Curvilinear Analysis in a Euclidean Space
Presented in a framework and notation customized for students and professionals
who are already familiar with Cartesian analysis in ordinary 3D physical engineering
space.
Rebecca M. Brannon
UNM SUPPLEMENTAL BOOK DRAFT
June 2004
Written by Rebecca Moss Brannon of Albuquerque NM, USA, in connection with adjunct
teaching at the University of New Mexico. This document is the intellectual property of Rebecca
Brannon.
Copyright is reserved.
June 2004
PREFACE
This document started out as a small set of notes assigned as supplemental reading
and exercises for the graduate students taking my continuum mechanics course at the
University of New Mexico (Fall of 1999). After the class, I posted the early versions of the
manuscript on my UNM web page to solicit interactions from anyone on the internet
who cared to comment. Since then, I have regularly fielded questions and added
enhancements in response to the encouragement (and friendly hounding) that I received
from numerous grad students and professors from all over the world.
Perhaps the most important acknowledgement I could give for this work goes to
Prof. Howard “Buck” Schreyer, who introduced me to curvilinear coordinates when I
was a student in his Continuum Mechanics class back in 1987. Buck’s daughter, Lynn
Betthenum (sp?) has also contributed to this work by encouraging her own grad stu-
dents to review it and send me suggestions and corrections.
Although the vast majority of this work was performed in the context of my Univer-
sity appointment, I must acknowledge the professional colleagues who supported the
continued refinement of the work while I have been a staff member and manager at San-
dia National Laboratories in New Mexico.
Rebecca Brannon
rmbrann@me.unm.edu
June 2004
rmbrann@me.unm.edu
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DRAFT June 17, 2004
1
Curvilinear Analysis in a Euclidean Space
Rebecca M.Brannon, University of New Mexico, First draft: Fall 1998, Present draft: 6/17/04.
1. Introduction
This manuscript is a student’s introduction on the mathematics of curvilinear coordinates,
but can also serve as an information resource for practicing scientists. Being an introduction,
we have made every effort to keep the analysis well connected to concepts that should be
familiar to anyone who has completed a first course in regular-Cartesian-coordinate
1
(RCC)
vector and tensor analysis. Our principal goal is to introduce
engineering
specialists to the
mathematician’s
language of general curvilinear vector and tensor analysis. Readers who are
already well-versed in functional analysis will probably find more rigorous manuscripts
(such as [14]) more suitable. If you are completely new to the subject of general curvilinear
coordinates or if you seek guidance on the basic machinery associated with non-orthonormal
base vectors, then you will probably find the approach taken in this report to be unique and
(comparatively) accessible. Many engineering students presume that they can get along in
their careers just fine without ever learning any of this stuff. Quite often, that’s true. Nonethe-
less, there will undoubtedly crop up times when a system operates in a skewed or curved
coordinate system, and a basic knowledge of curvilinear coordinates makes life a lot easier.
Another reason to learn curvilinear coordinates — even if you never explicitly apply the
knowledge to any practical problems — is that you will develop a far deeper understanding
of Cartesian tensor analysis.
Learning the basics of curvilinear analysis is an essential first step to reading much of the
older materials modeling literature, and the theory is still needed today for non-Euclidean
surface and quantum mechanics problems. We added the proviso “older” to the materials
modeling literature reference because more modern analyses are typically presented using
structured notation
(also known as Gibbs, symbolic, or direct notation) in which the highest-
level fundamental meaning of various operations are called out by using a notation that does
not
explicitly suggest the procedure for actually performing the operation. For example,
would be the structure notation for the vector dot product whereas
would be
the procedural notation that clearly shows how to compute the dot product but has the disad-
1. Here, “regular” means that the basis is right, rectangular, and normalized. “Right” means the basis forms a right-handed
system (i.e., crossing the first base vector into the second results in a third vector that has a positive dot product with the
third base vectors). “Rectangular” means that the base vectors are mutually perpendicular. “Normalized” means that the
base vectors are dimensionless and of unit length. “Cartesian” means that all three coordinates have the same physical
units [12, p90]. The last “C” in the RCC abbreviation stands for “coordinate” and its presence implies that the basis is
itself defined in a manner that is coupled to the coordinates. Specifically, the basis is always tangent to the coordinate
grid. A goal of this paper is to explore the implications of removing the constraints of RCC systems. What happens when
the basis is not rectangular? What happens when coordinates of different dimensions are used? What happens when the
basis is selected independently from the coordinates?
a
˜
b
˜
•
a
1
b
1
a
2
b
2
a
3
b
3
+
+
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DRAFT June 17, 2004
2
vantage of being applicable only for RCC systems. The
same
operation would be computed
differently in a non-RCC system — the fundamental operation itself doesn’t change; instead
the method for computing it changes depending on the system you adopt. Operations such as
the dot and cross products are known to be invariant when expressed using combined com-
ponent+basis notation. Anyone who
chooses
to perform such operations using Cartesian com-
ponents will obtain the same result as anyone else who opts to use general curvilinear
components — provided that both researchers understand the connections between the two
approaches! Every now and then, the geometry of a problem clearly calls for the use of non-
orthonormal or spatially varying base vectors. Knowing the basics of curvilinear coordinates
permits analysts to choose the approach that most simplifies their calculations. This manu-
script should be regarded as providing two services: (1) enabling students of Cartesian analy-
sis to solidify their knowledge by taking a foray into curvilinear analysis and (2) enabling
engineering professionals to read older literature wherein it was (at the time) considered
more “rigorous” or stylish to present all analyses in terms of general curvilinear analysis.
In the field of materials modeling, the stress tensor is regarded as a function of the strain
tensor and other material state variables. In such analyses, the material often contains certain
“preferred directions” such as the direction of fibers in a composite matrix, and curvilinear
analysis becomes useful if those directions are not orthogonal. For plasticity modeling, the
machinery of non-orthonormal base vectors can be useful to understand six-dimensional
stress space, and it is especially useful when analyzing the response of a material when the
stress resides at a so-called yield surface “vertex”. Such a vertex is defined by the convergence
of two or more surfaces having different and generally non-orthogonal orientation normals,
and determination of whether or not a trial elastic stress rate is progressing into the “cone of
limiting normals” becomes quite straightforward using the formal mathematics of non-
orthonormal bases.
This manuscript is broken into three key parts: Syntax, Algebra, and Calculus. Chapter 2
introduces the most common coordinate systems and iterates the distinction between irregu-
lar bases and curvilinear coordinates; that chapter introduces the several fundamental quanti-
ties (such as metrics) which appear with irresistible frequency throughout the literature of
generalized tensor analysis. Chapter 3 shows how Cartesian formulas for basic vector and
tensor operations must be altered for non-Cartesian systems. Chapter 4 covers basis and coor-
dinate transformations, and it provides a gentle introduction to the fact that base vectors can
vary with position.
The fact that the underlying base vectors might be non-normalized, non-orthogonal, and/
or non-right-handed is the essential focus of Chapter 4. By contrast, Chapter 5 focuses on how
extra terms must appear in gradient expressions (in addition to the familiar terms resulting
from spatial variation of scalar and vector components); these extra terms account for the fact
that the coordinate base vectors vary in space. The fact that different base vectors can be used
at different points in space is an essential feature of curvilinear coordinates analysis.
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DRAFT June 17, 2004
3
The vast majority of engineering applications use one of the coordinate systems illustrated
in Fig. 1.1. Of these, the rectangular Cartesian coordinate system is the most popular choice.
For all three systems in Fig. 1.1, the base vectors are unit vectors. The base vectors are also
mutually perpendicular, and the ordering is “right-handed” (
i.e.,
the third base vector is
obtained by crossing the first into the second). Each base vector points in the direction that the
position vector moves if one coordinate is increased, holding the other two coordinates con-
stant; thus, base vectors for spherical and cylindrical coordinates vary with position. This is a
crucial concept: although the coordinate system has only one origin, there can be an infinite
number base vectors because the base vector orientations can depend on position.
Most practicing engineers can get along just fine without ever having to learn the theory
behind general curvilinear coordinates. Naturally, every engineer must, at some point, deal
with cylindrical and spherical coordinates, but they can look up whatever formulas they need
in handbook tables. So why bother learning about generalized curvilinear coordinates? Dif-
ferent people have different motivations for studying general curvilinear analysis. Those
dealing with general relativity, for example, must be able to perform tensor analysis on four
dimensional curvilinear manifolds. Likewise, engineers who analyze shells and membranes
in 3D space greatly benefit from general tensor analysis. Reading the literature of continuum
mechanics — especially the older work — demands an understanding of the notation. Finally,
the topic is just plain interesting in its own right. James Simmonds [7] begins his book on ten-
sor analysis with the following wonderful quote:
The magic of this theory will hardly fail to impose itself on anybody who has truly understood it; it
represents a genuine triumph of the method of absolute differential calculus, founded by Gauss, Rie-
mann, Ricci, and Levi-Civita.
—Albert Einstein
1
An important message articulated in this quote is the suggestion that, once you have mas-
x
1
x
2
x
3
e
˜
1
e
˜
2
e
˜
3
e
˜
r
e
˜
θ
e
˜
z
x
1
x
2
x
3
θ
r
z
φ
θ
e
˜
r
e
˜
θ
e
˜
φ
(orthogonal)
Cartesian coordinates
(orthogonal)
Cylindrical coordinates
(orthogonal)
Spherical coordinates
FIGURE 1.1 The most common engineering coordinate systems.
Note that all three systems are orthogonal because
the associated base vectors are mutually perpendicular. The cylindrical and spherical coordinate systems are
inhomogeneous because the base vectors vary with position. As indicated,
depends on
θ
for cylindrical
coordinates and
depends on both
θ
and
ψ
for spherical coordinates.
e
˜
r
e
˜
r
r
x
1
x
2
x
3
x
1
x
2
x
3
, ,
{
}
r
θ
z
, ,
{
}
r
θ φ
, ,
{
}
x
˜
x
˜
x
˜
x
˜
r
e
˜
r
θ φ
,
(
)
=
x
˜
r
e
˜
r
θ
( )
z
e
˜
z
+
=
x
˜
x
1
e
˜
1
x
2
e
˜
2
x
3
e
˜
3
+
+
=
(a)
(c)
(b)
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DRAFT June 17, 2004
4
tered tensor analysis, you will begin to recognize its basic concepts in many other seemingly
unrelated fields of study. Your knowledge of tensors will therefore help you master a broader
range of subjects.
1. From: “Contribution to the Theory of General Relativity,” 1915; as quoted and translated by C. Lanczos in The Einstein
Decade, p213.
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DRAFT June 17, 2004
5
This document is a teaching and learning tool. To assist with this goal, you will note that the
text is color-coded as follows:
Please direct comments to
rmbrann@me.unm.edu
1.1 Vector and Tensor Notation
The tensorial order of quantities will be indicated by the number of underlines. For exam-
ple, is a scalar, is a vector, is a second-order tensor, is a third order tensor,
etc.
We fol-
low Einstein’s summation convention where repeated indices are to be summed (this rule will
be later clarified for curvilinear coordinates).
You, the reader, are presumed familiar with basic operations in Cartesian coordinates (dot
product, cross-product, determinant, etc.). Therefore, we may define our
structured
terminol-
ogy and notational conventions by telling you their meanings in terms ordinary Cartesian
coordinates. A principal purpose of this document is to show how these
same
structured oper-
ations must be computed using
different procedures
when using non-RCC systems. In this sec-
tion, where we are merely explaining the meanings of the non-indicial notation structures, we
will use standard RCC conventions that components of vectors and tensors are identified by
subscripts that take on the values 1, 2, and 3. Furthermore, when exactly two indices are
repeated in a single term, they are understood to be summed from 1 to 3. Later on, for non-
RCC systems, the conventions for subscripts will be generalized.
The term “array” is often used for any matrix having one dimension equal to 1. This docu-
ment focuses exclusively on ordinary 3D physical space. Thus, unless otherwise indicated, the
word “array” denotes either a
or a
matrix. Any array of three numbers may be
expanded as the sum of the array components times the corresponding primitive
basis
arrays:
(1.1)
Everyday engineering problems typically characterize vectors using only the regular Carte-
sian (orthonormal right-handed) laboratory basis,
. Being orthonormal, the base
vectors have the property that
, where
is the Kronecker delta and the indices
( and ) take values from 1 to 3. Equation (1.1) is the matrix-notation equivalent of the usual
expansion of a vector as a sum of components times base vectors:
(1.2)
BLUE
definition
⇒
RED
important concept
⇒
s
v
˜
T
˜˜
ξ
˜˜˜
3
1
×
1
3
×
v
1
v
2
v
3
v
1
1
0
0
v
2
0
1
0
v
3
0
0
1
+
+
=
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
e
˜
i
e
˜
j
•
δ
ij
=
δ
ij
i
j
v
˜
v
1
e
˜
1
v
2
e
˜
2
v
3
e
˜
3
+
+
=
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DRAFT June 17, 2004
6
More compactly,
(1.3)
Many engineering problems (e.g., those with spherical or cylindrical symmetry) become
extremely complicated when described using the orthonormal laboratory basis, but they sim-
plify superbly when phrased in terms of some other basis,
. Most of the time, this
“other” basis is also a “regular” basis, meaning that it is orthonormal (
) and
right handed (
) — the only difference is that the new basis is oriented differ-
ently than the laboratory basis. The best choice for this other, more convenient, basis might
vary in space. Note, for example, that the bases for spherical and cylindrical coordinates illus-
trated in Fig. 1.1 are orthonormal and right-handed (and therefore “regular”) even though a
different set of base vectors is used at each point in space. This harks back to our earlier com-
ment that the properties of being orthonormal and curvilinear are distinct — one does not
imply or exclude the other.
Generalized curvilinear coordinates show up when studying quantum mechanics or shell
theory (or even when interpreting a material deformation from the perspective of a person
who translates, rotates, and stretches along with the material). For most of these advanced
physics problems, the governing equations are greatly simplified when expressed in terms of
an “irregular” basis (i.e., one that is not orthogonal, not normalized, and/or not right-
handed). To effectively study curvilinear coordinates and irregular bases, the reader must
practice constant vigilance to keep track of what
particular basis
a set of components is refer-
enced to
.
When working with irregular bases, it is customary to construct a complementary or
“dual” basis that is intimately related to the original irregular basis. Additionally, even
though it is might not be convenient for the application at hand, the regular laboratory basis
still exists. Sometimes a quantity is most easily interpreted using yet other bases. For example,
a tensor is usually described in terms of the laboratory basis or some “applications” basis, but
we all know that the tensor is particularly simplified if it is expressed in terms of its
principal
basis. Thus, any engineering problem might involve the simultaneous use of many different
bases. If the basis is changed, then the components of vectors and tensors must change too. To
emphasize the inextricable interdependence of components and bases, vectors are routinely
expanded in the form of components times base vectors
.
What is it that distinguishes vectors from simple
arrays of numbers? The answer is
that the component array for a vector is determined by the underlying basis and this compo-
nent array must change is a very particular manner when the basis is changed. Vectors have
(by definition) an invariant quality with respect to a change of basis. Even though the compo-
nents themselves change when a basis changes, they must change in a very specific way — it
they don’t change that way, then the thing you are dealing with (whatever it may be) is not a
vector. Even though components change when the basis changes, the
sum
of the components
times the base vectors remains the same. Suppose, for example, that
is the regular
v
˜
v
k
e
˜
k
=
E
˜
1
E
˜
2
E
˜
3
,
,
{
}
E
˜
i
E
˜
j
•
δ
ij
=
E
˜
3
E
˜
1
E
˜
2
×
=
v
˜
v
1
e
˜
1
v
2
e
˜
2
v
3
e
˜
3
+
+
=
3
1
×
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
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DRAFT June 17, 2004
7
laboratory basis and
is some alternative orthonormal right-handed basis. Let the
components of a vector with respect to the lab basis be denoted by and let denote the
components with respect to the second basis. The invariance of vectors requires that the basis
expansion of the vector must give the same result regardless of which basis is used. Namely,
(1.4)
The relationship between the components and the components can be easily character-
ized as follows:
Dotting both sides of (1.4) by
gives:
(1.5a)
Dotting both sides of (1.4) by
gives:
(1.5b)
Note that
. Similarly,
. We can define
a set of nine numbers (known as direction cosines)
. Therefore,
and
, where we have used the fact that the dot product is commutative
(
for any vectors and ). With these observations and definitions, Eq. (1.5)
becomes
(1.6a)
(1.6b)
These relationships show how the { } components are related to the
components. Satisfy-
ing these relationships is often the identifying characteristic used to identify whether or not
something really is a vector (as opposed to a simple collection of three numbers). This discus-
sion was limited to changing from one regular (i.e., orthonormal right-handed) to another.
Later on, the concepts will be revisited to derive the change of component formulas that apply
to irregular bases. The key point (which is exploited throughout the remainder of this docu-
ment) is that,
although the vector components themselves change with the basis, the
sum
of
components times base vectors in invariant
.
The statements made above about vectors also have generalizations to tensors. For exam-
ple, the analog of Eq. (1.1) is the expansion of a
matrix into a sum of individual compo-
nents times base tensors:
(1.7)
Looking at a tensor in this way helps clarify why tensors are often treated as nine-dimen-
sional vectors: there are nine components and nine associated “base tensors.” Just as the inti-
mate relationship between a vector and its components is emphasized by writing the vector in
the form of Eq. (1.4), the relationship between a tensor’s components and the underlying basis
is emphasized by writing tensors as the sum of components times “basis dyads”. Specifically
E
˜
1
E
˜
2
E
˜
3
,
,
{
}
v
˜
v
k
v
˜
k
v
k
e
˜
k
v
˜
k
E
˜
k
=
v
˜
k
v
k
E
˜
m
v
k
e
˜
k
E
˜
m
•
(
)
v
˜
k
E
˜
k
E
˜
m
•
(
)
=
e
˜
m
v
k
e
˜
k
e
˜
m
•
(
)
v
˜
k
E
˜
k
e
˜
m
•
(
)
=
v
˜
k
E
˜
k
E
˜
m
•
(
)
v
˜
k
δ
km
v
˜
m
=
=
v
k
e
˜
k
e
˜
m
•
(
)
v
k
δ
km
v
m
=
=
L
ij
e
˜
i
E
˜
j
•
≡
e
˜
k
E
˜
m
•
L
km
=
E
˜
k
e
˜
m
•
L
mk
=
a
˜
b
˜
•
b
˜
a
˜
•
=
a
˜
b
˜
v
k
L
km
v
˜
m
=
v
m
v
˜
k
L
mk
=
v
k
v
˜
m
3
3
×
A
11
A
12
A
13
A
21
A
22
A
23
A
31
A
32
A
33
A
11
1 0 0
0 0 0
0 0 0
A
12
0 1 0
0 0 0
0 0 0
…
A
33
0 0 0
0 0 0
0 0 1
+
+
+
=
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DRAFT June 17, 2004
8
in direct correspondence to Eq. (1.7) we write
.
(1.8)
Two vectors written side-by-side are to be multiplied
dyadically
, for example,
is a sec-
ond-order tensor
dyad
with Cartesian
ij
components
. Any tensor can be expressed as a
linear combination of the nine possible
basis dyads
. Specifically the dyad
corresponds to
an RCC component matrix that has zeros everywhere except “1” in the position, which was
what enabled us to write Eq. (1.7) in the more compact form of Eq. (1.8). Even a dyad
itself
can be expanded in terms of basis dyads as
. Dyadic multiplication is often
alternatively denoted with the symbol
⊗
. For example,
means the same thing as
.
We prefer using no “
⊗
” symbol for dyadic multiplication because it allows more appealing
identities such as
.
Working with third-order tensors requires introduction of
triads
, which are denoted struc-
turally by three vectors written side-by-side. Specifically,
is a third-order tensor with
RCC
ijk
components
. Any third-order tensor can always be expressed as a linear com-
bination of the fundamental basis triads. The concept of dyadic multiplication extends simi-
larly to fourth and higher-order tensors.
Using our summation notation that repeated indices are to be summed, the standard com-
ponent-basis expression for a tensor (Eq. 1.8) can be written
.
(1.9)
The components of a tensor change when the basis changes, but the sum of components times
basis dyads remains invariant. Even though a tensor comprises many components and basis
triads, it is this
sum
of individual parts that’s unique and physically meaningful.
A
raised single dot
is the
first-order inner product
. For example, in terms of a Cartesian
basis,
. When applied between tensors of higher or mixed orders, the single dot
continues to denote the first order inner product; that is, adjacent vectors in the basis dyads
are dotted together so that
.
Here
is the Kronecker delta, defined to equal 1 if
and 0 otherwise. The common opera-
tion,
denotes a first order vector whose Cartesian component is
. If, for exam-
ple,
, then the RCC components of may be found by the matrix multiplication:
implies (for RCC)
, or
(1.10)
Similarly,
, where the superscript “T” denotes the tensor
transpose
(
i.e.,
). Note that the effect of the raised single dot is to sum adjacent indi-
A
˜˜
A
11
e
˜
1
e
˜
1
A
12
e
˜
1
e
˜
2
…
A
33
e
˜
3
e
˜
3
+
+
+
=
a
˜
b
˜
a
i
b
j
e
˜
i
e
˜
j
ij
a
˜
b
˜
a
˜
b
˜
a
i
b
i
e
˜
i
e
˜
j
=
a
˜
b
˜
⊗
a
˜
b
˜
a
˜
b
˜
(
)
c
˜
•
a
˜
b
˜
c
˜
•
(
)
=
u
˜
v
˜
w
˜
u
i
v
j
w
k
A
˜˜
A
ij
e
˜
i
e
˜
j
=
u
˜
v
˜
•
u
k
v
k
=
A
˜˜
B
˜˜
•
A
ij
e
˜
i
e
˜
j
(
)
B
pq
e
˜
p
e
˜
q
(
)
•
A
ij
B
pq
δ
jp
e
˜
i
e
˜
q
(
)
A
ij
B
jq
e
˜
i
e
˜
q
=
=
=
δ
ij
i
=
j
A
˜˜
u
˜
•
i
th
A
ij
u
j
w
˜
A
˜˜
u
˜
•
=
w
˜
w
˜
A
˜˜
u
˜
•
=
w
i
A
ij
u
j
=
w
1
w
2
w
3
A
11
A
12
A
13
A
21
A
22
A
23
A
31
A
32
A
33
u
1
u
2
u
3
=
v
˜
B
˜˜
•
v
k
B
km
e
˜
m
B
˜˜
T
v
˜
•
=
=
B
ij
T
B
ji
=
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DRAFT June 17, 2004
9
ces. Applying similar heuristic notational interpretation, the reader can verify that
must be a scalar computed in RCC by
.
A first course in tensor analysis, for example, teaches that the
cross-product
between two
vectors
is a new vector obtained in RCC by
(1.11)
or, more compactly,
,
(1.12)
where
is the permutation symbol
= 1 if
= 123, 231, or 312
= –1 if
= 321, 132, or 213
= 0 if any of the indices , , or are equal.
(1.13)
Importantly,
(1.14)
This permits us to alternatively write Eq. (1.12) as
(1.15)
We will employ a self-defining notational structure for
all
conventional vector operations. For
example, the expression
can be immediately inferred to mean
(1.16)
The “
triple scalar-valued product
” is denoted with square brackets around a list of three
vectors and is defined
. Note that
(1.17)
We denote the
second-order inner product
by a
“double dot” colon
. For rectangular Carte-
sian components, the second-order inner product sums adjacent
pairs
of components. For
example,
,
, and
. Caution: many authors
insidiously use the term “inner product” for the similar looking scalar-valued operation
, but this operation is not an inner product because it fails the positivity axiom required
for any inner product.
u
˜
C
˜˜
v
˜
•
•
u
i
C
ij
v
j
v
˜
w
˜
×
v
˜
w
˜
×
v
2
w
3
v
3
w
2
–
(
)
e
˜
1
v
3
w
1
v
1
w
3
–
(
)
e
˜
2
v
1
w
2
v
2
w
1
–
(
)
e
˜
3
+
+
=
v
˜
w
˜
×
ε
ijk
v
j
w
k
e
˜
i
=
ε
ijk
ε
ijk
ijk
ε
ijk
ijk
ε
ijk
i j
k
e
˜
i
e
˜
j
×
ε
ijk
e
˜
k
=
v
˜
w
˜
×
v
j
w
k
e
˜
j
e
˜
k
×
(
)
=
C
˜˜
v
˜
×
C
˜˜
v
˜
×
C
ij
e
˜
i
e
˜
i
v
k
e
˜
k
×
C
ij
v
k
e
˜
i
e
˜
i
e
˜
k
×
C
ij
v
k
ε
mik
e
˜
m
=
=
=
u
˜
v
˜
w
˜
, ,
[
]
u
˜
v
˜
w
˜
×
(
)
•
≡
ε
ijk
e
˜
i
e
˜
j
e
˜
k
, ,
[
]
=
A
˜˜
:B
˜˜
A
ij
B
ij
=
ξ
˜˜˜
:C
˜˜
ξ
ijk
C
jk
e
˜
i
=
ξ
˜˜˜
:a
˜
b
˜
ξ
ijk
a
j
b
k
e
˜
i
=
A
ij
B
ji
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DRAFT June 17, 2004
10
1.2 Homogeneous coordinates
A coordinate system is called “
homoge-
neous
” if the associated base vectors are
the same throughout space. A basis is
“
orthogonal
” (or “
rectangular
”) if the
base vectors are everywhere mutually
perpendicular. Most authors use the term
“Cartesian coordinates” to refer to the
conventional
orthonormal
homogeneous
right-handed system of Fig. 1.1a. As seen
in Fig. 1.2b, a homogeneous system is not required to be orthogonal. Furthermore, no coordi-
nate system is required to have unit base vectors. The opposite of homogeneous is “curvilin-
ear,” and Fig. 1.3 below shows that a coordinate system can be both curvilinear and
orthogonal. In short,
the properties of being “orthogonal” or “homogeneous” are indepen-
dent
(one does not imply or exclude the other).
