Discussion:
The need of geodesics in physics
Mike
2003-10-26 15:54:37 UTC
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.
davidoff404
2003-10-26 19:51:04 UTC
Post by Mike
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
2003-10-26 20:48:17 UTC
Post by davidoff404
Post by Mike
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
2003-10-26 20:26:45 UTC
Post by Mike
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.
Post by Mike
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!
Mike
2003-10-26 21:18:07 UTC
Post by Uncle Al
Post by Mike
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.
Richard
2003-10-26 23:37:01 UTC
Post by Mike
Post by Uncle Al
Post by Mike
Using Stoke's theorem, one can show that the closed line integral must
be
Post by Uncle Al
Post by Mike
conserved from one end of a world-tube to the other end in a scalar
field.
Post by Uncle Al
Post by Mike
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
Post by Uncle Al
Post by Mike
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
Mike
2003-10-27 00:17:15 UTC
Post by Richard
Post by Mike
Post by Uncle Al
Post by Mike
Using Stoke's theorem, one can show that the closed line integral must
be
Post by Uncle Al
Post by Mike
conserved from one end of a world-tube to the other end in a scalar
field.
Post by Uncle Al
Post by Mike
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
Post by Uncle Al
Post by Mike
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?
Uncle Al
2003-10-27 01:03:27 UTC
Post by Mike
Post by Richard
Post by Mike
Post by Uncle Al
Post by Mike
Using Stoke's theorem, one can show that the closed line integral
must
Post by Richard
Post by Mike
be
Post by Uncle Al
Post by Mike
conserved from one end of a world-tube to the other end in a scalar
field.
Post by Uncle Al
Post by Mike
If the line integral must be conserve from one closed curve to the
next,
Post by Richard
Post by Mike
Post by Uncle Al
Post by Mike
then does this result in the need of a geodesic surface from one end
of
Post by Richard
Post by Mike
the
Post by Uncle Al
Post by Mike
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
Post by Richard
Post by Mike
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!
Ken S. Tucker
2003-10-27 16:54:55 UTC
Post by Uncle Al
Post by Mike
Post by Richard
Post by Mike
Post by Uncle Al
Post by Mike
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.
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
Post by Uncle Al
All metric theory
predictions are included in affine theories. Geodesics are a
convenience.
This may be semantics, but geodesics are the
*solution* of the motion equation, and need
to be provided in a manner demonstrable
in any CS. Maybe a brief example from
Newtons 1st law could be provided?
Post by Uncle Al
(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.
I think Einstein would agree, and so followed his
antisymmetrical metrics.
Post by Uncle Al
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
Good Luck
Ken S. Tucker
Gregory L. Hansen
2003-10-27 21:20:38 UTC
Post by Ken S. Tucker
Post by Uncle Al
Metric models are geometric and have geodesics. Affine models are
electromagnetic in form and have no geodesics.
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
I'm always tempted to make the connections non-scalar. If the geodesic
equation is

a_i = -{i,jk} v_j v_k

with a=d^2(x)/dt^2, v=dx/dt, {i,jk} an ASCII-compatible symbol for the
connection, you can multiply through by an m and call -m*{i,jk} the force
of gravity. So I always want to put some column vectors in there, like

a_i = -(1) {i,jk1} v_j v_k - (0) {i,jk2} v_j v_k
(0) (1)

and then multiply through by (m,q).

But maybe that's just dumb.
--
"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
Igor
2003-10-28 06:22:30 UTC
On Mon, 27 Oct 2003 21:20:38 +0000 (UTC),
Post by Gregory L. Hansen
Post by Ken S. Tucker
Post by Uncle Al
Metric models are geometric and have geodesics. Affine models are
electromagnetic in form and have no geodesics.
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
I'm always tempted to make the connections non-scalar. If the geodesic
equation is
a_i = -{i,jk} v_j v_k
with a=d^2(x)/dt^2, v=dx/dt, {i,jk} an ASCII-compatible symbol for the
connection, you can multiply through by an m and call -m*{i,jk} the force
of gravity. So I always want to put some column vectors in there, like
a_i = -(1) {i,jk1} v_j v_k - (0) {i,jk2} v_j v_k
(0) (1)
and then multiply through by (m,q).
But maybe that's just dumb.
Yeah, I see what you're trying to do (even though your column vectors
got a little skewed). The first term would be gravitation and the
second the Lorentz, but how you'd ever get a cross product out of a
symmetric connection is a bit of a stretch.
Ken S. Tucker
2003-10-28 19:25:37 UTC
Post by Igor
On Mon, 27 Oct 2003 21:20:38 +0000 (UTC),
Post by Gregory L. Hansen
Post by Ken S. Tucker
Post by Uncle Al
Metric models are geometric and have geodesics. Affine models are
electromagnetic in form and have no geodesics.
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
I'm always tempted to make the connections non-scalar. If the geodesic
equation is
a_i = -{i,jk} v_j v_k
with a=d^2(x)/dt^2, v=dx/dt, {i,jk} an ASCII-compatible symbol for the
connection, you can multiply through by an m and call -m*{i,jk} the force
of gravity.
I have no objection provided m is small relative to
the gravitating body, such as Mercury is to the Sun,
but check out Dover's P of R, A.E.'s GR Eq.(45),
and read geodesic is "the motion of the point".
I figure this is the solution from G_uv =0 (with the
energy density being zilch, by the specification of a
point) as opposed to the G_uv =k*T_uv, in the case
of a massive particle creating nonlinear *feed-back*
Post by Igor
Post by Gregory L. Hansen
So I always want to put some column vectors in there, like
a_i = -(1) {i,jk1} v_j v_k - (0) {i,jk2} v_j v_k
(0) (1)
and then multiply through by (m,q).
Very interesting how the indexes 1,2 are added
within your connections, and may very well be
ok, is a significant departure from conventional
mathematics.
I think QT excludes continuous acceleration of
charged particles. This is why Lorentz's
f_u = q*F_uv *U^v =0
may be reasonable.

