I've read that the metric tensor g_ab is symmetric. Can someone explain
to me why the metric tensor has to be symmetric? I tried to derive it
from the fact the metric tensor can be used to give the squared length
of a vector, but I got stuck with that way of proving the symmetry.

Amantine

Two vectors, A and B, the dot product A.B = B.A

If you consider a general (0,2) one form 'G' taking two vector arguments,
A=A_i, B=B_j such that G(A,B) = G_ij = A_i,B_j then its is a requirement
that G(A,B)=G(B,A) which is equivalent to saying G has to be symmetric

Schutz, A first Course in General Relativity, is a good source

Richard Miller

"Amantine" wrote in message
i.nl...
I've read that the metric tensor g_ab is symmetric. Can someone explain
to me why the metric tensor has to be symmetric? I tried to derive it
from the fact the metric tensor can be used to give the squared length
of a vector, but I got stuck with that way of proving the symmetry.

Amantine

richard miller wrote:
Two vectors, A and B, the dot product A.B = B.A

If you consider a general (0,2) one form 'G' taking two vector arguments,
A=A_i, B=B_j such that G(A,B) = G_ij = A_i,B_j then its is a requirement
that G(A,B)=G(B,A) which is equivalent to saying G has to be symmetric

Schutz, A first Course in General Relativity, is a good source

Richard Miller

"Amantine" wrote in message
i.nl...

I've read that the metric tensor g_ab is symmetric. Can someone explain
to me why the metric tensor has to be symmetric? I tried to derive it
from the fact the metric tensor can be used to give the squared length
of a vector, but I got stuck with that way of proving the symmetry.

Amantine




Thanks for the answer. I see now what I did wrong.

Amantine

Amantine wrote in message li.nl...
richard miller wrote:
Two vectors, A and B, the dot product A.B = B.A

If you consider a general (0,2) one form 'G' taking two vector arguments,
A=A_i, B=B_j such that G(A,B) = G_ij = A_i,B_j then its is a requirement
that G(A,B)=G(B,A) which is equivalent to saying G has to be symmetric

Schutz, A first Course in General Relativity, is a good source

Richard Miller

"Amantine" wrote in message
i.nl...

I've read that the metric tensor g_ab is symmetric. Can someone explain
to me why the metric tensor has to be symmetric? I tried to derive it
from the fact the metric tensor can be used to give the squared length
of a vector, but I got stuck with that way of proving the symmetry.

Amantine




Thanks for the answer. I see now what I did wrong.

Amantine

If you're serious about relativity and metrics, don't
become to ingrained with the idea that g_ab is symmetric.
In the study of simple static curved surfaces that
Reimann uses the g_ab is always symmetric as far as I
know. But relativity deals with motion on surfaces and
that subltely changes the logical application of g_ab.

Einstein investigated asymmetrical metrics and more
recently Moffat, it's not cranky although it is difficult.

In physics, the actualization of measurement uses the
effect of photons on charged particles, and that introduces
aspects of reality into the conventional tensor analysis
that requires respect for tensor dynamics. The conventional
tensor analysis was designed for static situations and was
rather crudely adopted to physics by GR.

Regards
Ken S. Tucker

PS: Suppose we choose the differential dx^u to describe
displacement. An immediate problem occurs when interfacing
to Quantum Theory and thus ElectroMagnetic theory. QT allows
only discrete lengths, but relativity (PoR) demands relative
motion.

Ok let's postulate

s^2 = g^uv x_u x_v is for finites (QT)

ds^2 = g_uv dx^u dx^v is for infinitesmals (PoR)


In a weak field limit and in the absence of velocity, dynamic
issues vanish so let's use the Kronecker delta as a metric and
set,

x^2 = d^uv x_u x_v }
} Orthogonal relations
dx^2 = d_uv dx^u dx^v }

In these last two equations the covariant/contravariant
notation is demonstrative since

d^uv == d_uv == d^u_v = Kronecker delta

Set S = X + Q , (physically Q is charge)

where X.Q=0 then define

S.S = X.X + Q.Q = s^2 = x^2 + q^2

where q^2 = +/- 1.

Use an EM potential*distance like O^u x_u to define

q = O^u x_u then

s^2 = x^2 + q^2 = d^uv x_u x_v + O^u O^v x_u x_v

= g^uv x_u x_v providing,

g^uv = d^uv + O^u O^v.

Physically suppose s^2 is the distance from charge "a" to "b"
and the potentials of those charges are A^u and B^v then,

g^uv = d^uv + A^u B^v

Here is where the physical (relativistic) basis for
the asymmetry of the metric tensor g^uv becomes apparent,
and necessary for applications to EM and QT. The term

A^u B^v = - A'^v B'^u

is a common asymmetry in PoR and EM. In simple terms
you and I are in opposite directions when sitting and
talking, likewise the relative direction of communication
relating charges "a" and "b" are relatively opposite and
a metric used to define that physical distance must
necessarily include that relativity, because neither
charge "a" or "b" is able to define distance, only the
relative relationship of the two can do that.
Within that common relation, neither "a" nor "b" is
preferred, hence the metric used to define the distance
includes both asymmetrically.
KST

Amantine wrote in message li.nl...
I've read that the metric tensor g_ab is symmetric. Can someone explain
to me why the metric tensor has to be symmetric? I tried to derive it
from the fact the metric tensor can be used to give the squared length
of a vector, but I got stuck with that way of proving the symmetry.

Amantine


AFAIK, there is no particular reason why the metric tensor absolutely
has to be symmetric. In fact, Einstein generalized the metric tensor
to a nonsymmetric form in one of his many attempts at constructing a
unified field theory of EM and inertial forces. But, although his
nonsymmetric theory failed as far as observable consequences, the math
behind it is still consistent, though far removed from traditional
differential geometries.