1.3 Curvilinear coordinates
The
coordinate grid
is the family of
lines along which only one coordinate var-
ies. If the grid has at least some curved
lines, the coordinate system is called “
cur-
vilinear
,” and, as shown in Fig. 1.3, the
associated base vectors (tangent to the
grid lines) necessarily change with posi-
tion, so
curvilinear systems are always
inhomogeneous
. The system in Fig. 1.3a
has base vectors that are everywhere orthogonal, so it is simultaneously curvilinear and
orthogonal. Note from Fig. 1.1 that conventional cylindrical and spherical coordinates are
both orthogonal and curvilinear. Incidentally, no matter what type of coordinate system is
used,
base vectors need not be of unit length; they only need to point in the direction that the
g
˜
1
g
˜
2
g
˜
1
g
˜
2
g
˜
1
g
˜
2
(b)
e
˜
1
e
˜
2
e
˜
1
e
˜
2
e
˜
1
e
˜
2
Orthogonal (or rectangular)
homogeneous coordinates
Nonorthogonal
homogeneous coordinates
FIGURE 1.2 Homogeneous coordinates.
The base vectors
are the same at all points in space. This condition is possible
only if the coordinate grid is formed by straight lines.
(a)
g
˜
1
g
˜
2
g
˜
2
g
˜
1
g
˜
2
g
˜
1
(a)
orthogonal
curvilinear coordinates
g
˜
1
g
˜
2
g
˜
2
g
˜
1
g
˜
1
g
˜
2
nonorthogonal
curvilinear coordinates
FIGURE 1.3 Curvilinear coordinates.
The base vectors are still
tangent to coordinate lines. The left system is curvilinear and
orthogonal (the coordinate lines always meet at right angles).
(b)
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DRAFT June 17, 2004
11
position vector would move when changing the associated coordinate, holding others con-
stant
.
1
We will call a basis “
regular
” if it consists of a right-handed orthonormal triad. The sys-
tems in Fig. 1.3 have irregular associated base vectors. The system in Fig 1.3a can be
“regularized” by normalizing the base vectors. Cylindrical and spherical systems are exam-
ples of regularized curvilinear systems.
In Section 2, we introduce mathematical tools for both irregular homogeneous and irregu-
lar curvilinear coordinates first deals with the possibility that the base vectors might be non-
orthogonal, non-normalized, and/or non-right-handed. Section 3 shows that the component
formulas for many operations such as the dot product take on forms that are different from
the regular (right-handed orthonormal) formulas. The distinction between homogeneous and
curvilinear coordinates becomes apparent in Section 5, where the derivative of a vector or
higher order tensor requires additional terms to account for the variation of curvilinear base
vectors with position. By contrast, homogeneous base vectors do not vary with position, so
the tensor calculus formulas look very much like their Cartesian counterparts, even if the
associated basis is irregular.
1.4 Difference between Affine (non-metric) and Metric spaces
As discussed by Papastavridis [12], there are situations where the axes used to define a
space don’t have the same physical dimensions, and there is no possibility of comparing the
units of one axis against the units of another axis. Such spaces are called “affine” or “non-met-
ric.” The apropos example cited by Papastavridis is “thermodynamic state space” in which
the pressure, volume, and temperature of a fluid are plotted against one another. In such a
space, the concept of lengths (and therefore angles) between two points becomes meaning-
less. In affine geometries, we are only interested in properties that remain invariant under
arbitrary scale and angle changes of the axes.
The remainder of this document is dedicated to
metric
spaces such as the ordinary physical
3D space that we all (hopefully) live in.
2. Dual bases for irregular bases
Suppose there are compelling physical reasons to use an irregular basis
.
Here, “irregular” means the basis might be nonorthogonal, non-normalized, and/or non-
right-handed. In this section we develop tools needed to derive modified component formu-
las for tensor operations such as the dot product.
For tensor
algebra
, it is irrelevant whether the
basis is homogeneous or curvilinear; all that matters is the possibility that the base vectors
1. Strictly speaking, it is not necessary to require that the base vectors have any relationship whatsoever with the coordinate
lines. If desired, for example, we
could
use arbitrary curvilinear coordinates while taking the basis to be everywhere
aligned with the laboratory basis. In this document, however, the basis is
always
assumed tangent to coordinate lines.
Such a basis is called the “associated” basis.
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
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DRAFT June 17, 2004
12
might not be orthogonal and/or might not be of unit length and/or might not form a right-
handed system.
Again, keep in mind that we will be deriving new procedures for computing
the operations, but the ultimate result and meanings for the operations will be unchanged. If,
for example, you had two vectors expressed in terms of an irregular basis, then you could
always transform those vectors into conventional RCC expansions in order to compute the
dot product. The point of this section is to deduce
faster
methods that permit you to obtain the
same result
directly from the irregular vector components without having to transform to
RCC.
To simplify the discussion, we will assume that the underlying space is our ordinary 3D
physical Euclidean space.
1
Whenever needed, we may therefore assume there exists a
right-
handed
orthonormal
laboratory basis
,
where .
This is particularly
convenient because we can then claim that
there exists a transformation tensor such that
. (2.1)
If this transformation tensor is written in component form with respect to the laboratory
basis, then
the column of the matrix [F] contains the components of the base vector with
respect to the laboratory basis.
In terms of the
lab
components of [F], Eq. (2.1) can be written
(2.2)
instead of
. After all, Eq. (1.10) appears to be tell-
ing us that adjacent indices should be summed, but Eq. (2.2) shows the summation index
being summed with the farther (first) index on the tensor. There’s a subtle and important phe-
nomenon here that needs careful attention whenever you deal with equations like (2.1) that
really represent
three separate equations
for each value of
i
from 1 to 3. To unravel the mystery,
let’s start by changing the symbol used for the free index in Eq. (2.1) by writing it equivalently
by
. Now, applying Eq. (1.10) gives
. Any vector, , can be
expanded as
. Applying this identity with replaced by gives
, or,
. The expression
represents the lab component of , so it must equal
. Consequently,
, which is equivalent to Eq. (2.2).
Incidentally, the transformation tensor may be written in a purely dyadic form as
(2.3)
1. To quote from Ref. [7], “Three-dimensional Euclidean space,
, may be characterized by a set of axioms that expresses
relationships among primitive, undefined quantities called points, lines, etc. These relationships so closely correspond to
the results of ordinary measurements of distance in the physical world that, until the appearance of general relativity, it
was thought that Euclidean geometry was
the
kinematic model of the universe.”
E
3
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
e
˜
i
e
˜
j
•
δ
ij
=
F
˜˜
g
˜
i
F
˜˜
e
˜
i
•
=
i
th
g
i
g
˜
i
F
ji
e
˜
j
=
F
ji
F
ij
j
g
˜
k
F
˜˜
e
˜
k
•
=
g
˜
k
( )
i
F
ij
e
˜
k
( )
j
=
v
˜
v
˜
v
i
e
˜
i
=
v
˜
g
˜
k
g
˜
k
g
˜
k
( )
i
e
˜
i
=
g
˜
k
F
ij
e
˜
k
( )
j
e
˜
i
=
e
˜
k
( )
j
j
th
e
˜
k
δ
kj
g
˜
k
F
ij
δ
kj
e
˜
i
F
ik
e
˜
i
=
=
F
˜˜
F
˜˜
g
˜
k
e
˜
k
=
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DRAFT June 17, 2004
13
Partial Answer:
(a) The term “regular” is defined on page 10.
(b) In terms of the lab basis, the component array of the first base vector is
, so this must be the first
column of the [F] matrix.
This transformation tensor is defined for the
specific
irregular basis of interest as it relates to
the laboratory basis. The transformation tensor for a different pair of bases will be different.
This does not imply that is not a tensor. Readers who are familiar with continuum mechan-
ics may be wondering whether our basis transformation tensor has anything to do with the
deformation gradient tensor used to describe continuum motion. The answer is “no.” In
general, the tensor in this document merely represents the relationship between the labora-
tory basis and the irregular basis. Even though our tensor is generally unrelated to the
deformation gradient tensor from continuum mechanics, it’s still interesting to consider the
special case in which these two tensors
are
the same. If a deforming material is conceptually
“painted” with an orthogonal grid in its reference state, then this grid will deform with the
material, thereby providing a natural “embedded” curvilinear coordinate system with an
associated “natural” basis that is everywhere tangent to the painted grid lines. When this
“natural” embedded basis is used, our transformation tensor will be identical to the defor-
mation gradient tensor . The component forms of many constitutive material models
become intoxicatingly simple in structure when expressed using an embedded basis (it
remains a point of argument, however, whether or not simple structure implies intuitiveness).
The embedded basis
co-varies
with the grid lines — in other words, these vectors stay always
tangent to the grid lines and they stretch in proportion with the stretching of the grid lines.
For this reason, the embedded basis is called the
covariant
basis. Later on, we will introduce a
companion triad of vectors, called the
contravariant
basis, that does not move with the grid
lines; instead we will find that the contravariant basis moves in a way that it remains always
perpendicular to material planes that do co-vary with the deformation. When a plane of parti-
cles moves with the material, its normal does not generally move with the material!
Study Question 2.1
Consider the following irregular base
vectors expressed in terms of the laboratory basis:
.
(a) Explain why this basis is irregular.
(b) Find the 3
×
3 matrix of components of the transformation
tensor with respect to the laboratory basis.
g
˜
2
g
˜
1
e
˜
1
e
˜
2
g
˜
1
e
˜
1
2
e
˜
2
+
=
g
˜
2
e
˜
1
–
e
˜
2
+
=
g
˜
3
e
˜
3
=
F
˜˜
1
2
0
F
˜˜
F
˜˜
F
˜˜
F
˜˜
F
˜˜
F
˜˜
F
˜˜
F
˜˜
F
˜˜
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DRAFT June 17, 2004
14
Our tensor can be seen as characterizing a transformation operation that will take you
from the orthonormal laboratory base vectors to the irregular base vectors. The three irregular
base vectors,
form a triad, which in turn defines a parallelepiped. The volume of
the parallelepiped is given by the
Jacobian
of the transformation tensor , defined by
.
(2.4)
Geometrically, the Jacobian
J
in Eq. (2.4) equals the volume of the parallelepiped formed by
the covariant base vectors
. To see why this triple scalar product is identically
equal to the determinant of the transformation tensor , we now introduce the
direct notation
definition of a determinant
:
The determinant, det[ ] (also called the
Jacobian
), of a tensor is the unique scalar
satisfying
[
] = det[ ]
for all vectors
.
(2.5)
Geometrically this strange-looking definition of the determinate states that if a parallelepiped
is formed by three vectors,
, and a transformed parallelepiped is formed by the
three transformed vectors
, then the ratio of the transformed volume to the
original volume will have a unique value, regardless what three vectors are chosen to form
original the parallelepiped! This volume ratio is the determinant of the transformation tensor.
Since Eq. (2.5) must hold for all vectors,
it must hold for any particular choices of those vectors
.
Suppose we choose to identify
with the underlying orthonormal basis
.
Then, recalling from Eq. (2.4) that
is denoted by the Jacobian , Eq. (2.5) becomes
. The underlying Cartesian basis
is orthonor-
mal and right-handed, so
. Recalling from Eq. (2.1) that the covariant basis is
obtained by the transformation
, we get
,
(2.6)
which completes the proof that
the Jacobian
J
can be computed by taking the determinant of
the Cartesian transformation tensor or by simply taking the triple scalar product of the covari-
ant base vectors, whichever method is more convenient:
.
(2.7)
The set of vectors
forms a basis if and only if
is invertible — i.e., the Jacobian
must be nonzero.
By choice, the laboratory basis
is regular and therefore right-
handed. Hence,
the irregular basis
is
right-handed if
left-handed if
.
(2.8)
F
˜˜
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
F
˜˜
J
det
F
˜˜
[ ]
≡
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
F
˜˜
F
˜˜
F
˜˜
F
˜˜
u
˜
•
F
˜˜
v
˜
•
F
˜˜
w
˜
•
,
,
F
˜˜
u
˜
v
˜
w
˜
, ,
[
]
u
˜
v
˜
w
˜
, ,
{
}
u
˜
v
˜
w
˜
, ,
{
}
F
˜˜
u
˜
•
F
˜˜
v
˜
•
F
˜˜
w
˜
•
,
,
{
}
u
˜
v
˜
w
˜
, ,
{
}
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
det
F
˜˜
[ ]
J
F
˜˜
e
˜
1
•
F
˜˜
e
˜
2
•
F
˜˜
e
˜
3
•
,
,
[
]
=
J
e
˜
1
e
˜
2
e
˜
3
,
,
[
]
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
e
˜
1
e
˜
2
e
˜
3
,
,
[
]
=1
g
˜
i
F
˜˜
e
˜
i
•
=
g
˜
1
g
˜
2
g
˜
3
,
,
[
]
J
=
J
det
F
˜˜
[ ]
g
˜
1
g
˜
2
g
˜
3
×
(
)
•
g
˜
1
g
˜
2
g
˜
3
,
,
[
]
≡
=
=
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
F
˜˜
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
J
0
>
J
0
<
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DRAFT June 17, 2004
15
2.1 Modified summation convention
Given that
is a basis, we know there exist unique coefficients
such that any vector can be written
. Using Einstein’s summation
notation, you may write this expansion as
(2.9)
By convention, components with respect to an irregular basis
are
identified
with
superscripts
,
1
rather than subscripts. Summations always occur on different levels — a
superscript is always paired with a subscript in these implied summations. The summation
convention rules for an irregular basis are:
1. An index that appears exactly once in any term is called a “
free index
,” and it must ap-
pear exactly once in
every
term in the expression.
2. Each particular free index must appear at the same level in every term. Distinct free in-
dices may permissibly appear at different levels.
3. Any index that appears exactly twice in a given term is called a
dummy sum index
and
implies summation from 1 to 3. No index may appear more than twice in a single term.
4. Given a dummy sum pair, one index must appear at the
upper “contravariant” level
,
and one must appear at the
lower “covariant” level
.
5. Exceptions to the above rules must be clearly indicated whenever the need arises.
•
Exceptions of rule #1 are extremely rare in tensor analysis because rule #1 can never be violated in
any well-formed tensor expression. However, exceptions to rule #1 do regularly appear in non-ten-
sor (matrix) equations. For example, one might define a matrix
with components given by
. Here, both and are free indices, and the right-hand-side of this equation violates
rule #1 because the index occurs exactly once in the first term but not in the second term. This
definition of the
numbers is certainly well-defined in a matrix sense
2
, but the equation is a vio-
lation of tensor index rule #1. Consequently if you
really do
wish to use the equation
to define some matrix
, then you should include a parenthetical comment that the tensor index
conventions are not to be applied — otherwise your readers will think you made a typo.
•
Exceptions of rules #2 and #4 can occur when working with a regular (right-handed orthonormal)
basis because it turns out that there is no distinction between covariant and contravariant components
when the basis is regular. For example, is identically equal to when the basis is regular. That’s
why indicial expressions in most engineering publications show
all
components using only sub-
scripts.
•
Exceptions of rule #3 sometimes occur when the indices are actually referenced to a
particular
basis
and are not intended to apply to
any
basis. Consider, for example, how you would need to handle an
exception to rule #3 when defining the
principle direction
and eigenvalue associated with
some tensor . You would have to write something like “
”
in order to call attention to the fact that the index is supposed to be a
free
index, not summed. An-
other exception to rule #3 occurs when an index appears only once, but you
really do
wish for a sum-
mation over that index. In that case you must explicitly show the summation sign in front of the
equation. Similarly, if you
really do
wish for an index to appear more than twice, then you must ex-
plicitly indicate whether that index is free or summed.
1. The superscripts are only indexes, not exponents. For example,
is the
second
contravariant component of a vector
— it is not the square of some quantity . If your work
does
involve some scalar quantity “ ”, then you should typeset
its square as
whenever there is any chance for confusion.
2. This equation is not well defined as an indicial
tensor
equation because it will not transform properly under a basis
change. The concept of what constitutes a well-formed tensor operation will be discussed in more detail later.
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
a
1
a
2
a
3
,
,
{
}
a
˜
a
˜
a
1
g
˜
1
a
2
g
˜
2
a
3
g
˜
3
+
+
=
a
˜
a
i
g
˜
i
=
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
a
2
a
˜
a
a
a
( )
2
A
[ ]
A
ij
v
i
v
j
+
=
i
j
i
A
ij
A
ij
v
i
v
j
+
=
A
[ ]
v
i
v
i
i
th
p
˜
i
λ
i
T
˜˜
T
˜˜
p
˜
i
•
λ
i
p
˜
i
(no sum over index i)
=
i
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DRAFT June 17, 2004
16
We arbitrarily elected to place the index on the lower level of our basis
, so
(recalling rule #4) we call it the
“covariant” basis
. The coefficients
have the index
on the upper level and are therefore called the
contravariant components
of the vector .
Later on, we will define
co
variant components
with respect to a carefully defined
complementary
“
contra
variant” basis
. We will then have
two ways to write the
vector:
. Keep in mind: we have not yet indicated how these contra- and co-
variant components are computed, or what they mean physically, or why they are useful. For
now, we are just introducing the standard “high-low” notation used in the study of irregular
bases.
(You may find the phrase “co-go-below” helpful to remember the difference between co- and con-
tra-variant.)
We will eventually show
there are
four
ways to write a second-order tensor
. We will intro-
duce contravariant components
and covariant components
, such that
. We will also introduce “mixed” components
and
such that
. Note the use of a “dot” to serve as a place holder to indicate the
order of the indices (the order of the indices is dictated by the order of the dyadic basis pair).
As shown in Section 3.4, use of a “dot” placeholder is necessary only for
nonsymmetric
tensors.
(namely, we will find that symmetric tensor components satisfy the property that
,
so the placement of the “dot” is inconsequential for symmetric tensors.)
. In professionally typeset
manuscripts, the dot placeholder might not be necessary because, for example,
can be
typeset in a manner that is clearly distinguishable from
. The dot placeholders are more
frequently used in handwritten work, where individuals have unreliable precision or clarity
of penmanship. Finally, the number of dot placeholders used in an expression is typically
kept to the minimum necessary to clearly demark the order of the indices. For example,
means the same thing as
. Either of these expressions clearly show that the indices are sup-
posed to be ordered as “ followed by ,” not vice versa. Thus, only one dot is enough to
serve the purpose of indicating order. Similarly,
means the same thing as
, but the dots
serve no clarifying purpose for this case when all indices are on the same level (thus, they are
omitted). The importance of clearly indicating the
order
of the indices is inadequately empha-
sized in some texts.
1
1. For example, Ref. [4] fails to clearly indicate index ordering. They use neither well-spaced typesetting nor dot placehold-
ers, which can be confusing.
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
a
1
a
2
a
3
,
,
{
}
a
˜
a
1
a
2
a
3
,
,
{
}
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
a
˜
a
i
g
˜
i
a
i
g
˜
i
=
=
T
ij
T
ij
T
˜˜
T
ij
g
˜
i
g
˜
j
T
ij
g
˜
i
g
˜
j
˜
=
=
T
•
j
i
T
i
•
j
T
˜˜
T
•
j
i
g
˜
i
g
˜
j
T
i
•
j
g
˜
i
g
˜
j
=
=
T
•
j
i
T
j
•
i
=
T
j
i
T
j
i
T
•
j
i
•
T
•
j
i
i
j
T
••
ij
T
ij
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DRAFT June 17, 2004
17
IMPORTANT:
As proved later (Study question 2.6),
there is no difference between cova-
riant and contravariant components whenever the basis is orthonormal
. Hence, for example,
is the same as
. Nevertheless, in order to always satisfy rules #2 and
#4 of the sum conventions, we rewrite all familiar orthonormal formulas so that the summed
subscripts are on different levels. Furthermore, throughout this document, the following are
all equivalent symbols for the
Kronecker delta
:
=
.
(2.10)
BEWARE: as discussed in Section 3.5,
the set of
values should be regarded as indexed
symbols as defined above, not as components of any particular tensor. Yes, it’s true that
are components of the identity tensor
with respect to the underlying rectangular Cartesian
basis
, but they are
not
the contravariant components of the identity tensor with
respect to an irregular
basis
. Likewise,
are
not
the covariant components of
the identity tensor with respect to the irregular basis. Interestingly, the mixed-level Kronecker
delta components, , do turn out to be the mixed components of the identity tensor with
respect to either basis! Most of the time, we will be concerned only with the components of
tensors with respect to the irregular basis. The Kronecker delta is important in its own right.
This is one reason why we denote the identity tensor by a symbol different from its compo-
nents. Later on, we will note the importance of the permutation symbol
(which equals +1
if
ijk
={123, 231, or 312}, -1 if
ijk
={321, 213, or 132}, and zero otherwise). The permutation sym-
bol represents the components of the alternating tensor with respect to the
any
regular (
i.e.,
right-handed orthonormal basis), but not with respect to an irregular basis. Consequently, we
will represent the alternating
tensor
by a different symbol so that we can continue to use the
permutation
symbol
as an independent indexed quantity. Tracking the basis to which
components are referenced is one the most difficult challenges of curvilinear coordinates.
Important notation glitch
Square brackets [ ] will be used to indicate a
matrix, and
braces { } will indicate a
matrix containing vector components. For example,
denotes the
matrix that contains the contravariant components of a vector . Similarly,
is the matrix that contains the covariant components of a second-order tensor , and
will be used to denote the
matrix
containing the contravariant components of . Any
indices appearing inside a matrix merely indicate the co/contravariant nature of the matrix —
they are not interpreted in the same way as indices in an indicial expression. The indices
merely to indicate the (high/low/mixed)
level
of the matrix components. The rules on page 15
apply only to proper indicial equations, not to equations involving matrices. We will later
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
δ
ij
δ
i
j
δ
ij
,
,
1 if
i
=
j
0 if
i j
≠
δ
ij
δ
ij
I
˜˜
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
δ
ij
δ
i
j
I
˜˜
ε
ijk
ξ
˜˜˜
ε
ijk
3
3
×
3
1
×
v
i
{ }
3
1
×
v
˜
T
ij
[
]
T
˜˜
T
ij
[
]
T
˜˜
ij
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DRAFT June 17, 2004
18
prove, for example, that the determinant
can be computed by the determinant of the
mixed components of . Thus, we might write
. The brackets around
indicate that this equation involves the
matrix
of mixed components, so the rules on page 15
do not apply to the indices. It’s okay that and don’t appear on the left-hand side.
2.2 The metric coefficients and the dual contravariant basis
For reasons that will soon become apparent, we introduce a symmetric set of numbers
,
called the “
metric coefficients
,” defined
.
(2.11)
When the space is Euclidean, the
base vectors can be expressed as linear combi-
nations of the underlying orthonormal laboratory basis and the above set of dot products can
be computed using the ordinary orthonormal basis formulas.
1
We also introduce a
dual “contravariant” basis
defined such that
.
(2.12)
Geometrically, Eq. (2.12) requires that the first contravariant base vector
must be perpen-
dicular to both
and
, so it must be of the form
. The
as-yet
undeter-
mined scalar
is determined by requiring that
equal unity.
(2.13)
In the second-to-last step, we recognized the triple scalar product,
, to be the
Jacobian defined in Eq. (2.7). In the last step we asserted that the result must equal unity.
Consequently, the scalar is merely the reciprocal of the Jacobian:
(2.14)
All three contravariant base vectors can be determined similarly to eventually give the final
result:
,
, .
(2.15)
where
.
