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M1.5 Vector representations 21
properly, the meaning of any mathematical expression will not change. Let us
consider the expression
E = q
r
q
r
= g
rm
g
rn
q
m
q
n
= δ
n
m
q
m
q
n
(M1.85)
Application of (M1.82) and (M1.84) gives the expression to the right of the second
equality sign. For comparison purposes we have also added one of the forms of
(M1.80) as the final expression in (M1.85). It should be carefully observed that the
summation indices m, n, r occur twice only.
To take full advantage of the reciprocal systems we perform a scalar multiplica-
tion first by the contravariant basis vector q
i
and then by the covariant basis vector
q
j
, yielding
(E · q
i
) · q
j
= g
rm
g
rn
δ
i
n
δ
m
j
= δ
n
m
δ
i
n
δ
m
j
(M1.86)
from which it follows immediately that
g
rj
g
ri
= δ
i
j
(M1.87a)
By interchanging i and j , observing the symmetry of the fundamental quantities,
we find
g
ir
g
rj
= δ
j
i
or (g
ij
)(g
ij
) =
100
010
001
=⇒ (g
ij
) = (g
ij
)
−1
(M1.87b)
Hence, the matrices (g
ij
)and(g
ij
) are inverse to each other. Owing to the symmetry
properties of the metric fundamental quantities, i.e. g
ij
= g
ji
and g
ij
= g
ji
,we
need six elements only to specify either metric tensor. In case of an orthogonal
system g
ij
= 0, g
ij
= 0fori = j so that (M1.87a) reduces to
g
ii
g
ii
= 1(M1.88)
thus verifying equation (M1.61). At this point we must recall the rule that we do
not sum over repeated free indices i, j, k, l.
Next we wish to show that, in an orthonormal system, there is no difference
between contravariant and covariant basis vectors. The proof is very simple:
e
i
=
q
i
√
g
ii
=
g
in
√
g
ii
q
n
=
√
g
ii
q
i
=
q
i
g
ii
= e
i
(M1.89)
Here use of the raising rule has been made. With the help of (M1.89) it is easy
to show that there is no difference between contravariant and covariant physical
measure numbers. Utilizing (M1.71) we find
A · e
i
= A · e
i
=⇒ A
*
n
e
n
· e
i
= A
*
n
e
n
· e
i
=⇒ A
*
i
= A
*
i
(M1.90)
22 Algebra of vectors
M1.6 Products of vectors in general coordinate systems
There are various ways to express the dyadic product of vector A with vector B by
employing covariant and contravariant basis vectors:
AB = A
m
B
n
q
m
q
n
= A
m
B
n
q
m
q
n
= A
m
B
n
q
m
q
n
= A
m
B
n
q
m
q
n
(M1.91)
This yields four possibilities for formulating the scalar product A · B:
A · B = A
m
B
n
q
m
· q
n
= A
m
B
n
g
mn
= A
m
B
n
q
m
· q
n
= A
m
B
n
g
mn
= A
m
B
n
q
m
· q
n
= A
m
B
m
= A
m
B
n
q
m
· q
n
= A
m
B
m
(M1.92)
1
The last two forms with mixed basis vectors (covariant and contravariant) are
more convenient since the sums involve the evaluation of only three terms. In
contrast, nine terms are required for the first two forms since they involve the
metric fundamental quantities.
There are two useful forms in which to express the vector product A × B.From
the basic definition (M1.28) and the properties of the reciprocal systems (M1.55)
we obtain
A × B = A
m
q
m
× B
n
q
n
=
√
g
q
1
q
2
q
3
A
1
A
2
A
3
B
1
B
2
B
3
(M1.93)
where all measure numbers are of the contravariant type. If it is desirable to
express the vector product in terms of covariant measure numbers we use (M1.53)
and (M1.57). Thus, we find
A × B = A
m
q
m
× B
n
q
n
=
1
√
g
q
1
q
2
q
3
A
1
A
2
A
3
B
1
B
2
B
3
(M1.94)
The two forms involving mixed basis vectors are not used, in general.
