Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

TENSORS


Using (26.76), deduce the way in which the quantitiesΓkijtransform under a general
coordinate transformation, and hence show that they do not form the components of a
third-order tensor.

In a new coordinate system


Γ′kij=e′k·

∂e′i
∂u′j

,


but from (26.69) and (26.67) respectively we have, on reversing primed and unprimed
variables,


e′k=

∂u′k
∂un

en and e′i=

∂ul
∂u′i

el.

Therefore in the new coordinate system the quantities Γ′kijare given by


Γ′kij=

∂u′k
∂un

en·


∂u′j

(


∂ul
∂u′i

el

)


=


∂u′k
∂un

en·

(


∂^2 ul
∂u′j∂u′i

el+

∂ul
∂u′i

∂el
∂u′j

)


=


∂u′k
∂un

∂^2 ul
∂u′j∂u′i

en·el+

∂u′k
∂un

∂ul
∂u′i

∂um
∂u′j

en·

∂el
∂um

=


∂u′k
∂ul

∂^2 ul
∂u′j∂u′i

+


∂u′k
∂un

∂ul
∂u′i

∂um
∂u′j

Γnlm, (26.78)

where in the last line we have used (26.76) and the reciprocity relationen·el=δnl.From
(26.78), because of the presence of the first term on the right-hand side, we conclude
immediately that the Γkijdo not form the components of a third-order tensor.


In a given coordinate system, in principle we may calculate the Γkij using

(26.76). In practice, however, it is often quicker to use an alternative expression,


which we now derive, for the Christoffel symbol in terms of the metric tensorgij


and its derivatives with respect to the coordinates.


Firstly we note that the Christoffel symbol Γkijis symmetric with respect to

the interchange of its two subscriptsiandj. This is easily shown: since


∂ei
∂uj

=

∂^2 r
∂uj∂ui

=

∂^2 r
∂ui∂uj

=

∂ej
∂ui

,

it follows from (26.75) that Γkijek=Γkjiek. Taking the scalar product witheland


using the reciprocity relationek·el=δklgives immediately that


Γlij=Γlji.

To obtain an expression for Γkijwe then usegij=ei·ej and consider the

derivative


∂gij
∂uk

=

∂ei
∂uk

·ej+ei·

∂ej
∂uk
=Γlikel·ej+ei·Γljkel

=Γlikglj+Γljkgil, (26.79)
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