(2.16)
1. Note: If the space is
not
Euclidean, then an orthonormal basis does
not
exist, and the metric coefficients
must be spec-
ified
a priori
. Such a space is called Riemannian. Shell and membrane theory deals with 2D curved Riemannian mani-
folds embedded in 3D space. The geometry of general relativity is that of a
four-dimensional
Riemannian manifold. For
further examples of Riemannian spaces, see, e.g., Refs. [5, 7].
det
T
˜˜
T
˜˜
det
T
˜˜
det
T
•
j
i
[
]
=
T
•
j
i
ij
i
j
g
ij
g
ij
g
˜
i
g
˜
j
•
≡
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
g
ij
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
g
˜
i
g
˜
j
•
δ
j
i
=
g
˜
1
g
˜
2
g
˜
3
g
˜
1
α
g
˜
2
g
˜
3
×
(
)
=
α
g
˜
1
g
˜
1
•
g
˜
1
g
˜
1
•
α
g
˜
2
g
˜
3
×
(
)
[
]
g
˜
1
•
α
g
˜
1
g
˜
2
g
˜
3
×
(
)
•
α
J
1
=
=
=
=
“set”
g
˜
1
g
˜
2
g
˜
3
×
(
)
•
J
α
α
1
J
---
=
g
˜
1
1
J
---
g
˜
2
g
˜
3
×
(
)
=
g
˜
2
1
J
---
g
˜
3
g
˜
1
×
(
)
=
g
˜
3
1
J
---
g
˜
1
g
˜
2
×
(
)
=
J
g
˜
1
g
˜
2
g
˜
3
×
(
)
•
g
˜
1
g
˜
2
g
˜
3
,
,
[
]
≡
=
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DRAFT June 17, 2004
19
An alternative way to obtain the dual contravariant basis is to assert that
is in
fact a basis; we may therefore demand that coefficients
must exist such that each covariant
base vector can be written as a linear combination of the contravariant basis:
.
Dotting both sides with and imposing Eqs. (2.11) and (2.12) shows that the transformation
coefficients must be identical to the covariant metric coefficients: .
Thus
.
(2.17)
This equation may be solved for the contravariant basis. Namely,
,
(2.18)
where
the matrix of
contravariant
metric components
is obtained by inverting the covari-
ant metric matrix
. Dotting both sides of Eq. (2.18) by we note that
,
(2.19)
which is similar in form to Eq. (2.11).
In later analyses,
keep in mind that
is the inverse of the
matrix. Furthermore, both
metric matrices are symmetric. Thus, whenever these are multiplied together with a con-
tracted index, the result is the Kronecker delta:
.
(2.20)
Another quantity that will appear frequently in later analyses is the determinant of the
covariant
metric matrix and the determinant of the contravariant
metric matrix:
and
.
(2.21)
Recalling that the
matrix is the inverse of the
matrix, we note that
.
(2.22)
Furthermore, as shown in Study Question 2.5,
is related to the Jacobian
J
from Eqs. (2.4)
,
(2.23)
Thus
.
(2.24)
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
L
ik
g
˜
i
g
˜
i
L
ik
g
˜
k
=
g
˜
k
L
ik
g
ik
=
g
˜
i
g
ik
g
˜
k
=
g
˜
i
g
ik
g
˜
k
=
g
ij
g
ij
[ ]
g
˜
j
g
ij
g
˜
i
g
˜
j
•
=
g
ij
g
ij
g
ik
g
kj
g
ki
g
kj
g
ik
g
jk
g
ki
g
jk
δ
j
i
=
=
=
=
g
o
g
ij
g
o
g
ij
g
o
det
g
11
g
12
g
13
g
21
g
22
g
23
g
31
g
32
g
33
≡
g
o
det
g
11
g
12
g
13
g
21
g
22
g
23
g
31
g
32
g
33
≡
g
ij
[
]
g
ij
[ ]
g
o
1
g
o
-----
=
g
o
g
o
J
2
=
g
o
1
J
2
-----
=
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DRAFT June 17, 2004
20
Non-trivial lemma
1
Note that Eq. (2.21) shows that may be regarded as a function of the
nine components of the
matrix. Taking the partial derivative of with respect to a partic-
ular
component gives a result that is identical to the signed subminor (also called the
cofactor) associated with that component. The “subminor” is a number associated with each
matrix position that is equal to the determinant of the
submatrix obtained by striking
out the row and column of the original matrix. To obtain the cofactor (i.e., the
signed
subminor
) associated with the
ij
position, the subminor is multiplied by
. Denoting this
cofactor by
, we have
(2.25)
Any book on matrix analysis will include a proof that the
inverse
of a matrix
can be
obtained by taking the transpose of the cofactor matrix and dividing by the determinant:
(2.26)
From which it follows that
. Applying this result to the case that
is
the
symmetric
matrix
, and recalling that
is denoted , and also recalling that
is given by
(2.27)
Similarly,
(2.28)
With these equations, we are introduced for the first time to a new index notation rule: sub-
script indices that appear in the “denominator” of a derivative should be regarded as super-
script indices in the expression as a whole. Similarly, superscripts in the denominator should
be regarded as subscripts in the derivative as a whole. With this convention, there is no viola-
tion of the index rule that requires free indices to be on the same level in all terms.
Recalling Eq. (2.23), we can apply the chain rule to note that
(2.29)
Equation (2.27) permits us to express the left-hand-side of this equation in terms of the
contra-
1. This side-bar can be skipped without impacting your ability to read subsequent material.
g
o
g
ij
g
o
g
ij
2
2
×
i
th
j
th
1
–
(
)
i
j
+
g
ij
C
∂
g
o
∂
g
ij
---------
g
ij
C
=
A
[ ]
A
[ ]
1
–
A
[ ]
CT
det
A
[ ]
----------------
=
A
[ ]
C
det
A
[ ]
(
)
A
[ ]
T
–
=
A
[ ]
g
ij
[ ]
det
g
ij
[ ]
g
o
g
ij
[ ]
1
–
g
ij
[
]
∂
g
o
∂
g
ij
---------
g
o
g
ij
=
∂
g
o
∂
g
ij
---------
g
o
g
ij
=
∂
g
o
∂
g
ij
---------
2
J
∂
J
∂
g
ij
---------
=
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DRAFT June 17, 2004
21
variant
metric. Thus, we may solve for the derivative on the right-hand-side to obtain
(2.30)
where we have again recalled that
.
Partial Answer:
(a) (b)
(c)
,
.
(d)
=
.
∂
J
∂
g
ij
---------
1
2
---
Jg
ij
=
g
o
J
2
=
Study Question 2.2
Let
represent the ordinary orthonormal laboratory basis.
Consider the following irregular base vectors:
.
(a) Construct the metric coefficients
.
(b) Construct
by inverting
.
(c) Construct the contravariant (dual) basis by
directly using the formula of Eq. (2.15). Sketch
the dual basis in the picture at right and visually
verify that it satisfies the condition of Eq. (2.12).
(d) Confirm that the formula of Eq. (2.18)
gives the same result as derived in part (c).
(e) Redo parts (a) through (d) if
is now replaced by
.
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
g
˜
2
g
˜
1
e
ˆ
˜
1
e
ˆ
˜
2
g
˜
1
e
˜
1
2
e
˜
2
+
=
g
˜
2
e
˜
1
–
e
˜
2
+
=
g
˜
3
e
˜
3
=
g
ij
g
ij
[
]
g
ij
[ ]
g
˜
3
g
˜
3
5
e
˜
3
=
g
11
=5
g
13
=0
g
22
=2
,
,
g
11
=2 9
⁄
g
33
=1
g
21
= 1
–
9
⁄
,
,
J
=3
g
˜
2
=
1
3
---
g
˜
3
g
˜
1
×
(
)
= –
2
3
---
e
˜
1
+
1
3
---
e
˜
2
g
˜
2
g
21
g
˜
1
g
22
g
˜
2
g
23
g
˜
3
+
+
=
1
9
---
g
˜
1
–
5
9
---
g
˜
2
+
2
3
---
e
ˆ
˜
1
–
1
3
---
e
ˆ
˜
2
+
=
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DRAFT June 17, 2004
22
Partial Answer:
(a) yes -- note the direction of the third base vector (b)
A faster way to get the metric coefficients:
Suppose you already have the
with
respect to the regular laboratory basis. We now prove that the
matrix can by obtained by
. Recall that
. Therefore
,
(2.31)
The left hand side is
, and the right hand side represents the
ij
components of
with
respect to the regular laboratory basis, which completes the proof that
the covariant metric
coefficients
are equal to the
ij
laboratory
components of the tensor
Study Question 2.3
Let
represent the ordinary orthonormal laboratory basis.
Consider the following irregular base vectors:
.
(a) is this basis right handed?
(b) Compute
and
.
(c) Construct the contravariant (dual) basis.
(d) Prove that
points in the direction of the
outward normal to the upper surface of the
shaded parallelogram.
(d) Prove that (see drawing label) is the
reciprocal of the length of .
(e) Prove that the area of the face of the parallelepiped whose normal is parallel to is
given by the Jacobian times the magnitude of .
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
g
˜
2
g
˜
1
e
ˆ
˜
1
e
ˆ
˜
2
g
˜
1
θ
h
1
h
2
g
˜
1
e
˜
1
2
e
˜
2
+
=
g
˜
2
3
e
˜
1
e
˜
2
+
=
g
˜
3
7
e
˜
3
–
=
g
ij
[
]
g
ij
[ ]
g
˜
1
h
k
g
˜
k
g
˜
k
J
g
˜
k
F
ij
[
]
g
ij
[ ]
F
[ ]
T
F
[ ]
g
˜
i
F
˜˜
e
˜
i
•
=
g
˜
i
g
˜
j
•
F
˜˜
e
˜
i
•
(
)
F
˜˜
e
˜
j
•
(
)
•
e
˜
i
F
˜˜
T
•
(
)
F
˜˜
e
˜
j
•
(
)
•
e
˜
i
F
˜˜
T
F
˜˜
•
(
)
e
˜
j
•
•
=
=
=
g
ij
F
[ ]
T
F
[ ]
g
ij
F
˜˜
T
F
˜˜
•
Study Question 2.4
Using the [F] matrix from Study Question 2.1, verify that
gives the same matrix for
F
[ ]
T
F
[ ]
g
ij
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DRAFT June 17, 2004
23
Partial Answer:
(a) Substitute Eqs (2.1) and (2.32) into Eq. (2.12) and use the fact that
. (b) Follow logic similar to Eq. (2.31). (c) A correct proof must use the fact that is
invertible. Reminder: a tensor
is positive definite iff
. The dot product
positivity property states that
. (d) Easy: the determinant of a product is the product
of the determinants.
(a) By definition,
, so the fact that these dot products equal the Kronecker delta says,
for example, that
(i.e., it is a unit vector) and
is perpendicular to
, etc. (b) There is no
info about the handedness of the system. (c) By definition, the first contravariant base vector must be
Study Question 2.5
Recall that the covariant basis may be regarded as a transformation
of the right-handed orthonormal laboratory basis. Namely,
.
(a) Prove that the
contravariant
basis is obtained by the inverse transpose transformation:
,
(2.32)
where is the same as .
(b)
Prove that the contravariant metric coefficients
are equal to the
ij
laboratory
compo-
nents of the tensor
.
(c) Prove that, for any invertible tensor , the tensor
is positive definite. Explain
why this implies that the
and
matrices are positive definite.
(d) Use the result from part (b) to prove that the determinant of the covariant metric
matrix
equals the square of the Jacobian .
g
˜
i
F
˜˜
e
˜
i
•
=
g
˜
i
F
˜˜
T
–
e
˜
i
•
=
e
˜
i
e
˜
i
g
ij
F
˜˜
1
–
F
˜˜
T
–
•
S
˜˜
S
˜˜
T
S
˜˜
•
g
ij
[ ]
g
ij
g
o
g
ij
J
2
e
˜
i
e
˜
j
•
δ
j
i
=
S
˜˜
A
˜˜
u
˜
A
˜˜
u
˜
•
•
0
>
u
˜
0
˜
≠
∀
v
˜
v
˜
•
> 0 if
v
˜
0
˜
≠
= 0 iff
v
˜
=
0
˜
Study Question 2.6
SPECIAL CASE (orthonormal systems)
Suppose that the metric coefficients happen to equal the Kronecker delta:
(2.33)
(a) Explain why this condition implies that the covariant base vectors are orthonormal.
(b) Does the above equation tell us anything about the handedness of the systems?
(c) Explain why
orthonormality implies that contravariant base vectors are identically
equal to the covariant base vectors
.
g
ij
δ
ij
=
g
ij
g
˜
i
g
˜
j
•
≡
g
˜
1
1
=
g
˜
1
g
˜
2
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DRAFT June 17, 2004
24
perpendicular to the second and third covariant vectors, and it must satisfy
, which (with
some thought) leads to the conclusion that
.
Super-fast way to get the dual (contravariant) basis and metrics
the covariant basis can be connected to the lab basis through the equation
(2.34)
When we introduced this equation, we explained that the column of the lab component
matrix [F] would contain the lab components of . Later, in Study Question 2.5, Eq. (2.34), we
asserted that
,
(2.35)
Consequently, we may conclude that the column of
must contain the lab compo-
nents of . This means that the
row
of
must contain the contravariant base vec-
tor.
Partial Answer:
(a) (b)
(c), .
(d)
(e) All results the same. This way was computationally faster!
g
˜
1
g
˜
1
•
1
=
g
˜
1
g
˜
1
=
g
˜
i
F
˜˜
e
˜
i
•
=
i
th
g
˜
i
g
˜
i
F
˜˜
T
–
e
˜
i
•
=
i
th
F
[ ]
T
–
g
˜
i
i
th
F
[ ]
1
–
i
th
Study Question 2.7
Let
represent the ordinary orthonormal laboratory basis.
Consider the following irregular base vectors:
.
(a) Construct the
matrix by putting the lab compo-
nents of into the column.
(b) Find
(c) Find from the row of
.
(d) Directly compute
from the result of part (c).
(e) Compare the results with those found in earlier study questions and comment on
which method was fastest.
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
g
˜
2
g
˜
1
e
ˆ
˜
1
e
ˆ
˜
2
g
˜
1
e
˜
1
2
e
˜
2
+
=
g
˜
2
e
˜
1
–
e
˜
2
+
=
g
˜
3
e
˜
3
=
F
[ ]
g
˜
i
i
th
F
[ ]
1
–
g
˜
i
i
th
F
[ ]
1
–
g
ij
F
[ ]
1 1
– 0
2 1 0
0 0 1
=
F
[ ]
1
–
1 3
⁄
1 3
⁄
0
2 3
⁄
–
1 3
⁄
0
0
0 1
=
g
˜
2
= –
2
3
---
e
˜
1
+
1
3
---
e
˜
2
g
12
g
˜
1
g
˜
2
•
dot product of 1st and 2nd rows
1
9
---
–
=
=
=
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DRAFT June 17, 2004
25
Partial Answer:
(a) We know that
because it’s the coefficient of
in the expression
. (b) Draw a path from the origin to the tip of the vector such that the path consists of
straight line segments that are always parallel to one of the base vectors
or
. You can thereby
demonstrate graphically that
. Hence
,
(c) You’re on your own. (d) The
length of the second segment equals the magnitude of
, or
. (e) In general, when a vector is
broken into segments parallel to the covariant basis, then the length of the segment parallel to must
equal
, with no implied sum on i. Lesson here? Because the base vectors are not necessarily of
unit length, the meaning of the contravariant component must never be confused with the length of
the associated line segment! The contravariant component is merely the coefficient of in the lin-
ear expansion
. The component’s magnitude equals the length of the segment divided by the
length of the base vector!
Study Question 2.8
Consider the same irregular base vectors in Study Question 2.2.
Namely, , ,
.
Now consider three vectors, , , and , in
the 1-2 plane as shown.
(a) Referring to the sketch, graphically
demonstrate that
gives a vector
identically equal to . Thereby explain
why the contravariant components of
are: , ,
and .
(b) Using similar geometrical arguments,
find the contravariant components of .
(c) Find the contravariant components of .
(d) The path from the tail to the tip of a vec-
tor can be decomposed into parts that are parallel to the base vectors. For example, the vec-
tor can be viewed as a segment equal to
plus a segment equal to
. What are the
lengths of each of these individual segments?
(e) In general, when any given vector is broken into segments parallel to the covariant
base vectors {
, how are the lengths of these segments related (if at all) to the
contravariant components
of the vector?
g
˜
1
e
˜
1
2
e
˜
2
+
=
g
˜
2
e
˜
1
–
e
˜
2
+
=
g
˜
3
e
˜
3
=
a
˜
b
˜
g
˜
2
g
˜
1
e
ˆ
˜
1
e
ˆ
˜
2
c
˜
a
˜
b
˜
c
˜
g
˜
1
g
˜
2
–
a
˜
a
˜
a
1
=1
a
2
= 1
–
a
3
=0
b
˜
c
˜
b
˜
g
˜
1
2
g
˜
2
u
˜
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
u
1
u
2
u
3
,
,
{
}
a
1
=1
g
˜
1
a
˜
g
˜
1
g
˜
2
–
=
b
˜
g
˜
1
g
˜
2
b
˜
g
˜
1
2
g
˜
2
+
=
b
1
=1
b
2
= etc
2
g
˜
2
2 5
u
˜
g
˜
i
u
i
g
ii
u
i
u
i
g
˜
i
u
˜
u
i
g
˜
i
=
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DRAFT June 17, 2004
26
2.3 Transforming components by raising and lower indices.
At this point, we have not yet identified a procedural means of determining the contravar-
iant components
or covariant components
— we merely assert that
they
exist
. The sets
and
are each bases, so we know that we may
express any vector as
.
(2.36)
Dotting both sides of this equation by gives
.
Equation (2.12) allows us to simplify the second term as
. Using
Eq. (2.19), the third term simplifies as
, where the last step utilized
the fact that the
matrix is symmetric. Thus, we conclude
(2.37)
.
(2.38)
Similarly, by dotting both sides of Eq. (2.36) by you can show that
(2.39)
.
(2.40)
These very important equations provide a means of transforming between types of compo-
nents.
In Eq. (2.40), note how the
matrix effectively “lowers” the index on , changing it
to an “
i
”. Similarly, in Eq. (2.38), the
matrix “raises” the index on , changing it to an “
i
”.
Thus the metric coefficients serve a role that is very similar to that of the Kronecker delta in
orthonormal theory.
Incidentally, note that Eqs. (2.17) and (2.18) are also examples of raising
a
1
a
2
a
3
,
,
{
}
a
1
a
2
a
3
,
,
{
}
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
a
˜
a
˜
a
k
g
˜
k
a
k
g
˜
k
=
=
g
˜
i
a
˜
g
˜
i
•
a
k
g
˜
k
g
˜
i
•
a
k
g
˜
k
g
˜
i
•
=
=
a
k
g
˜
k
g
˜
i
•
a
k
δ
k
i
a
i
=
=
a
k
g
˜
k
g
˜
i
•
a
k
g
ki
g
ik
a
k
=
=
g
ik
a
i
a
˜
g
˜
i
•
=
a
i
g
ik
a
k
=
g
˜
i
a
i
a
˜
g
˜
i
•
=
a
i
g
ik
a
k
=
g
ik
a
k
g
ik
a
k
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DRAFT June 17, 2004
27
and lowering indices.
Partial Answer:
For the first one, the index “m” is repeated. The
allows you to lower the super-
script “m” on
so that it becomes an “l” when the metric coefficient is removed. The final simplified
result for the first expression is . Be certain that your final simplified results still have the same free
indices on the same level as they were in the original expression. The very last expression has a twist:
recognizing the repeated index , we can lower the index on
and turn it into an “i” when we
remove
so that a simplified expression becomes
. Alternatively, if we prefer to remove
the
, we could raise the index on
, making it into a “k” so that an alternative simplification of
the very last expression becomes
. Either method gives the same result. The final simplifi-
cation comes from recalling that the mixed metric components are identically equal to the Kronecker
delta. Thus,
, which is merely one of the expressions given in Eq. (2.20).
Finding contravariant vector components — classical method
we used geometrical methods to find the contravariant components of several vectors. Eqs
(2.37) through (2.40) provide us with an
algebraic
means of determining the contravariant and
covariant components of vectors. Namely, given a basis
and a vector , the con-
tra- and co-variant components of the vector are determined as follows:
STEP 1. Compute the covariant metric coefficients
.
STEP 2. Compute the contravariant metric coefficients
by inverting
.
STEP 3. Compute the contravariant basis
.
STEP 4. Compute the covariant components
.
STEP 5. Compute the contravariant components
. Alternatively,
.
Finding contravariant vector components — accelerated method
If the lab components
of the vector are available, then you can quickly compute the covariant and contravariant
components by noting that
and, similarly,
. The steps are as follows:
STEP 1. Construct the
matrix by putting the lab components of into the
column.
STEP 2. The covariant components are found from
, using lab components.
STEP 3. The contravariant components are found from
, using lab components.
Study Question 2.9
Simplify the following expressions so that there are no metric coeffi-
cients:
, , , , ,
(2.41)
a
m
g
ml
g
pk
u
p
f
n
g
ni
r
i
g
ij
s
j
g
ij
b
k
g
kj
g
ij
g
jk
g
ml
a
m
a
l
j
j
g
jk
g
ij
g
ij
g
jk
g
i
k
=
g
jk
j
g
ij
g
ij
g
jk
g
i
k
=
g
ij
g
jk
δ
i
k
=
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
a
˜
g
ij
g
˜
i
g
˜
j
•
=
g
ij
g
ij
[ ]
g
˜
i
g
ij
g
˜
j
=
a
i
a
˜
g
˜
i
•
=
a
i
a
˜
g
˜
i
•
=
a
i
g
ij
a
j
=
a
˜
a
i
a
˜
g
˜
i
•
a
˜
F
˜˜
e
˜
i
•
•
F
˜˜
T
a
˜
•
(
)
e
˜
i
•
=
=
=
a
i
F
˜˜
1
–
a
˜
•
(
)
e
˜
i
•
=
F
[ ]
g
˜
i
i
th
F
[ ]
T
a
{ }
F
[ ]
1
–
a
{ }
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DRAFT June 17, 2004
28
Partial Answer:
Classical method: First get the contra- and co-variant base vectors, then apply the
formulas:
, and so on giving
.
, and so on giving
.
ACCELERATED METHOD:
covariant components:
. . . agrees!
contravariant components:
. . . agrees!
By applying techniques similar to those used to derive Eqs. (2.37) through (2.40), the follow-
ing analogous results can be shown to hold for tensors:
(2.42)
(2.43a)
(2.43b)
(2.43c)
.
(2.43d)
Again, observe how the metric coefficients play a role similar to that of the Kronecker delta
for orthonormal bases. For example, in Eq. (2.43a), the expression
“becomes”
as
follows: upon seeing
, you look for a subscript
i
or
m
on
T
. Finding the subscript
m
, you
replace it with an
i
and change its level from a subscript to a superscript. The metric coeffi-
cient
similarly raises the subscript
n
on
to a become a superscript
j
on
.
Incidentally, if the
matrix and lab components
are available for , then
, , ,
and
(2.44)
These are nifty quick-answer formulas, but for general discussions, Eqs. (2.43) are really more
meaningful and those equations are applicable even when you
don’t
have lab components.
Study Question 2.10
Using the basis and vectors from Study Question 2.8, apply the clas-
sical step-by-step algorithm to compute the covariant and contravariant components of
the vectors, and . Check whether the accelerated method gives the same answer. Be
sure to verify that your contravariant components agree with those determined geometri-
cally in Question 2.8.
a
˜
b
˜
a
2
a
˜
g
˜
2
•
2
e
˜
1
e
˜
2
+
(
)
e
˜
1
–
e
˜
2
+
(
)
•
1
–
=
=
=
a
1
a
2
a
3
,
,
{
}
4
1
–
0
,
,
{
}
=
a
2
a
˜
g
˜
2
•
2
e
˜
1
e
˜
2
+
(
)
2
3
---
e
˜
1
–
1
3
---
e
˜
2
+
(
)
•
1
–
=
=
=
a
1
a
2
a
3
,
,
{
}
1
1
–
0
,
,
{
}
=
F
[ ]
T
a
{ }
lab
1 2 0
1
– 1 0
0 0 1
2
1
0
4
1
–
0
=
=
F
[ ]
1
–
a
{ }
lab
1 3
⁄
1 3
⁄
0
2 3
⁄
–
1 3
⁄
0
0
0 1
2
1
0
1
1
–
0
=
=
T
˜˜
T
ij
g
˜
i
g
˜
j
T
ij
g
˜
i
g
˜
j
T
•
j
i
g
˜
i
g
˜
j
T
i
•
j
g
˜
i
g
˜
j
=
=
=
=
T
ij
g
˜
i
T
˜˜
g
˜
j
•
•
T
mn
g
im
g
jn
T
•
k
i
g
kj
T
k
•
j
g
ki
=
=
=
=
T
ij
g
˜
i
T
˜˜
g
˜
j
•
•
T
mn
g
mi
g
nj
T
•
j
k
g
ki
T
i
•
k
g
kj
=
=
=
=
T
•
j
i
g
˜
i
T
˜˜
g
˜
j
•
•
T
ik
g
kj
T
kj
g
ki
T
m
•
n
g
mi
g
nj
=
=
=
=
T
i
•
j
g
˜
i
T
˜˜
g
˜
j
•
•
T
kj
g
ki
T
ik
g
kj
T
•
n
m
g
mi
g
nj
=
=
=
=
T
mn
g
im
g
jn
T
ij
g
im
g
jn
T
mn
T
ij
F
[ ]
T
[ ]
lab
T
˜˜
T
ij
[ ]
F
[ ]
1
–
T
[ ]
lab
F
[ ]
T
–
=
T
ij
[ ]
F
[ ]
T
T
[ ]
lab
F
[ ]
=
T
•
j
i
[
]
F
[ ]
1
–
T
[ ]
lab
F
[ ]
=
T
i
•
j
[
]
F
[ ]
T
T
[ ]
lab
F
[ ]
T
–
=
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DRAFT June 17, 2004
29
Our final index changing properties involve the
mixed
Kronecker delta itself.