Igor:
Amantine wrote in message news:
I've read that the metric tensor g_ab is symmetric. Can someone explain
to me why the metric tensor has to be symmetric? I tried to derive it
from the fact the metric tensor can be used to give the squared length
of a vector, but I got stuck with that way of proving the symmetry.

Amantine


AFAIK, there is no particular reason why the metric tensor absolutely
has to be symmetric. In fact, Einstein generalized the metric tensor
to a nonsymmetric form in one of his many attempts at constructing a
unified field theory of EM and inertial forces. But, although his
nonsymmetric theory failed as far as observable consequences, the math
behind it is still consistent, though far removed from traditional
differential geometries.

In general relativity, the metric tensor has to be symmetric. The
metric has to be locally the lorentz metric, which is real and diagonal.
In order to have real eigenvalues and be anti-symmetric, the metric
would have to hermitian to be diagnolizable into real eigenvalues
and the transformation would be unitary, not orthogonal.

(Bilge) wrote in message ...

Igor:
AFAIK, there is no particular reason why the metric tensor absolutely
has to be symmetric. In fact, Einstein generalized the metric tensor
to a nonsymmetric form in one of his many attempts at constructing a
unified field theory of EM and inertial forces. But, although his
nonsymmetric theory failed as far as observable consequences, the math
behind it is still consistent, though far removed from traditional
differential geometries.

In general relativity, the metric tensor has to be symmetric. The
metric has to be locally the lorentz metric, which is real and diagonal.

Not really. The metric you're talking about cannot extend
beyond a point, however the metric was designed for
calculating finite distances, recall surveying, and needs
to hold over finite distances, such as the curvature of
the Earth.
Consider surveying Earth from the north pole to the equator
using lasers, and taking into account Coriollos acceleration,
will the path from the north pole to lat/long (0,0) on the
equator be the same as the path from (0,0) to the north pole?

Simplify to a rotating disk. Use a laser from the origin
and reflected from the circumference, well the paths out
and in will be different mapped by observers everywhere
on the disk.
Now reverse the direction of disk rotation, the laser
follows an asymmetrical path, that is, the original out/
in path becomes the in/out path.

((I learned this while intoxicated trying to play a
record backward to hear "Freddy is the devil", and then
examining the scratchs on the record))

I think Einstein described that geometry using

g^12 = - g^21

with g^12 == (dx/dt)*y/r - (dy/dt)*x/r.

When applied to unified fields, place a charge on the
circumference of the disk, and the rotation generates
a Magnetic field B(z) at the center of the disk. When
the rotation of the disk is reversed the magnetic field
is -B(z). Then define B(z) = F^12 = -F^21. Of course
the magnetic field at the origin is a consequence of
EM force/radiation, and should behave as light does.
Regards
Ken S. Tucker






















In order to have real eigenvalues and be anti-symmetric, the metric
would have to hermitian to be diagnolizable into real eigenvalues
and the transformation would be unitary, not orthogonal.

Ken S. Tucker:
(Bilge) wrote:

Igor:
AFAIK, there is no particular reason why the metric tensor absolutely
has to be symmetric. In fact, Einstein generalized the metric tensor
to a nonsymmetric form in one of his many attempts at constructing a
unified field theory of EM and inertial forces. But, although his
nonsymmetric theory failed as far as observable consequences, the math
behind it is still consistent, though far removed from traditional
differential geometries.

In general relativity, the metric tensor has to be symmetric. The
metric has to be locally the lorentz metric, which is real and diagonal.

Not really.

Yes, it does, or it isn't general relativity.

(Bilge) wrote in message ...
Ken S. Tucker:
(Bilge) wrote:

Igor:
AFAIK, there is no particular reason why the metric tensor absolutely
has to be symmetric. In fact, Einstein generalized the metric tensor
to a nonsymmetric form in one of his many attempts at constructing a
unified field theory of EM and inertial forces. But, although his
nonsymmetric theory failed as far as observable consequences, the math
behind it is still consistent, though far removed from traditional
differential geometries.

In general relativity, the metric tensor has to be symmetric. The
metric has to be locally the lorentz metric, which is real and diagonal.

Not really.

Yes, it does, or it isn't general relativity.

Is GR valid on a rotating disk?

(Bilge) wrote in message ...
Igor:
Amantine wrote in message news:
I've read that the metric tensor g_ab is symmetric. Can someone explain
to me why the metric tensor has to be symmetric? I tried to derive it
from the fact the metric tensor can be used to give the squared length
of a vector, but I got stuck with that way of proving the symmetry.

Amantine


AFAIK, there is no particular reason why the metric tensor absolutely
has to be symmetric. In fact, Einstein generalized the metric tensor
to a nonsymmetric form in one of his many attempts at constructing a
unified field theory of EM and inertial forces. But, although his
nonsymmetric theory failed as far as observable consequences, the math
behind it is still consistent, though far removed from traditional
differential geometries.

In general relativity, the metric tensor has to be symmetric. The
metric has to be locally the lorentz metric, which is real and diagonal.
In order to have real eigenvalues and be anti-symmetric, the metric
would have to hermitian to be diagnolizable into real eigenvalues
and the transformation would be unitary, not orthogonal.

You make a very good point there. As I seem to recall, Einstein's
main disappointment here was that the anti-symmetric part of the Ricci
tensor could not be put into the form of a curl. Frankly, I don't
ever recall any account of him addressing the issue of
diagonalizablity, but that would appear to be a primary objection.
I'm also aware that he proposed a theory involving a complex Hermitian
metric, but gave up when he realized that it acted on the real vector
components in a reducible way.