On performing the scalar triple product operation (M1.33) we find
[A, B, C] = A
m
q
m
· (B
n
q
n
× C
r
q
r
)
= A
m
√
gq
m
·
q
1
q
2
q
3
B
1
B
2
B
3
C
1
C
2
C
3
=
√
g
A
1
A
2
A
3
B
1
B
2
B
3
C
1
C
2
C
3
(M1.95)
1
For the scalar product A ·A we usually write A
2
.
M1.7 Problems 23
which is in agreement with (M1.38). All measure numbers are of the contravari-
ant type. If covariant measure numbers of the three vectors are desired, utilizing
(M1.94) we obtain
[A, B, C] = A
m
q
m
· (B
n
q
n
× C
r
q
r
) =
1
√
g
A
1
A
2
A
3
B
1
B
2
B
3
C
1
C
2
C
3
(M1.96)
In this expression the covariant measure numbers may be replaced with the help of
(M1.76) by A
i
= A · q
i
,B
i
= B · q
i
,C
i
= C · q
i
, yielding
[A, B, C][q
1
, q
2
, q
3
] =
A · q
1
A · q
2
A · q
3
B · q
1
B · q
2
B · q
3
C · q
1
C · q
2
C · q
3
(M1.97)
Finally, if in this equation we replace the basis vectors q
1
, q
2
, q
3
by the arbitrary
vectors D, F, G, we obtain
[A, B, C][D, F, G] =
A · DA· FA· G
B · DB· FB· G
C · DC· FC· G
(M1.98)
On setting in this expression D = A, F = B,andG = C we obtain the Gram
determinant, which was already stated without proof as equation (M1.41).
M1.7 Problems
M1.1: ArethethreevectorsA = 2q
1
+ 6q
2
− 2q
3
, B = 3q
1
+ q
2
+ 2q
3
,and
C = 8q
1
+ 16q
2
− 3q
3
linearly dependent?
M1.2:
(a) Are the three vectors A = 4q
1
− q
2
+ 5q
3
, B =−2q
1
+ 3q
2
+ q
3
,and
C =−2q
1
− 2q
2
− 6q
3
linearly dependent?
(b) Use Cartesian basis vectors to show that the vectors F = A × B, G = B × C,
and H = C × A are collinear.
(c) The vectors A and B are the same as before but now C = 8q
1
+ 7q
2
− 5q
3
.
Is the new set of vectors linearly dependent? Show that now F is orthogonal to G
and H.
24 Algebra of vectors
M1.3: Decompose the vector A = A
1
q
1
+ A
2
q
2
into the two components A
1
=
A
1
q
1
and A
2
= A
2
q
2
by assuming that (i) A
1
deviates from A by 15
◦
and (ii) the
length
|
A
1
|
=
1
3
|
A
|
. Determine the length
|
A
2
|
and the angle between A
2
and A.
M1.4: The vector A = 3i
1
+ 2i
2
+ 4i
3
is given.
(a) Find the measure numbers A
i
of A = A
1
q
1
+ A
2
q
2
+ A
3
q
3
if the basis vectors
are given by q
1
= i
1
+ 2i
2
, q
2
= 2i
2
+ i
3
,andq
3
= 2i
3
.
(b) By employing the reciprocal basis vectors q
i
find the measure numbers A
i
of A.
M1.5: By direct transformation of the contravariant measure numbers in (M1.38),
show that
[A, B, C] =
1
√
g
A
1
A
2
A
3
B
1
B
2
B
3
C
1
C
2
C
3
M1.6: Show that the lowering and raising rules do not apply to physical measure
numbers.
M1.7: Show that
(A × B) · (B × C) × (C × A) = (A · B × C)
2
M1.8: The vectors
A =
B × C
[A, B, C]
,
B =
C × A
[A, B, C]
,
C =
A × B
[A, B, C]
with [A, B, C] = 0 are given. Find [
A,
B,
C].
M1.9: The unit vector e is perpendicular to the plane defined by the vectors B and
C. Show that [e, B, C] =
|
B × C
|
.