The mixed
Kronecker delta changes the index symbol
without
changing its level.
For example,
(2.45)
etc.
(2.46)
Contrast this with the metric coefficients
and
, which change both the symbol and the
level of the sub/superscripts.
Partial Answer:
(a) Seeing the
, you can get rid of it if you can find a subscript “i” or “m.” Find-
ing “i” as a subscript on H, you raise its level and change it to an “m,” leaving a “dot” placeholder
where the old subscript lived. After similarly eliminating the
, the final simplified result is
.
(b)
(c) For the first and last “g” factor, you can lower the “i” to make it an “m” and then raise
the “m” to make it a “p.” Alternatively, you can recognize the opportunity to apply Eq. (2.20) so that
, and then Eq. (2.46) allows you to change the original superscript “i” on the H tensor to
a “p” leaving its level unchanged. (d) The i and j are already on the upper level, so you leave them
alone. To raise the “k” subscript, you multiply by
. The “p” is raised similarly so that the final
result is
. (g) Yes, of course!
δ
i
j
v
i
δ
i
j
v
j
=
v
i
δ
j
i
v
j
=
T
ij
δ
i
k
T
kj
=
δ
j
n
T
i
•
j
δ
m
i
T
m
•
n
=
g
ij
g
ij
Study Question 2.11
Let denote a fourth-order tensor. Use the method of raising and
lowering indices to simplify the following expressions so that there are no metric coeffi-
cients or Kronecker deltas (
,
, or ) present.
(a)
(b)
(c)
Multiply the following covariant, or mixed components by appropriate combinations of the
metric coefficients to convert them all to pure contravariant components (
i.e.,
all super-
scripts).
(d)
(e)
(f)
(g) Are we having fun yet?
H
~
~
~
~
g
ij
g
ij
δ
i
j
H
ijkl
g
im
g
jn
δ
m
q
H
ipqn
g
is
g
nj
H
•
j
•
l
i
•
k
•
g
im
g
jn
g
mp
H
••
kp
ij
••
H
ijkl
H
•
j
•
l
i
•
k
•
g
im
g
jn
H
••
kl
mn
••
H
•
pm
•
s
••
j
g
im
g
mp
δ
i
p
=
g
km
H
ijmn
H
••
kp
ij
••
g
km
g
pn
=
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DRAFT June 17, 2004
30
3. Tensor algebra for general bases
A general basis is one that is not restricted to be orthogonal, normalized, or right-handed.
Throughout this section, it is important to keep in mind that
structured (direct) notation for-
mulas remain unchanged
. Here we show how the
component
formulas take on different forms
for general bases that are permissibly nonorthogonal, non-normalized, and/or non-right-
handed.
3.1 The vector inner (“dot”) product for general bases.
Presuming that the contravariant components of two vectors and are known, we may
expand the direct notation for the dot product as
.
(3.1)
Substituting the definition (2.11) into Eq. (3.1), the formula for the dot product becomes
= .
(3.2)
This result can be written in other forms by raising and lowering indices. Noting, for example,
that
, we obtain a much simpler formula for the dot product
= .
(3.3)
The simplicity of this formula and its close similarity to the familiar formula from orthonor-
mal theory is one of the principal reasons for introducing the dual bases.
We can lower the
index on by writing it as
so we obtain another formula for the dot product:
= .
(3.4)
Finally, we recognize the index raising combination
, to obtain yet another for-
mula:
= .
(3.5)
Whenever new results are derived for a general basis, it is always advisable to ensure that
they all reduce to a familiar form whenever the basis is orthonormal. For the special case that
the basis
happens to be orthonormal, we note that
. Furthermore, for an
orthonormal basis, there is no difference between contravariant and covariant components.
Thus Eqs. 3.2, 3.3, 3.4, and 3.5 do indeed all reduce to the usual orthonormal formula.
a
˜
b
˜
a
˜
b
˜
•
a
i
g
˜
i
(
)
b
j
g
˜
j
(
)
•
a
i
b
j
g
˜
i
g
˜
j
•
(
)
=
=
a
˜
b
˜
•
a
i
b
j
g
ij
=
a
1
b
1
g
11
a
1
b
2
g
12
a
1
b
3
g
13
a
2
b
1
g
21
…
a
3
b
3
g
33
+
+
+
+
+
b
j
g
ij
b
i
=
a
˜
b
˜
•
a
i
b
i
=
a
1
b
1
a
2
b
2
a
3
b
3
+
+
a
i
g
ki
a
k
a
˜
b
˜
•
a
k
b
i
g
ki
=
a
1
b
1
g
11
a
1
b
2
g
12
a
1
b
3
g
13
a
2
b
1
g
21
…
a
3
b
3
g
33
+
+
+
+
+
b
i
g
ki
b
k
=
a
˜
b
˜
•
a
i
b
i
=
a
1
b
1
a
2
b
2
a
3
b
3
+
+
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
g
ij
=
δ
ij
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DRAFT June 17, 2004
31
Partial Answer:
(a)
(b)
(c) yes, of course it agrees!
Index contraction
The vector dot product is a special case of a more general operation,
which we will call index contraction. Given two free indices in an expression, we say that we
“contract” them when they are turned into dummy summation indices according to the fol-
lowing rules:
1. if the indices started at the same level, multiply them by a metric coefficient
2. if the indices are at different levels, multiply by a Kronecker delta (and simplify).
In both cases, the indices on the metric coefficient or Kronecker delta must be the same sym-
bol as the indices being contracted, but at different levels so that they become dummy sum-
mation symbols upon application of the summation conventions. The symbol
denotes
contraction of the first and second indices;
would denote contraction of the second and
eighth indices (assuming, of course, that the expression has eight free indices to start with!).
Index contraction is
actually
basis contraction and it is the ordering of the basis that dictates
the contraction.
Consider, for example, the expression
. This expression has six free indices, and
we will suppose that it therefore corresponds to a sixth-order tensor with the associated base
vectors being ordered the same as the free indices
. To contract the first and sec-
ond indices ( and ), which lie on different levels, we must dot the first and second base vec-
tors into themselves, resulting in the Kronecker delta . Thus, contracting the first and
Study Question 3.1
Consider the following nonorthonormal base vectors:
and
Consider two vectors, and , lying in the
1-2 plane as drawn to scale in the figure.
(a) Express and in terms of the lab
basis.
(b) Compute
in the ordinary familiar
manner by using lab components.
(c) Use Eqs. (3.2) and (3.3) to compute
.
Does the result agree with part (b)?
a
˜
b
˜
g
˜
2
g
˜
1
e
˜
1
e
˜
2
g
˜
1
e
˜
1
2
e
˜
2
+
=
g
˜
2
e
˜
1
–
e
˜
2
+
=
g
˜
3
e
˜
3
=
a
˜
b
˜
a
˜
b
˜
e
˜
a
˜
b
˜
•
a
˜
b
˜
•
a
˜
2
e
˜
1
e
˜
2
+
=
a
˜
b
˜
•
2
=
C
1
2
C
2
8
v
i
W
jk
Z
lmn
g
˜
i
g
˜
j
g
˜
k
g
˜
l
g
˜
m
g
˜
n
i
j
δ
j
i
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DRAFT June 17, 2004
32
second indices of
gives
, which simplifies to
. Note that the
contraction operation has reduced the order of the result from a sixth-order tensor down to a
fourth-order tensor. To compute the contraction
, we must dot the fourth base vector
into the sixth base vector
, which results in
. Thus, contracting the fourth and sixth indi-
ces of
gives
. To summarize,
Index contraction is equivalent to dotting together the base vectors associated with the identified indices.
denotes contraction of the
and
base vectors.
No matter what basis is used to expand a tensor, the result of a contraction operation will be
the same. In other words, the contraction operation is invariant under a change in basis. By
this, we mean that you can apply the contract and then change the basis or vice versa — the
result will be the same.
Incidentally, note that the vector dot product
can be expressed in terms of a contrac-
tion as
operating on the
dyad
. The formal
notation for index contraction is rarely
used, but the
phrase
“index contraction” is very common.
3.2 Other dot product operations
Using methods similar to those in Section 3.1, indicial forms of the operation
are
found to be
etc.
In all these formulas, the dot product results in a summation between adjacent indices on
opposite levels.
If you know only
and , you must first raise an index to apply the dot
product. For example,
.
(3.6)
In general, the correct indicial expression for any operation involving dot products can be
derived by starting with the direct notation and expanding each argument in whatever form
happens to be available.
This approach typically leads to the opportunity to apply one or
more of Eqs. (2.11), (2.12), or (2.19). For example, suppose you know two tensors in the forms
and
. Then
.
(3.7)
Note the change in dummy sum indices from
ij
to
mn
required to avoid violation of the sum-
mation conventions. For the final step, we merely applied Eq. (2.11). If desired, the above
result may be simplified by using the
to lower the “
j
” superscript on
(changing it to an
v
i
W
jk
Z
lmn
δ
j
i
v
i
W
jk
Z
lmn
v
i
W
ik
Z
lmn
C
4
6
g
˜
l
g
˜
n
g
ln
v
i
W
jk
Z
lmn
g
ln
v
i
W
jk
Z
lmn
C
α
β
α
th
β
th
u
˜
v
˜
•
C
1
2
u
˜
v
˜
C
α
β
T
˜˜
v
˜
•
T
˜˜
v
˜
•
T
ij
v
j
g
˜
i
T
ij
v
j
g
˜
i
T
•
j
i
v
j
g
˜
i
=
=
=
T
ij
v
k
T
˜˜
v
˜
•
T
ij
v
k
g
kj
g
˜
i
=
A
˜˜
A
ij
g
˜
i
g
˜
j
=
B
˜˜
B
•
j
i
g
˜
i
g
˜
j
=
A
˜˜
B
˜˜
•
A
ij
g
˜
i
g
˜
j
(
)
B
•
n
m
g
˜
m
g
˜
n
(
)
•
=
A
ij
B
•
n
m
g
˜
i
g
˜
j
g
˜
m
•
(
)
g
˜
n
=
A
ij
B
•
n
m
g
jm
g
˜
i
g
˜
n
=
g
jm
A
ij
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DRAFT June 17, 2004
33
“
m
”) so that
.
(3.8)
Alternatively, we could have used the
to lower the “
m
” superscript on
B
(changing it to
a “
j
”) so that
.
(3.9)
These are all valid expressions for the composition of two tensors. Note that
the final result in
the last three equations involved components times the basis dyad
. Hence those compo-
nents represent the mixed “
” components of
.
3.3 The transpose operation
The transpose
of a tensor
is defined in direct notation such that
for all vectors and .
(3.10)
Since this must hold for all vectors, it must hold for any particular choice. Taking
and
, we see that
.
(3.11)
Similarly, you can show that
. (3.12)
A
˜˜
B
˜˜
•
A
•
m
i
B
•
n
m
g
˜
i
g
˜
n
=
g
jm
A
˜˜
B
˜˜
•
A
ij
B
jn
g
˜
i
g
˜
n
=
g
˜
i
g
˜
n
i
•
n
A
˜˜
B
˜˜
•
Study Question 3.2
Complete the following table, which shows the indicial components
for all of the
sixteen possible ways to express the operation
, depending on what
type of components are available for and . The shaded cells have the simplest form
because they do not involve metric coefficients.
A
˜˜
B
˜˜
•
A
˜˜
B
˜˜
B
nj
g
˜
n
g
˜
j
B
nj
g
˜
n
g
˜
j
B
•
j
n
g
˜
n
g
˜
j
B
n
•
j
g
˜
n
g
˜
j
A
im
g
˜
i
g
˜
m
A
im
B
nj
g
mn
g
˜
i
g
˜
j
A
im
B
mj
g
˜
i
g
˜
j
A
im
B
n
•
j
g
mn
g
˜
i
g
˜
j
A
im
g
˜
i
g
˜
m
A
im
B
mj
g
˜
i
g
˜
j
A
im
B
•
j
n
g
mn
g
˜
i
g
˜
j
A
•
m
i
g
˜
i
g
˜
m
A
•
m
i
B
•
j
m
g
˜
i
g
˜
j
A
•
m
i
B
n
•
j
g
mn
g
˜
i
g
˜
j
A
i
•
m
g
˜
i
g
˜
m
A
i
•
m
B
mj
g
˜
i
g
˜
j
A
i
•
m
B
nj
g
mn
g
˜
i
g
˜
j
B
˜˜
T
B
˜˜
u
˜
B
˜˜
T
v
˜
•
•
v
˜
B
˜˜
u
˜
•
•
=
u
˜
v
˜
u
˜
=
g
˜
i
v
˜
=
g
˜
j
B
˜˜
T
( )
ij
B
ji
=
B
˜˜
T
( )
ij
B
ji
=
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DRAFT June 17, 2004
34
The mixed components of the transpose are more different.
Namely, taking
and
,
.
(3.13)
Note that the high/low level of the indices remains unchanged. Only their ordering is
switched.
Similarly,
.
(3.14)
All of the above relations could have been derived in the following alternative manner: The
transpose
of any dyad is simply
. Therefore, knowing that the transpose of a sum is
the sum of the transposes and that any tensor can be written as a sum of dyads, we can write:
.
(3.15)
The coefficient of
is
. Thus,
A tensor
is symmetric only if it equals its own transpose. Therefore, referring to
Eqs. the components of a symmetric tensor satisfy
, , .
(3.16)
The last relationship shows that the “dot placeholder” is unnecessary for symmetric tensors,
and we may write simply without ambiguity.
3.5 The identity tensor for a general basis.
The identity tensor
is the unique symmetric tensor for which
for all vectors and .
(3.17)
Since this must hold for all vectors, it must hold for any particular choice. Taking
and
, we find that
. The left-hand side represents the
ij
covariant compo-
nents of the identity, and the right hand side is
. By taking
and
, it can be simi-
larly shown that the contravariant components of the identity are
. By taking
and
, the mixed components of the identity are found to equal . Thus,
(3.18)
This result further validates raising and lowering indices by multiplying by appropriate met-
ric coefficients — such an operation merely represents dotting by the identity tensor.
u
˜
=
g
˜
i
v
˜
=
g
˜
j
B
˜˜
T
( )
i
•
j
B
•
i
j
=
B
˜˜
T
( )
•
j
i
B
j
•
i
=
u
˜
v
˜
(
)
T
v
˜
u
˜
B
˜˜
T
B
ij
g
˜
i
g
˜
j
(
)
T
B
ij
g
˜
i
g
˜
j
(
)
T
B
ij
g
˜
j
g
˜
i
=
=
=
g
˜
i
g
˜
j
B
ij
B
˜˜
T
(
)
ji
B
ij
=
A
˜˜
A
ij
A
ji
=
A
ij
A
ji
=
A
•
j
i
A
j
•
i
=
A
i
j
I
˜˜
u
˜
I
˜˜
v
˜
•
•
u
˜
v
˜
•
=
u
˜
v
˜
u
˜
=
g
˜
i
v
˜
=
g
˜
j
g
˜
i
I
˜˜
g
˜
j
•
•
g
˜
i
g
˜
j
•
=
g
ij
u
˜
=
g
˜
i
u
˜
=
g
˜
j
g
ij
u
˜
=
g
˜
i
u
˜
=
g
˜
j
δ
i
j
I
˜˜
g
ij
g
˜
i
g
˜
j
g
ij
g
˜
i
g
˜
j
δ
i
j
g
˜
i
g
˜
j
δ
j
i
g
˜
i
g
˜
j
=
=
=
=
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DRAFT June 17, 2004
35
3.6 Eigenproblems and similarity transformations
The
eigenproblem
for a tensor
requires the determination of all eigenvectors and
corresponding eigenvalues
λ
such that
.
(3.19)
In this form, the eigenvector is called the
“right” eigenvector
. We can also define a
“left”
eigenvector
such that
.
(3.20)
In other words, the
left eigenvectors are the right eigenvectors of
. The characteristic
equation for
is the same as that for
, so
the eigenvalues are the same for both the
right and left eigenproblems
. Consider a particular eigenpair
:
.
(no sum on
k
)
(3.21)
Dot from the left by a left eigenvector
, noting that
(no sum on
m
). Then
.
(no sum on
k
)
(3.22)
Rearranging,
. (no
sums)
(3.23)
From this result we conclude that the
left and right eigenvectors corresponding distinct
(
) eigenvalues are orthogonal
. This motivates
renaming the eigenvectors using dual
basis notation
:
Rename .
Rename .
(3.24)
The magnitudes of eigenvectors are arbitrary, so we can
select normalization such that the
renamed vectors are truly dual bases:
.
(3.25)
Nonsymmetric tensors might not possess a complete set of eigenvectors (i.e., the geometric
multiplicity of eigenvalues might be less than their algebraic multiplicity). If, however, the
tensor
happens to possess a complete (or “
spanning
”) set of eigenvectors, then those
eigenvectors form an acceptable basis.
The mixed components of
with respect to this
principal basis are then
.
(no sum on
j
)
(3.26)
Ah!
Whenever a tensor
possesses a complete set of eigenvectors, it is diagonal in its
mixed
T
˜˜
u
˜
T
˜˜
u
˜
•
λ
u
˜
=
u
˜
v
˜
v
˜
T
˜˜
•
λ
v
˜
=
T
˜˜
T
T
˜˜
T
T
˜˜
λ
k
u
˜
k
,
(
)
T
˜˜
u
˜
k
•
λ
k
u
˜
k
=
v
˜
m
v
˜
m
T
˜˜
•
λ
m
v
˜
m
=
λ
m
v
˜
m
u
˜
k
•
λ
k
v
˜
m
u
˜
k
•
=
λ
m
λ
k
–
(
)
v
˜
m
u
˜
k
•
(
)
0
=
λ
m
λ
k
≠
u
˜
k
p
˜
k
=
v
˜
m
p
˜
m
=
p
˜
m
p
˜
k
•
δ
k
m
=
T
˜˜
T
˜˜
T
•
j
i
p
˜
i
T
˜˜
p
˜
j
•
•
p
˜
i
λ
j
p
˜
j
(
)
•
λ
j
δ
j
i
=
=
=
T
˜˜
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DRAFT June 17, 2004
36
principal basis!
Stated differently,
.
(3.27)
or
.
(3.28)
In matrix analysis books, this result is usually presented as a similarity transformation. As
we can invoke the existence of a basis transformation tensor
, such that
and
.
(3.29)
The columns of the matrix of
with respect to the laboratory
basis are simply
the right eigenvectors expressed in the lab basis:
.
(3.30)
,
(3.31)
where
.
(3.32)
T
•
j
i
[
]
λ
1
0 0
0
λ
2
0
0 0
λ
3
p
˜
i
p
˜
j
=
T
˜˜
λ
k
p
˜
k
p
˜
k
k
1
=
3
∑
=
F
˜˜
p
˜
k
F
˜˜
e
˜
k
•
=
p
˜
k
F
˜˜
T
–
e
˜
k
•
=
F
˜˜
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
F
˜˜
[ ]
e
˜
e
˜
p
˜
1
{ }
e
˜
p
˜
2
{ }
e
˜
p
˜
3
{ }
e
˜
[
]
=
T
˜˜
λ
k
F
˜˜
e
˜
k
•
(
)
F
˜˜
T
–
e
˜
k
•
(
)
k
1
=
3
∑
=
F
˜˜
Λ
˜˜
F
˜˜
1
–
•
•
Λ
˜˜
λ
k
e
˜
k
e
˜
k
k
1
=
3
∑
λ
1
0 0
0
λ
2
0
0 0
λ
3
e
˜
k
e
˜
k
→
≡
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DRAFT June 17, 2004
37
Partial Answer:
.
The above discussion focused on eigenproblems. We now mention the more general relation-
ship between similar components and similarity transformations.
When two tensors
and
possess the same components but with respect to different
mixed
bases, i.e., when
and
,
(identical
mixed
components)
(3.34)
then the tensors are similar. In other words, there exists a transformation tensor
such that
and
therefore
,
(3.35)
In the context of continuum mechanics, the transformation tensor
is typically the defor-
mation gradient tensor.
It is important to have at least a vague notion of this concept in order
to communicate effectively with researchers who prefer to do all analyses in general curvilin-
ear coordinates.
To them, the discovery that two tensors have identical mixed components
with respect to different bases has great significance, whereas you might find it more mean-
ingful to recognize this situation as merely a similarity transformation.
Study Question 3.3
Consider a tensor having components with respect to the orthonor-
mal laboratory basis given by
.
(3.33)
Prove that this tensor has a spanning set of eigenvectors
. Find the labora-
tory components of the basis transformation tensor
, such that
. Also
is similar to a tensor that is diagonal in the laboratory basis,
T
˜˜
[ ]
2 1
2
–
1 2
2
–
0 0 2
e
˜
k
e
˜
k
=
p
˜
1
p
˜
2
p
˜
3
,
,
{
}
F
˜˜
p
˜
k
F
˜˜
e
˜
k
•
=
T
˜˜
T
˜˜
F
˜˜
[ ]
1 2 1
1
–
2 1
0 1 0
=
T
˜˜
S
˜˜
T
˜˜
α
•
j
i
g
˜
i
g
˜
j
=
S
˜˜
α
•
j
i
G
˜
i
G
˜
j
=
F
˜˜
g
˜
k
F
˜˜
G
˜
k
•
=
T
˜˜
F
˜˜
S
˜˜
F
˜˜
1
–
•
•
=
F
˜˜
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DRAFT June 17, 2004
38
3.7 The alternating tensor
When using a nonorthonormal or non-right-handed basis, it is a good idea to use a differ-
ent symbol for the alternating tensor so that the permutation symbol
may retain its
usual meaning. As we did for the Kronecker delta, we will assign the same meaning to the
permutation symbol
regardless of the contra/covariant level of the indices.
Namely,
=
.
(3.39)
Important: These quantities are all defined the same regardless of the contra- or co- level at
which the indices are placed.
The above defined permutation symbols are the components of
the alternating tensor with respect to any regular
right-handed
orthonormal laboratory basis,
so they do not transform co/contravariant level via the metric tensors. It is not allowed to
raise or lower indices on the ordinary permutation symbol with the metric tensors.
A similar
situation was encountered in connection with the Kronecker delta of Eq. (2.10).
Through the use of the permutation symbol, the three formulas written explicitly in Eq.
(2.15) can be expressed compactly as a
single
indicial equation:
.
(3.40)
In terms of an orthonormal right-handed basis, the alternating tensor is
.
(3.41)
Study Question 3.4
Suppose two tensors
and
possess the
same
contravariant
components with respect to
different
bases.
In other words,
and
(same contravariant components)
(3.36)
Demonstrate by direct substitution that
,
(3.37)
where
is a basis transformation tensor defined such that
.
(3.38)
T
˜˜
S
˜˜
T
˜˜
β
ij
g
˜
i
g
˜
j
=
S
˜˜
β
ij
G
˜
i
G
˜
j
=
T
˜˜
F
˜˜
S
˜˜
F
˜˜
T
•
•
=
F
˜˜
g
˜
k
F
˜˜
G
˜
k
•
=
ξ
˜˜˜
ε
ijk
ε
ijk
ε
ijk
ε
ij
•
••
k
etc.
,
,
,
+1 if
ijk
123 231 312
,
,
=
–1 if
ijk
321 213 132
,
,
=
0 otherwise
ε
ijm
g
˜
m
1
J
---
g
˜
i
g
˜
j
×
(
)
=
ξ
˜˜˜
ε
ijk
e
˜
i
e
˜
j
e
˜
k
=
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DRAFT June 17, 2004
39
In terms of a general basis, the basis form for the alternating tensor
is
,
(3.42)
where
, , , etc.
(3.43)
.
(3.44)
Thus, simplifying the last expression,
.
(3.45)
Hence, the covariant alternating tensor components simply equal to the permutation symbol
times the Jacobian. This result could have been derived in the following alternative way: Sub-
stituting Eq. (2.1) into Eq. (3.43) and using the direct notation definition of determinant gives
= [
,
,
] =
.
(3.46)
ξ
˜˜˜
ξ
ijk
g
˜
i
g
˜
j
g
˜
k
ξ
ijk
g
˜
i
g
˜
j
g
˜
k
ξ
ij
•
••
k
g
˜
i
g
˜
j
g
˜
k
etc.
