| Topic: |
Science > Physics |
| User: |
"mathlover" |
| Date: |
08 Nov 2006 02:15:59 AM |
| Object: |
the symmetric lower indices of Christoffel symbol |
Dear all,
I'm self-studying general relativity and I read from the book that
\Gamma^i_{jk} = \Gamma^i_{kj} where the indices are from 0 to 3.
But I read from the Wikipedia that if the space had torsion, then
the symmetry of the Christoffel symbol may be broken.
http://en.wikipedia.org/wiki/Christoffel_symbols
My problem is, I can't figure out where the proof is wrong in my GR
book(Schutz) if the torsion were added in the space.
I present the arguments of the book below:
In Cartesian coordinates, we have:
\Phi_{,\beta, \alpha} = \Phi_{,\alpha, \beta} since the partial
derivatives commute.
But if a tensor is symmetric in one basis it is symmetric in all bases.
Therefore:
\Phi_{,\beta ;\alpha} = \Phi_{,\alpha ;\beta} (where the lower index
preceded with a comma means covariant derivative.) in any basis. Using
the definition of covariant derivative, we have:
\Phi_{,\beta, \alpha} - \Phi_{,\mu}\Gamma^\mu_{\beta\alpha} =
\Phi_{,\alpha, \beta} - \Phi_{,\mu}\Gamma^\mu_{\alpha\beta}
But \Phi_{,\beta, \alpha} = \Phi_{,\alpha, \beta} , so we have
\Gamma^\mu_{\alpha\beta} = \Gamma^\mu_{\beta\alpha}
Thanks in advance! Sincerely
.
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| User: "" |
|
| Title: Re: the symmetric lower indices of Christoffel symbol |
08 Nov 2006 04:53:35 PM |
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mathlover <CPPandC@gmail.com> wrote:
Dear all,
I'm self-studying general relativity and I read from the book that
\Gamma^i_{jk} = \Gamma^i_{kj} where the indices are from 0 to 3.
But I read from the Wikipedia that if the space had torsion, then
the symmetry of the Christoffel symbol may be broken.
http://en.wikipedia.org/wiki/Christoffel_symbols
My problem is, I can't figure out where the proof is wrong in my GR
book(Schutz) if the torsion were added in the space.
I present the arguments of the book below:
In Cartesian coordinates, we have:
\Phi_{,\beta, \alpha} = \Phi_{,\alpha, \beta} since the partial
derivatives commute.
But if a tensor is symmetric in one basis it is symmetric in all bases.
Therefore:
\Phi_{,\beta ;\alpha} = \Phi_{,\alpha ;\beta} (where the lower index
preceded with a comma means covariant derivative.) in any basis.
This assumes that Cartesian coordinates exist, and that the covariant
derivative is equal to the partial derivative in those coordinates.
If spacetime is not flat, the first is not true; if the torsion tensor
is not zero, the second is not true even in a locally freely falling
frame.
(Nonzero torsion violates the principle of equivalence, in the sense that
even by going to a freely falling frame one cannot transform the connection
away completely -- the torsion tensor is an honest tensor, and is nonzero
in any frame if it is nonzero in one. But for many reasonable proposals
for coupling torsion to matter, the violation is too small to conflict
with observation.)
Steve Carlip
.
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| User: "mathlover" |
|
| Title: Re: the symmetric lower indices of Christoffel symbol |
08 Nov 2006 08:56:47 PM |
|
|
Excuse me, allow me reply an article to check if my concepts are
correct.
carlip-nospam@physics.ucdavis.edu =BCg=B9D=A1G
mathlover <CPPandC@gmail.com> wrote:
Dear all,
I'm self-studying general relativity and I read from the book that
\Gamma^i_{jk} =3D \Gamma^i_{kj} where the indices are from 0 to 3.
But I read from the Wikipedia that if the space had torsion, then
the symmetry of the Christoffel symbol may be broken.
http://en.wikipedia.org/wiki/Christoffel_symbols
My problem is, I can't figure out where the proof is wrong in my GR
book(Schutz) if the torsion were added in the space.
I present the arguments of the book below:
In Cartesian coordinates, we have:
\Phi_{,\beta, \alpha} =3D \Phi_{,\alpha, \beta} since the partial
derivatives commute.
But if a tensor is symmetric in one basis it is symmetric in all bases.
Therefore:
\Phi_{,\beta ;\alpha} =3D \Phi_{,\alpha ;\beta} (where the lower ind=
ex
preceded with a comma means covariant derivative.) in any basis.
This assumes that Cartesian coordinates exist, and that the covariant
derivative is equal to the partial derivative in those coordinates.
If spacetime is not flat, the first is not true; if the torsion tensor
is not zero, the second is not true even in a locally freely falling
frame.
The covariant derivative of a vector field is a type(1,1) tensor. A
tensor is a geometric object which is independent of coordinates. Thus,
a tensorial equation holds in every coordinates.