((Knowing Mr. Hansen has R.C. Tolmans,
"Relativity, Thermodynamics and Cosmology",
he may want to glance Eq. (103.1) and (103.2).))
Post by Igor
Post by Gregory L. Hansen
But maybe that's just dumb.
Well it's more complex than I usually do.
Post by Igor
Yeah, I see what you're trying to do (even though your column vectors
got a little skewed). The first term would be gravitation and the
second the Lorentz, but how you'd ever get a cross product out of a
symmetric connection is a bit of a stretch.
The "cross-product" is primarily a magnetic effect,
from relative charge motion considered from a rest
frame. If I'm not mistaken, a direct proportionality
exists between the Magnetic field and the angular
momentum of the mass the charge creating the current
and B-field is attached (at rest) to.
Wouldn't this suggest that the metric describing
angular momentum and it's direct analog be equal?
Regards
Ken S. Tucker
Igor
2003-10-29 05:12:36 UTC
Post by Ken S. Tucker
Post by Igor
On Mon, 27 Oct 2003 21:20:38 +0000 (UTC),
Post by Gregory L. Hansen
Post by Ken S. Tucker
Post by Uncle Al
Metric models are geometric and have geodesics. Affine models are
electromagnetic in form and have no geodesics.
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
I'm always tempted to make the connections non-scalar. If the geodesic
equation is
a_i = -{i,jk} v_j v_k
with a=d^2(x)/dt^2, v=dx/dt, {i,jk} an ASCII-compatible symbol for the
connection, you can multiply through by an m and call -m*{i,jk} the force
of gravity.
I have no objection provided m is small relative to
the gravitating body, such as Mercury is to the Sun,
but check out Dover's P of R, A.E.'s GR Eq.(45),
and read geodesic is "the motion of the point".
I figure this is the solution from G_uv =0 (with the
energy density being zilch, by the specification of a
point) as opposed to the G_uv =k*T_uv, in the case
of a massive particle creating nonlinear *feed-back*
Post by Igor
Post by Gregory L. Hansen
So I always want to put some column vectors in there, like
a_i = -(1) {i,jk1} v_j v_k - (0) {i,jk2} v_j v_k
(0) (1)
and then multiply through by (m,q).
Very interesting how the indexes 1,2 are added
within your connections, and may very well be
ok, is a significant departure from conventional
mathematics.
I think QT excludes continuous acceleration of
charged particles. This is why Lorentz's
f_u = q*F_uv *U^v =0
may be reasonable.
((Knowing Mr. Hansen has R.C. Tolmans,
"Relativity, Thermodynamics and Cosmology",
he may want to glance Eq. (103.1) and (103.2).))
Post by Igor
Post by Gregory L. Hansen
But maybe that's just dumb.
Well it's more complex than I usually do.
Post by Igor
Yeah, I see what you're trying to do (even though your column vectors
got a little skewed). The first term would be gravitation and the
second the Lorentz, but how you'd ever get a cross product out of a
symmetric connection is a bit of a stretch.
The "cross-product" is primarily a magnetic effect,
from relative charge motion considered from a rest
frame. If I'm not mistaken, a direct proportionality
exists between the Magnetic field and the angular
momentum of the mass the charge creating the current
and B-field is attached (at rest) to.
Wouldn't this suggest that the metric describing
angular momentum and it's direct analog be equal?
Regards
Ken S. Tucker
I'm not completely sure I understand what you are trying to say here.
All I was saying was that we have two completely different types of
mathematical objects here. The magnetic force is a cross product of
two vectors, sometimes called a bi-vector or a second rank
antisymmetric tensor. The affine connection, on the other hand, has a
second rank symmetric structure in it's covariant indices. I don't
understand how those two patterns could match up.

is only true for spin and not angular momentum in general.
Ken S. Tucker
2003-10-29 16:19:47 UTC
Post by Igor
Post by Ken S. Tucker
Post by Igor
On Mon, 27 Oct 2003 21:20:38 +0000 (UTC),
Post by Gregory L. Hansen
Post by Ken S. Tucker
Post by Uncle Al
Metric models are geometric and have geodesics. Affine models are
electromagnetic in form and have no geodesics.
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
I'm always tempted to make the connections non-scalar. If the geodesic
equation is
a_i = -{i,jk} v_j v_k
with a=d^2(x)/dt^2, v=dx/dt, {i,jk} an ASCII-compatible symbol for the
connection, you can multiply through by an m and call -m*{i,jk} the force
of gravity.
I have no objection provided m is small relative to
the gravitating body, such as Mercury is to the Sun,
but check out Dover's P of R, A.E.'s GR Eq.(45),
and read geodesic is "the motion of the point".
I figure this is the solution from G_uv =0 (with the
energy density being zilch, by the specification of a
point) as opposed to the G_uv =k*T_uv, in the case
of a massive particle creating nonlinear *feed-back*
Post by Igor
Post by Gregory L. Hansen
So I always want to put some column vectors in there, like
a_i = -(1) {i,jk1} v_j v_k - (0) {i,jk2} v_j v_k
(0) (1)
and then multiply through by (m,q).
Very interesting how the indexes 1,2 are added
within your connections, and may very well be
ok, is a significant departure from conventional
mathematics.
I think QT excludes continuous acceleration of
charged particles. This is why Lorentz's
f_u = q*F_uv *U^v =0
may be reasonable.
((Knowing Mr. Hansen has R.C. Tolmans,
"Relativity, Thermodynamics and Cosmology",
he may want to glance Eq. (103.1) and (103.2).))
Post by Igor
Yeah, I see what you're trying to do (even though your column vectors
got a little skewed). The first term would be gravitation and the
second the Lorentz, but how you'd ever get a cross product out of a
symmetric connection is a bit of a stretch.
The "cross-product" is primarily a magnetic effect,
from relative charge motion considered from a rest
frame. If I'm not mistaken, a direct proportionality
exists between the Magnetic field and the angular
momentum of the mass the charge creating the current
and B-field is attached (at rest) to.
Wouldn't this suggest that the metric describing
angular momentum and it's direct analog be equal?
Regards
Ken S. Tucker
I'm not completely sure I understand what you are trying to say here.
All I was saying was that we have two completely different types of
mathematical objects here. The magnetic force is a cross product of
two vectors, sometimes called a bi-vector or a second rank
antisymmetric tensor. The affine connection, on the other hand, has a
second rank symmetric structure in it's covariant indices.
Agreed, what you have stated is rigorously correct
using the concept of the Magnetic field, but the
magnetic force could be calculated from,

f=qE*(1-v^2), with magnetic force basically,

f(m) = -q*E*v^2 = -q*Q/r * v^2/r
Post by Igor
I don't understand how those two patterns could match up.
Well setting energy (=mass) = q*Q/r and a
centrifugal acceleration term from the affine
to be v^2/r reproduces basic f(m).
Mr. Hansen is very familiar with centrifugal metrics
including spacetime metrics like g_0i. We discussed
these before, so I have some *very* basic notions,
Post by Igor
is only true for spin and not angular momentum in general.
Thanks, BTW way I read with interest, awhile ago
your 2 stage process to generate a photons.
Ken S. Tucker
Shmuel (Seymour J.) Metz
2003-10-29 21:35:10 UTC
Post by Igor
The magnetic force is a cross product of
two vectors, sometimes called a bi-vector or a second rank
antisymmetric tensor.
In non-Relativistic EM, yes. But in a Relativistic context, the
electrical and magnetic fields are subsumed in a single antisymmetric
tensor, and the force is given by taking the dot product with the
4-current.
Post by Igor
The affine connection, on the other hand, has a
second rank symmetric structure in it's covariant indices. I don't
understand how those two patterns could match up.
They don't; you take derivatives of the connection to get a tensor.
Both curvature and torsion are tensors.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