=
=
=
=
ξ
ijk
g
˜
i
g
˜
j
g
˜
k
, ,
[
]
=
ξ
ijk
g
˜
i
g
˜
j
g
˜
k
, ,
[
]
=
ξ
ij
•
••
k
g
˜
i
g
˜
j
g
˜
k
, ,
[
]
=
ξ
ijk
g
˜
i
g
˜
j
×
(
)
g
˜
k
•
J
ε
ijm
g
˜
m
(
)
g
˜
k
•
J
ε
ijm
δ
k
m
=
=
=
ξ
ijk
J
ε
ijk
=
ξ
ijk
F
˜˜
e
˜
i
•
F
˜˜
e
˜
j
•
F
˜˜
e
˜
k
•
J
e
˜
i
e
˜
j
e
˜
k
, ,
[
]
J
ε
ijk
=
Study Question 3.5
Use Eq. (2.32), in Eq. (3.43) to prove that
(3.47)
ξ
ijk
1
J
---
ε
ijk
=
Study Question 3.6
Consider a basis that is orthonormal but left-handed (see Eq. 2.8).
Prove that
.
(3.48)
This is why some textbooks claim to “define” the permutation symbol to be its negative
when the basis is left-handed. We do
not
adopt this practice. We keep the permutation
symbol
unchanged. For a left-handed basis, the permutation symbol stays the same, but
the components of the alternating
tensor
change sign.
ξ
ijk
ε
ijk
–
=
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DRAFT June 17, 2004
40
3.8 Vector valued operations
CROSS PRODUCT
In direct notation, the cross product is defined
.
(3.49)
Hence,
.
(3.50)
Alternatively,
.
(3.51)
Axial vectors
Some textbooks choose to identify certain vectors such as angular velocity as
“different” because, according to these books, they take on different form in a left-handed
basis. This viewpoint is objectionable since it intimates that physical quantities are affected by
the choice of bases. Those textbooks handle these supposedly different “axial” vectors by
redefining the left-handed cross-product to be negative of the right-hand definition. Malvern
suggests alternatively redefining the permutation symbol to be its negative for left-hand
basis. Malvern’s suggested sign change is handled automatically by defining the alternating
tensor as we have above. Specifically Eqs. (3.44) and (2.8) show that the alternating tensor
components
will automatically change sign for left-handed systems. Hence, with this
approach, there is no need for special formulas for left-hand bases.
3.9 Scalar valued operations
TENSOR INNER (double dot) PRODUCT
The inner product between two dyads,
and
is a scalar defined
(3.52)
When applied to base vectors, this result tells us that
etc.
(3.53)
The “etc” stands for the many other ways we could possibly mix up the level of the indices;
the basic trend should be clear.
The inner product between two tensors, and , is defined to be distributive over addi-
tion. In other words, each tensor can be expanded in terms of known components times base
vectors and then Eq. (3.53) can be applied to the base vectors. If, for example, the mixed
u
˜
v
˜
×
ξ
:u
˜
v
˜
=
~
~
~
u
˜
v
˜
×
ξ
ijk
u
j
v
k
g
˜
i
=
u
˜
v
˜
×
ξ
ijk
u
j
v
k
g
˜
i
=
ξ
ijk
a
˜
b
˜
r
˜
s
˜
a
˜
b
˜
(
)
:
r
˜
s
˜
( )
a
˜
r
˜
•
(
)
b
˜
s
˜
•
(
)
≡
g
˜
i
g
˜
j
(
)
: g
˜
m
g
˜
n
(
)
g
im
g
jn
=
g
˜
i
g
˜
j
(
)
: g
˜
m
g
˜
n
(
)
δ
i
m
δ
j
n
=
g
˜
i
g
˜
j
(
)
: g
˜
m
g
˜
n
(
)
δ
m
i
δ
n
j
=
g
˜
i
g
˜
j
(
)
: g
˜
m
g
˜
n
(
)
g
im
g
jn
=
g
˜
i
g
˜
j
(
)
: g
˜
m
g
˜
n
(
)
g
im
δ
n
j
=
g
˜
i
g
˜
j
(
)
: g
˜
m
g
˜
n
(
)
δ
i
m
g
jn
=
A
˜˜
B
˜˜
A
•
j
i
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DRAFT June 17, 2004
41
components of and the contravariant
components of are known, then the inner
product between and is
(3.54)
Stated differently, to find the inner product between two tensors, and , you should con-
tract
1
the adjacent indices
pairwise
so that the first index in is contracted with the first index
in and the second index in is contracted with the second index in . To clarify, here are
some expressions for the tensor inner product for various possibilities of which components
of each tensor are available:
(3.55)
Note that
(3.56)
where “tr” is the trace operation.
TENSOR MAGNITUDE
The magnitude of a tensor is defined
(3.57)
Based on the result of Eq. (3.55), note that the magnitude of a tensor is
not
found by simply
summing the squares of the tensor components (and rooting the result). Instead, the tensor
magnitude is computed by rooting the summation running over each component of multi-
plied by its
dual
component. If, for example, the components
are known, then the dual
components
must be computed by
so that
(3.58)
TRACE
The trace an operation in which the two base vectors of a second order tensor are
contracted
2
together, resulting in a scalar. In direct notation, the trace of a tensor can be
defined
. There are four ways to write the tensor :
.
(3.59)
The double dot operation is distributive over addition. Furthermore, for any vectors and ,
. Therefore, the trace is found by contracting the base vectors to give
,
(3.60)
1. The definition of index “contraction” is given on page 31.
2. The definition of index “contraction” is given on page 31.
A
˜˜
B
mn
B
˜˜
A
˜˜
B
˜˜
A
˜˜
:B
˜˜
A
•
j
i
g
˜
i
g
˜
j
(
)
:
B
mn
g
˜
m
g
˜
n
(
)
A
•
j
i
B
mn
g
˜
i
g
˜
j
(
)
: g
˜
m
g
˜
n
(
)
A
•
j
i
B
mn
g
im
δ
n
j
A
•
j
i
B
mj
g
im
=
=
=
=
A
˜˜
B
˜˜
A
˜˜
B
˜˜
A
˜˜
B
˜˜
A
˜˜
:B
˜˜
A
ij
B
ij
A
ij
B
ij
A
•
j
i
B
mj
g
im
A
•
j
i
B
i
•
j
A
ij
B
mn
g
im
g
jn
…
=
=
=
=
=
=
A
˜˜
:B
˜˜
tr
A
˜˜
T
B
˜˜
•
(
)
tr
A
˜˜
B
˜˜
T
•
(
)
=
=
A
˜˜
A
˜˜
A
˜˜
:A
˜˜
=
A
˜˜
A
•
j
i
A
i
•
j
A
i
•
j
g
im
A
•
n
m
g
jn
=
A
˜˜
:A
˜˜
A
•
j
i
A
i
•
j
A
•
j
i
A
•
n
m
(
)
g
im
g
jn
=
=
B
˜˜
tr
B
˜˜
I
˜˜
:B
˜˜
=
B
˜˜
B
˜˜
B
ij
g
˜
i
g
˜
j
B
ij
g
˜
i
g
˜
j
B
•
j
i
g
˜
i
g
˜
j
B
i
•
j
g
˜
i
g
˜
j
=
=
=
=
u
˜
v
˜
I
˜˜
: u
˜
v
˜
(
)
u
˜
v
˜
•
=
tr
B
˜˜
B
ij
g
˜
i
g
˜
j
•
B
ij
g
˜
i
g
˜
j
•
B
•
j
i
g
˜
i
g
˜
j
•
B
i
•
j
g
˜
i
g
˜
j
•
=
=
=
=
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DRAFT June 17, 2004
42
or
.
(3.61)
Note that the last two expressions are the closest in form to the familiar formula for the trace
in orthonormal bases.
TRIPLE SCALAR-VALUED VECTOR PRODUCT
In direct notation,
.
(3.62)
.
(3.63)
DETERMINANT
The determinant of a tensor is defined in direct notation as
for all vectors
.
(3.64)
Recalling that the triple scalar product is defined
we may apply
previous formulas for the dot and cross product to conclude that
,
(3.65)
Recalling Eqs. (3.44) and (3.47), we may introduce the ordinary permutation symbol to write
this result as
.
(3.66)
Now that we are using the ordinary permutation symbol, we may interpret this result as a
matrix equation. Specifically, the left hand side represents the determinant of the contravari-
ant component matrix
times the permutation symbol
. Therefore, the above equation
implies that
.
(3.67)
Similarly, it can be shown that
.
(3.68)
In other words, if you have the matrix of covariant components, you compute the determi-
nant of the tensor by finding the determinant of the
matrix and multiplying the result
by .
Suppose the components of are known in
mixed
.
(3.69)
tr
B
˜˜
B
ij
g
ij
B
ij
g
ij
B
•
k
k
B
k
•
k
=
=
=
=
u
˜
v
˜
w
˜
, ,
[
]
u
˜
v
˜
w
˜
×
(
)
•
≡
u
˜
v
˜
w
˜
, ,
[
]
ξ
ijk
u
i
v
j
w
k
=
T
˜˜
T
˜˜
u
˜
•
T
˜˜
v
˜
•
T
˜˜
w
˜
•
,
,
[
]
det
T
˜˜
u
˜
v
˜
w
˜
, ,
[
]
=
u
˜
v
˜
w
˜
, ,
{
}
u
˜
v
˜
w
˜
, ,
[
]
u
˜
v
˜
w
˜
×
(
)
•
=
ξ
ijk
T
ip
T
jq
T
kr
det
T
˜˜
(
)ξ
pqr
=
ε
ijk
T
ip
T
jq
T
kr
det
T
˜˜
J
2
-----------
ε
pqr
=
T
ij
ε
pqr
det
T
ij
[
]
det
T
˜˜
(
)
J
2
⁄
=
det
T
˜˜
g
o
det
T
ij
[
]
=
det
T
˜˜
g
o
det
T
ij
[
]
=
T
˜˜
T
ij
[
]
g
o
T
˜˜
ξ
ijk
T
•
p
i
T
•
q
j
T
•
r
k
det
T
˜˜
ξ
pqr
=
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DRAFT June 17, 2004
43
Recalling Eqs. (3.45), we may introduce the ordinary permutation symbol. Noting that the
J
’s
on each side cancel, the above result as
.
(3.70)
Therefore, the determinant of the tensor is equal to the determinant of the matrix of
mixed
components:
.
(3.71)
Likewise, it can be shown that
.
(3.72)
3.10 Tensor-valued operations
TRANSPOSE
We have already discussed one tensor-valued operation, the transpose. Spe-
cifically, consider a tensor expressed in one of its four possible ways:
.
(3.73)
In Section 3.3, we showed that the transpose may be obtained by transposing the matrix base
dyads.
.
(3.74)
Most people are more accustomed to thinking of transposing components rather than base
vectors. Referring to the above result, we note that
, or
(3.75a)
, or
(3.75b)
, or
(3.75c)
, or
(3.75d)
Using our
matrix
notational conventions, these equations would be written
.
(3.76a)
.
(3.76b)
.
(3.76c)
.
(3.76d)
Here, we have intentionally used different indices ( and ) on the left-hand-side to remind
the reader that matrix equations are not subject to the same index rules. The indices are
present within the matrix brackets only to indicate to the reader which components (covari-
ant, contravariant, or mixed) are contained in the matrix.
ε
ijk
T
•
p
i
T
•
q
j
T
•
r
k
det
T
˜˜
ε
pqr
=
T
˜˜
det
T
˜˜
det
T
•
j
i
[
]
=
det
T
˜˜
det
T
i
•
j
[
]
=
T
˜˜
T
ij
g
˜
i
g
˜
j
T
ij
g
˜
i
g
˜
j
T
•
j
i
g
˜
i
g
˜
j
T
i
•
j
g
˜
i
g
˜
j
=
=
=
=
T
˜˜
T
T
ij
g
˜
j
g
˜
i
T
ij
g
˜
j
g
˜
i
T
•
j
i
g
˜
j
g
˜
i
T
i
•
j
g
˜
j
g
˜
i
=
=
=
=
T
˜˜
T
(
)
ji
T
ij
=
g
˜
j
T
˜˜
T
g
˜
i
•
•
g
˜
i
T
˜˜
g
˜
j
•
•
=
T
˜˜
T
(
)
ji
T
ij
=
g
˜
j
T
˜˜
T
g
˜
i
•
•
g
˜
i
T
˜˜
g
˜
j
•
•
=
T
˜˜
T
(
)
j
•
i
T
•
j
i
=
g
˜
j
T
˜˜
T
g
˜
i
•
•
g
˜
i
T
˜˜
g
˜
j
•
•
=
T
˜˜
T
(
)
•
i
j
T
i
•
j
=
g
˜
j
T
˜˜
T
g
˜
i
•
•
g
˜
i
T
˜˜
g
˜
j
•
•
=
T
˜˜
T
(
)
mn
[
]
T
ij
[
]
T
=
T
˜˜
T
(
)
mn
[
]
T
ij
[
]
T
=
T
˜˜
T
(
)
m
•
n
[
]
T
•
j
i
[
]
T
=
T
˜˜
T
(
)
•
n
m
[
]
T
i
•
j
[
]
T
=
m
n
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DRAFT June 17, 2004
44
Equation (3.76a) says that the
matrix
of contravariant components of
are obtained by
taking the transpose of the matrix of contravariant components of the original tensor
.
Similarly, Eq. (3.76b) says that the covariant component matrix for
is just the transpose of
the covariant matrix for .
The
mixed
component formulas are more tricky. Equation (3.76c) states that the
low-high
mixed components of
are obtained by taking the transpose of the
high-low
components
of . Likewise, Eq. (3.76d) states that the
high-low
mixed components of
are obtained by
taking the transpose of the
low-high
components of .
INVERSE
Consider a tensor . Let
. Then
.
(3.77)
Hence
.
(3.78)
Hence, the
contra
variant components of
are obtained by inverting the
matrix of
covari-
ant
components.
Alternatively note a different component form of Eq. (3.77) is
.
(3.79)
This result states that the
high-low
mixed components of
are obtained by inverting the
high-low
mixed
matrix.
The formulas for inverses may be written in our matrix notation as
.
(3.80a)
.
(3.80b)
.
(3.80c)
.
(3.80d)
COFACTOR
Without proof, we claim that similar methods can be used to show that the
matrix of
high-low
mixed components of
are the cofactor of the
low-high
matrix.
Note that the components come from the matrix. The contravariant matrix for
will
equal the cofactor of the
covariant
matrix for
times the metric scale factor
. The full set of
T
˜˜
T
[
]
T
˜˜
[ ]
T
˜˜
T
T
˜˜
T
T
T
˜˜
T
˜˜
T
T
˜˜
T
˜˜
U
˜˜
T
˜˜
1
–
=
T
˜˜
U
˜˜
•
I
˜˜
=
T
ik
U
kj
δ
i
j
=
T
˜˜
1
–
T
ij
T
•
k
i
U
•
j
k
δ
j
i
=
•
j
i
T
˜˜
1
–
T
•
j
i
[
]
T
˜˜
1
–
(
)
mn
[
]
T
ij
[
]
1
–
=
T
˜˜
1
–
(
)
mn
[
]
T
ij
[
]
1
–
=
T
˜˜
1
–
(
)
m
•
n
[
]
T
˜˜
1
–
(
)
i
•j
[
]
1
–
=
T
˜˜
1
–
(
)
•
n
m
[
]
T
˜˜
1
–
(
)
•
j
i
[
]
1
–
=
•
j
i
T
˜˜
C
T
i
•
j
[
]
•
j
i
i
•
j
T
˜˜
C
T
˜˜
g
o
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DRAFT June 17, 2004
45
formulas is
.
(3.81a)
.
(3.81b)
.
(3.81c)
.
(3.81d)
DYAD
By direct expansion,
. Thus,
.
(3.82)
Similarly,
.
(3.83)
.
(3.84)
.
(3.85)
T
˜˜
C
(
)
mn
[
]
g
o
T
ij
[
]
C
=
T
˜˜
C
(
)
mn
[
]
g
o
T
ij
[
]
C
=
T
˜˜
C
(
)
m
•
n
[
]
T
•
j
i
[
]
C
=
T
˜˜
C
(
)
•
n
m
[
]
T
i
•
j
[
]
C
=
u
˜
v
˜
u
i
g
˜
i
(
)
v
j
g
˜
j
(
)
u
i
v
j
g
˜
i
g
˜
j
=
=
u
˜
v
˜
(
)
ij
u
i
v
j
=
u
˜
v
˜
(
)
ij
u
i
v
j
=
u
˜
v
˜
(
)
•
j
i
u
i
v
j
=
u
˜
v
˜
(
)
i
•
j
u
i
v
j
=
rmbrann@me.unm.edu
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DRAFT June 17, 2004
46
4. Basis and Coordinate transformations
This chapter discusses how vector and tensor components transform when the basis changes.
This chapter may be skipped without any negative impact on the understandability of later chapters.
Chapter 3 had focused exclusively on the implications of using an irregular basis. For vec-
tor and tensor algebra (Chapter 3), all that matters is the possibility that the basis might be
non-normalized and/or non-right-handed and/or non-orthogonal. For vector and tensor
algebra, it doesn’t matter whether or not the basis changes from one point in space to another
— that’s because algebraic operations are always applied
at a single location
. By contrast, dif-
ferential operations such as the spatial gradient (discussed later in Chapter 5) are dramatically
affected by whether or not the basis is fixed or spatially varying. This present chapter pro-
vides a gentle transition between the topics of Chapter 3 to the topics of Chapter 4 by outlin-
ing the transformation rules that govern how components of a vector with respect to one
irregular basis will change if a different basis is used
at that same point in space
.
Throughout this document, we have asserted that any vector can (and should) be
regarded as invariant in the sense that the vector itself is unchanged upon a change in basis.
The vector can be expanded as a sum of components times corresponding base vectors :
(4.1)
It’s true that
the individual components will change if the basis is changed, but the
sum
of
components times base vectors will be invariant
. Consequently,
a vector’s components must
change in a very specific way if a different basis is used
.
Basis transformation discussions are complicated and confusing because you have to con-
sider two different systems at the same time. Since each system has both a covariant and a
contravariant basis, talking about two different systems entails keeping track of
four
different
basis triads. To help with this book-keeping nightmare, we will now refer to the first system
as the “A” system and the other system as the “B” system. Each contravariant index (which is
a superscript) will be accompanied by a subscript (either A or B) to indicate which system the
index refers to. Likewise, each covariant index (a subscript) will now be accompanied by a
superscript (A or B) to indicate the associated system. You should regard the system indicator
(A or B) to serve the same sort of role as the “dot placeholder” discussed on page 16 — they
are not indices. With this convention, we may now say
are the covariant base vectors for system-A
(4.2)
are the contravariant base vectors for system-A
(4.3)
are the covariant base vectors for system-B
(4.4)
are the contravariant base vectors for system-B
(4.5)
v
˜
v
i
g
˜
i
v
˜
v
i
g
˜
i
=
g
˜
1
A
g
˜
2
A
g
˜
3
A
,
,
{
}
g
˜
A
1
g
˜
A
2
g
˜
A
3
,
,
{
}
g
˜
1
B
g
˜
2
B
g
˜
3
B
,
,
{
}
g
˜
B
1
g
˜
B
2
g
˜
B
3
,
,
{
}
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DRAFT June 17, 2004
47
The system flag, A or B, acts as a “dot-placeholder” and moves along with its associated index
in operations. Now that we are dealing with up to four basis triads, the components of vectors
must also be flagged to indicate the associated basis. Hence, depending on which of the above
four bases are used, the basis expansion of a vector can be any of the following:
(4.6)
(4.7)
(4.8)
(4.9)
Keep in mind that the system label (A or B) is to be regarded as a “dot-placeholder,” not an
implicitly summed index. The metric coefficients for any given system are defined in the
usual way and the Kronecker delta relationship between contravariant and covariant base
vectors
within a single system
still holds. Specifically,
(4.10)
(4.11)
(4.12)
(4.13)
In previous chapters (which dealt with only a single system), we showed that the matrix con-
taining the
components could be obtained by inverting the
matrix. This relationship
still holds within each
individual
system (A or B) listed above. Namely,
(4.14)
(4.15)
Now that we are considering two systems simultaneously, we can further define new
matrices that inter-relate the two systems:
(4.16a)
(4.16b)
(4.16c)
(4.16d)
v
˜
v
A
i
g
˜
i
A
=
v
˜
v
i
A
g
˜
A
i
=
v
˜
v
B
i
g
˜
i
B
=
v
˜
v
B
i
g
˜
i
B
=
g
ij
AA
g
˜
i
A
g
˜
j
A
•
=
g
iA
Aj
g
˜
i
A
g
˜
A
j
•
δ
i
j
=
=
g
Aj
iA
g
˜
A
i
g
˜
j
A
•
δ
j
i
=
=
g
AA
ij
g
˜
A
i
g
˜
A
j
•
=
g
ij
BB
g
˜
i
B
g
˜
j
B
•
=
g
iB
Bj
g
˜
i
B
g
˜
B
j
•
δ
i
j
=
=
g
Bj
iB
g
˜
B
i
g
˜
j
B
•
δ
j
i
=
=
g
BB
ij
g
˜
B
i
g
˜
B
j
•
=
g
ij
g
ij
[ ]
g
ik
AA
g
AA
kj
g
iA
Aj
δ
i
j
=
=
g
ik
BB
g
BB
kj
g
iB
Bj
δ
i
j
=
=
g
ij
AB
g
˜
i
A
g
˜
j
B
•
=
g
AB
ij
g
˜
A
i
g
˜
B
j
•
=
g
ij
BA
g
˜
i
B
g
˜
j
A
•
=
g
BA
ij
g
˜
B
i
g
˜
A
j
•
=
g
Aj
iB
g
˜
A
i
g
˜
j
B
•
=
g
iB
Aj
g
˜
i
A
g
˜
B
j
•
=
g
Bj
iA
g
˜
B
i
g
˜
j
A
•
=
g
iA
Bj
g
˜
i
B
g
˜
A
j
•
=
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DRAFT June 17, 2004
48
Partial Answer:
(a) ,
, ,
.
(b) , , ,
(c) , ,
,
,
, ,
,
From the commutativity of the dot product, Eq. (4.16) implies that
, ,
,
and .
(4.17)
System metrics (i.e., “g” matrices that involve AA or BB) are components of a tensor (the iden-
tity tensor). Coupling matrix components (i.e., ones that involve A and B) are not components
of a tensor — instead, a coupling matrix characterizes interrelationship between two bases.
Study Question 4.1
Consider the following irregular base vectors (the A-system):
and
Additionally consider a second B-system:
and
(a) Find the contravariant and covariant met-
rics for each individual system.
(b) Find the contravariant basis for each indi-
vidual system.
(c) Directly apply Eq. (4.16) to find the cou-
pling matrices. Then verify that
.
g
˜
1
A
e
˜
1
e
˜
2
g
˜
1
B
g
˜
2
B
g
˜
2
A
g
˜
1
A
e
˜
1
2
e
˜
2
+
=
g
˜
2
A
e
˜
1
–
e
˜
2
+
=
g
˜
3
A
e
˜
3
=
g
˜
1
B
2
e
˜
1
e
˜
2
+
=
g
˜
2
B
e
˜
1
–
4
e
˜
2
+
=
g
˜
3
B
e
˜
3
=
g
ij
AB
[
]
g
ij
BA
[
]
T
=
g
ij
AA
[
]
5 1 0
1 2 0
0 0 1
=
g
AA
ij
[
]
2 9
⁄
1 9
⁄
–
0
1 9
⁄
–
5 9
⁄
0
0
0
1
=
g
ij
BB
5 2 0
2 17 0
0 0 1
=
g
ij
AA
17 81
⁄
1 81
⁄
–
0
1 81
⁄
–
5 81
⁄
0
0
0
1
=
g
˜
A
1
1
3
---
e
˜
1
1
3
---
e
˜
2
+
=
g
˜
A
2
2
–
3
------
e
˜
1
1
3
---
e
˜
2
+
=
g
˜
B
1
4
9
---
e
˜
1
1
9
---
e
˜
2
+
=
g
˜
B
2
1
–
9
------
e
˜
1
2
9
---
e
˜
2
+
=
g
ij
AB
[
]
4 7 0
1
– 5 0
0 0 1
=
g
ij
BA
[
]
4 1
– 0
7 5 0
0 0 1
=
g
Bj
iA
[
]
2 3
⁄
1 3
⁄
–
0
1 3
⁄
1 3
⁄
0
0
0
1
=
g
iB
Aj
[
]
2 3
⁄
1 3
⁄
0
1 3
⁄
–
1 3
⁄
0
0
0 1
=
g
AB
ij
[
]
5 27
⁄
1 27
⁄
0
7 27
⁄
–
4 27
⁄
0
0
0
1
=
g
BA
ij
[
]
5 27
⁄
7 27
⁄
–
0
1 27
⁄
4 27
⁄
0
0
0
1
=
g
Aj
iB
[
]
1 1 0
1
– 2 0
0 0 1
=
g
iA
Bj
[
]
1 1
– 0
1 2 0
0 0 1
=
g
ij
AB
g
ji
BA
=
g
AB
ij
g
BA
ij
=
g
Aj
iB
g
jA
Bi
=
g
i B
A j
g
Bi
jA
=
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DRAFT June 17, 2004
49
The first relationship in Eq. (4.17) does not say that the matrix containing
is symmet-
ric.