M2
Vector functions
M2.1 Basic definitions and operations
In general scalars and vectors depend on the position coordinates q
i
or q
i
and on
the time t. Therefore, we have to deal with expressions of the type
B = B
n
(q
1
,q
2
,q
3
,t)q
n
(q
1
,q
2
,q
3
,t) = B
n
(q
1
,q
2
,q
3
,t)q
n
(q
1
,q
2
,q
3
,t)(M2.1)
While in stationary Cartesian coordinates the basis vectors are independent of
position, in the general case q
i
and q
i
are functions of the corresponding position
coordinates. In rotating coordinate systems they are functions of time also.
Of particular interest to our studies are linear vector functions defined by
B(r
(1)
+ r
(2)
) = B(r
(1)
) + B(r
(2)
), B(λr) = λB(r)(M2.2)
where λ is a scalar and r
(1)
and r
(2)
are position vectors. An example of a linear
vector function is given by
B(r) = B(q
n
q
n
) = B(q
n
)q
n
= B
n
q
n
with B
i
= B(q
i
)(M2.3)
Since r = q
n
q
n
we have q
i
= q
i
·r and therefore
B(r) = B
n
q
n
= B
n
q
n
·r = B·r (M2.4)
This is the defining equation for the complete dyadic
B = B
n
q
n
. In our studies the
sum will always be restricted to three terms. By expressing the position vector in
B(r)asr = q
n
q
n
, we obtain analogously
B(r) = B(q
n
q
n
) = B(q
n
)q
n
= B
n
q
n
= B
n
q
n
·r = B·r (M2.5)
with q
i
= q
i
·r. Hence, the complete dyadic B may be written as
B = B
n
q
n
= B
n
q
n
(M2.6)
25
26 Vector functions
Scalar multiplication of a vector by a dyadic yields a vector. In special but
very important cases the summation is terminated after two terms. The resulting
planar dyadic will be of great interest in the following work. Dyadics consisting
of only one term are called dyads; those consisting of two terms only are known as
singular dyadics. Whenever two vectors of a dyadic, such as B
i
or q
i
(i = 1, 2, 3),
are linearly dependent then the complete dyadic transforms to a singular dyadic.
As we know already, the nine components of a three-term dyadic can be arranged
as a square matrix. Therefore, the rules of matrix algebra can be applied to perform
operations with second-order dyadics. Unless specifically stated otherwise, we will
be dealing with complete dyadics.
Let us now think of the dyadic
B as representing an operator. Scalar multiplication
of the dyadic by the original vector r results in a new vector r
, which is called the
image vector:
r
= B·r = (B
m
q
m
)·(q
n
q
n
) = B
m
q
n
δ
m
n
= B
m
q
m
= (B
m
q
m
)·(q
n
q
n
) = B
m
q
n
δ
n
m
= B
m
q
m
(M2.7)
There are several ways to represent a complete dyadic. Some important results
are given below. As will be seen, various dyadic measure numbers occur, which will
now be discussed. Let us first consider the form in which the dyadic
B is expressed
with the help of the covariant vectorial measure numbers B
n
and contravariant basis
vectors q
n
,i.e.B = B
n
q
n
. Scalar multiplication of B by q
i
gives
B·q
i
= B
n
q
n
·q
i
= B
n
δ
n
i
= B
i
(M2.8)
The vector B
i
may be represented in the two equivalent forms
B
i
= B
n
i
q
n
= B
ni
q
n
(M2.9)
Repeating this procedure by expressing
B in the form B = B
n
q
n
yields analogously
B·q
i
= B
n
δ
i
n
= B
i
, B
i
= B
ni
q
n
= B
i
n
q
n
(M2.10)
From (M2.9) and (M2.10) we obtain four possibilities for representing the dyadic
B:
B = B
m
n
q
m
q
n
= B
mn
q
m
q
n
= B
mn
q
m
q
n
= B
n
m
q
m
q
n
(M2.11)
While B
i
j
and B
j
i
are called the mixed measure numbers of B,thetermsB
ij
and
B
ij
are the covariant and contravariant measure numbers of B, respectively. At
each measure number the positions of the subscripts and superscripts indicate not
M2.1 Basic definitions and operations 27
only the kind of the corresponding basis vectors (covariant or contravariant) of the
dyadic but also the order in which they appear.