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Igor
2003-10-28 06:14:35 UTC
Post by Ken S. Tucker
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
The problem with that is that both sides of that equation are tensors.
If they vanish in one frame, then they must vanish in all frames.
Geodesics are based on affine connections, which are not tensors, so
they have non-homogeneous transformation properties. Moreover,
geodesic acceleration should be independent of mass, which applies to
forces such as centrifugal, coriolis, and of course, gravitation.
Acceleration due to Lorentz force will be inversely dependent on mass,
making it a no go.
Post by Ken S. Tucker
Post by Uncle Al
All metric theory
predictions are included in affine theories. Geodesics are a
convenience.
This may be semantics, but geodesics are the
*solution* of the motion equation, and need
to be provided in a manner demonstrable
in any CS. Maybe a brief example from
Newtons 1st law could be provided?
It's actually a bit more than semantics. Not all motion is geodesic
motion. See remarks above.
Ken S. Tucker
2003-10-28 18:28:28 UTC
Post by Igor
Post by Ken S. Tucker
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
The problem with that is that both sides of that equation are tensors.
If they vanish in one frame, then they must vanish in all frames.
Geodesics are based on affine connections, which are not tensors, so
they have non-homogeneous transformation properties.
Ah, but the geodesic equation is the absolute derivative
of the 4 velocity U^u = dx^u/ds, ie

DU^u/ds = U^u;v * U^v =0.
Post by Igor
Moreover,
geodesic acceleration should be independent of mass, which applies to
forces such as centrifugal, coriolis, and of course, gravitation.
True for acceleration of point masses, (but not true
for relatively large masses, that's why acceleration
is a better term than force in GR, otherwise you'll
need to include the term k*t_uv >0 we figuring G_uv).
Post by Igor
Acceleration due to Lorentz force will be inversely dependent on mass,
making it a no go.
But I wrote the Lorentz force vanishes. Charged particle
motion must respect Quantum Theory, Lorentz force does
not, so the stated objection evaporates.
Post by Igor
Post by Ken S. Tucker
Post by Uncle Al
All metric theory
predictions are included in affine theories. Geodesics are a
convenience.
This may be semantics, but geodesics are the
*solution* of the motion equation, and need
to be provided in a manner demonstrable
in any CS. Maybe a brief example from
Newtons 1st law could be provided?
It's actually a bit more than semantics. Not all motion is geodesic
motion. See remarks above.
Classically true, but Lorentz force vanishes in view
of QT, meaning all motion must be geodesical,
otherwise absolute acceleration exists.

Ken S. Tucker
Igor
2003-10-29 05:03:19 UTC
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
The problem with that is that both sides of that equation are tensors.
If they vanish in one frame, then they must vanish in all frames.
Geodesics are based on affine connections, which are not tensors, so
they have non-homogeneous transformation properties.
Ah, but the geodesic equation is the absolute derivative
of the 4 velocity U^u = dx^u/ds, ie
DU^u/ds = U^u;v * U^v =0.
True, but how does that relate to the LF?
Post by Ken S. Tucker
Post by Igor
Moreover,
geodesic acceleration should be independent of mass, which applies to
forces such as centrifugal, coriolis, and of course, gravitation.
True for acceleration of point masses, (but not true
for relatively large masses, that's why acceleration
is a better term than force in GR, otherwise you'll
need to include the term k*t_uv >0 we figuring G_uv).
Post by Igor
Acceleration due to Lorentz force will be inversely dependent on mass,
making it a no go.
But I wrote the Lorentz force vanishes. Charged particle
motion must respect Quantum Theory, Lorentz force does
not, so the stated objection evaporates.
But why are you making that assumption? I can make that statement
certainly, but I'm excluding a large variety of systems by doing it.
So I don't think it is very realistic.
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
Post by Uncle Al
All metric theory
predictions are included in affine theories. Geodesics are a
convenience.
This may be semantics, but geodesics are the
*solution* of the motion equation, and need
to be provided in a manner demonstrable
in any CS. Maybe a brief example from
Newtons 1st law could be provided?
It's actually a bit more than semantics. Not all motion is geodesic
motion. See remarks above.
Classically true, but Lorentz force vanishes in view
of QT, meaning all motion must be geodesical,
otherwise absolute acceleration exists.
Ken S. Tucker
Okay, I don't fully comprehend that last statement about LF vanishing
in QT. Aren't you saying that subatomic particles experience no EM
force?
Ken S. Tucker
2003-10-31 17:05:10 UTC
Post by Igor
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
The problem with that is that both sides of that equation are tensors.
If they vanish in one frame, then they must vanish in all frames.
Geodesics are based on affine connections, which are not tensors, so
they have non-homogeneous transformation properties.
Ah, but the geodesic equation is the absolute derivative
of the 4 velocity U^u = dx^u/ds, ie
DU^u/ds = U^u;v * U^v =0.
True, but how does that relate to the LF?
Well multiply by mass (scalar invariant) then,

m*DU^u/ds =0

but QT requires m to change discretely, and
since dm >0 requires a continuous change in m,
dm=0, therefore set P^u = m*U^u (the 4-
momentum) and find

DP^u/ds =0 (P^u = m*U^u, dm=0).

Is true generally, and applies to LF, so

f^u=DP^u/ds =0.
Post by Igor
Post by Ken S. Tucker
Post by Igor
Moreover,
geodesic acceleration should be independent of mass, which applies to
forces such as centrifugal, coriolis, and of course, gravitation.
True for acceleration of point masses, (but not true
for relatively large masses, that's why acceleration
is a better term than force in GR, otherwise you'll
need to include the term k*t_uv >0 we figuring G_uv).
Post by Igor
Acceleration due to Lorentz force will be inversely dependent on mass,
making it a no go.
But I wrote the Lorentz force vanishes. Charged particle
motion must respect Quantum Theory, Lorentz force does
not, so the stated objection evaporates.
But why are you making that assumption? I can make that statement
certainly, but I'm excluding a large variety of systems by doing it.
So I don't think it is very realistic.
Consider, for example f_0 = q*F_0i*U^i,
which in vectors is approximately
f_0 = q*E_i (dot) V_i.

