_ij is the scalar mean curvature. The
generalization to higher dimensional manifolds is then pretty
straightforward.

davidoff

Mike wrote:

Using Stoke's theorem, one can show that the closed line integral must be
conserved from one end of a world-tube to the other end in a scalar field.
If the line integral must be conserve from one closed curve to the next,
then does this result in the need of a geodesic surface from one end of the
tube to the other?

Metric theories of gravitation have geodesic paths. Affine theories
of gravitation do not have geodesic paths. Since affine theories can
give identical predictions vs. metric theories, I'd posit geodesics
were not required to describe gravitation.

I'm thinking in terms of classical string theory and
trying to justify the Euler-Lagrange equations. If an extremal surface can
be justified for other reasons, then perhaps the Euler-Lagrange equation can
be derived from that. I am told that the Euler-Lagrange vector is normal to
the surface and that it is zero for an extremal surface which must then be a
geodesic surface. But if the surface must be extremal for other reasons,
then perhaps we can derive the meaning of energy and momentum as some
geometric features on this surface. The surface may need to be extremal, for
example, because the most probable path is one along equi-potential paths
where the gradient in the direction of the normal is zero.

I would appreciate some comments, thank you.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!

"davidoff404" wrote in message
...
Mike wrote:
Using Stoke's theorem, one can show that the closed line integral must
be
conserved from one end of a world-tube to the other end in a scalar
field.
If the line integral must be conserve from one closed curve to the next,
then does this result in the need of a geodesic surface from one end of
the
tube to the other? I'm thinking in terms of classical string theory and
trying to justify the Euler-Lagrange equations. If an extremal surface
can
be justified for other reasons, then perhaps the Euler-Lagrange equation
can
be derived from that. I am told that the Euler-Lagrange vector is normal
to
the surface and that it is zero for an extremal surface which must then
be a
geodesic surface. But if the surface must be extremal for other reasons,
then perhaps we can derive the meaning of energy and momentum as some
geometric features on this surface. The surface may need to be extremal,
for
example, because the most probable path is one along equi-potential
paths
where the gradient in the direction of the normal is zero.

I would appreciate some comments, thank you.

Mike.


Try looking at the simplest possible case of what you're talking about.

You're talking about minimizing the length of a 1-dimensional line here. But
there are two perspectives. One, we do this procedure because it is a
mathematically elegant way of formulating what we already know. And two,
there is something inherent in this geometry that leads to the physical
characteristics of energy and momentum. I am working on the second approach.
I'm wondering, if we can find that certain conservations law are simply
required (as is the case of line integrals in a scalar field), is it
possible to calculate mass and momentum from the resulting geometry (or at
least what looks like mass and momentum). It might help if we could describe
physical quantities in terms of the inherent features of a geodesic line. If
only it were true that the kinetic energy were the tangential component of
the geodesic line and potential energy were the normal component, for
example. Then adding them up and integrating would seem to be a more natural
characteristic of the geometry.

"Uncle Al" wrote in message
...
Mike wrote:

Using Stoke's theorem, one can show that the closed line integral must
be
conserved from one end of a world-tube to the other end in a scalar
field.
If the line integral must be conserve from one closed curve to the next,
then does this result in the need of a geodesic surface from one end of
the
tube to the other?

Metric theories of gravitation have geodesic paths. Affine theories
of gravitation do not have geodesic paths. Since affine theories can
give identical predictions vs. metric theories, I'd posit geodesics
were not required to describe gravitation.

We can formulate physics with either Lagrangian dynamics or Hamiltonian
mathematics. If one is equivalent to the other, then it is fair to ask about
the geometrics of the one.

Mike wrote:

"Uncle Al" wrote in message
...
Mike wrote:

Using Stoke's theorem, one can show that the closed line integral must
be
conserved from one end of a world-tube to the other end in a scalar
field.
If the line integral must be conserve from one closed curve to the next,
then does this result in the need of a geodesic surface from one end of
the
tube to the other?

Metric theories of gravitation have geodesic paths. Affine theories
of gravitation do not have geodesic paths. Since affine theories can
give identical predictions vs. metric theories, I'd posit geodesics
were not required to describe gravitation.

We can formulate physics with either Lagrangian dynamics or Hamiltonian
mathematics. If one is equivalent to the other, then it is fair to ask about
the geometrics of the one.

And this has what to do with the OP's question? I think it is 'fair' to
say that Al provided the only correct answer, anything more would be
redundant. There is nothing left to elaborate on except the math, and
only Al can cite all of that in one breath. And he will if you provoke
him

Richard Perry

"Richard" wrote in message
...


Mike wrote:

"Uncle Al" wrote in message
...
Mike wrote:

Using Stoke's theorem, one can show that the closed line integral
must
be
conserved from one end of a world-tube to the other end in a scalar
field.
If the line integral must be conserve from one closed curve to the
next,
then does this result in the need of a geodesic surface from one end
of
the
tube to the other?

Metric theories of gravitation have geodesic paths. Affine theories
of gravitation do not have geodesic paths. Since affine theories can
give identical predictions vs. metric theories, I'd posit geodesics
were not required to describe gravitation.