Instead this equation is saying that the matrix containing
may be obtained by the
transpose of the
different
matrix that contains
. Given a “g”
matrix
of any type, you can
transform it immediately to other types of g-matrices by using the following rules:
• To exchange system labels (A or B) left-right, apply a transpose.
• To exchange system labels (A or B) up-down, apply an inverse.
• To exchange system labels (A or B) along a diagonal, apply an inverse transpose.
For example, if
is available, then
, (4.18)
Similarly, if
is available, then
, (4.19)
Here, and in all upcoming matrix equations involving “g” components, a star (*) is inserted
where indices normally reside in indicial equations. These equations show how to move two
system labels simultaneously.
To move only one system label to a new location, you need a matrix product. In matrix
products, abutting system labels must be on opposite levels (the system labels are
not
summed — index rules apply only to indicial expressions, not to matrix expressions. For
example, what operation would you need to move only the “B” label in
from the bot-
tom to the top? The answer is that you need to use the B-system metric as follows:
(this is the matrix form of
)
(4.20)
This behavior also applies to transforming bases. For example,
(4.21)
All
of the “g” matrices may be obtained only from knowledge of one system metric (same sys-
tem labels) and one coupling matrix (different system labels).
g
ij
AB
g
ij
AB
g
ij
BA
g
ij
AB
g
**
BA
[
]
g
**
AB
[
]
T
=
left-right
g
AB
**
g
**
AB
[
]
1
–
=
up-down
g
BA
**
g
**
AB
[
]
T
–
=
diagonal
g
iB
Aj
g
B
*
*
A
[
]
g
*
B
A
*
[
]
T
=
left-right
g
A
*
*
B
g
*
B
A
*
[
]
1
–
=
up-down
g
*
A
B
*
g
*
B
A
*
[
]
T
–
=
diagonal
g
B
*
*
A
[
]
g
**
BA
[
]
g
**
BB
[
]
g
B
*
*
A
[
]
=
g
ij
BA
g
ik
BB
g
Bj
kA
=
g
˜
i
A
g
iB
Ak
g
˜
k
B
=
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DRAFT June 17, 2004
50
Partial Answer:
(a) We are given a regular metric and a mixed coupling matrix. Start by expressing
as the product of any AB matrix times a BB metric times a BA matrix:
.
We now need to get
expressed in terms of the given coupling matrix
. In other words, we
need to swap A and B along the diagonal, which requires an inverse-transpose: .
Next,
we need to express the coupling matrix
in terms of the given coupling matrix
. This requires
only an up-down motion of the system labels, so that’s just an inverse:
. Putting it all
back in the first equation gives
(b)
. The matrix in this relation is
. Therefore
, ,
and
.
(c) is similar.
Study Question 4.2
Suppose you are studying a problem involving two bases, A and
B, and you know one system metric
and one coupling matrix
as follows:
and
(4.22)
(a) Find
and
(b) Express the
base vectors in terms of the
base vectors.
(b) Express the
base vectors in terms of the
base vectors.
g
ij
BB
g
iA
Bj
g
**
BB
[
]
4 2 3
2 4
1
–
3 1
–
5
=
g
*
A
B
*
[
]
1 2
4
–
0 1
–
2
3 0
2
–
=
g
**
AA
[
]
g
AA
**
[
]
g
˜
i
A
g
˜
k
B
g
˜
A
i
g
˜
k
B
g
**
AA
[
]
g
**
AA
[
]
g
*
B
A
*
[
]
g
**
BB
[
]
g
B
*
*
A
[
]
=
g
*
B
A
*
[
]
g
*
A
B
*
[
]
g
*
B
A
*
[
]
g
*
A
B
*
[
]
T
–
=
g
B
*
*
A
[
]
g
*
A
B
*
[
]
g
B
*
*
A
[
]
g
*
A
B
*
[
]
1
–
=
g
**
AA
[
]
g
*
A
B
*
[
]
T
–
g
**
BB
[
]
g
*
A
B
*
[
]
1
–
1 2
4
–
0 1
–
2
3 0
2
–
T
–
4 2 3
2 4
1
–
3 1
–
5
1 2
4
–
0 1
–
2
3 0
2
–
19
2
------ 5
2
--- 57
2
------
5 11 49
9
2
---
–
1
2
---
15
2
------
–
=
=
=
g
˜
i
A
g
iB
Ak
g
˜
k
B
=
g
*
B
A
*
[
]
g
*
A
B
*
[
]
T
–
1 3
3
2
---
2 5 3
0 1
–
1
2
---
–
=
=
g
˜
1
A
g
˜
1
B
3
g
˜
2
B
3
2
---
g
˜
3
B
+
+
=
g
˜
2
A
2
g
˜
1
B
5
g
˜
2
B
3
g
˜
3
B
+
+
=
g
˜
3
A
g
˜
2
B
–
1
2
---
g
˜
3
B
–
=
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DRAFT June 17, 2004
51
We showed in previous sections how the covariant components of a vector are related to
the contravariant components
from the same system
. For example, we showed that
when considering only one system. When considering more than one basis at a time, this
result needs to explicitly show the system (A or B) flags:
(4.23)
(4.24)
These equations show how to transform components
within a single system.
To transform com-
ponents from one system to another, the formulas are
(4.25)
(4.26)
Equivalently,
(4.27)
(4.28)
Tensor transformations are similar. For example
and
(4.29)
These formulas are constructed one index position at a time. In the last formula, for example,
the first free index position on the
T
is the pair . On the other side of the equation, the first
index pair on the
T
is . Because is a dummy summed index, the “g” must have . Thus,
the first “
g
” is a combination of the free index pair on the left side and the “flipped” index pair
on the right
. The “
T
” in the above equations may be moved to the middle if you like it
better that way:
and
(4.30)
If you wanted to write these as matrix equations, you can write them in a way that
dummy summed indices are abutting, keeping in mind that the
g
components are unchanged
when indices
along with their system label
are moved left-to-right. The
g
components require an
inverse on their matrix forms to move system labels up-down. Thus, the matrix forms for
these transformation formulas are
and
(4.31)
Numerous matrix versions of component transformation formulas may be written. In this
case, we wrote the matrix expressions on the assumption that the known coupling and metric
matrices were
and
.
v
i
g
ij
v
j
=
v
i
A
g
ij
AA
v
A
j
=
v
i
B
g
ij
BB
v
B
j
=
v
i
A
g
iB
Ak
v
k
B
=
v
i
A
g
ik
AB
v
B
k
=
v
A
i
g
AB
ik
v
k
B
=
v
A
i
g
Ak
iB
v
B
k
=
v
i
A
v
k
B
g
Bi
kA
=
v
i
A
v
B
k
g
ki
BA
=
v
A
i
v
k
B
g
BA
ki
=
v
A
i
v
B
k
g
kA
Bi
=
T
ij
AA
T
pq
BB
g
Bi
pA
g
Bj
qA
=
T
Bj
iB
T
AA
pq
g
pB
Ai
g
qj
AB
=
i
B
p
A
p
A
p
g
pB
Ai
T
ij
AA
g
Bi
pA
T
pq
BB
g
Bj
qA
=
T
Bj
iB
g
pB
Ai
T
AA
pq
g
qj
AB
=
T
**
AA
[
]
g
B
*
*
A
[
]
T
T
**
BB
[
]
g
B
*
*
A
[
]
=
T
Bj
iB
g
B
*
*
A
[
]
T
T
AA
**
[
]
g
B
*
*
A
[
]
T
–
g
**
AA
[
]
=
g
B
*
*
A
[
]
g
**
AA
[
]
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DRAFT June 17, 2004
52
Partial Answer:
(a) , , , , , , , .
(b) The first third and fourth results of part (a) imply that
. Drawing grid
lines parallel to and
, the path from the origin to the tip of vector may be reached by first tra-
versing along a line segment equal to
, and then moving anti-parallel to
to reach the tip.
(c) Using results from study question 4.1,
becomes in matrix form
.
Multiplying this out verifies that it is true.
(d) The answer is
.
Study Question 4.3
Consider the same irregular base vectors in Study Question 4.1.
Namely,
and
and
and
Let .
(a) Determine the covariant and contravari-
ant components of with respect to sys-
tems A and B. (i.e., find
).
(b) Demonstrate graphically that your
answers to part (a) make sense.
(c) Verify the following:
(d) Fill in the question marks to make the following equation true:
e
ˆ
˜
1
e
ˆ
˜
2
v
˜
g
˜
1
A
g
˜
1
B
g
˜
2
B
g
˜
2
A
g
˜
1
A
e
˜
1
2
e
˜
2
+
=
g
˜
2
A
e
˜
1
–
e
˜
2
+
=
g
˜
3
A
e
˜
3
=
g
˜
1
B
2
e
˜
1
e
˜
2
+
=
g
˜
2
B
e
˜
1
–
4
e
˜
2
+
=
g
˜
3
B
e
˜
3
=
v
˜
3
e
˜
1
3
e
˜
2
+
=
v
˜
v
i
A
v
A
i
v
i
B
v
B
i
,
,
,
v
i
A
g
ik
AB
v
B
k
=
v
i
A
g
iB
Ak
v
k
B
=
v
A
i
g
Ak
iB
v
B
k
=
v
A
i
g
AB
ik
v
k
B
=
v
i
B
g
??
??
v
?
A
=
v
1
A
9
=
v
2
A
0
=
v
A
1
2
=
v
A
2
1
–
=
v
1
B
9
=
v
2
B
9
=
v
B
1
5
3
---
=
v
B
2
1
3
---
=
v
˜
v
A
1
g
˜
1
A
v
A
2
g
˜
2
A
+
2
g
˜
1
A
g
˜
2
A
–
=
=
g
˜
1
A
g
˜
2
A
v
˜
2
g
˜
1
A
g
˜
2
A
v
i
A
g
ik
AB
v
B
k
=
9
0
0
4 7 0
1
– 5 0
0 0 1
5 3
⁄
1 3
⁄
0
=
v
i
B
g
iA
Bk
v
k
A
=
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DRAFT June 17, 2004
53
Coordinate transformations
A principal goal of this chapter is to show how the
compo-
nents must be related to the
components.
We are seeking
component
transformations
resulting from a change of
basis
. Strictly speaking, coordinates may always be selected inde-
pendently from the basis. This, however, is rarely done. Usually the basis is taken to be
defined to be the one that is naturally defined by the coordinate system grid lines. The natural
covariant basis co-varies with the coordinates — i.e., each covariant base vector always points
tangent to the coordinate a grid line and has a length that varies in proportion to the grid
spacing density. The natural contravariant basis contra-varies with the grid — i.e., each con-
travariant base vector points normal to a surface of constant coordinate value. When people
talk about component transformations resulting from a change in
coordinates
, they implicitly
assuming that the basis is coupled to the choice of coordinates.
There is an unfortunate
implicit assumption in many publications that vector and tensor components
must
be coupled
to the spatial coordinates. In fact, it’s fine to use a basis that is selected entirely independently
from the coordinates. For example, to study loads on a ferris wheel, you might decide to use a
fixed Cartesian basis that is aligned with gravity, while also using cylindrical
coordi-
nates
to identify points in space.
Spatial
coordinates
are any three numbers that uniquely define a location in space. For
example, Cartesian coordinates are frequently denoted
, cylindrical coordinates are
, and so on.
Any quantity that is known to vary in space can be written as a function
of the coordinates. Coordinates uniquely define the position vector , but
coordinates
are not
the same thing as
components
of the position vector.
For example, the position vector in cylin-
drical coordinates is
. Notice that position vector is
not
.
The
components of the position vector are
, not
. The position vector has only
two
nonzero components. It has no
term. Does this mean that the position vector depends
only on and ? Nope. The position vector’s dependence on is buried
implicitly
in the
dependence of on .
Whether the basis is chosen to be coupled to the coordinates is entirely up to you. If conve-
nient to do so (as in the ferris wheel example), you might choose to use cylindrical coordinates
but a
Cartesian
basis so that the position vector would be written
. In this case, the chosen basis is the Cartesian laboratory
basis
, which is entirely uncoupled from the coordinates
.
To speak in generality, we will denote the three spatial coordinates by
. If, for
example, cylindrical coordinates are used, then
,
, and
. If spherical
coordinates are used, then
,
, and
. If Cartesian coordinates are used,
then , ,
and .
v
B
i
v
A
i
r
θ
z
, ,
(
)
x y z
, ,
{
}
r
θ
z
, ,
{
}
x
˜
x
˜
r
e
˜
r
z
e
˜
z
+
=
r
e
˜
r
θ
e
˜
θ
z
e
˜
z
+
+
r
zero
z
,
,
{
}
r
θ
z
, ,
{
}
θ
e
˜
θ
r
z
θ
e
˜
r
θ
x
˜
r
θ
cos
(
)
e
˜
x
r
θ
sin
(
)
e
˜
y
z
e
˜
z
+
+
=
e
˜
x
e
˜
y
e
˜
z
, ,
{
}
r
θ
z
, ,
{
}
η
1
η
2
η
3
,
,
{
}
η
1
r
=
η
2
θ
=
η
3
z
=
η
1
r
=
η
2
θ
=
η
3
ψ
=
η
1
x
=
η
2
y
=
η
3
z
=
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DRAFT June 17, 2004
54
The position vector is truly a function of all three
coordinates. The basis can
be selected independently from the choice of coordinates, However, we can always define the
natural
basis
is
associated with the choice of coordinates by
(4.32)
The natural covariant base vector equals the derivative of the position vector with
respect to the coordinate. Hence, this base vector points in the direction of increasing
and it is tangent to the grid line along which the other two coordinates remain constant.
Examples of the natural basis for various coordinate systems are provided in Section 5.3. The
natural basis defined in Eq. (4.32) is clearly coupled to the coordinates themselves. Conse-
quently, a change in coordinates will result in a change of the natural basis. Thus, the compo-
nents of any vector expressed using the
natural
basis will change upon a change in
coordinates. Having the basis be coupled to the coordinates is a
choice
. One can alternatively
choose the basis to be uncoupled from the coordinates (see the discussion of non-holonomic-
ity in Ref. [12]).
Each particular value of the coordinates
defines a unique location in space.
Conversely, each location in space corresponds to a unique set of
coordinates.
That means the
coordinates themselves can be regarded as functions of the position vector,
and we can therefore take the spatial gradients of the coordinates
. As further explained in Sec-
contravariant
natural base vector
is defined to equal these coordinate gradients:
(4.33)
Being the gradient of , the contravariant base vector must point normal to surfaces of
constant .
Proving that these are indeed the contravariant base vectors associated with the
covariant natural basis defined in Eq. (4.32) is simply a matter of applying the chain rule to
demonstrate that
comes out to equal the Kronecker delta:
(4.34)
The metric coefficients corresponding to Eq. (4.32) are
(4.35)
Many textbooks present this result in pure component form by writing the vector in terms
of its cartesian components as
so that the above expression becomes
for the natural basis
(4.36)
η
1
η
2
η
3
,
,
{
}
g
˜
i
∂
x
˜
∂η
i
--------
≡
i
th
g
˜
i
x
˜
i
th
η
i
η
1
η
2
η
3
,
,
{
}
x
˜
η
1
η
2
η
3
,
,
{
}
g
˜
j
∂η
j
∂
x
˜
--------
≡
η
j
g
˜
j
η
j
g
˜
i
g
˜
j
•
g
˜
i
g
˜
j
•
∂
x
˜
∂η
i
--------
∂η
j
∂
x
˜
--------
•
∂η
j
∂η
i
--------
δ
i
j
=
=
=
g
ij
g
˜
i
g
˜
j
•
∂
x
˜
∂η
i
--------
∂
x
˜
∂η
j
--------
•
=
=
x
˜
x
˜
x
k
e
˜
k
=
g
ij
∂
x
k
∂η
i
--------
∂
x
k
∂η
j
--------
k
1
=
3
∑
=
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DRAFT June 17, 2004
55
When written this way, it’s important to recognize that are the
Cartesian
components of the
position vector (not the covariant components with respect to an irregular basis). That’s why
we introduced the summation in Eq. (4.36) -- otherwise, we would be violating the high-low
rule of the summation conventions.
Similarly, the contravariant metric coefficients are
for the natural basis
(4.37)
These are the expressions for the metric coefficients when the natural basis is used. Let us reit-
erate that there is no law that says the basis used to described vectors and tensors must neces-
sarily be coupled to the choice of spatial coordinates in any way. We cite these relationships to
assist those readers who wish to make connections between our own present formalism and
other published discourses on curvilinear coordinates.
In the upcoming discussion of coordinate transformations, we will
not
assume that the
metric coefficients are given by the above expressions. Because our transformation equations
will be expressed in terms of metrics only, they will apply even when the chosen basis is
not
the natural basis. When we derive the component transformation rules, we will include both
the general basis transformation expression and its specialization to the case that the basis is
chosen to be the
natural
basis.
4.1 What is a vector? What is a tensor?
The bane of many rookie graduate students is the common qualifier question: “what is a
vector?” The frustration stems, in part, from the fact that the “correct” answer depends on
who is doing the asking. Different people have different answers to this question. More than
likely, the professor follows up with the even harder question “what is a tensor?”
Definition #0 (used for undergraduates):
A college freshman is typically told that a vector
is something with length and direction, and nothing more is said (certainly no mention is
made of tensors!). The idea of a tensor is tragically withheld from most undergraduates.
Definition #1 (classical, but our least favorite):
Some professors merely want their stu-
dent to say that there are two kinds of vectors: A contravariant vector is a set of three numbers
that transform according to
when changing from an “A-basis” to a
“B-basis”, whereas a covariant vector is a set of three numbers
that transform
such that
. These professors typically wish for their students to define
four
differ-
ent kinds of tensors, the contravariant tensor being a set of nine numbers that transform
x
k
g
ij
∂η
i
∂
x
k
--------
∂η
j
∂
x
k
--------
k
1
=
3
∑
=
v
1
v
2
v
3
, ,
{
}
v
B
k
v
A
i
g
iB
Ak
=
v
1
v
2
v
3
, ,
{
}
v
k
B
v
i
A
g
Ak
iB
=
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DRAFT June 17, 2004
56
according to
, the mixed tensors being ones whose components transform
according to
or
, etc.
Definition #2 (our preference for ordinary engineering applications):
We regard a vec-
tor to be an entity that exists independent of our act of observing it — independent of our act
of representing it quantitatively. Of course, to decide if something is an engineering vector,
we demand that it must nevertheless lend itself to a quantitative representation as a sum of
three components
times
1
three base vectors
, and we
test
whether or
not we have a vector by verifying that
holds upon a change of basis. The base
vectors themselves are regarded as abstractions defined in terms of geometrical fundamental
primitives. Namely, the base vectors are regarded as directed line segments between points in
space, which are presumed (by axiom) to exist. The key distinction between definition #2 and
#1 is that definition #2 makes no distinction between covariant vectors and contravariant vec-
tors. Definition #2 instead uses the terms “covariant components” and “contravariant compo-
nents” — these are components of the
same
vector.
Recall that definition of a
second-order tensor
under our viewpoint required introduction of
new objects, called dyads, that could be constructed from vectors. From there, we asserted
that any collection of nine numbers could be assembled as a sum of these numbers times basis
dyads
. A tensor would then be defined as any linear combination of tensor dyads for
which the coefficients of these dyads (i.e., the tensor components) follow the transformation
rules outlined in the preceding section.
Definition #3 (for mathematicians or for advanced engineering)
Mathematicians define
vectors as being “members of a vector space.” A vector space must comprise certain basic components:
A1. A field
R
must exist.
(For engineering applications, the field is the set of reals.)
A2. There must be a discerning definition of membership in a set
V
.
A3. There must be a rule for multiplying a scalar times a member of
V
. Furthermore, this
multiplication rule must be
proved
closed in :
If
and
then
A4. There must be a rule for adding two members of
V
.
Furthermore, this vector addition rule must be
proved
closed in :
If
and
then
A5. There must be a well defined process for determining whether two members of are equal.
A6. The multiplication and addition rules must satisfy the following rules:
•
and
•
•There must exist a zero vector
such that
.
1. The act of multiplying a scalar times a vector is presumed to be well defined and to satisfy the rules outlined in
definition #3.
T
BB
ij
T
AA
mn
g
Bm
iA
g
Bn
jA
=
T
iB
Bj
T
mA
An
g
im
BA
g
BA
jn
=
T
Bj
iB
T
An
mA
g
BA
im
g
jn
BA
=
v
1
v
2
v
3
, ,
{
}
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
v
B
k
v
A
i
g
iB
Ak
=
g
˜
i
g
˜
j
α
v
˜
V
α
R
∈
v
˜
V
∈
α
v
˜
V
∈
V
v
˜
V
∈
w
˜
V
∈
v
˜
w
˜
+
V
∈
V
v
˜
w
˜
+
w
˜
v
˜
+
=
α
v
˜
v
˜
α
=
u
˜
v
˜
w
˜
+
(
)
+
u
˜
v
˜
+
(
)
w
˜
+
=
0
˜
V
∈
v
˜
0
˜
+
v
˜
=
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DRAFT June 17, 2004
57
Unfortunately, textbooks seem to fixate on item A6, completely neglecting the far more subtle and dif-
ficult items A2, A3, A4, and A5. Engineering vectors are more than something with length and direc-
tion. Likewise, engineering vectors are more than simply an array of three numbers. When people
define vectors according to the way their components change upon a change of basis (definitions #1 and
#2), they are implicitly addressing axiom A2. Our “definition #2” is a special case of this more general
definition. In general, axiom A2 is the most difficult axiom to satisfy when discussing
specific
vector
spaces.
Whenever possible, vector spaces are typically supplemented with the definition of an inner product
(here denoted
), which is a scalar-valued binary
1
operation between two vectors, and , that
must satisfy the following rules:
A7.
A8.
if
and
only if
.
An inner product space is just a vector space that has an inner product.
Partial Answer:
(a) You may assume that the set of reals constitutes an adequate field, but you
may obtain a careful definition of the term “field” in Reference [6]. In order to deal with axiom A2, you
will need to locate a carefully crafted definition of what it means to be a “real valued continuous func-
1.The term “binary” is just an fancy way of saying that the function has
two
arguments.
a
˜
b
˜
,
(
)
a
˜
b
˜
a
˜
b
˜
,
(
)
b
˜
a
˜
,
(
)
=
a
˜
a
˜
,
(
)
0
>
a
˜
0
˜
≠
a
˜
a
˜
,
(
)
0
˜
=
a
˜
0
˜
=
Study Question 4.4
(a) Apply definition #3 to prove that the set of all real-valued continuous functions of
one variable (on a domain being reals) is a vector space.
(b) Suppose that and are members of this vector space. Prove that the scalar-valued
functional
(4.38)
is an acceptable definition of the inner product,
.
(c) Explain what a Taylor-series expansion
has in common with a basis expansion
(
) of a vector. A Fourier-series expansion is related to the Taylor series in a
way that is analogous to what operation from ordinary vector analysis in 3D space?
(d) Explain how the binary function
is analogous to the dyad
from conven-
tional vector/tensor analysis.
(e) Explain why the dummy variables and play roles similar to the indices and
(f) Explain why a general binary function
is analogous to a conventional tensor
components .
(g) Assuming that the goal of this document is to give you greater insight into vector
and tensor analysis in 3D space, do you think this problem contributes toward that
goal? Explain?
f
g
f x
( )
g x
( )
x
d
∞
–
∞
∫
f g
,
(
)
f x
( )
f
˜
f
i
g
i
=
f x
( )
g y
( )
f
˜
g
˜
x
y
i
j
a x y
,
(
)
A
ij
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DRAFT June 17, 2004
58
tion of one (real) variable” [see References 8, 9, or 10 if you are stumped]. (b) Demonstrate that axioms
A7 and A8 hold true. (c) The Taylor series expansion writes a function as a linear combination of more
simple functions, namely, the powers , for ranging from 0 to . The Fourier series expansion
writes a function as a linear combination of trigonometric functions. (d) The dyad
has components
, where the indices and range from 1 to 3. The function f(x)g(y) has values defined for values of
the independent variables and ranging from
to . A dyad
is the most primitive of tensors,
and it is defined such that it has no meaning (other than book-keeping) unless it operates on an arbi-
trary vector so that
means
. Likewise, we can define the most primitive function of two
variables, f(x)g(y) so that it takes on meaning when appearing in an operation on a general function
so that f(x)g(y) is defined so that it means
. (e) see pre-
vious answer (f).
When addressing the question of whether or not something is indeed a tensor, you must
commit yourself to which of the definitions discussed on page 55 you wish to use. When we
cover the topic of curvilinear calculus, we will encounter the Christoffel symbols
and
.