If q moves in direction of E_i then it's
potential energy will vary continuously and
this is prohibited by QT, therefore
E_i (dot) V_i =0 and thus f_0 =0.

Simply stated, charge q's V_i is always
perpendicular to E_i.
Post by Igor
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
Post by Uncle Al
All metric theory
predictions are included in affine theories. Geodesics are a
convenience.
This may be semantics, but geodesics are the
*solution* of the motion equation, and need
to be provided in a manner demonstrable
in any CS. Maybe a brief example from
Newtons 1st law could be provided?
It's actually a bit more than semantics. Not all motion is geodesic
motion. See remarks above.
Classically true, but Lorentz force vanishes in view
of QT, meaning all motion must be geodesical,
otherwise absolute acceleration exists.
Okay, I don't fully comprehend that last statement about LF vanishing
in QT. Aren't you saying that subatomic particles experience no EM
force?
Yes thats exactly right! But this does not exclude
higher and lower energy levels. It means these
levels are varied by quanta (like photon emission
or absorption).
I imagine that one must add the EM field of the
photon to the nuclear field to create an electron
path (geodesic) that preserves E_i (dot) V_i
where E_i is the Electric field of the photon plus
the nuclear field relative to an electrons charge q,
when changing orbits.

Our notions of forces have been largely obsoleted
by GR and geodesical motion. Why should it be a
surprise to apply GR to LF in view of the evidence
of QT and find electric (quantum) geodesics?
Regards
Ken S. Tucker
Shmuel (Seymour J.) Metz
2003-11-03 11:40:40 UTC
Post by Ken S. Tucker
but QT requires m to change discretely,
No. QT allows both discrete and continuous spectra. Further, even in
systems with discrete spectra, QT is based on differential equations
involving continuous variables. The Devil is in the details; you can't
understand what is going on in QT by reading a popularization.
Post by Ken S. Tucker
changing orbits
That would be the old QT, obsolete even before QED came along.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

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the right to publicly post or ridicule any abusive E-mail.

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Mark
2003-11-04 00:40:38 UTC
Post by Shmuel (Seymour J.) Metz
Post by Ken S. Tucker
changing orbits
That would be the old QT, obsolete even before QED came along.
The equations of motion for r(t), v(t) are still the same
r'(t) = v(t), v'(t) = -kr(t)/|r(t)|^3
only now with the boundary condition
[r(0), v(0)] = i h-bar/m

The orbital parameters
h = r x v, e = v x h/k - r/|r|
are still there. They're just q-numbers now. In fact, the
commutation relations are those
[h_1, e_1] = [h_2, e_2] = [h_3, e_3] = 0
[h_1, h_2] = i h-bar h_3 & cyc.
[h_1, e_2] = i h-bar e_3 & cyc.
[e_1, e_2] = i h-bar K h_3 & cyc. (K depends on orbit type)
are just those of SO(4), Galilei(2) or SL(2, C) depending whether the
orbit is elliptical (|e| < 1), parabolic (|e| = 1) or hyperbolic (|e| > 1).

The bound states are just the (k,k) representatives of the Lie algebra
so(4) (the restriction on the range of representations arising from
the constraint e.h = 0).

Hence, the term "Hydrogen SO(4)".
Shmuel (Seymour J.) Metz
2003-11-06 05:32:09 UTC
Post by Mark
The equations of motion for r(t), v(t) are still the same
r'(t) = v(t), v'(t) = -kr(t)/|r(t)|^3
No, they are replaced by Schrodinger's Equation, and variables in the
potential are replaced with the corresponding operators. A physical
state may be a superposition of distinct eigenstates, which doesn't
correspond to any solution of the classical equations.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

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the right to publicly post or ridicule any abusive E-mail.

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tb+ (Thomas Bushnell, BSG)
2003-11-06 06:35:01 UTC
Post by Shmuel (Seymour J.) Metz
No, they are replaced by Schrodinger's Equation, and variables in the
potential are replaced with the corresponding operators. A physical
state may be a superposition of distinct eigenstates, which doesn't
correspond to any solution of the classical equations.
Almost right. A physical state *is* a superposition of distinct
eigenstates.

Remember, there is no privileged basis. What is not a superposition
in one basis will be in another basis.

And, for any physical state, there is some basis for which it is an
eigenstate and not a superposition at all. But, that eigenstate might
not correspond to any solution of the classical equations.

Thomas
Igor
2003-11-04 09:42:05 UTC
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
The problem with that is that both sides of that equation are tensors.
If they vanish in one frame, then they must vanish in all frames.
Geodesics are based on affine connections, which are not tensors, so
they have non-homogeneous transformation properties.
Ah, but the geodesic equation is the absolute derivative
of the 4 velocity U^u = dx^u/ds, ie
DU^u/ds = U^u;v * U^v =0.
True, but how does that relate to the LF?
Well multiply by mass (scalar invariant) then,
m*DU^u/ds =0
but QT requires m to change discretely, and
since dm >0 requires a continuous change in m,
dm=0, therefore set P^u = m*U^u (the 4-
momentum) and find
DP^u/ds =0 (P^u = m*U^u, dm=0).
I don't know where you got that notion from. QT says nothing about m
changing discretely. In fact, the rules of QT maintain continuous
change of mechanical variables (of course under the restrictions of
the operator algebra). What is discrete in QT are the quantized
values of the so-called stationary states, but these tend to be the
exception rather than the rule. Continuous fluctuation tends to occur
everywhere in the quantum world.
Post by Ken S. Tucker
Is true generally, and applies to LF, so
f^u=DP^u/ds =0.
Post by Igor
Post by Ken S. Tucker
Post by Igor
Moreover,
geodesic acceleration should be independent of mass, which applies to
forces such as centrifugal, coriolis, and of course, gravitation.
True for acceleration of point masses, (but not true
for relatively large masses, that's why acceleration
is a better term than force in GR, otherwise you'll
need to include the term k*t_uv >0 we figuring G_uv).
Post by Igor
Acceleration due to Lorentz force will be inversely dependent on mass,
making it a no go.
But I wrote the Lorentz force vanishes. Charged particle
motion must respect Quantum Theory, Lorentz force does
not, so the stated objection evaporates.
But why are you making that assumption? I can make that statement
certainly, but I'm excluding a large variety of systems by doing it.
So I don't think it is very realistic.
Consider, for example f_0 = q*F_0i*U^i,
which in vectors is approximately
f_0 = q*E_i (dot) V_i.
If q moves in direction of E_i then it's
potential energy will vary continuously and
this is prohibited by QT, therefore
E_i (dot) V_i =0 and thus f_0 =0.
Simply stated, charge q's V_i is always
perpendicular to E_i.
But potential still varies continuously in QT and its changes are very
easily tracked in the position representation. I don't know how you
came to that conclusion.
Mark
2003-11-05 03:01:01 UTC
Post by Igor
I don't know where you got that notion from. QT says nothing about m
changing discretely.
Mass is quantized on a discrete spectrum in QFT, but the origin of
the mass matrix and its spectrum is unknown.

