26.8 The tensorsδijandijk
Nsubscripts, with an arbitraryNth-order tensor (i.e. one having independently
variable components) and determining whether the result is a scalar.
Use the quotient law to show that the elements ofT, equation (26.24), are the components
of a second-order tensor.The outer productxixjis a second-order tensor. Contracting this with theTijgiven in
(26.24) we obtain
Tijxixj=x^22 x^21 −x 1 x 2 x 1 x 2 −x 1 x 2 x 2 x 1 +x^21 x^22 =0,which is clearly invariant (a zeroth-order tensor). Hence by the quotient theoremTijmust
also be a tensor.
26.8 The tensorsδijandijkIn many places throughout this book we have encountered and used the two-
subscript quantityδijdefined by
δij={
1ifi=j,
0otherwise.Let us now also introduce the three-subscriptLevi–Civita symbolijk, the value
of which is given by
ijk=
+1 ifi, j, kis an even permutation of 1, 2 ,3,−1ifi, j, kis an odd permutation of 1, 2 ,3,
0otherwise.We will now show thatδijandijkare respectively the components of a second-and a third-order Cartesian tensor. Notice that the coordinatesxido not appear
explicitly in the components of these tensors, their components consisting entirely
of 0 and 1.
In passing, we also note thatijkis totally antisymmetric, i.e. it changes signunder the interchange of any pair of subscripts. In factijk, or any scalar multiple
of it, is theonlythree-subscript quantity with this property.
Treatingδijfirst, the proof that it is a second-order tensor is straightforwardsince if, from (26.16), we consider the equation
δkl′=LkiLljδij=LkiLli=δkl,we see that the transformation ofδijgenerates the same expression (a pattern
of 0’s and 1’s) as does the definition ofδij′in the transformed coordinates. Thus
δijtransforms according to the appropriate tensor transformation law and is
therefore a second-order tensor.
Turning now toijk, we have to consider the quantity′lmn=LliLmjLnkijk. (26.28)