So, Schutz(The book) finds the equation \Phi_{,\beta, \alpha} =3D
\Phi_{,\alpha, \beta} in a locally inertial frame (it exists in torsion
free GR), then he can extend the equation to arbitrary coordinates, so
he has: \Phi_{,\beta ;\alpha} =3D \Phi_{,\alpha ;\beta}, then the
symmetry on the two lower indices is established.
But the key point is that the locally inertial frames do not exist with
torsion. So, Schutz(the book) can't write down the first equation, thus
the symmetry on the two lower indices is broken. Am I right?
(Nonzero torsion violates the principle of equivalence, in the sense that
even by going to a freely falling frame one cannot transform the connecti=
on
away completely -- the torsion tensor is an honest tensor, and is nonzero
in any frame if it is nonzero in one. But for many reasonable proposals
for coupling torsion to matter, the violation is too small to conflict
with observation.)
Thanks for your clear demonstration of the concepts of torsion. I
learned much. Thanks you!
=20
Steve Carlip
Sincerely
.
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| User: "" |
|
| Title: Re: the symmetric lower indices of Christoffel symbol |
09 Nov 2006 12:45:50 PM |
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mathlover <CPPandC@gmail.com> wrote:
Excuse me, allow me reply an article to check if my concepts are
correct.
[...]
The covariant derivative of a vector field is a type(1,1) tensor. A
tensor is a geometric object which is independent of coordinates. Thus,
a tensorial equation holds in every coordinates.
So, Schutz(The book) finds the equation \Phi_{,\beta, \alpha} =
\Phi_{,\alpha, \beta} in a locally inertial frame (it exists in torsion
free GR), then he can extend the equation to arbitrary coordinates, so
he has: \Phi_{,\beta ;\alpha} = \Phi_{,\alpha ;\beta}, then the
symmetry on the two lower indices is established.
But the key point is that the locally inertial frames do not exist with
torsion. So, Schutz(the book) can't write down the first equation, thus
the symmetry on the two lower indices is broken. Am I right?
Right.
(The definition of "locally inertial" that you're using is a standard
one. Note, though, that even if torsion is present, one can find a
local frame in which freely falling objects move in straight lines,
because the antisymmetric part of the connection doesn't come into the
geodesic equation. But other laws of physics will be different in a
freely falling frame than in flat space; there will be additional
couplings to torsion. One key question is whether such couplings would
be observable. You might look at http://arxiv.org/abs/gr-qc/0608121
for an analysis of one class of theories with torsion.)
Steve Carlip
.
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| User: "Ken S. Tucker" |
|
| Title: Re: the symmetric lower indices of Christoffel symbol |
09 Nov 2006 02:18:32 PM |
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mathlover wrote:
Dear all,
I'm self-studying general relativity and I read from the book that
\Gamma^i_{jk} = \Gamma^i_{kj} where the indices are from 0 to 3.
Yes
But I read from the Wikipedia that if the space had torsion, then
the symmetry of the Christoffel symbol may be broken.
http://en.wikipedia.org/wiki/Christoffel_symbols
IMO
It comes down to the covariant derivative of
the Kronecker Delta == &x^u/&x^v == d^u_v
using "&" as partial.
There's some mathematical philosophy applied,
for example, those partials specify the use of
independent dimensions, such that a point
sliding along one has no displacement along
any of the others and thus are so -called
independent axes, and that holds true even
when the axes are non-orthogonal.
From what I understand, the mathematicians
specified the use of independent axes and thus
d^u_v = {0,1}, with a zero covariant derivative in
tensor analysis. I'd regard that as the basis of
known tensor analysis, and a departure from
d^u_v = {0,1} would require a formidable new
mathematical structure.
The hard thing is that a non-symmetric Gamma
as you questioned above, might permit the covariant
derivative of the Kronecker to be non-zero, and
really *rocks-the-boat*, i.e. needs lots of math.
My problem is, I can't figure out where the proof is wrong in my GR
book(Schutz) if the torsion were added in the space.
Perhaps a pseudo-tensor shouldn't contain
invariant information, after-all it's a Christoffel.
Best Regards
and an excellent quest, I'd like to learn more
too.
Ken
PS: Steve Carlips study is always good.
I present the arguments of the book below:
In Cartesian coordinates, we have:
\Phi_{,\beta, \alpha} = \Phi_{,\alpha, \beta} since the partial
derivatives commute.
But if a tensor is symmetric in one basis it is symmetric in all bases.