What *is* known only furthers the mystery: the mass operator doesn't
commute with the SU(2) symmetry generators corresponding to the W and
anti-W.

The mystery only deepens, considering that [anti-]W can only interact with
particles in the left helicity state (and anti-particles in the right
state). But for a particle with mass, the distinction between left
and right helicity states is relative. But interacting vs. not
interacting can't be relative too! Therefore, the helicity states
can't be either, and the particles can't actually have any mass.
Which means, in turn, that their mass is not intrinsic, but dynamic
in origin, arising from an interaction with an otherwise unknown or
unseen/undiscovered field. The mass matrix, then, just describes
the coupling of the particles with this field.

Were it not for this extra kink, all the charge states of all the
particles would exist with a 3-fold degeneracy that would be
completely indistinguishable from one another. The mass matrix
mixes between these groups, and the mass eigenstates for each
charge state are actually mixtures across the 3 groups with 3
separate eigenvalues.

The mystery deepens yet further when you step back and loop at the
charge states. They form a pattern: they reside at the points of
a 6 dimensional Cartesian lattice forming within it a 5-dimensional
hypercube! Indeed, you can write down an orthonormal basis
(X,A,B,C,D,E), and the 32 vertices are precisely those for the
32 combinations of vectors
xX + (x-a)A + (x-b)B + (x-c)C + (x-d)D + (x-e)E
for which a,b,c,d,e = +/- 1/2 with x undetermined (possibly with
one set for each of the 3 groups).

And the mystery deepens yet further when you consider what the
W and anti-W actually do; and what the 6 (charged) gluons in SU(3)
do. The W switches a (+a,-b) to a (-a,+b); anti-W the other way.
The gluons switch a (+c,-d) to a (-c,+d); and the other 5 combinations
involving c, d and e. Left and right are distinguished solely by
one of the a or b, say, by +b vs. -b, with left being (+,-) and (-,+)
and right being (+,+) or (-,-); so (+,-) & (+,+) are the left & right
states of one particle; (-,+) & (-,-) the left & right states of
another.
Ken S. Tucker
2003-11-05 17:18:59 UTC
Post by Igor
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
The problem with that is that both sides of that equation are tensors.
If they vanish in one frame, then they must vanish in all frames.
Geodesics are based on affine connections, which are not tensors, so
they have non-homogeneous transformation properties.
Ah, but the geodesic equation is the absolute derivative
of the 4 velocity U^u = dx^u/ds, ie
DU^u/ds = U^u;v * U^v =0.
True, but how does that relate to the LF?
Well multiply by mass (scalar invariant) then,
m*DU^u/ds =0
but QT requires m to change discretely, and
since dm >0 requires a continuous change in m,
dm=0, therefore set P^u = m*U^u (the 4-
momentum) and find
DP^u/ds =0 (P^u = m*U^u, dm=0).
I don't know where you got that notion from. QT says nothing about m
changing discretely. In fact, the rules of QT maintain continuous
change of mechanical variables (of course under the restrictions of
the operator algebra). What is discrete in QT are the quantized
values of the so-called stationary states, but these tend to be the
exception rather than the rule. Continuous fluctuation tends to occur
everywhere in the quantum world.
We should make sure we're talking about the
same thing. It is *invariant* energy which can
only vary discretely and NOT continuously.
A simple atom emits and absorbs energy
by photons.
Post by Igor
Post by Ken S. Tucker
Is true generally, and applies to LF, so
f^u=DP^u/ds =0.
Post by Igor
Post by Ken S. Tucker
Post by Igor
Moreover,
geodesic acceleration should be independent of mass, which applies to
forces such as centrifugal, coriolis, and of course, gravitation.
True for acceleration of point masses, (but not true
for relatively large masses, that's why acceleration
is a better term than force in GR, otherwise you'll
need to include the term k*t_uv >0 we figuring G_uv).
Post by Igor
Acceleration due to Lorentz force will be inversely dependent on mass,
making it a no go.
But I wrote the Lorentz force vanishes. Charged particle
motion must respect Quantum Theory, Lorentz force does
not, so the stated objection evaporates.
But why are you making that assumption? I can make that statement
certainly, but I'm excluding a large variety of systems by doing it.
So I don't think it is very realistic.
Consider, for example f_0 = q*F_0i*U^i,
which in vectors is approximately
f_0 = q*E_i (dot) V_i.
If q moves in direction of E_i then it's
potential energy will vary continuously and
this is prohibited by QT, therefore
E_i (dot) V_i =0 and thus f_0 =0.
Simply stated, charge q's V_i is always
perpendicular to E_i.
But potential still varies continuously in QT and its changes are very
easily tracked in the position representation.
We're talking about Lorentz force not potential.
Post by Igor
I don't know how you came to that conclusion.
Fair enough. Just to make clear definitions, define
invariant energy P by P*P = P_u*P^u ok?

An invariant force would be f = dP/ds, and
dP = f*ds ok?

The Lorentz force f_u *dx^u = f*ds, ok?

f_u*dx^u = q*F_uv *(dx^v/ds)* dx^u ok?

Since dx^u*dx^v is symmetrical and F_uv is
antisymmetrical F_uv*dx^u*dx^v =0 when
summed ok?

Then f*ds = 0 = dP ok?

That's it!

To summarize: The Lorentz Force Equation
predicts dP=0. But we know P is not a constant,
therefore P varies discretely, ie by quanta.

Best Regards
Ken S. Tucker
PS: Snippable comment - That LF predicts and is
compatible with QT is excellent. It forms a great
bridge between GR and QT embodied in

DP^u/ds = 0 (absolute derivative).

This equation is derived w.r.t LF, QT (therefore
EM) and GR, without any added assumptions, ie
straight from working in the system.
It can be called a *generalized geodesic*,
(I'd hestitate to call it THE *universal geodesic*).

