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Igor
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 Posted: Sun Oct 05, 2008 11:25 am Post subject: Question about the contracted Christoffel Symbols |
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Are the contracted Christoffel Symbols vectors? A quick glance at the
transformation properties of the Christoffel Symbols seems to suggest
that the non-homogeneous part of the transformation would vanish under
contraction by simply applying the chain rule. However, I have not
been able to find a verification of this in any textbooks or online.
Moreover, by applying the transformation rules to the gradient of the
log of the metric tensor, it appears that there will be an additional
term related to the derivative of the Jacobian determinant. But I>m
unable to determine whether this term vanishes or not. Any help would
be tremendously appreciated. |
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a student
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| Posted: Tue Oct 07, 2008 3:14 pm Post subject: Re: Question about the contracted Christoffel Symbols |
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On Oct 5, 10:25 pm, Igor <thoov...@excite.com> wrote:
[quote]Are the contracted Christoffel Symbols vectors? A quick glance at the
transformation properties of the Christoffel Symbols seems to suggest
that the non-homogeneous part of the transformation would vanish under
contraction by simply applying the chain rule. However, I have not
been able to find a verification of this in any textbooks or online.
Moreover, by applying the transformation rules to the gradient of the
log of the metric tensor, it appears that there will be an additional
term related to the derivative of the Jacobian determinant. But I>m
unable to determine whether this term vanishes or not. Any help would
be tremendously appreciated.
[/quote]
The easiest way to see that it cannot be a vector is to note that the
covariant
divergence of an arbitrary vector V^a is given by
(V^a);a = (V^a),a + G_a V^a ,
where G_a is the contracted Christoffel symbol. If G_a was a vector,
it
would immediately follow from this formula that the first term on the
righthand side - (V^a),a - was a scalar, which is not the case (as is
easily seen by writing down how it transforms). Therefore
the derivative of the logarithm of the Jacobian determinant does not
vanish in general.
i |
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Igor Khavkine
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| Posted: Tue Oct 07, 2008 3:14 pm Post subject: Re: Question about the contracted Christoffel Symbols |
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On Oct 5, 7:25 am, Igor <thoov...@excite.com> wrote:
[quote]Are the contracted Christoffel Symbols vectors? A quick glance at the
transformation properties of the Christoffel Symbols seems to suggest
that the non-homogeneous part of the transformation would vanish under
contraction by simply applying the chain rule. However, I have not
been able to find a verification of this in any textbooks or online.
Moreover, by applying the transformation rules to the gradient of the
log of the metric tensor, it appears that there will be an additional
term related to the derivative of the Jacobian determinant. But I>m
unable to determine whether this term vanishes or not. Any help would
be tremendously appreciated.
[/quote]
The identity used to compute the contracted Christoffel symbol is
G_b = G^a_ab = @_b log sqrt(-g),
where @_b is the partial coordinate derivative and g is the
determinant of the metric tensor in the same coordinate system.
If G_b is a vector, it would transform as G_b -> J^b_b' G_b under a
coordinate transformation whose Jacobian is J^a_b. On the other hand,
the determinant g also transforms as g -> g J^2, where J is now the
determinant of J^a_b. This introduces an extra term of the form J^b_b'
@_b log |J|, which has no reason to vanish and is not captured by the
standard vector transformation rule. Therefore, G_b is not a vector.
The underlying reason is that sqrt(-g) is not a mere function on the
manifold, it is a density, which transforms with with a change in
coordinates.
Hope this helps.
Igor |
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Igor
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| Posted: Thu Oct 09, 2008 5:51 am Post subject: Re: Question about the contracted Christoffel Symbols |
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On Oct 7, 11:14 am, Igor Khavkine <igor...@gmail.com> wrote:
[quote]On Oct 5, 7:25 am, Igor <thoov...@excite.com> wrote:
Are the contracted Christoffel Symbols vectors? A quick glance at the
transformation properties of the Christoffel Symbols seems to suggest
that the non-homogeneous part of the transformation would vanish under
contraction by simply applying the chain rule. However, I have not
been able to find a verification of this in any textbooks or online.
Moreover, by applying the transformation rules to the gradient of the
log of the metric tensor, it appears that there will be an additional
term related to the derivative of the Jacobian determinant. But I>m
unable to determine whether this term vanishes or not. Any help would
be tremendously appreciated.
