Mathematical Methods for Physics and Engineering - Matematica.NET

TENSORS◮Using (26.76), deduce the way in which the quantities Γ k ijtransform under a generalcoordinate transformation, and hence show that they do not form the components of athird-order tensor.In a new coordinate systemΓ ′k ij = e ′k · ∂e′ i∂u , ′jbut from (26.69) and (26.67) respectively we have, on reversing primed and unprimedvariables,e ′k = ∂u′k∂u n en and e ′ i = ∂ul∂u e l.′iTherefore in the new coordinate system the quantities Γ ′k ij are given byΓ ′k ij = ∂u′k∂u n en ·= ∂u′k∂u n en ·( )∂ ∂ul∂u ′j ∂u e ′i l( )∂ 2 u l∂u ′j ∂u e ′i l + ∂ul ∂e l∂u ′i ∂u ′j= ∂u′k ∂ 2 u le n · e∂u n ∂u ′j ∂u ′i l + ∂u′k ∂u l ∂u me n · ∂e l∂u n ∂u ′i ∂u ′j ∂u m= ∂u′k ∂ 2 u l+ ∂u′k ∂u l ∂u m∂u l ∂u ′j ∂u ′i ∂u n ∂u ′i ∂u ′j Γn lm, (26.78)where in the last line we have used (26.76) and the reciprocity relation e n · e l = δl n.From(26.78), because of the presence of the first term on the right-hand side, we concludeimmediately that the Γ k ijdo not form the components of a third-order tensor. ◭In a given coordinate system, in principle we may calculate the Γ k ijusing(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 tensor g ijand its derivatives with respect to the coordinates.Firstly we note that the Christoffel symbol Γ k ijis symmetric with respect tothe interchange of its two subscripts i and j. This is easily shown: since∂e i∂u j =∂2 r∂u j ∂u i =∂2 r∂u i ∂u j = ∂e j∂u i ,it follows from (26.75) that Γ k ij e k =Γ k ji e k. Taking the scalar product with e l andusing the reciprocity relation e k · e l = δk l gives immediately thatΓ l ij =Γ l ji.To obtain an expression for Γ k ijwe then use g ij = e i · e j and consider thederivative∂g ij∂u k = ∂e i∂u k · e j + e i · ∂e j∂u k=Γ l ike l · e j + e i · Γ l jke l=Γ l ikg lj +Γ l jkg il , (26.79)966