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Second and higherorder Cartesian tensors

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Second and higherorder Cartesian tensors
26.5 SECOND- AND HIGHER-ORDER CARTESIAN TENSORS
If vi are the components of a first-order tensor, show that ∇ · v = ∂vi /∂xi is a zero-order
tensor.
In the rotated coordinate system ∇ · v is given by
∂vk
∂vi
∂v ∂xj ∂
= i =
(Lik vk ) = Lij Lik
,
∂xi
∂xi
∂xi ∂xj
∂xj
since the elements Lij are not functions of position. Using the orthogonality relation (26.6)
we then find
∂vj
∂vk
∂vk
∂vi
= Lij Lik
= δjk
=
.
∂xi
∂xj
∂xj
∂xj
Hence ∂vi /∂xi is invariant under rotation of the axes and is thus a zero-order tensor; this
was to be expected since it can be written in a coordinate-free way as ∇ · v. 26.5 Second- and higher-order Cartesian tensors
Following on from scalars with no subscripts and vectors with one subscript,
we turn to sets of quantities that require two subscripts to identify a particular
element of the set. Let these quantities by denoted by Tij .
Taking (26.9) as a guide we define a second-order Cartesian tensor as follows:
the Tij form the components of such a tensor if, under the same conditions as
for (26.9),
and
Tij = Lik Ljl Tkl
(26.16)
Tij = Lki Llj Tkl .
(26.17)
At the same time we may define a Cartesian tensor of general order as follows.
The set of expressions Tij···k form the components of a Cartesian tensor if, for all
rotations of the axes of coordinates given by (26.4) and (26.5), subject to (26.6),
the expressions using the new coordinates, Tij···k
are given by
and
Tij···k
= Lip Ljq · · · Lkr Tpq···r
(26.18)
.
Tij···k = Lpi Lqj · · · Lrk Tpq···r
(26.19)
It is apparent that in three dimensions, an Nth-order Cartesian tensor has 3N
components.
Since a second-order tensor has two subscripts, it is natural to display its
components in matrix form. The notation [Tij ] is used, as well as T, to denote
the matrix having Tij as the element in the ith row and jth column.§
We may think of a second-order tensor T as a geometrical entity in a similar
way to that in which we viewed linear operators (which transform one vector into
§
We can also denote the column matrix containing the elements vi of a vector by [vi ].
935
TENSORS
another, without reference to any coordinate system) and consider the matrix
containing its components as a representation of the tensor with respect to a
particular coordinate system. Moreover, the matrix T = [Tij ], containing the
components of a second-order tensor, behaves in the same way under orthogonal
transformations T = LTLT as a linear operator.
However, not all linear operators are second-order tensors. More specifically,
the two subscripts in a second-order tensor must refer to the same coordinate
system. In particular, this means that any linear operator that transforms a vector
into a vector in a different vector space cannot be a second-order tensor. Thus,
although the elements Lij of the transformation matrix are written with two
subscripts, they cannot be the components of a tensor since the two subscripts
each refer to a different coordinate system.
As examples of sets of quantities that are readily shown to be second-order
tensors we consider the following.
(i) The outer product of two vectors. Let ui and vi , i = 1, 2, 3, be the components
of two vectors u and v, and consider the set of quantities Tij defined by
Tij = ui vj .
(26.20)
The set Tij are called the components of the the outer product of u and v. Under
rotations the components Tij become
Tij = ui vj = Lik uk Ljl vl = Lik Ljl uk vl = Lik Ljl Tkl ,
(26.21)
which shows that they do transform as the components of a second-order tensor.
Use has been made in (26.21) of the fact that ui and vi are the components of
first-order tensors.
The outer product of two vectors is often denoted, without reference to any
coordinate system, as
T = u ⊗ v.
(26.22)
(This is not to be confused with the vector product of two vectors, which is itself
a vector and is discussed in chapter 7.) The expression (26.22) gives the basis to
which the components Tij of the second-order tensor refer: since u = ui ei and
v = vi ei , we may write the tensor T as
T = ui ei ⊗ vj ej = ui vj ei ⊗ ej = Tij ei ⊗ ej .
(26.23)
Moreover, as for the case of first-order tensors (see equation (26.10)) we note
that the quantities Tij are the components of the same tensor T, but referred to
a different coordinate system, i.e.
T = Tij ei ⊗ ej = Tij ei ⊗ ej .
These concepts can be extended to higher-order tensors.
936
26.5 SECOND- AND HIGHER-ORDER CARTESIAN TENSORS
(ii) The gradient of a vector. Suppose vi represents the components of a vector;
let us consider the quantities generated by forming the derivatives of each vi ,
i = 1, 2, 3, with respect to each xj , j = 1, 2, 3, i.e.
Tij =
∂vi
.
∂xj
These nine quantities form the components of a second-order tensor, as can be
seen from the fact that
Tij =
∂vk
∂vi
∂(Lik vk ) ∂xl
=
= Lik
Ljl = Lik Ljl Tkl .
∂xj
∂xl ∂xj
∂xl
In coordinate-free language the tensor T may be written as T = ∇v and hence
gives meaning to the concept of the gradient of a vector, a quantity that was not
discussed in the chapter on vector calculus (chapter 10).
A test of whether any given set of quantities forms the components of a secondorder tensor can always be made by direct substitution of the xi in terms of the
xi , followed by comparison with the right-hand side of (26.16). This procedure is
extremely laborious, however, and it is almost always better to try to recognise
the set as being expressible in one of the forms just considered, or to make
alternative tests based on the quotient law of section 26.7 below.
Show that the Tij given by
T = [Tij ] =
x22
−x1 x2
−x1 x2
x21
(26.24)
are the components of a second-order tensor.
Again we consider a rotation θ about the e3 -axis. Carrying out the direct evaluation first
we obtain, using (26.7),
T11
= x2 2 = s2 x21 − 2scx1 x2 + c2 x22 ,
= −x1 x2 = scx21 + (s2 − c2 )x1 x2 − scx22 ,
T12
T21
= −x1 x2 = scx21 + (s2 − c2 )x1 x2 − scx22 ,
T22
= x1 2 = c2 x21 + 2scx1 x2 + s2 x22 .
Now, evaluating the right-hand side of (26.16),
T11
= ccx22 + cs(−x1 x2 ) + sc(−x1 x2 ) + ssx21 ,
= c(−s)x22 + cc(−x1 x2 ) + s(−s)(−x1 x2 ) + scx21 ,
T12
= (−s)cx22 + (−s)s(−x1 x2 ) + cc(−x1 x2 ) + csx21 ,
T21
= (−s)(−s)x22 + (−s)c(−x1 x2 ) + c(−s)(−x1 x2 ) + ccx21 .
T22
After reorganisation, the corresponding expressions are seen to be the same, showing, as
required, that the Tij are the components of a second-order tensor.
The same result could be inferred much more easily, however, by noting that the Tij
are in fact the components of the outer product of the vector (x2 , −x1 ) with itself. That
(x2 , −x1 ) is indeed a vector was established by (26.12) and (26.13). 937
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