26.5 Second- and higher-order Cartesian tensors
Ifviare the components of a first-order tensor, show that∇·v=∂vi/∂xiis a zero-order
tensor.
In the rotated coordinate system∇·vis given by
(
∂vi
∂xi
)′
=
∂vi′
∂x′i
=
∂xj
∂x′i
∂
∂xj
(Likvk)=LijLik
∂vk
∂xj
,
since the elementsLijare not functions of position. Using the orthogonality relation (26.6)
we then find
∂v′i
∂x′i
=LijLik
∂vk
∂xj
=δjk
∂vk
∂xj
=
∂vj
∂xj
.
Hence∂vi/∂xiis 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 byTij.
Taking (26.9) as a guide we define asecond-order Cartesian tensoras follows:
theTijform the components of such a tensor if, under the same conditions as
for (26.9),
Tij′=LikLjlTkl (26.16)
and
Tij=LkiLljTkl′. (26.17)
At the same time we may define a Cartesian tensor of general order as follows.
The set of expressionsTij···kform 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′···kare given by
Tij′···k=LipLjq···LkrTpq···r (26.18)
and
Tij···k=LpiLqj···LrkTpq′···r. (26.19)
It is apparent that in three dimensions, anNth-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 asT, to denote
the matrix havingTijas the element in theith row andjth column.§
We may think of a second-order tensorTas 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 elementsviof a vector by [vi].