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The algebra of tensors

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The algebra of tensors
TENSORS
Physical examples involving second-order tensors will be discussed in the later
sections of this chapter, but we might note here that, for example, magnetic
susceptibility and electrical conductivity are described by second-order tensors.
26.6 The algebra of tensors
Because of the similarity of first- and second-order tensors to column vectors and
matrices, it would be expected that similar types of algebraic operation can be
carried out with them and so provide ways of constructing new tensors from old
ones. In the remainder of this chapter, instead of referring to the Tij (say) as the
components of a second-order tensor T, we may sometimes simply refer to Tij
as the tensor. It should always be remembered, however, that the Tij are in fact
just the components of T in a given coordinate system and that Tij refers to the
components of the same tensor T in a different coordinate system.
The addition and subtraction of tensors follows an obvious definition; namely
that if Vij···k and Wij···k are (the components of) tensors of the same order, then
their sum and difference, Sij···k and Dij···k respectively, are given by
Sij···k = Vij···k + Wij···k ,
Dij···k = Vij···k − Wij···k ,
for each set of values i, j, . . . , k. That Sij···k and Dij···k are the components of
tensors follows immediately from the linearity of a rotation of coordinates.
It is equally straightforward to show that if the Tij···k are the components of
a tensor, then so is the set of quantities formed by interchanging the order of (a
pair of) indices, e.g. Tji···k .
If Tji···k is found to be identical with Tij···k then Tij···k is said to be symmetric
with respect to its first two subscripts (or simply ‘symmetric’, for second-order
tensors). If, however, Tji···k = −Tij···k for every element then it is an antisymmetric
tensor. An arbitrary tensor is neither symmetric nor antisymmetric but can always
be written as the sum of a symmetric tensor Sij···k and an antisymmetric tensor
Aij···k :
Tij···k = 12 (Tij···k + Tji···k ) + 12 (Tij···k − Tji···k )
= Sij···k + Aij···k .
Of course these properties are valid for any pair of subscripts.
In (26.20) in the previous section we had an example of a kind of ‘multiplication’
of two tensors, thereby producing a tensor of higher order – in that case two
first-order tensors were multiplied to give a second-order tensor. Inspection of
(26.21) shows that there is nothing particular about the orders of the tensors
involved and it follows as a general result that the outer product of an Nth-order
tensor with an Mth-order tensor will produce an (M + N)th-order tensor.
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