Therefore:
\Phi_{,\beta ;\alpha} = \Phi_{,\alpha ;\beta} (where the lower index
preceded with a comma means covariant derivative.) in any basis. Using
the definition of covariant derivative, we have:
\Phi_{,\beta, \alpha} - \Phi_{,\mu}\Gamma^\mu_{\beta\alpha} =
\Phi_{,\alpha, \beta} - \Phi_{,\mu}\Gamma^\mu_{\alpha\beta}
But \Phi_{,\beta, \alpha} = \Phi_{,\alpha, \beta} , so we have
\Gamma^\mu_{\alpha\beta} = \Gamma^\mu_{\beta\alpha}
Thanks in advance! Sincerely
.
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| User: "FrediFizzx" |
|
| Title: Re: the symmetric lower indices of Christoffel symbol |
09 Nov 2006 03:59:08 PM |
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"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1163103512.442262.251270@b28g2000cwb.googlegroups.com...
mathlover wrote:
Dear all,
I'm self-studying general relativity and I read from the book that
\Gamma^i_{jk} = \Gamma^i_{kj} where the indices are from 0 to 3.
Yes
But I read from the Wikipedia that if the space had torsion, then
the symmetry of the Christoffel symbol may be broken.
http://en.wikipedia.org/wiki/Christoffel_symbols
IMO
It comes down to the covariant derivative of
the Kronecker Delta == &x^u/&x^v == d^u_v
using "&" as partial.
There's some mathematical philosophy applied,
for example, those partials specify the use of
independent dimensions, such that a point
sliding along one has no displacement along
any of the others and thus are so -called
independent axes, and that holds true even
when the axes are non-orthogonal.
From what I understand, the mathematicians
specified the use of independent axes and thus
d^u_v = {0,1}, with a zero covariant derivative in
tensor analysis. I'd regard that as the basis of
known tensor analysis, and a departure from
d^u_v = {0,1} would require a formidable new
mathematical structure.
The hard thing is that a non-symmetric Gamma
as you questioned above, might permit the covariant
derivative of the Kronecker to be non-zero, and
really *rocks-the-boat*, i.e. needs lots of math.
My problem is, I can't figure out where the proof is wrong in my GR
book(Schutz) if the torsion were added in the space.
Perhaps a pseudo-tensor shouldn't contain
invariant information, after-all it's a Christoffel.
Best Regards
and an excellent quest, I'd like to learn more
too.
Ken
PS: Steve Carlip's study is always good.
Yep, and I read that the Gravity Probe B results are supposed to be out
next April. Wonder if there will be any leakage of results before then?
;-)
http://einstein.stanford.edu/
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com
.
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| User: "Ken S. Tucker" |
|
| Title: Re: the symmetric lower indices of Christoffel symbol |
10 Nov 2006 12:51:00 PM |
|
|
FrediFizzx wrote:
"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1163103512.442262.251270@b28g2000cwb.googlegroups.com...
mathlover wrote:
Dear all,
I'm self-studying general relativity and I read from the book that
\Gamma^i_{jk} = \Gamma^i_{kj} where the indices are from 0 to 3.
Yes
But I read from the Wikipedia that if the space had torsion, then
the symmetry of the Christoffel symbol may be broken.
http://en.wikipedia.org/wiki/Christoffel_symbols
IMO
It comes down to the covariant derivative of
the Kronecker Delta == &x^u/&x^v == d^u_v
using "&" as partial.
There's some mathematical philosophy applied,
for example, those partials specify the use of
independent dimensions, such that a point
sliding along one has no displacement along
any of the others and thus are so -called
independent axes, and that holds true even
when the axes are non-orthogonal.
From what I understand, the mathematicians
specified the use of independent axes and thus
d^u_v = {0,1}, with a zero covariant derivative in
tensor analysis. I'd regard that as the basis of
known tensor analysis, and a departure from
d^u_v = {0,1} would require a formidable new
mathematical structure.
The hard thing is that a non-symmetric Gamma
as you questioned above, might permit the covariant
derivative of the Kronecker to be non-zero, and
really *rocks-the-boat*, i.e. needs lots of math.
My problem is, I can't figure out where the proof is wrong in my GR
book(Schutz) if the torsion were added in the space.
Perhaps a pseudo-tensor shouldn't contain
invariant information, after-all it's a Christoffel.
Best Regards
and an excellent quest, I'd like to learn more
too.
Ken
PS: Steve Carlip's study is always good.
Yep, and I read that the Gravity Probe B results are supposed to be out
next April. Wonder if there will be any leakage of results before then?
;-)
http://einstein.stanford.edu/
FrediFizzx
So far as I know the *naked* data is classified,
with the USA via NASA claiming a propriety use
of the data that is available on a preferential basis.
I think that policy prevents other groups doing
work on the data, that in science, would be helpful,
and saves time and money.
The experiment *basically* questions the relation
of GR to Mach's Principle, for a review see AE's
"The Meaning of Relativity" pg 100, and years on
Weinberg's "G&C" pg 240.
I figure "frame-dragging" is a CS artifact, maybe
time for us to write something up about that?
Regards
Ken S. Tucker
.
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