We can formulate physics with either Lagrangian dynamics or Hamiltonian
mathematics. If one is equivalent to the other, then it is fair to ask
about
the geometrics of the one.

And this has what to do with the OP's question? I think it is 'fair' to
say that Al provided the only correct answer, anything more would be
redundant. There is nothing left to elaborate on except the math, and
only Al can cite all of that in one breath. And he will if you provoke
him


So what you are saying is that geodesics are a luxury in physics?

In article ,
Uncle Al wrote:
Mike wrote:

Using Stoke's theorem, one can show that the closed line integral must be
conserved from one end of a world-tube to the other end in a scalar field.
If the line integral must be conserve from one closed curve to the next,
then does this result in the need of a geodesic surface from one end of the
tube to the other?

Metric theories of gravitation have geodesic paths. Affine theories
of gravitation do not have geodesic paths. Since affine theories can
give identical predictions vs. metric theories, I'd posit geodesics
were not required to describe gravitation.

If an affine theory gives the same predictions as a metric theory, doesn't
that make it a metric theory? As I understand it, a metric space means
distances can be defined on it. So if a geometrical theory makes
predictions in terms of rulers and clocks*c, it's metric.
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé

Mike wrote:

"Richard" wrote in message
...


Mike wrote:

"Uncle Al" wrote in message
...
Mike wrote:

Using Stoke's theorem, one can show that the closed line integral
must
be
conserved from one end of a world-tube to the other end in a scalar
field.
If the line integral must be conserve from one closed curve to the
next,
then does this result in the need of a geodesic surface from one end
of
the
tube to the other?

Metric theories of gravitation have geodesic paths. Affine theories
of gravitation do not have geodesic paths. Since affine theories can
give identical predictions vs. metric theories, I'd posit geodesics
were not required to describe gravitation.

We can formulate physics with either Lagrangian dynamics or Hamiltonian
mathematics. If one is equivalent to the other, then it is fair to ask
about
the geometrics of the one.

And this has what to do with the OP's question? I think it is 'fair' to
say that Al provided the only correct answer, anything more would be
redundant. There is nothing left to elaborate on except the math, and
only Al can cite all of that in one breath. And he will if you provoke
him


So what you are saying is that geodesics are a luxury in physics?

Metric models are geometric and have geodesics. Affine models are
electromagnetic in form and have no geodesics. All metric theory
predictions are included in affine theories. Geodesics are a
convenience. (Use Hamiltonian or Lagrangian formalisms. They are
proven to be equivalent. One is usually much easier to handle for a
given problem).

http://arXiv.org/abs/gr-qc/0304106
Affine (teleparallel) gravitation.

Metric and affine theories have a slim non-interesection of disjoint
predictions. The only way to tell which is The One is to look at the
disagreements, not at the agreements. Einstein made no geometric
mistakes, neither did Euclid. Euclid missed hyperbolic and elliptic
geometries by having a weak Fifth Postulate. Euclid is incomplete by
trivial demonstration. You cannot deep sea navigate using Euclid, nor
can you survey large tracts of land.

Einstein postulated the Equivalence Prnciple and effected a tensor
theory of gravitation that is symmetric to parity transformation.
Affine theories ignore the Equivalence Principle and can possess
gravitational stress-energy pseudotensors that are anti-symmetric to
parity transformation. Einstein may be incomplete.

THE OBVIOUS THING TO DO IS TO TEST THE EQUIVALENCE PRINCIPLE AGAINST
PARITY-TRANSFORMED TEST MASSES.

http://www.mazepath.com/uncleal/qz.pdf

This point seems to have been lost on physics despite its trivially
easy and inexpensive implimentation in existing apparatus. The math
is rigorous, both as explicit calculation as as a pure geometric model
with only two parameters- intrinsic scale and one measured empirical
angle.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!

"Gregory L. Hansen" wrote:

In article ,
Uncle Al wrote:
Mike wrote:

Using Stoke's theorem, one can show that the closed line integral must be
conserved from one end of a world-tube to the other end in a scalar field.
If the line integral must be conserve from one closed curve to the next,
then does this result in the need of a geodesic surface from one end of the
tube to the other?

Metric theories of gravitation have geodesic paths. Affine theories
of gravitation do not have geodesic paths. Since affine theories can
give identical predictions vs. metric theories, I'd posit geodesics
were not required to describe gravitation.

If an affine theory gives the same predictions as a metric theory, doesn't
that make it a metric theory? As I understand it, a metric space means
distances can be defined on it. So if a geometrical theory makes
predictions in terms of rulers and clocks*c, it's metric.

Metric theories are DEFINITELY NOT affine theories,

"Gravitation without the equivalence principle,"
http://arXiv.org/abs//gr-qc/0304106

Poincaré group gauge theory can be equivalent to Einstein-Cartan
gravitation theory. Einstein-Cartan theory operates in Riemann-Cartan
spacetime U^4. A curvature and a torsion tensor can be obtained on
U^4.

1) If the torsion tensor vanishes, V^4 pseudo-Riemannian spacetime
obtains (metric);
2) If the curvature tensor vanishes, A^4 Weitzenböck spacetime
obtains (affine/teleparallel);
3) If both tensors vanish, M^4 Minkowski spacetime obtains.

Pick your vierbein, but don't confuse them.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!