These three-index quantities characterize how the
natural
curvilinear basis varies in space.
Their definition is based upon your choice of basis,
. Naturally it stands to reason
that choosing some other basis will still permit you to construct Christoffel symbols for that
system. Any review of the literature will include a statement that the Christoffel symbols “are
not third-order tensors.” This statement merely means that
(4.39)
Note that it is perfectly acceptable for you to construct a
distinct
tensors defined
and (4.40)
Equation (4.39) tells us that these two tensors are not equal. That is,
(4.41)
Stated differently, for each basis, there exists a
basis-dependent
Christoffel tensor. This should
not be disturbing. After all, the base vectors
themselves
are, by definition, basis-dependent, but
that doesn’t mean they aren’t vectors. Changing the basis will change the associated Christof-
fel tensor. A particular choice for the basis is itself (of course) basis dependent -- change the
basis, and you will (obviously) change the base vectors themselves. You can always construct
other vectors and tensors from the basis, but you would naturally expect these tensors to
change upon a change of basis if the new definition of the tensor is defined the same as the old
definition except that the new base vectors are used. What’s truly remarkable is that there
exist certain combinations of base vectors and basis-referenced definitions of components that
turn out to be
invariant
under a change of basis.
x
n
n
∞
f
˜
g
˜
f
i
g
j
i
j
x
y
∞
–
∞
f
˜
g
˜
f
˜
g
˜
v
˜
•
f
˜
g
˜
v
˜
•
(
)
v y
( )
f x
( )
g y
( )
v y
( )
y
d
∞
–
∞
∫
f x
( )
g y
( )
v y
( )
y
d
∞
–
∞
∫
=
Γ
ij
k
Γ
kij
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
Γ
ijB
BBk
Γ
pqA
AAr
g
ip
BA
g
jq
BA
g
BA
kr
≠
Γ
˜˜˜
A
Γ
ijA
AAk
g
˜
A
m
g
˜
A
n
g
˜
p
A
=
Γ
˜˜˜
B
Γ
ijB
BBk
g
˜
B
m
g
˜
B
n
g
˜
p
B
=
Γ
˜˜˜
A
Γ
˜˜˜
B
≠
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59
Consider two tensors constructed from the metrics
and
:
and (4.42)
These two tensors, it turns out, are equal even though they were constructed using
different
bases and
different
components! Even though
and even though
, it turns
out that the differences cancel each other out in the combinations defined in Eq. (4.42) so that
both tensors are equal. In fact, as was shown in Eq. (3.18), these tensors are simply the identity
tensor!
(4.43)
This is an exceptional situation. More often than not, when you construct a tensor by mul-
tiplying an ordered array of numbers by corresponding base vectors in different systems, you
will end up with two different tensors. To clarify, suppose that is a function (scalar, vector,
or tensor valued) which can be constructed from the basis. Presume that the
same
function can
be applied to any basis, and furthermore presume that the results will be different, so we
must denote the results by different symbols.
and if
(4.44)
In the older literature, the components of were called tensor if and only if
. In
more modern treatments, the
function
is called basis invariant if
, whereas the
function
f
is basis-
dependent
if
. These two tensors are equal only if the components of
with respect to one basis happen to equal the components of
with respect to the
same
basis. It’s not uncommon for two distinct tensors to have the same components with respect
to
different
bases — such a result would
not
imply equality of the two tensors.
4.2 Coordinates are not the same thing as components
All base vectors in this section are to be regarded as the natural basis so that they are
related to the coordinates
by
. Recall that we have typeset the three coordi-
nates as
. Despite this typesetting, these numbers are
not
contravariant components of any
vector . By this we mean that if you compare a different set of components
from a second
system, they are
not
generally related to the original set of components via a vector transfor-
mation rule. We may only presume that transformation rules exist such that each
can be
expressed as a function of the
coordinates. If coordinates were also vector components,
then they would have to be related by a transformation formula of the form
valid for homogeneous, but not curvilinear coordinates
(4.45)
A coordinate transformation of this form is a
linear
, which holds only for homogenous (i.e.,
non
-curvilinear) coordinate systems. Even though the
coordinates are not generally com-
g
ij
AA
g
ij
BB
G
˜˜
AA
g
ij
AA
g
˜
A
i
g
˜
A
j
=
G
˜˜
BB
g
ij
BB
g
˜
B
i
g
˜
B
j
=
g
˜
A
i
g
˜
B
i
≠
g
ij
AA
g
ij
BB
≠
G
˜˜
AA
G
˜˜
BB
I
˜˜
=
=
f
F
A
f
g
˜
1
A
g
˜
2
A
g
˜
3
A
,
,
(
)
=
F
B
f
g
˜
1
B
g
˜
2
B
g
˜
3
B
,
,
(
)
=
F
A
F
A
F
B
=
f
F
A
F
B
=
F
A
F
B
≠
F
A
F
B
η
k
g
˜
k
∂
x
˜
∂η
k
⁄
=
η
k
η
˜
η
k
η
k
η
m
η
k
∂η
k
∂η
m
----------
η
m
=
←
η
k
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60
ponents of a vector, their
increments
do transform like vectors. That is,
this is true for both homogenous and curvilinear coordinates!
(4.46)
For curvilinear systems, the transformation formulas that express coordinates
in terms of
are
nonlinear
functions, but their
increments
are linear for the same reason that the incre-
ment of a nonlinear function
is an expression that is linear with respect to incre-
ments: .
Geometrically,
at any point on a curvy line, we can construct a
straight
tangent to that line.
4.3 Do there exist a “complementary” or “dual” coordinates?
Recall that the coordinates are typeset as
. We’ve explained that the coordinate
incre-
ments
transform as vectors even though the coordinates themselves are not components
of vectors. The
increments are the components of the position increment vector
. In
other words,
(4.47)
To better emphasize the order of operation, we should write this as
(4.48)
On the left side, we have the
component of the vector differential, whereas on the right
side we have the differential of the
coordinate.
The covariant component of the position increment is well-defined:
(4.49)
Looking back at Eq. (4.48), natural question would be: is it possible to define complementary
or dual coordinates that are related to the baseline coordinates such that
this is NOT possible in general
(4.50)
We will now prove (by contradiction) that such dual coordinates do
not
generally exist. If
such coordinates did exist, then it would have to be true that
(4.51)
For the to exist, this equation would need to be integrable. In other words, the expression
would have to be an exact differential, which means that the metrics would have to
be given by
(4.52)
d
η
k
d
η
k
∂η
k
∂η
m
----------
d
η
m
=
←
η
k
η
m
y
f x
( )
=
dy
f
′
x
( )
dx
=
η
k
d
η
k
d
η
k
d
x
˜
d
x
˜
(
)
k
d
η
k
=
d
x
˜
(
)
k
d
η
k
( )
=
k
th
k
th
d
x
˜
(
)
i
g
ik
d
x
˜
(
)
k
=
η
i
η
i
d
x
˜
(
)
i
d
η
i
( )
=
←
d
η
i
( )
g
ij
d
η
j
( )
=
η
i
g
ij
d
η
j
( )
g
ij
∂η
i
∂η
j
--------
=
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61
and therefore, the second partials would need to be interchangeable:
(4.53)
This constraint is not satisfied by curvilinear systems. Consider, for example, cylindrical coor-
dinates,
, where the metric matrix is
(4.54)
Note that
, but
(4.55)
Hence, since these are not equal, the integrability condition of Eq. (4.53) is violated and there-
fore there do
not
exist dual coordinates.
∂
g
ij
∂η
k
---------
∂
2
η
i
∂η
j
∂η
k
------------------
∂
2
η
i
∂η
k
∂η
j
------------------
∂
g
ik
∂η
j
----------
=
=
=
η
1
=
r
η
2
=
θ η
3
=
z
,
,
(
)
g
ij
[ ]
1 0 0
0
r
2
0
0 0 1
=
∂
g
22
∂η
1
----------
∂
g
22
∂
r
----------
2
r
=
=
∂
g
21
∂η
2
----------
∂
g
21
∂θ
----------
0
=
=
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62
5. Curvilinear calculus
Chapter 3 focused on algebra, where the only important issue was the possibility that the
basis might be irregular (i.e., nonorthogonal, nonnormalized, and/or non-right-handed). In
that chapter, it made no difference whether the coordinates are homogeneous or curvilinear.
For homogeneous coordinates (where the base vectors are the same at all points in space), ten-
sor calculus formulas for the gradient are similar to the formulas for ordinary rectangular
Cartesian coordinates. By contrast, when the coordinates are curvilinear, there are new addi-
tional terms in gradient formulas to account for the variation of the base vectors with posi-
tion.
5.1 A introductory example
Before developing the general curvilinear theory, it is useful to first show how to develop
the formulas
without
using the full power of curvilinear calculus. Suppose an engineering
problem is most naturally described using cylindrical coordinates, which are related to the
laboratory Cartesian coordinates by
, ,
.
(5.1)
The base vectors for cylindrical coordinates are related to the lab base vectors by
.
(5.2)
Suppose you need the gradient of a scalar,
. If
s
is written as a function of , then
the familiar Cartesian formula applies:
.
(5.3)
Because curvilinear coordinates were selected, the function
is probably simple in
form. If you were a sucker for punishment, you could substitute the inverse relationships,
, ,
.
(5.4)
Then you’d have the function
, with which you could directly apply Eq. (5.3). This
approach is unsatisfactory for several reasons. Strictly speaking, Eq. (5.4) is incorrect because
the arctangent has
two
solutions in the range from 0 to
; the correct formula would have to
be the two-argument arctangent. Furthermore, actually computing the derivative would be
tedious, and the final result would be expressed in terms of the lab Cartesian basis instead of
the cylindrical basis.
x
1
r
θ
cos
=
x
2
r
θ
sin
=
x
3
z
=
e
˜
r
θ
cos
e
˜
1
θ
sin
e
˜
2
+
=
e
˜
θ
θ
sin
–
e
˜
1
θ
cos
e
˜
2
+
=
e
˜
z
e
˜
3
=
ds d
x
˜
⁄
x
i
ds
d
x
˜
------
∂
s
∂
x
1
--------
e
˜
1
∂
s
∂
x
2
--------
e
˜
2
∂
s
∂
x
3
--------
e
˜
3
+
+
=
s r
θ
z
, ,
(
)
r
x
1
2
x
1
2
+
=
θ
tan
1
–
x
2
x
1
-----
=
z
x
3
=
s x
1
x
2
x
3
, ,
(
)
2
π
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63
The better approach is to look at the problem from a more academic slant by applying the
chain rule. Given that
then
,
(5.5)
where the subscripts indicate which variables are held constant in the derivatives. The deriva-
tives of the cylindrical coordinates with respect to can be computed
a priori.
For example,
the quantity
is the gradient of . Physically, we know that the gradient of a quantity is
perpendicular to surfaces of constant values of that quantity. Surfaces of constant are cylin-
ders of radius , so we know that
must be perpendicular to the cylinder. In other
words, we know that
must be parallel to . The gradients of the cylindrical coordi-
nates are obtained by applying the ordinary Cartesian formula:
.
(5.6)
We must determine these derivatives
implicitly
by using Eqs. (5.1) to first compute the
derivatives of the Cartesian coordinates with respect to the cylindrical coordinates. We can
arrange the nine possible derivatives of Eqs. (5.1) in a matrix:
.
(5.7)
The inverse derivatives are obtained inverting this matrix to give
.
(5.8)
s
s r
θ
z
, ,
(
)
=
ds
d
x
˜
------
∂
s
∂
r
-----
θ
z
,
dr
d
x
˜
------
∂
s
∂θ
------
r z
,
d
θ
d
x
˜
------
∂
s
∂
z
-----
r
θ
,
dz
d
x
˜
------
+
+
=
x
˜
dr d
x
˜
⁄
r
r
r
dr d
x
˜
⁄
dr d
x
˜
⁄
e
˜
r
dr
d
x
˜
------
∂
r
∂
x
1
--------
e
˜
1
∂
r
∂
x
2
--------
e
˜
2
∂
r
∂
x
3
--------
e
˜
3
+
+
=
d
θ
d
x
˜
------
∂θ
∂
x
1
--------
e
˜
1
∂θ
∂
x
2
--------
e
˜
2
∂θ
∂
x
3
--------
e
˜
3
+
+
=
dz
d
x
˜
------
∂
z
∂
x
1
--------
e
˜
1
∂
z
∂
x
2
--------
e
˜
2
∂
z
∂
x
3
--------
e
˜
3
+
+
=
∂
x
1
∂
r
--------
∂
x
1
∂θ
--------
∂
x
1
∂
z
--------
∂
x
2
∂
r
--------
∂
x
2
∂θ
--------
∂
x
2
∂
z
--------
∂
x
3
∂
r
--------
∂
x
3
∂θ
--------
∂
x
3
∂
z
--------
θ
cos
r
θ
sin
–
0
θ
sin
r
θ
cos
0
0
0
1
=
∂
r
∂
x
1
--------
∂
r
∂
x
2
--------
∂
r
∂
x
3
--------
∂θ
∂
x
1
--------
∂θ
∂
x
2
--------
∂θ
∂
x
3
--------
∂
z
∂
x
1
--------
∂
z
∂
x
2
--------
∂
z
∂
x
3
--------
θ
cos
θ
sin
0
θ
sin
r
-----------
–
θ
cos
r
------------ 0
0
0
1
=
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64
Substituting these derivatives into Eq. (5.6) gives
.
(5.9)
Referring to Eq. (5.2) we conclude that
This result was rather tedious to derive, but it was a one-time-only task. For any coordinate
system, it is always a good idea to derive the coordinate gradients and save them for later use.
Now that we have the coordinate gradients, Eq. (5.5) becomes
,
(5.11)
where the commas are a shorthand notation for derivatives. This is the formula for the gradi-
ent of a scalar that you would find in a handbook.
The above analysis showed that high-powered curvilinear theory is not necessary to
derive the formula for the gradient of a scalar function of the cylindrical coordinates. All
dr
d
x
˜
------
θ
cos
e
˜
1
θ
sin
e
˜
2
+
=
d
θ
d
x
˜
------
θ
sin
r
-----------
–
e
˜
1
θ
cos
r
------------
e
˜
2
+
=
dz
d
x
˜
------
e
˜
3
=
(5.10)
dr
d
x
˜
------
e
˜
r
=
d
θ
d
x
˜
------
e
˜
θ
r
-----
=
dz
d
x
˜
------
e
˜
3
=
ds
d
x
˜
------
s
,
r
e
˜
r
s
,
θ
r
------
e
˜
θ
s
,
z
e
˜
3
+
+
=
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65
that’s needed is knowledge of tensor analysis in Cartesian coordinates. However, for more
complicated coordinate systems, a fully developed curvilinear theory is indispensable.
5.2 Curvilinear coordinates
In what follows, three coordinates
are pre-
sumed to identify the position of a point in space. These
coordinates are identified with superscripts merely by con-
vention. As sketched in Figure 5.1, the associated base vectors
at any point in space are tangent to the grid lines at that point. The base vector
points in
the direction of increasing
. Importantly, the position vector is
not
generally equal to
. For example, the position vector for spherical coordinates is simply
(
not
); for spherical coordinates, the dependence of the position vector on the
coordinates and is hidden in dependence of on and .
5.3 The “associated” curvilinear covariant basis
Given three coordinates
that identify the location in space, the associated
covariant base vectors are defined to be tangent to the grid lines along which only one coordi-
nate varies (the others being held constant). By definition, the three coordinates
identify the position in space:
;
i.e.,
is a function of
,
, and
.
(5.12)
The
i
th
covariant base vector is defined to point in the direction that moves when
is
O
g
˜
1
g
˜
2
g
˜
3
x
˜
FIGURE 5.1 Curvilinear covariant basis.
The covariant base
vectors are tangent to the grid lines associated with
. Different
base vectors are used at different locations in space because the grid
itself varies in space. The lengths of the covariant base vectors “co-
vary” with the coordinate. Each unit change in the coordinate
corresponds to a specific change in the position vector, and the
covariant basis has a length equal to the change in position per unit
change in coordinate.
η
k
η
1
η
2
x
˜
g
˜
1
g
˜
2
η
1
η
2
η
3
,
,
{
}
x
˜
g
˜
k
η
k
x
˜
η
k
g
˜
k
x
˜
r
e
˜
r
=
x
˜
r
e
˜
r
θ
e
˜
θ
φ
e
˜
ψ
+
+
=
θ
φ
e
˜
r
θ
φ
η
1
η
2
η
3
,
,
{
}
η
1
η
2
η
3
,
,
{
}
x
˜
x
˜
η
1
η
2
η
3
,
,
(
)
=
x
˜
η
1
η
2
η
3
x
˜
η
k
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66
increased, holding the other coordinates constant. Hence, the natural definition of the
i
th
cova-
riant basis is the partial derivative of with respect to :
.
(5.13)
Note the following new summation convention: the superscript on is interpreted as a
sub-
script
in the final result because appears in the “denominator” of the derivative.
Note that the natural covariant base vectors are not necessarily unit vectors. Furthermore,
because the coordinates
do not necessarily have the same physical units, the nat-
ural base vectors themselves will have different physical units. For example, cylindrical coor-
dinates
have dimensions of length, radians, and length, respectively. Consequently,
two of the base vectors (
and
) are dimensionless, but
has dimensions of length. Such
a situation is not unusual and should not be alarming. We will find that the
components
of a
vector will have physical dimensions appropriate to ensure that the each term in the sum of
components times base vectors will have the same physical dimensions as the vector itself.
Again this point harks back to the fact that neither components nor base vectors are invariant,
but the
sum
of components times base vectors is invariant.
Equation (5.12) states that the coordinates uniquely identify the position in space. Con-
versely, any position in space corresponds to a unique set of coordinates. That is, each coordi-
nate may be regarded as a single-valued function of position vector:
.
(5.14)
By the chain rule, note that
.
(5.15)
Therefore, recalling Eq. (5.13), the contravariant dual basis must be given by
.
(5.16)
This derivative is the spatial gradient of . Hence, as
sketched in Fig. 5.2,
each contravariant base vector is
normal to surfaces of constant , and it points in the
direction of increasing
.
Starting with Eq. (5.12), the increment in the position
vector is given by
,
(5.17)
x
˜
η
i
g
˜
i
∂
x
˜
∂η
i
--------
≡
η
i
η
i
η
1
η
2
η
3
,
,
{
}
r
θ
z
, ,
{
}
g
˜
1
g
˜
3
g
˜
2
η
j
η
j
x
˜
( )
=
d
η
j
d
x
˜
--------
∂
x
˜
∂η
i
--------
•
∂η
j
∂η
i
--------
δ
i
j
=
=
η
1
η
2
FIGURE 5.2 Curvilinear contravariant
basis.
The contravariant base vectors are
normal to surfaces of constant
.
η
k
x
˜
g
˜
1
g
˜
2
g
˜
j
∂η
j
∂
x
˜
--------
≡
η
j
g
˜
j
η
j
η
j
d
x
˜
∂
x
˜
∂η
k
---------
d
η
k
=
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67
.
(5.18)
Now we note a key difference between homogeneous and curvilinear coordinates:
ix.
For homogeneous coordinates, each of the base vectors is same
throughout space — they are independent of the coordinates
. Hence,
for homogeneous coordinates, Eq. (5.18) can be
integrated to show that
.
x.
For curvilinear coordinates, each varies in space — they depend on
the coordinates
. Hence,
for curvilinear coordinates,
Eq. (5.18) does
not
imply that
.
EXAMPLE
Consider cylindrical coordinates:
, ,
.
Figure 5.3 illustrates how, for simple enough coordinate systems, you can determine the cova-
riant base vectors graphically. Recall that
and therefore that
is the
coefficient of
when is varied by differentially changing
holding
and
con-
stant. For cylindrical coordinates, when the radial coordinate,
, is varied holding the oth-
ers constant, the position vector moves such
, and therefore
must equal
because it is the coefficient of
. Similarly,
must be equal to
since that is the coeffi-
cient of
in
when the second coordinate,
, is varied holding the others constant.
Summarizing,
, ,
and
.
(5.19)
To derive these results analytically (rather than geometrically), we utilize the underlying rect-
d
x
˜
d
η
k
g
˜
k
=
g
˜
k
η
1
η
2
η
3
,
,
{
}
x
˜
η
k
g
˜
k
=
g
˜
k
η
1
η
2
η
3
,
,
{
}
x
˜
η
k
g
˜
k
=
η
1
=
r
η
2
=
θ η
3
=
z
e
˜
r
e
˜
θ
e
˜
z
θ
r
z
x
1
x
2
x
3
x
˜
x
˜
r
e
˜
r
θ
( )
z
e
˜
z
+
=
x
˜
d
x
˜
=d
r
e
˜
r
x
˜
d
x
˜
=
r
d
θ
e
˜
θ
Here we vary
r
FIGURE 5.3 Covariant basis for cylindrical coordinates.
The views down the -axis show how the covariant base
vectors can be determined graphically by varying one coordinate, holding the others constant.
z
holding
θ
and
z
constant
Here we vary
θ
holding
r
and
z
constant
x
1
x
1
x
2
x
2
g
˜
1
∂
x
˜
∂η
1
⁄
(
)
η
2
η
3
,
=
g
˜
1
d
η
1
x
˜
η
1
η
2
η
3
η
1
=
r
x
˜
d
x
˜
d
r
e
˜
r
=
g
˜
1
e
˜
r
d
r
g
˜
2
r
e
˜
θ
d
θ
d
x
˜
η
2
=
θ
g
˜
1
e
˜
r
=
g
˜
2
r
e
˜
θ
=
g
˜
3
e
˜
z
=
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DRAFT June 17, 2004
68
angular Cartesian basis
to write the position vector as
,
(5.20)
where
(5.21a)
(5.21b)
.
(5.21c)
Then
(5.22a)
(5.22b)
.
(5.22c)
which we note is equivalent to the graphically derived Eqs. (5.19). Also note that
and
are dimensionless, whereas
has physical dimensions of length.
The metric coefficients
for cylindrical coordinates are derived in the usual way:
(5.23a)
(5.23b)
,
(5.23c)
or, in matrix form,
.
←
cylindrical
coordinates
(5.24)
Whenever the metric tensor comes out to be diagonal as it has here, the coordinate system is
orthogonal and the base vectors at each point in space are mutually perpendicular.
Inverting
the covariant
matrix gives the contravariant metric coefficients:
.
←
cylindrical
coordinates
(5.25)
e
˜
1
e
˜
2
e
˜
3
,
,
{
}
x
˜
x
1
e
˜
1
x
2
e
˜
2
x
3
e
˜
3
+
+
=
x
1
r
θ
cos
=
x
2
r
θ
sin
=
x
3
z
=
g
˜
1
∂
x
˜
∂
r
------
θ
z
,
∂
x
1
∂
r
--------
θ
z
,
e
˜
1
∂
x
2
∂
r
--------
θ
z
,
e
˜
2
∂
x
3
∂
r
--------
θ
z
,
e
˜
3
+
+
θ
cos
e
˜
1
θ
e
˜
2
sin
+
=
=
=
g
˜
2
∂
x
˜
∂θ
------
r z
,
∂
x
1
∂θ
--------
r z
,
e
˜
1
∂
x
2
∂θ
--------
r z
,
e
˜
2
∂
x
3
∂θ
--------
r z
,
e
˜
3
+
+
r
θ
sin
e
˜
1
–
r
θ
cos
e
˜
2
+
=
=
=
g
˜
3
∂
x
˜
∂
z
------
r
θ
,
∂
x
1
∂
z
--------
r
θ
,
e
˜
1
∂
x
2
∂
z
--------
r
θ
,
e
˜
2
∂
x
3
∂
z
--------
r
θ
,
e
˜
3
+
+
e
˜
3
=
=
=
g
˜
1
g
˜
3
g
˜
2
g
ij
g
11
g
˜
1
g
˜
1
•
1
=
=
g
12
g
˜
1
g
˜
2
•
0
=
=
g
13
g
˜
1
g
˜
3
•
0
=
=
g
12
g
˜
2
g
˜
1
•
0
=
=
g
22
g
˜
2
g
˜
2
•
r
2
=
=
g
23
g
˜
2
g
˜
3
•
0
=
=
g
31
g
˜
3
g
˜
1
•
0
=
=
g
32
g
˜
3
g
˜
2
•
0
=
=
g
33
g
˜
3
g
˜
3
•
1
=
=
g
ij
[ ]
1 0 0
0
r
2
0
0 0 1
=
g
ij
[ ]
g
ij
[
]
1
0
0
0 1
r
2
⁄
0
0
0
1
=
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69
The contravariant dual basis can be derived in the usual way by applying the formula of
(5.26a)
(5.26b)
.
(5.26c)
With considerably more effort, we can alternatively derive these results by directly applying
Eq. (5.16). To do this, we must express the coordinates
in terms of
:
(5.27a)
(5.27b)
.