One thing that is of interest : since P<>0 the
Limited Einstein Field Equation G_uv =0 is not
applicable (as it applies to massless points) so
the EFE G_uv = k*T_uv would be required.
L8R
KST
Igor
2003-11-06 01:20:30 UTC
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
The problem with that is that both sides of that equation are tensors.
If they vanish in one frame, then they must vanish in all frames.
Geodesics are based on affine connections, which are not tensors, so
they have non-homogeneous transformation properties.
Ah, but the geodesic equation is the absolute derivative
of the 4 velocity U^u = dx^u/ds, ie
DU^u/ds = U^u;v * U^v =0.
True, but how does that relate to the LF?
Well multiply by mass (scalar invariant) then,
m*DU^u/ds =0
but QT requires m to change discretely, and
since dm >0 requires a continuous change in m,
dm=0, therefore set P^u = m*U^u (the 4-
momentum) and find
DP^u/ds =0 (P^u = m*U^u, dm=0).
I don't know where you got that notion from. QT says nothing about m
changing discretely. In fact, the rules of QT maintain continuous
change of mechanical variables (of course under the restrictions of
the operator algebra). What is discrete in QT are the quantized
values of the so-called stationary states, but these tend to be the
exception rather than the rule. Continuous fluctuation tends to occur
everywhere in the quantum world.
We should make sure we're talking about the
same thing. It is *invariant* energy which can
only vary discretely and NOT continuously.
A simple atom emits and absorbs energy
by photons.
Post by Igor
Post by Ken S. Tucker
Is true generally, and applies to LF, so
f^u=DP^u/ds =0.
Post by Igor
Post by Ken S. Tucker
Post by Igor
Moreover,
geodesic acceleration should be independent of mass, which applies to
forces such as centrifugal, coriolis, and of course, gravitation.
True for acceleration of point masses, (but not true
for relatively large masses, that's why acceleration
is a better term than force in GR, otherwise you'll
need to include the term k*t_uv >0 we figuring G_uv).
Post by Igor
Acceleration due to Lorentz force will be inversely dependent on mass,
making it a no go.
But I wrote the Lorentz force vanishes. Charged particle
motion must respect Quantum Theory, Lorentz force does
not, so the stated objection evaporates.
But why are you making that assumption? I can make that statement
certainly, but I'm excluding a large variety of systems by doing it.
So I don't think it is very realistic.
Consider, for example f_0 = q*F_0i*U^i,
which in vectors is approximately
f_0 = q*E_i (dot) V_i.
If q moves in direction of E_i then it's
potential energy will vary continuously and
this is prohibited by QT, therefore
E_i (dot) V_i =0 and thus f_0 =0.
Simply stated, charge q's V_i is always
perpendicular to E_i.
But potential still varies continuously in QT and its changes are very
easily tracked in the position representation.
We're talking about Lorentz force not potential.
Post by Igor
I don't know how you came to that conclusion.
Fair enough. Just to make clear definitions, define
invariant energy P by P*P = P_u*P^u ok?
An invariant force would be f = dP/ds, and
dP = f*ds ok?
The Lorentz force f_u *dx^u = f*ds, ok?
f_u*dx^u = q*F_uv *(dx^v/ds)* dx^u ok?
Since dx^u*dx^v is symmetrical and F_uv is
antisymmetrical F_uv*dx^u*dx^v =0 when
summed ok?
Then f*ds = 0 = dP ok?
That's it!
To summarize: The Lorentz Force Equation
predicts dP=0. But we know P is not a constant,
therefore P varies discretely, ie by quanta.
Best Regards
Ken S. Tucker
Yeah, I think I see what you're trying to say. The quantity f ds is
actually invariant work. And your result threw me for a few minutes
since magnetic fields do no work, but electric fields do. Then I
realized that it could be broken down as f_i * d x^i + f _0 * d x^0
= 0. The first term is the work done by the electric field and the
second term is non-zero so they will balance out. Note that f_0 is
essentially E dot v and will be related to power.

I'm still not sure where you're going with this, but I think I
understand your argument a little better. Thanks.
Ken S. Tucker
2003-11-06 19:08:58 UTC
Post by Igor
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
Post by Igor
Post by Ken S. Tucker
That's a bit confusing, one can set the Lorentz
Force to generally vanish ie.
f_u = q*F_uv *U^v =0
and in effect decribe EM geodesics, then put
q*F_uv in the metric forming a non-symmetrical
metric.
The problem with that is that both sides of that equation are tensors.
If they vanish in one frame, then they must vanish in all frames.
Geodesics are based on affine connections, which are not tensors, so
they have non-homogeneous transformation properties.
Ah, but the geodesic equation is the absolute derivative
of the 4 velocity U^u = dx^u/ds, ie
DU^u/ds = U^u;v * U^v =0.
True, but how does that relate to the LF?
Well multiply by mass (scalar invariant) then,
m*DU^u/ds =0
but QT requires m to change discretely, and
since dm >0 requires a continuous change in m,
dm=0, therefore set P^u = m*U^u (the 4-
momentum) and find
DP^u/ds =0 (P^u = m*U^u, dm=0).
I don't know where you got that notion from. QT says nothing about m
changing discretely. In fact, the rules of QT maintain continuous
change of mechanical variables (of course under the restrictions of
the operator algebra). What is discrete in QT are the quantized
values of the so-called stationary states, but these tend to be the
exception rather than the rule. Continuous fluctuation tends to occur
everywhere in the quantum world.
We should make sure we're talking about the
same thing. It is *invariant* energy which can
only vary discretely and NOT continuously.
A simple atom emits and absorbs energy
by photons.
Post by Igor
Post by Ken S. Tucker
Is true generally, and applies to LF, so
f^u=DP^u/ds =0.
Post by Igor
Post by Ken S. Tucker
Post by Igor
Moreover,
geodesic acceleration should be independent of mass, which applies to
forces such as centrifugal, coriolis, and of course, gravitation.
True for acceleration of point masses, (but not true
for relatively large masses, that's why acceleration
is a better term than force in GR, otherwise you'll
need to include the term k*t_uv >0 we figuring G_uv).
Post by Igor
Acceleration due to Lorentz force will be inversely dependent on mass,
making it a no go.
But I wrote the Lorentz force vanishes. Charged particle
motion must respect Quantum Theory, Lorentz force does
not, so the stated objection evaporates.
But why are you making that assumption? I can make that statement
certainly, but I'm excluding a large variety of systems by doing it.
So I don't think it is very realistic.
Consider, for example f_0 = q*F_0i*U^i,
which in vectors is approximately
f_0 = q*E_i (dot) V_i.
If q moves in direction of E_i then it's
potential energy will vary continuously and
this is prohibited by QT, therefore
E_i (dot) V_i =0 and thus f_0 =0.
Simply stated, charge q's V_i is always
perpendicular to E_i.
But potential still varies continuously in QT and its changes are very
easily tracked in the position representation.
We're talking about Lorentz force not potential.
Post by Igor
I don't know how you came to that conclusion.
Fair enough. Just to make clear definitions, define
invariant energy P by P*P = P_u*P^u ok?
An invariant force would be f = dP/ds, and
dP = f*ds ok?
The Lorentz force f_u *dx^u = f*ds, ok?
f_u*dx^u = q*F_uv *(dx^v/ds)* dx^u ok?
Since dx^u*dx^v is symmetrical and F_uv is
antisymmetrical F_uv*dx^u*dx^v =0 when
summed ok?
Then f*ds = 0 = dP ok?
That's it!
To summarize: The Lorentz Force Equation
predicts dP=0. But we know P is not a constant,
therefore P varies discretely, ie by quanta.
Best Regards Ken S. Tucker
Yeah, I think I see what you're trying to say. The quantity f ds is
actually invariant work.
Right (and thanks). Perhaps I might check out the
following with you to ascertain co-understanding,
(\$ = integral)

