Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1
26.8 THE TENSORSδijANDijk

Write the following as contracted Cartesian tensors:a·b,∇^2 φ,∇×v,∇(∇·v),∇×(∇×v),
(a×b)·c.

The corresponding (contracted) tensor expressions are readily seen to be as follows:


a·b=aibi=δijaibj,

∇^2 φ=

∂^2 φ
∂xi∂xi

=δij

∂^2 φ
∂xi∂xj

,


(∇×v)i=ijk

∂vk
∂xj

,


[∇(∇·v)]i=


∂xi

(


∂vj
∂xj

)


=δjk

∂^2 vj
∂xi∂xk

,


[∇×(∇×v)]i=ijk


∂xj

(


klm

∂vm
∂xl

)


=ijkklm

∂^2 vm
∂xj∂xl

,


(a×b)·c=δijcijklakbl =iklciakbl.

An important relationship between the-andδ- tensors is expressed by the

identity


ijkklm=δilδjm−δimδjl. (26.30)

To establish the validity of this identity between two fourth-order tensors (the


LHS is a once-contracted sixth-order tensor) we consider the various possible


cases.


The RHS of (26.30) has the values

+1 ifi=landj=m=i, (26.31)

−1ifi=mandj=l=i, (26.32)

0 for any other set of subscript valuesi, j, l, m. (26.33)

In each product on the LHSkhas the same value in both factors and for a


non-zero contribution none ofi, l, j, mcan have the same value ask. Since there


are only three values, 1, 2 and 3, that any of the subscripts may take, the only


non-zero possibilities arei=landj=mor vice versa but not all four subscripts


equal (since then eachfactor is zero, as it would be ifi=jorl=m). This


reproduces (26.33) for the LHS of (26.30) and also the conditions (26.31) and


(26.32). The values in (26.31) and (26.32) are also reproduced in the LHS of


(26.30) since


(i) ifi=landj=m,ijk=lmk=klmand, whetherijkis +1 or−1, the
product of the two factors is +1; and
(ii) ifi=mandj=l,ijk=mlk=−klmand thus the productijkklm(no
summation) has the value−1.

This concludes the establishment of identity (26.30).

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