By utilizing the property (M1.51) of reciprocal systems as well as the lower-
ing and raising rules for the basis vectors, it is a simple task to obtain relations
between the different types of measure numbers of a dyadic. This will be clar-
ified by the following two examples. Application of the raising rule (M1.82)
to the first expression of (M2.11) yields, together with the second expression
of (M2.11),
B = B
r
n
q
r
q
n
= B
r
n
g
rm
q
m
q
n
= B
mn
q
m
q
n
(M2.12)
Scalar multiplication of (M2.12) first by the covariant basis vector q
j
and then by
q
i
results in
(B·q
j
)·q
i
= B
r
n
g
rm
δ
n
j
δ
m
i
= B
mn
δ
n
j
δ
m
i
(M2.13a)
Applying the lowering rule (M1.84) to the basis vector q
r
of the dyadic B =
B
n
r
q
r
q
n
and multiplying the result from the right-hand side first by q
j
andthenby
q
i
yields
(
B·q
j
)·q
i
= B
n
r
g
rm
δ
j
n
δ
i
m
= B
mn
δ
j
n
δ
i
m
(M2.13b)
The final evaluation of the expressions of (M2.13a) and (M2.13b) gives the raising
rule and the lowering rule for the measure numbers of a dyadic:
B
ij
= g
ri
B
r
j
,B
ij
= g
ri
B
j
r
(M2.14)
Since in Cartesian coordinate systems there is no difference between covariant
and contravariant basis vectors, from (M2.11) it may easily be seen that, in these
systems, the different measure numbers of a dyadic are identical:
B = B
m
n
i
m
i
n
= B
mn
i
m
i
n
= B
mn
i
m
i
n
= B
n
m
i
m
i
n
with B
i
j
= B
ij
= B
ij
= B
j
i
(M2.15)
In the following sections, for the sake of brevity, we will not list all possible
representations of the dyadic, but will usually limit ourselves to the form
= B
n
q
n
= B
m
n
q
m
q
n
(M2.16)
This does not imply that a preference should be given to this particular form.
28 Vector functions
M2.2 Special dyadics
Whenever the antecedents and the consequents of the dyadic
= B
m
q
m
form
linearly independent systems, then is called a complete dyadic. If the antecedents
or the consequents are coplanar then the complete dyadic reduces to a planar dyadic
of only two terms.
M2.2.1 The conjugate dyadic
A special but very important dyadic known as the conjugate dyadic is obtained by
interchanging the positions of the first vector (antecedent) and the second vector
(consequent) of the dyadic
B = B
m
q
m
:
B = q
m
B
m
(M2.17)
Thus, the original antecedent becomes the consequent. The conjugate dyadic will
be identified by means of the tilde. An important relationship involving the original
and the conjugate dyadic is
D·B = D·B
m
q
m
= q
m
D·B
m
= q
m
B
m
·D =
B·D
(M2.18)
M2.2.2 The symmetric dyadic
A dyadic
is called symmetric if interchanging the positions of the two vectors
does not change the dyadic:
=
(M2.19)
An important consequence is
·D = D·
= D·
(M2.20)
It is easy to show that the sum of an arbitrary dyadic B and the corresponding
conjugate dyadic
B is symmetric:
= B +
B = B
m
q
m
+ q
m
B
m
,
= q
m
B
m
+ B
m
q
m
= =
(M2.21)
For the Cartesian coordinate system the explicit form of a symmetric dyadic is
given by
= B
11
i
1
i
1
+ B
12
i