f = dP/ds =0 ok?

\$ f ds = constant (= invariant energy) ok?
Post by Igor
And your result threw me for a few minutes
since magnetic fields do no work, but electric fields do. Then I
realized that it could be broken down as f_i * d x^i + f _0 * d x^0
= 0. The first term is the work done by the electric field and the
second term is non-zero so they will balance out. Note that f_0 is
essentially E dot v and will be related to power.
Ah, but f_0 is an enigma, lets study it together.

1) If f_0 is non-zero then E dot v is non-zero.
If E dot v is non-zero then *spiral* orbits become
possible, for example, the charge q can move slowly
into the electric field E. (this is the old classical idea of
how electrons move in the nuclear field, and was
replaced by Plancks Quantum hypothesis (imho)).
So a non-zero f_0 violates QT.

2)You pointed out E dot v is "power", well consider
electrical current moving threw a resistor and this
produces power by the product Volts x Current
and in turn produces heat. The production of heat
is by quanta, ie. infared photons, hence power is
quantized. This can be solved and explained by,

\$ f_0 ds = q*\$ F_0i dx^i = constant.

For simplicity, work the RHS and get
q*E*x = q*Q/r = energy.=constant =quanta.
Post by Igor
I'm still not sure where you're going with this, but I think I
understand your argument a little better. Thanks.
You're very welcome, thank YOU.
I'm sorry my understanding is not simple, blame QT,
I (we) are working in the system. I think we agree
f=0, so the discussion is on the LF components f_0
and f_i being zero or not. (I'm arguing they are zero).

Best Regards
Ken S. Tucker
Shmuel (Seymour J.) Metz
2003-11-06 05:35:59 UTC
Post by Ken S. Tucker
We should make sure we're talking about the
same thing. It is *invariant* energy which can
only vary discretely and NOT continuously.
No. Energy is quantized only in a bound system. In an unbound system,
e.g., an atom that can be ionized, you need to deal with continuous
spectra.
--
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Ken S. Tucker
2003-11-06 19:25:21 UTC
Post by Shmuel (Seymour J.) Metz
Post by Ken S. Tucker
We should make sure we're talking about the
same thing. It is *invariant* energy which can
only vary discretely and NOT continuously.
No. Energy is quantized only in a bound system. In an unbound system,
e.g., an atom that can be ionized, you need to deal with continuous
spectra.
Shmuel (Seymour J.) Metz, SysProg and JOAT
Hi Seymour, I think this our first communication,
hope to have many more....

But the "spectra" itself is formed by photons, and the
photons themselves are quantum particles, so the atom
you mention is still varying by discrete energy quantities
called photons.
So far as I can tell the composition of the spectum cannot
produce continuous energy variation. (Unless the production
of indefinitely long waves is possible, but this is unlikely
given a finite transmitting dipole).
Yes?
Regards Ken S. Tucker
Shmuel (Seymour J.) Metz
2003-11-08 23:21:55 UTC
Post by Ken S. Tucker
But the "spectra" itself is formed by photons,
No. The term "continuous spectra" is not the same as the usage of
the term "spectrum" in optics. An observable in QM is a Hermitian
operator, and such an operator has a characteristic that is known in
the literature as its spectrum.

Read any good book on Functional Analysis for a explanation. Or read
"Principles of Quantum Mechanics" (P.A.M. Dirac) for an explanation
from a Physics perspective.

 But the spectrum in optics is related to the spectrum of the
Hamiltonian.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

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Mark
2003-11-10 00:11:54 UTC
Post by Shmuel (Seymour J.) Metz
Post by Ken S. Tucker
But the "spectra" itself is formed by photons,
No. The term "continuous spectra" is not the same as the usage of
the term "spectrum" in optics. An observable in QM is a Hermitian
operator, and such an operator has a characteristic that is known in
the literature as its spectrum.
Actually, the term "spectrum" is generic to linear algebras, and is not
specific to operator algebras. If A is an algebra over a field F, the
spectrum of an element a of A would be
spec(a) = { f in F: (a-f) has no two-sided inverse in A }.
Ken S. Tucker
2003-11-11 20:47:52 UTC
Post by Mark
Post by Shmuel (Seymour J.) Metz
Post by Ken S. Tucker
But the "spectra" itself is formed by photons,
Ken
No. The term "continuous spectra" is not the same as the usage of
the term "spectrum" in optics. An observable in QM is a Hermitian
operator, and such an operator has a characteristic that is known in
the literature as its spectrum.
Seymour
Actually, the term "spectrum" is generic to linear algebras, and is not
specific to operator algebras. If A is an algebra over a field F, the
spectrum of an element a of A would be
spec(a) = { f in F: (a-f) has no two-sided inverse in A }.
Mark
This is amusing, see this thread has been cross posted
to sci-math and sci-physics, so there are different
meanings to words. Anyway I'm the guy who is talking
about physics, but I'll try to make it interesting for the
higher class posters in sci-math, who may disagree.

An invariant has no units, specifically invariant mass
or energy. Two invariant masses can be equated on
a balance without resort to units.

I argue the invariant mass/energy can only change
by small but finite amounts, hence discontinuously.

Let E be the invariant then dE =0 as it has no
continuous function and derivative, although
E =/=constant.

Transactions are quantized to the nearest penny,
thus d(money) =0.

For mathematical masochists, one might suggest
a Fourier series to form a digital pulse (penny),
however, the wavelengths required exceed the
size of the penny to form the square pulse,
((SUM i, (n=1+2*i) 1/n sin n*w*t)).

So a wave mechanical solution is an
approximation.but is DOA mathematically.

