First and zeroorder Cartesian tensors

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First and zeroorder Cartesian tensors
26.4 First- and zero-order Cartesian tensors
Using the above example as a guide, we may consider any set of three quantities
vi , which are directly or indirectly functions of the coordinates xi and possibly
involve some constants, and ask how their values are changed by any rotation of
the Cartesian axes. The specific question to be answered is whether the specific
forms vi in the new variables can be obtained from the old ones vi using (26.4),
vi = Lij vj .
If so, the vi are said to form the components of a vector or first-order Cartesian
tensor v. By definition, the position coordinates are themselves the components
of such a tensor.The first-order tensor v does not change under rotation of the
coordinate axes; nevertheless, since the basis set does change, from e1 , e2 , e3 to
e1 , e2 , v3 , the components of v must also change. The changes must be such that
v = vi ei = vi ei
is unchanged.
Since the transformation (26.9) is orthogonal, the components of any such
first-order Cartesian tensor also obey a relation that is the inverse of (26.9),
vi = Lji vj .
We now consider explicit examples. In order to keep the equations to reasonable
proportions, the examples will be restricted to the x1 x2 -plane, i.e. there are
no components in the x3 -direction. Three-dimensional cases are no different in
principle – but much longer to write out.
Which of the following pairs (v1 , v2 ) form the components of a first-order Cartesian tensor
in two dimensions?:
(i) (x2 , −x1 ),
(ii) (x2 , x1 ),
(iii) (x21 , x22 ).
We shall consider the rotation discussed in the previous example, and to save space we
denote cos θ by c and sin θ by s.
(i) Here v1 = x2 and v2 = −x1 , referred to the old axes. In terms of the new coordinates
they will be v1 = x2 and v2 = −x1 , i.e.
v1 = x2 = −sx1 + cx2
v2 = −x1 = −cx1 − sx2 .
Now if we start again and evaluate v1 and v2 as given by (26.9) we find that
v1 = L11 v1 + L12 v2 = cx2 + s(−x1 )
v2 = L21 v1 + L22 v2 = −s(x2 ) + c(−x1 ).
The expressions for v1 and v2 in (26.12) and (26.13) are the same whatever the values
of θ (i.e. for all rotations) and thus by definition (26.9) the pair (x2 , −x1 ) is a first-order
Cartesian tensor.
(ii) Here v1 = x2 and v2 = x1 . Following the same procedure,
v1 = x2 = −sx1 + cx2
v2 = x1 = cx1 + sx2 .
But, by (26.9), for a Cartesian tensor we must have
v1 = cv1 + sv2 = cx2 + sx1
v2 = (−s)v1 + cv2 = −sx2 + cx1 .
These two sets of expressions do not agree and thus the pair (x2 , x1 ) is not a first-order
Cartesian tensor.
(iii) v1 = x21 and v2 = x22 . As in (ii) above, considering the first component alone is
sufficient to show that this pair is also not a first-order tensor. Evaluating v1 directly gives
v1 = x1 = c2 x21 + 2csx1 x2 + s2 x22 ,
whilst (26.9) requires that
v1 = cv1 + sv2 = cx21 + sx22 ,
which is quite different. There are many physical examples of first-order tensors (i.e. vectors) that will be
familiar to the reader. As a straightforward one, we may take the set of Cartesian
components of the momentum of a particle of mass m, (mẋ1 , mẋ2 , mẋ3 ). This set
transforms in all essentials as (x1 , x2 , x3 ), since the other operations involved,
multiplication by a number and differentiation with respect to time, are quite
unaffected by any orthogonal transformation of the axes. Similarly, acceleration
and force are represented by the components of first-order tensors.
Other more complicated vectors involving the position coordinates more than
once, such as the angular momentum of a particle of mass m, namely J =
x × p = m(x × ẋ), are also first-order tensors. That this is so is less obvious in
component form than for the earlier examples, but may be verified by writing
out the components of J explicitly or by appealing to the quotient law to be
discussed in section 26.7 and using the Cartesian tensor ijk from section 26.8.
Having considered the effects of rotations on vector-like sets of quantities we
may consider quantities that are unchanged by a rotation of axes. In our previous
nomenclature these have been called scalars but we may also describe them as
tensors of zero order. They contain only one element (formally, the number of
subscripts needed to identify a particular element is zero); the most obvious nontrivial example associated with a rotation of axes is the square of the distance of
a point from the origin, r 2 = x21 + x22 + x23 . In the new coordinate system it will
have the form r 2 = x1 2 + x2 2 + x3 2 , which for any rotation has the same value as
x21 + x22 + x23 .
In fact any scalar product of two first-order tensors (vectors) is a zero-order
tensor (scalar), as might be expected since it can be written in a coordinate-free
way as u · v.
By considering the components of the vectors u and v with respect to two Cartesian
coordinate systems (related by a rotation), show that the scalar product u · v is invariant
under rotation.
In the original (unprimed) system the scalar product is given in terms of components by
ui vi (summed over i), and in the rotated (primed) system by
ui vi = Lij uj Lik vk = Lij Lik uj vk = δjk uj vk = uj vj ,
where we have used the orthogonality relation (26.6). Since the resulting expression in the
rotated system is the same as that in the original system, the scalar product is indeed
invariant under rotations. The above result leads directly to the identification of many physically important quantities as zero-order tensors. Perhaps the most immediate of these is
energy, either as potential energy or as an energy density (e.g. F · dr, eE · dr, D · E,
B · H, µ · B), but others, such as the angle between two directed quantities, are
As mentioned in the first paragraph of this chapter, in most analyses of physical
situations it is a scalar quantity (such as energy) that is to be determined. Such
quantities are invariant under a rotation of axes and so it is possible to work with
the most convenient set of axes and still have confidence in the results.
Complementing the way in which a zero-order tensor was obtained from two
first-order tensors, so a first-order tensor can be obtained from a zero-order tensor
(i.e. a scalar). We show this by taking a specific example, that of the electric field
E = −∇φ; this is derived from a scalar, the electrostatic potential φ, and has
Ei = −
Clearly, E is a first-order tensor, but we may prove this more formally by
considering the behaviour of its components (26.14) under a rotation of the
coordinate axes, since the components of the electric field Ei are then given by
Ei =
∂xj ∂φ
=− =− = Lij Ej ,
∂xi ∂xj
where (26.5) has been used to evaluate ∂xj /∂xi . Now (26.15) is in the form
(26.9), thus confirming that the components of the electric field do behave as the
components of a first-order tensor.
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