The identity used to compute the contracted Christoffel symbol is
G_b = G^a_ab = @_b log sqrt(-g),
where @_b is the partial coordinate derivative and g is the
determinant of the metric tensor in the same coordinate system.
If G_b is a vector, it would transform as G_b -> J^b_b' G_b under a
coordinate transformation whose Jacobian is J^a_b. On the other hand,
the determinant g also transforms as g -> g J^2, where J is now the
determinant of J^a_b. This introduces an extra term of the form J^b_b'
@_b log |J|, which has no reason to vanish and is not captured by the
standard vector transformation rule. Therefore, G_b is not a vector.
The underlying reason is that sqrt(-g) is not a mere function on the
manifold, it is a density, which transforms with with a change in
coordinates.
Hope this helps.
Igor
[/quote]
Thanks. I played around with it a bit and discovered that even in a
two dimensional system, there will still be the nonhomogeneous term
involving the derivative of the Jacobian determinant. By narrowing
the group, you can probably make it transform like a vector, but it
won>t behave that way under general orthogonal transformations. |
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Guest
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| Posted: Sun Oct 19, 2008 3:52 pm Post subject: Re: Question about the contracted Christoffel Symbols |
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On 9 Okt, 07:51, Igor <thoov...@excite.com> wrote:
[quote]Thanks. =A0I played around with it a bit and discovered that even in a
two dimensional system, there will still be the nonhomogeneous term
involving the derivative of the Jacobian determinant. =A0By narrowing
the group, you can probably make it transform like a vector, but it
won>t behave that way under general orthogonal transformations.-
[/quote]
Poincare group. If it transforms as a vector, it can consistently be set
to zero. But then you don>t need to covariantize derivatives anyway. |
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Melroy
Guest
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| Posted: Sun Oct 19, 2008 8:05 pm Post subject: Re: Question about the contracted Christoffel Symbols |
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On Oct 9, 12:51 am, Igor <thoov...@excite.com> wrote:
[quote]On Oct 7, 11:14 am, Igor Khavkine <igor...@gmail.com> wrote:
On Oct 5, 7:25 am, Igor <thoov...@excite.com> wrote:
Are the contractedChristoffelSymbolsvectors? A quick glance at the
transformation properties of theChristoffelSymbolsseems to suggest
that the non-homogeneous part of the transformation would vanish under
contraction by simply applying the chain rule. However, I have not
been able to find a verification of this in any textbooks or online.
Moreover, by applying the transformation rules to the gradient of the
log of the metric tensor, it appears that there will be an additional
term related to the derivative of the Jacobian determinant. But I>m
unable to determine whether this term vanishes or not. Any help would
be tremendously appreciated.
The identity used to compute the contractedChristoffelsymbol is
G_b = G^a_ab = @_b log sqrt(-g),
where @_b is the partial coordinate derivative and g is the
determinant of the metric tensor in the same coordinate system.
If G_b is a vector, it would transform as G_b -> J^b_b' G_b under a
coordinate transformation whose Jacobian is J^a_b. On the other hand,
the determinant g also transforms as g -> g J^2, where J is now the
determinant of J^a_b. This introduces an extra term of the form J^b_b'
@_b log |J|, which has no reason to vanish and is not captured by the
standard vector transformation rule. Therefore, G_b is not a vector.
The underlying reason is that sqrt(-g) is not a mere function on the
manifold, it is a density, which transforms with with a change in
coordinates.
Hope this helps.
Igor
Thanks. I played around with it a bit and discovered that even in a
two dimensional system, there will still be the nonhomogeneous term
involving the derivative of the Jacobian determinant. By narrowing
the group, you can probably make it transform like a vector, but it
won>t behave that way under general orthogonal transformations.
[/quote]
A followup question. What was the original motivation for introducing
Christoffel symbols? I presume they were invented much before
GE was thought of, right? |
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Pierre Asselin
Guest
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| Posted: Mon Oct 20, 2008 3:04 am Post subject: Re: Question about the contracted Christoffel Symbols |
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Melroy <melroysoares@hotmail.com> wrote:
[quote]A followup question. What was the original motivation for introducing
Christoffel symbols? I presume they were invented much before
GE was thought of, right?
[/quote]
There is a short historical passage at:
http://en.wikipedia.org/wiki/Affine_connection#Motivation_from_tensor_calculus
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