Returning to physics, I maintain we cannot
infinitely divide the two masses on the balance,
nor can we infinitesmally vary either mass.
Therefore
dE =0

is a Law of Physics and describes how invariant
energy varies. This might look like a peculiar law
but compare this to the GR geodesical equation
of motion, given by the absolute derivative

DU^u/ds =0.

(Absolute acceleration does not exist)

This equation is very successful.
Now 4-momentum is E*U^u so a quantum-
mechanical generally relativistic geodesic is,

DE^u/ds =0 .

(absolute force does not exist)

This equation rules out any force like f^u (or f_u).

So it sets all the components of Lorentz force to
zero to describe how a charge moves geodesically.
That's a powerful constraint, and *appears* to be
verified so far.

One might call DE^u =0 a unified field theory,
but I think it's only a requirement. The metric
that satisfies this equation is the field, and I
suspect the unified field equation will turn out
to be G_uv=kT_uv. Einstein has suggested
(in his later years) using anti-symmetrical metrics
to go from G_uv = kT_uv to DE^u=0, to unify.

Regards to sci - math & physics
Ken S. Tucker
Shmuel (Seymour J.) Metz
2003-11-11 23:11:28 UTC
Post by Mark
Actually, the term "spectrum" is generic to linear algebras, and is
not specific to operator algebras.
Correct, although I don't know whether the more general meaning is
ever used in QM. I'm certainly not aware of the field being other
than R, C or H.

 If you use the definition that requires commutativity, then omit
H.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

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Shmuel (Seymour J.) Metz
2003-10-28 23:06:24 UTC
Post by Uncle Al
Metric models are geometric and have geodesics. Affine models are
electromagnetic in form and have no geodesics.
I hope that you mean that the geodesics are not relevant to the
Physics, rather then that they don't exist. An affine connection is
all that you need to define geodesics; you don't need a metric.
Post by Uncle Al
Einstein made no geometric mistakes, neither did Euclid.
Euclid did; his axioms and postulates were not adequate to prove all
of the statements he listed as theorems.
Post by Uncle Al
Euclid is incomplete by trivial demonstration.
Yes, but not in the sense you mean.
Post by Uncle Al
You cannot deep sea navigate using Euclid, nor
can you survey large tracts of land.
Sure you can; Euclid covered Solid Geometry.
--
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Gregory L. Hansen
2003-10-27 00:37:05 UTC
Post by Uncle Al
Post by Mike
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
Uncle Al
2003-10-27 01:13:34 UTC
Post by Gregory L. Hansen
Post by Uncle Al
Post by Mike
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!
2003-10-30 02:31:25 UTC
Post by Uncle Al
Post by Mike
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've seen your page, but it doesn't explain to me what affine theories
are all about. What ARE they all about? Do you have a page that is
better than yours that I could look at?

Gregory L. Hansen
2003-10-30 15:42:05 UTC
Post by Uncle Al
Post by Mike
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've seen your page, but it doesn't explain to me what affine theories
are all about. What ARE they all about? Do you have a page that is
better than yours that I could look at?
When you move around in a space with arbitrary curvature (all this started
with the mathematics of curved surfaces in three-dimensional Euclidean
space), you need to take derivatives of your basis vectors, and that's
basically what the connections are, indicated by Christoffel symbols. The
geodesic equation can be found by parallel transporting a tangent vector,
and it looks something like

a_i = -{i,jk} v_j v_k

where a_i is the i'th component of the acceleration, and v_j and v_k are
components of velocity, and the summation convention is used.

A metric is a rule for finding distances between points along a path, it's
sort of a generalization of Pythagorean's theorem. A metric space is an
affine space where a metric can be defined, and when the metric is
defined you can relate that to the Christoffel symbols. So I presume an
affine theory is one with a manifold that doesn't have a metric. But I've
been having trouble working up a physical picture of that.
--
"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
2003-11-01 01:51:59 UTC
Post by Gregory L. Hansen
A metric is a rule for finding distances between points along a path, it's
sort of a generalization of Pythagorean's theorem. A metric space is an
affine space where a metric can be defined, and when the metric is
defined you can relate that to the Christoffel symbols. So I presume an
affine theory is one with a manifold that doesn't have a metric. But I've
been having trouble working up a physical picture of that.
No, none of this is outside Riemannian geometry.

First of all, connection doesn't mean Christoffel symbol, in general
in differntial geometry. Connections and metrics are completely
independent structures that a manifold can have. And connection
is also a MUCH more general concept that applies also to gauge groups.

The best way to understand the indices of the usual connection
Gamma_{mn}^p
are:
the _m is a spacetime index
the _n^p is the index for a transformation matric.
This gives you a GL(n) connection (and, when the connection
leaves the covariant derivative of the metric 0, an O(n) connection).
The corresponding 1-form is:
Omega^p_n = Gamma_{mn}^p dx^m
[summation convention used], which is a O(n)-valued 1-form.

You can also have connections, say, for groups like U(1), T(n)
the n-dimensional translation group (as here), SU(2), SU(3), etc.

Likewise, "curvature" is a property of connections, not of
the spacetime or metric, itself and every type of connection can
have its own curvature (or field strength). The corresponding
curvature 2-form is:
Theta = d Omega + Omega ^ Omegs.
So, for an ordinary connection, this gives you the O(n)-valued
2-form:
Theta^a_c
= d Omega^a_c + Omega^a_b ^ Omega^b_c
= 1/2 (@_m Gamma_{nc}^a - @_n Gamma_{mc}^a
+ Gamma_{mb}^a Gamma_{nc}^b - Gamma_{nb}^a Gamma_{mc}^b) dx^m ^ dx^n
whose coefficients are just the Riemann Curvature components
Theta^a_c = 1/2 R^a_{c mn} dx^m ^ dx^n

The T(4)-connection B_m^a has indices _m for spacetime index
and ^a the T(4) index (labelling the T(4) generators P_a).

A SEPARATE Levi-Civita torsion-free connection (i.e. Christoffel
symbol) resides in this affine space, which has 0 curvature.

The curvature of the T(4)-connection yields the equivalent of
a torsion, which can be added to the Levi-Civita connection to
give you a connection which is not torsion-free.

The curvature of the Levi-Civita connection is 0 in this theory.

So, it's just ordinary Minkowski space with a connection that
consists of a Levi-Civita part, plus a non-zero torsion. You
can equally well consider any combination of the two, including
the one Uncle Al didn't list ...

No torsion, no curvature: Locally Minkowski
No torsion, curvature: Ordinary Riemannian geometry
Torsion, no curvature: Locally Minkowski with an affine connection
Torsion, curvature: Riemannian geometry with an affine connection