1
i
2
+ B
13
i
1
i
3
+ B
12
i
2
i
1
+ B
22
i
2
i
2
+ B
23
i
2
i
3
+ B
13
i
3
i
1
+ B
23
i
3
i
2
+ B
33
i
3
i
3
(M2.22)
since B
ij
= B
ji
, which is the condition characterizing symmetric matrices. Since
g
ij
= g
ji
the unit dyadic is also symmetric:
E = g
mn
q
m
q
n
,
E = g
mn
q
n
q
m
= g
nm
q
n
q
m
= E (M2.23)
M2.2 Special dyadics 29
M2.2.3 The antisymmetric or skew-symmetric dyadic
A dyadic
is called antisymmetric if interchanging the positions of the two
vectors results in a change of sign of the dyadic:
=−
(M2.24)
From this definition we immediately see that
·D = D·
=−D·
(M2.25)
The difference between a dyadic
B and the conjugate dyadic
B is always antisym-
metric:
= B −
B = B
m
q
m
− q
m
B
m
,
= q
m
B
m
− B
m
q
m
=− (M2.26)
From (M2.21) and (M2.26) we conclude that any dyadic may be expressed as the
sum of a symmetric and an antisymmetric dyadic:
=
1
2
(B +
B) +
1
2
(B −
B) =
+
(M2.27)
The representation of an antisymmetric dyadic in the Cartesian coordinate system
is given by
= 0 + B
12
i
1
i
2
+ B
13
i
1
i
3
− B
12
i
2
i
1
+ 0 + B
23
i
2
i
3
− B
13
i
3
i
1
− B
23
i
3
i
2
+ 0
(M2.28)
The elements on the main diagonal must be zero to satisfy the condition of anti-
symmetry B
ij
=−B
ji
which characterizes every antisymmetric matrix.
In order to discover another property of an antisymmetric dyadic, we wish to
reconsider the vectorial triple product. Using the Grassmann rule (M1.44) we find
D × (A × B) = D·(BA − AB)(M2.29)
which is the scalar product of a vector D with an antisymmetric dyadic.
30 Vector functions
M2.2.4 The adjoint dyadic
The adjoint dyadic of
= B
s
q
s
is defined by
a
= (B
s
q
s
)
a
=
1
2
(q
m
× q
n
)(B
m
× B
n
)
(M2.30)
On expanding and rearranging the sums we obtain for the adjoint dyadic the useful
form
a
= (q
1
× q
2
)(B
1
× B
2
) + (q
2
× q
3
)(B
2
× B
3
) + (q
3
× q
1
)(B
3
× B
1
)(M2.31)
According to (M2.17) the conjugate of the adjoint dyadic may be written as
a
=
1
2
(B
m
× B
n
)(q
m
× q
n
)(M2.32)
A little reflection shows that
a
=
a
(M2.33)
We will now give the component form of the adjoint dyadic whose original defini-
tion (M2.30) involves the vector products q
j
× q
k
and B
j
× B
k
. With the help of
(M1.53) and (M1.93) we find
B
i
× B
j
= B
m
i
q
m
× B
n
j
q
n
= [q
1
, q
2
, q
3
]
q
1
q
2
q
3
B
1
i
B
2
i
B
3
i
B
1
j
B
2
j
B
3
j
= [q
1
, q
2
, q
3
]D
ij
(M2.34)
and
(q
1
× q
2
)(B
1
× B
2
) = q
3
[q
1
, q
2
, q
3
][q
1
, q
2
, q
3
]D
12
= q
3
D
12
(q
3
× q
1
)(B
3
× B
1
) = q
2
D
31
(q
2
× q
3
)(B
2
× B
3
) = q
1
D
23
(M2.35)
By adding the three terms, as required by (M2.31), we obtain the desired result
a
= q
1
D
23
+ q
2
D
31
+ q
3
D
12
= M
m
n
q
m
q
n
(M2.36)
The M
i
j
are elements of the adjoint matrix (M
i
j
) corresponding to the coefficient
matrix (B
i
j
) of the dyadic = B
m
n
q
m
q
n
. Thus, we obtain the adjoint dyadic by
replacing the elements B
i
j
by the corresponding elements of the adjoint matrix
which is obtained by transposing the cofactor matrix of (B
i
j
).