The tensors ij and

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The tensors ij and
26.8 THE TENSORS δij AND ijk
N subscripts, with an arbitrary Nth-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 of T, equation (26.24), are the components
of a second-order tensor.
The outer product xi xj is a second-order tensor. Contracting this with the Tij given in
(26.24) we obtain
Tij xi xj = x22 x21 − x1 x2 x1 x2 − x1 x2 x2 x1 + x21 x22 = 0,
which is clearly invariant (a zeroth-order tensor). Hence by the quotient theorem Tij must
also be a tensor. 26.8 The tensors δij and ijk
In many places throughout this book we have encountered and used the twosubscript quantity δij defined by
1 if i = j,
δij =
0 otherwise.
Let us now also introduce the three-subscript Levi–Civita symbol ijk , the value
of which is given by
+1 if i, j, k is an even permutation of 1, 2, 3,
ijk = −1 if i, j, k is an odd permutation of 1, 2, 3,
We will now show that δij and ijk are respectively the components of a secondand a third-order Cartesian tensor. Notice that the coordinates xi do not appear
explicitly in the components of these tensors, their components consisting entirely
of 0 and 1.
In passing, we also note that ijk is totally antisymmetric, i.e. it changes sign
under the interchange of any pair of subscripts. In fact ijk , or any scalar multiple
of it, is the only three-subscript quantity with this property.
Treating δij first, the proof that it is a second-order tensor is straightforward
since if, from (26.16), we consider the equation
δkl = Lki Llj δij = Lki Lli = δkl ,
we see that the transformation of δij generates the same expression (a pattern
of 0’s and 1’s) as does the definition of δij in the transformed coordinates. Thus
δij transforms according to the appropriate tensor transformation law and is
therefore a second-order tensor.
Turning now to ijk , we have to consider the quantity
lmn = Lli Lmj Lnk ijk .
Let us begin, however, by noting that we may use the Levi–Civita symbol to
write an expression for the determinant of a 3 × 3 matrix A,
|A|lmn = Ali Amj Ank ijk ,
which may be shown to be equivalent to the Laplace expansion (see chapter 8).§
Indeed many of the properties of determinants discussed in chapter 8 can be
proved very efficiently using this expression (see exercise 26.9).
Evaluate the determinant of the matrix
A= 3
0 .
Setting l = 1, m = 2 and n = 3 in (26.29) we find
|A| = ijk A1i A2j A3k
= (2)(4)(1) − (2)(0)(−2) − (1)(3)(1) + (−3)(3)(−2)
+ (1)(0)(1) − (−3)(4)(1) = 35,
which may be verified using the Laplace expansion method. We can now show that the ijk are in fact the components of a third-order tensor.
Using (26.29) with the general matrix A replaced by the specific transformation
matrix L, we can rewrite the RHS of (26.28) in terms of |L|
lmn = Lli Lmj Lnk ijk = |L|lmn .
Since L is orthogonal its determinant has the value unity, and so lmn = lmn .
Thus we see that lmn has exactly the properties of ijk but with i, j, k replaced by
l, m, n, i.e. it is the same as the expression ijk written using the new coordinates.
This shows that ijk is a third-order Cartesian tensor.
In addition to providing a convenient notation for the determinant of a matrix,
δij and ijk can be used to write many of the familiar expressions of vector
algebra and calculus as contracted tensors. For example, provided we are using
right-handed Cartesian coordinates, the vector product a = b × c has as its
ith component ai = ijk bj ck ; this should be contrasted with the outer product
T = b ⊗ c, which is a second-order tensor having the components Tij = bi cj .
This may be readily extended to an N × N matrix A, i.e.
|A|i1 i2 ···iN = Ai1 j1 Ai2 j2 · · · AiN jN j1 j2 ···jN ,
where i1 i2 ···iN equals 1 if i1 i2 · · · iN is an even permutation of 1, 2,. . ., N and equals −1 if it is an
odd permutation; otherwise it equals zero.
26.8 THE TENSORS δij AND ijk
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 = ai bi = δij ai bj ,
∂2 φ
∂2 φ
= δij
∂xi ∂xi
∂xi ∂xj
(∇ × v)i = ijk
∂2 vj
[∇(∇ · v)]i =
= δjk
∂xi ∂xj
∂xi ∂xk
∂2 vm
[∇ × (∇ × v)]i = ijk
= ijk klm
∂xj ∂xl
(a × b) · c = δij ci jkl ak bl = ikl ci ak bl . ∇2 φ =
An important relationship between the - and δ- tensors is expressed by the
ijk klm = δil δjm − δim δjl .
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
The RHS of (26.30) has the values
+1 if i = l and j = m = i,
−1 if i = m and j = l = i,
0 for any other set of subscript values i, j, l, m.
In each product on the LHS k has the same value in both factors and for a
non-zero contribution none of i, l, j, m can have the same value as k. Since there
are only three values, 1, 2 and 3, that any of the subscripts may take, the only
non-zero possibilities are i = l and j = m or vice versa but not all four subscripts
equal (since then each factor is zero, as it would be if i = j or l = 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) if i = l and j = m, ijk = lmk = klm and, whether ijk is +1 or −1, the
product of the two factors is +1; and
(ii) if i = m and j = l, ijk = mlk = −klm and thus the product ijk klm (no
summation) has the value −1.
This concludes the establishment of identity (26.30).
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