Question about the contracted Christoffel Symbols
Igor
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.
a student
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
Igor Khavkine
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
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.
thomas_l...@hotmail.com
> Thanks. =A0I played around with it a bit and discovered that even in a
> two dimensional system, there will still be the nonhomogeneous term
> the group, you can probably make it transform like a vector, but it
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.
Melroy
> On Oct 7, 11:14 am, Igor Khavkine <igor...@gmail.com> wrote:
>
>
>
> > On Oct 5, 7:25 am, Igor <thoov...@excite.com> wrote:
>
> > > 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.
>
>
> > 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.
A followup question. What was the original motivation for introducing
Christoffel symbols? I presume they were invented much before
GE was thought of, right?
Pierre Asselin
> A followup question. What was the original motivation for introducing
> Christoffel symbols? I presume they were invented much before
> GE was thought of, right?
There is a short historical passage at:
http://en.wikipedia.org/wiki/Affine_connection#Motivation_from_tensor_calculus
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