(5.27c)
Then applying Eq. (5.16) gives
(5.28a)
(5.28b)
.
(5.28c)
which agrees with the previous result in Eq. (5.26).
Because the curvilinear basis
associated with cylindrical coordinates is
orthogonal, the dual vectors
are in exactly the same directions but have different
magnitudes. It is common practice to use the associated ortho
normal
basis
as we
have in the above equations.
This practice also eliminates the sometimes confusing complica-
tion of dimensional base vectors. When a coordinate system is orthogonal, the shared
orthonormal basis is called the “
physical basis
.” Notationally, vector components
or
are defined
and
, respectively. Because cylindrical coor-
dinates are orthogonal, it is conventional to also define so-called “physical components” with
respect to the orthonormalized cylindrical basis.
For cylindrical coordinates, the physical components are denoted
. Whenever
you derive a formula in terms of the general covariant and contravariant vector components,
g
˜
1
g
11
g
˜
1
g
12
g
˜
2
g
13
g
˜
3
+
+
g
˜
1
e
˜
r
=
=
=
g
˜
2
g
21
g
˜
1
g
22
g
˜
2
g
23
g
˜
3
+
+
g
˜
2
r
2
-----
e
˜
θ
r
-----
=
=
=
g
˜
3
g
31
g
˜
1
g
32
g
˜
2
g
33
g
˜
3
+
+
g
˜
3
e
˜
z
=
=
=
r
θ
z
, ,
{
}
x
1
x
2
x
3
, ,
{
}
r
x
1
2
x
2
2
+
=
θ
tan
1
–
x
2
x
1
-----
=
z
x
3
=
g
˜
1
d
r
d
x
˜
------
∂
r
∂
x
1
--------
e
˜
1
∂
r
∂
x
2
--------
e
˜
2
∂
r
∂
x
3
--------
e
˜
3
+
+
x
1
r
-----
e
˜
1
x
2
r
-----
e
˜
2
+
θ
cos
e
˜
1
θ
sin
e
˜
2
+
e
˜
r
=
=
=
=
=
g
˜
2
d
θ
d
x
˜
------
∂θ
∂
x
1
--------
e
˜
1
∂θ
∂
x
2
--------
e
˜
2
∂θ
∂
x
3
--------
e
˜
3
+
+
x
2
–
x
1
2
--------
cos
2
θ
e
˜
1
1
x
1
-----
cos
2
θ
e
˜
2
+
1
r
---
e
˜
θ
=
=
=
=
g
˜
3
d
z
d
x
˜
------
∂
z
∂
x
1
--------
e
˜
1
∂
z
∂
x
2
--------
e
˜
2
∂
z
∂
x
3
--------
e
˜
3
+
+
e
˜
3
e
˜
z
=
=
=
=
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
e
˜
r
e
˜
θ
e
˜
z
,
,
{
}
v
1
v
2
v
3
, ,
{
}
v
1
v
2
v
3
, ,
{
}
v
k
g
˜
k
v
˜
•
≡
v
k
g
˜
k
v
˜
•
≡
v
r
v
θ
v
z
, ,
{
}
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70
it’s a good idea to convert the final result to physical coordinates and the physical basis. For
cylindrical coordinates, these conversion formulas are:
(5.29a)
(5.29b)
.
(5.29c)
v
1
g
˜
1
v
˜
•
e
˜
r
v
˜
•
v
r
=
=
=
v
1
g
˜
1
v
˜
•
e
˜
r
v
˜
•
v
r
=
=
=
v
2
g
˜
2
v
˜
•
r
e
˜
θ
(
)
v
˜
•
rv
θ
=
=
=
v
2
g
˜
2
v
˜
•
1
r
---
e
˜
θ
v
˜
•
v
θ
r
-----
=
=
=
v
3
g
˜
3
v
˜
•
e
˜
z
v
˜
•
v
z
=
=
=
v
3
g
˜
3
v
˜
•
e
˜
z
v
˜
•
v
z
=
=
=
Study Question 5.1
For spherical coordinates,
, the underlying rectan-
gular Cartesian coordinates are
(5.30a)
(5.30b)
.
(5.30c)
(a) Follow the above example to prove that the covariant basis for spherical coordinates is
, where
(5.31a)
,
where (5.31b)
where .
(5.31c)
(b) Prove that the dual contravariant basis for spherical coordinates is
, (5.32a)
,
(5.32b)
.
(5.32c)
(c) As was done for cylindrical coordinates in Eq. (5.29) show that the spherical covariant
and contravariant vector components are related to the spherical “physical” components by
(5.33a)
(5.33b)
.
(5.33c)
η
1
=
r
η
2
=
θ η
3
=
φ
,
,
{
}
x
1
r
θ
sin
φ
cos
=
x
2
r
θ
sin
φ
sin
=
x
3
r
θ
cos
=
g
˜
1
e
˜
r
=
e
˜
r
θ
sin
φ
cos
e
˜
1
θ
sin
φ
sin
e
˜
2
θ
cos
e
˜
3
+
+
=
g
˜
2
r
e
˜
θ
=
e
˜
θ
θ
cos
φ
cos
e
˜
1
θ
φ
sin
cos
e
˜
2
θ
sin
e
˜
3
–
+
=
g
˜
3
r
θ
sin
e
˜
φ
=
e
˜
φ
φ
sin
e
˜
1
–
φ
cos
e
˜
2
+
=
g
˜
1
e
˜
r
=
g
˜
2
1
r
---
e
˜
θ
=
g
˜
3
1
r
θ
sin
--------------
e
˜
φ
=
v
1
v
r
=
v
1
v
r
=
v
2
rv
θ
=
v
2
v
θ
r
-----
=
v
3
r
θ
sin
(
)
v
φ
=
v
3
v
φ
r
θ
sin
--------------
=
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71
5.4 The gradient of a scalar
Section 5.1 provided a simplified example for finding the formula for the gradient of a scalar
function of cylindrical coordinates. Now we outline the procedure for more complicated gen-
eral curvilinear coordinates. Recall Eq. (5.18):
,
(5.34)
where
.
(5.35)
Dotting both sides of Eq. (5.34) with shows that
.
(5.36)
This important equation is crucial in determining expressions for gradient operations. Con-
sider, for example, a scalar-valued field
.
(5.37)
The increment in this function is given by
,
(5.38)
,
(5.39)
which holds for all
. The direct notation definition of the gradient
of a scalar field is
.
(5.40)
Comparing the above two equations gives the
formula for the gradient of a scalar in curvilin-
ear coordinates:
.
(5.41)
Notice that this formula is very similar in form to the familiar formula for rectangular Carte-
sian coordinates. Gradient formulas won’t look significantly different until we compute vec-
tor gradients in the next section.
Example: cylindrical coordinates
Consider a scalar field
.
(5.42)
d
x
˜
d
η
k
g
˜
k
=
g
˜
k
∂
x
˜
∂η
k
---------
=
g
˜
i
d
η
i
g
˜
i
d
x
˜
•
=
s
s
η
1
η
2
η
3
,
,
(
)
=
d
s
∂
s
∂η
k
---------d
η
k
=
d
s
∂
s
∂η
k
---------
g
˜
k
d
x
˜
•
=
d
x
˜
d
s
d
x
˜
⁄
d
s
d
s
d
x
˜
------
d
x
˜
•
=
d
x
˜
∀
d
s
d
x
˜
------
∂
s
∂η
k
---------
g
˜
k
=
s
s r
θ
z
, ,
(
)
=
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72
Applying Eq. (5.41) with the contravariant basis of Eq. (5.26) gives
, (5.43)
which is the gradient formula typically found in math handbooks.
5.5 Gradient of a vector -- “simplified” example
Now let’s look at the gradient of a vector. If the vector is expressed in Cartesian coordi-
nates, the formula for the gradient is
.
(5.45)
Suppose a vector is expressed in the cylindrical coordinates:
.
(5.46)
The
components , ,
and
are presumably known as functions of the
coordi-
nates. Importantly,
the base vectors also vary with the coordinates!
First, recall the product rule for the gradient of a scalar times a vector:
.
(5.47)
Applying the product rule to each of the terms in Eq. (5.46), the gradient of
gives
+ .
(5.48)
Eq. (5.11) applies to the gradient of any scalar. Hence, the first three terms of Eq. (5.48) can be
d
s
d
x
˜
------
∂
s
∂
r
-----
e
˜
r
∂
s
∂θ
------
e
˜
θ
r
-----
∂
s
∂
z
-----
e
˜
z
+
+
=
Study Question 5.2
Follow the above example [using Eq. (5.31) in Eq. (5.41)] to prove that
the formula for the gradient of a scalar
s
in spherical coordinates is
.
(5.44)
d
s
d
x
˜
------
∂
s
∂
r
-----
e
˜
r
∂
s
∂θ
------
e
˜
θ
r
-----
∂
s
∂φ
------
e
˜
φ
r
θ
sin
--------------
+
+
=
d
v
˜
d
x
˜
------
∂
v
i
∂
x
j
-------
e
˜
i
e
˜
j
=
v
˜
v
˜
v
r
e
˜
r
v
θ
e
˜
θ
v
z
e
˜
z
+
+
=
v
r
v
θ
v
z
r
θ
z
, ,
{
}
d s
u
˜
(
)
d
x
˜
---------------
u
˜
ds
d
x
˜
------
s
d
u
˜
d
x
˜
-------
+
=
d
v
˜
d
x
˜
⁄
d
v
˜
d
x
˜
------
e
˜
r
dv
r
d
x
˜
--------
e
˜
θ
dv
θ
d
x
˜
--------
e
˜
z
dv
z
d
x
˜
--------
+
+
=
v
r
d
e
˜
r
d
x
˜
--------
v
θ
d
e
˜
θ
d
x
˜
---------
v
z
d
e
˜
z
d
x
˜
--------
+
+
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DRAFT June 17, 2004
73
written
.
(5.49)
Now we need formulas for the gradients of the base vectors. Applying the product rule to Eq.
(5.2), noting that the gradients of the Cartesian basis are all zero, gives
.
(5.50)
Substituting Eqs. (5.49) and (5.51) into (5.48) gives
+
+ .
(5.52)
dv
r
d
x
˜
--------
v
r
,
r
e
˜
r
v
r
,
θ
r
--------
e
˜
θ
v
r
,
z
e
˜
3
+
+
=
dv
θ
d
x
˜
--------
v
θ
,
r
e
˜
r
v
θ
,
θ
r
---------
e
˜
θ
v
θ
,
z
e
˜
3
+
+
=
dv
z
d
x
˜
--------
v
z
,
r
e
˜
r
v
z
,
θ
r
--------
e
˜
θ
v
z
,
z
e
˜
3
+
+
=
d
e
˜
r
d
x
˜
--------
θ
sin
–
e
˜
1
θ
cos
e
˜
2
+
(
)
d
θ
d
x
˜
------
e
˜
θ
d
θ
d
x
˜
------
=
=
d
e
˜
θ
d
x
˜
---------
θ
cos
–
e
˜
1
θ
sin
e
˜
2
–
(
)
d
θ
d
x
˜
------
e
˜
r
d
θ
d
x
˜
------
–
=
=
d
e
˜
z
d
x
˜
--------
0
˜˜
=
.
(5.51)
d
e
˜
r
d
x
˜
--------
1
r
---
e
˜
θ
e
˜
θ
=
d
e
˜
θ
d
x
˜
---------
1
r
---
e
˜
r
e
˜
θ
–
=
d
e
˜
z
d
x
˜
--------
0
˜˜
=
d
v
˜
d
x
˜
------
e
˜
r
v
r
,
r
e
˜
r
v
r
,
θ
r
--------
e
˜
θ
v
r
,
z
e
˜
3
+
+
=
e
˜
θ
v
θ
,
r
e
˜
r
v
θ
,
θ
r
---------
e
˜
θ
v
θ
,
z
e
˜
3
+
+
e
˜
z
v
z
,
r
e
˜
r
v
z
,
θ
r
--------
e
˜
θ
v
z
,
z
e
˜
3
+
+
v
r
1
r
---
e
˜
θ
e
˜
θ
v
θ
1
r
---
e
˜
r
e
˜
θ
–
+
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DRAFT June 17, 2004
74
Collecting terms gives
=
+
+
+ + +
+ +
+
.
(5.53)
This result is usually given in textbooks in matrix form with respect to the cylindrical basis:
.
(5.54)
There’s no doubt that this result required a considerable amount of effort to derive. Typically,
these kinds of formulas are compiled in the appendices of most tensor analysis reference
books. The appendix of R.B. Bird’s book on macromolecular hydrodynamics is particularly
well-organized and error-free. If, however, you use Bird’s appendix, you will notice that the
components given for the gradient of a vector seem to be the transpose of what we have pre-
sented above; that’s because Bird (and some others) define the gradient of a tensor to be the
transpose of our definition. Before using anyone’s gradient table, you should always ascertain
which definition the author uses.
Now we are going to perform the same sort of analysis to show how the gradient is deter-
mined for general curvilinear coordinates.
5.6 Gradient of a vector in curvilinear coordinates
The formula for the gradient of a scalar in curvilinear coordinates was not particularly tough
to derive and comprehend — it didn’t look profoundly different from the formula for rectan-
gular Cartesian coordinates. Taking gradients of vectors, however, begins a new nightmare.
Consider a vector field,
.
(5.55)
Each component of the vector is of course a function of the coordinates, but for general curvi-
d
v
˜
d
x
˜
------
v
r
,
r
(
)
e
˜
r
e
˜
r
v
r
,
θ
r
--------
v
θ
r
-----
–
e
˜
r
e
˜
θ
v
r
,
z
(
)
e
˜
r
e
˜
z
v
θ
,
r
(
)
e
˜
θ
e
˜
r
v
θ
,
θ
r
---------
v
r
r
----
+
e
˜
θ
e
˜
θ
v
θ
,
z
(
)
e
˜
θ
e
˜
z
v
z
,
r
(
)
e
˜
z
e
˜
r
v
z
,
θ
r
--------
e
˜
z
e
˜
θ
v
z
,
z
e
˜
z
e
˜
z
d
v
˜
d
x
˜
------
v
r
,
r
v
r
,
θ
-
v
θ
r
-----------------
v
r
,
z
v
θ
,
r
v
θ
,
θ
+
v
r
r
------------------
v
θ
,
z
v
z
,
r
v
z
,
θ
r
--------
v
z
,
z
=
v
˜
v
˜
η
1
η
2
η
3
,
,
(
)
=
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DRAFT June 17, 2004
75
linear coordinates, so are the base vectors
! Written out,
.
(5.56)
Therefore the increment
involves both increments
of the components and increments
of the base vectors:
.
(5.57)
Applying the chain rule and using Eq. (5.36), the component increments can be written
.
(5.58)
Similarly, the base vector increments are
.
(5.59)
Substituting these results into Eq. (5.57) and rearranging gives
,
(5.60)
which holds for all
. Recall the gradient
of a vector is defined in direct notation
such that
.
(5.61)
Comparing the above two equations gives us a formula for the gradient:
.
(5.62)
Incidentally, these equations serve as further examples of how a superscript in the “denomi-
nator” is regarded as a subscript in the summation convention.
Christoffel Symbols
Note that
the nine
vectors [
i.e.
, the coefficients of in the last
term of Eq. (5.62)] are strictly properties of the coordinate system and they may be computed
and tabulated
a priori
.
This family of
system-dependent
vectors is denoted
.
(5.63)
Recalling Eq. (5.35), note that
.
(5.64)
Thus,
only six of the nine
vectors are independent due to the symmetry in
. The
k
th
con-
v
˜
v
i
η
1
η
2
η
3
,
,
(
)
g
˜
i
η
1
η
2
η
3
,
,
(
)
=
d
v
˜
d
v
i
d
g
˜
i
d
v
˜
d
v
i
g
˜
i
v
i
d
g
˜
i
+
=
d
v
i
∂
v
i
∂η
j
--------d
η
j
∂
v
i
∂η
j
--------
g
˜
j
d
x
˜
•
=
=
d
g
˜
i
∂
g
˜
i
∂η
j
--------d
η
j
∂
g
˜
i
∂η
j
--------
g
˜
j
d
x
˜
•
=
=
d
v
˜
∂
v
i
∂η
j
--------
g
˜
i
g
˜
j
d
x
˜
•
v
i
∂
g
˜
i
∂η
j
--------
g
˜
j
d
x
˜
•
+
=
d
x
˜
d
v
˜
d
x
˜
⁄
d
v
˜
d
v
˜
d
x
˜
------
d
x
˜
•
=
d
x
˜
∀
d
v
˜
d
x
˜
------
∂
v
i
∂η
j
--------
g
˜
i
g
˜
j
v
i
∂
g
˜
i
∂η
j
--------
g
˜
j
+
=
∂
g
˜
i
∂η
j
⁄
v
i
Γ
˜
ij
∂
g
˜
i
∂η
j
--------
≡
Γ
˜
ij
∂
2
x
˜
∂η
j
∂η
i
-----------------
∂
2
x
˜
∂η
i
∂η
j
-----------------
Γ
˜
ji
=
=
=
Γ
˜
ij
ij
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DRAFT June 17, 2004
76
travariant component of
is obtained by dotting into
:
.
(5.65)
The quantity
is called the
Christoffel symbol of the second kind
.
Important comment about notation:
Even though the Christoffel symbols
have three
indices, they are
not
components of a third-order tensor. By this, we mean that a second basis
will have a different set of Christoffel symbols
and, as discussed below, they are that
are not obtainable via a simple tensor transformation of the Christoffel symbols of the original
system [i.e.,
]. Instead of the notation
, Christoffel
symbols are therefore frequently denoted in the literature with the odd-looking notation
.
The fact that Christoffel symbols are not components of a tensor is, of course, a strong justifi-
cation for avoiding typesetting Christoffel symbols in the same way as tensors. However, to
be 100% consistent, proponents of this notational ideology would — by the same arguments
— be compelled to
not
typeset
coordinates
using the notation
which erroneously makes it
look as though the
are contravariant components of some vector , even though they
aren’t. Being comfortable with the typesetting
, we are also comfortable with the typeset-
ting
. The key is for the analyst to recognize that neither of these symbols connote tensors.
Instead, they are “basis-intrinsic” quantities (i.e., indexed quantities whose meaning is
defined for a particular basis and whose connection with counterparts from a different basis
are
not
obtained via a tensor transformation rule). Of course, the base vectors themselves are
basis-intrinsic objects. Any new object that is constructed from basis-intrinsic quantities
should be
itself
regarded as basis-intrinsic until proven otherwise. For example, the metric
were initially regarded as basis-intrinsic because they were constructed from
basis-intrinsic objects (the base vectors), but it was proved that they turned out to also satisfy
the tensor transformation rule. Consequently, even though the metric matrix is constructed
from basis-intrinsic quantities, it turns out to
not
be basis intrinsic itself (the metric compo-
nents are components of the identity tensor).
On page 86, we define Christoffel symbols of the “first” kind, which are useful in Rieman-
nian spaces where there is no underlying rectangular Cartesian basis. In general,
if the term
“Christoffel symbol” is used by itself, it should be taken to mean the Christoffel symbol of the
second
kind defined above
. Christoffel symbols may appear rather arcane, but keep in mind
that
these quantities simply characterize how the base vectors vary in space
. Christoffel are
also sometimes called the “
affinities
Γ
˜
ij
g
˜
k
Γ
˜
ij
Γ
ij
k
g
˜
k
Γ
˜
ij
•
≡
Γ
ij
k
Γ
ij
k
g
˜
k
Γ
ij
k
Γ
ij
k
Γ
mn
p
g
˜
m
g
˜
m
•
(
)
g
˜
n
g
˜
n
•
(
)
g
˜
p
g
˜
k
•
(
)
≠
Γ
ij
k
k
ij
{ }
η
k
η
k
η
˜
η
k
Γ
ij
k
g
ij
g
˜
i
g
˜
j
•
≡
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DRAFT June 17, 2004
77
By virtue of Eq. (5.64), note that
.
(5.66)
Like the components
of the tensor , the Christoffel symbols
are properties of the
par-
ticular
coordinate system and its basis. Consequently, the Christoffel symbols are not the com-
ponents of a third-order tensor.
Specifically, if we were to consider some
different
curvilinear
coordinate system and compute the Christoffel symbols for that system, the result would not
be the same as would be obtained by a tensor transformation of the Christoffel symbols of the
first coordinate system to the second system. This is in contrast to the metric coefficients
which
do
happen to be components of a tensor (namely, the identity tensor).
Recalling that
is the
k
th
contravariant component of
and by Eq. (5.63)
, we conclude that the variation of the base vectors in space is completely
characterized by the Christoffel symbols. Namely,
.
(5.67)
Increments in the base vectors
By the chain rule, the increment in the covariant base vec-
tor can always be written
(5.68)
or, using the notation introduced in Eq. (5.67)
(5.69)
Manifold torsion
Recall that the Christoffel symbols are not components of basis-indepen-
dent tensors. Consider, however [12], the anti-symmetric part of the Christoffel symbols:
(5.70)
As long as Eq. (5.66) holds, then the manifold torsion will be zero. For a non-holomomic sys-
tem, it’s possible that the manifold torsion will be nonzero, but it will turn out to be a basis
independent (i.e., “free” vector). Henceforth, we assume that Eq. (5.66) holds true, so no fur-
ther mention will be made of the manifold torsion.
Γ
ij
k
Γ
ji
k
=
F
ij
F
˜˜
Γ
ij
k
g
ij
Γ
ij
k
Γ
˜
ij
Γ
˜
ij
∂
g
˜
i
∂η
j
⁄
=
∂
g
˜
i
∂η
j
--------
Γ
ij
k
g
˜
k
=
d
g
˜
i
∂
g
˜
i
∂η
j
--------
d
η
j
=
d
g
˜
i
Γ
ij
k
g
˜
k
d
η
j
=
2
Γ
ij
[ ]
k
Γ
ij
k
Γ
ji
k
–
≡
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DRAFT June 17, 2004
78
EXAMPLE: Christoffel symbols for cylindrical coordinates
In terms of the underlying
rectangular Cartesian basis, the covariant base vectors from Eq. (5.19) can be written
(5.71a)
(5.71b)
.
(5.71c)
Therefore, applying Eq. (5.63), using
(5.72a)
(5.72b)
.
(5.72c)
Noting that
is the coefficient of in the expression for
, we find that only three of the
27 Christoffel symbols are nonzero. Namely, noting that
is the coefficient of
in
and noting that
the coefficient of
in
,
and
(all
other
).
(5.73)
If you look up cylindrical Christoffel symbols in a handbook, you will probably find the sub-
scripts (1,2,3) replaced with the coordinate symbols (
r,
θ
,z
) for clarity so they are listed as
and
(all
other
).
(5.74)
g
˜
1
θ
cos
e
˜
1
θ
sin
e
˜
2
+
=
g
˜
2
r
θ
sin
–
e
˜
1
r
θ
e
˜
2
cos
+
=
g
˜
3
e
˜
3
=
η
1
=
r
η
2
=
θ η
3
=
z
,
,
{
}
Γ
˜
11
∂
g
˜
1
∂
r
---------
0
˜
=
=
Γ
˜
12
∂
g
˜
1
∂θ
---------
θ
sin
–
e
˜
1
θ
e
˜
2
cos
+
1
r
---
g
˜
2
=
=
=
Γ
˜
13
∂
g
˜
1
∂
z
---------
0
˜
=
=
Γ
˜
21
Γ
˜
12
=
Γ
˜
22
∂
g
˜
2
∂θ
---------
r
θ
cos
–
e
˜
1
r
θ
sin
e
˜
2
–
r
g
˜
1
–
=
=
=
Γ
˜
23
∂
g
˜
2
∂
z
---------
0
˜
=
=
Γ
˜
31
Γ
˜
13
=
Γ
˜
32
Γ
˜
23
=
Γ
˜
33
∂
g
˜
3
∂
z
---------
0
˜
=
=
Γ
ij
k
g
˜
k
Γ
˜
ij
Γ
12
2
g
˜
2
Γ
˜
12
Γ
22
1
g
˜
1
Γ
˜
22
Γ
12
2
Γ
21
2
1
r
---
=
=
Γ
22
1
r
–
=
Γ
ij
k
0
=
Γ
r
θ
θ
Γ
θ
r
θ
1
r
---
=
=
Γ
θθ
r
r
–
=
Γ
ij
k
0
=
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DRAFT June 17, 2004
79
Partial Answer:
To find the vector
, first differentiate
from Eq. (5.31a) with respect to
(i.e., with respect to
θ
). You then recognize from (5.31b) that the result is
. Then t
is
found by dotting
by
.
Covariant differentiation of contravariant components
Eq. (5.62) gives
,
(5.77)
or, changing the dummy summation indices so that the basis dyads will have the same sub-
scripts in both terms,
.
(5.78)
We now introduce a compact notation called
covariant vector differentiation
:
.
(5.79)
The notation for covariant differentiation varies widely: the slash in
is also often denoted
with a comma
although many writers use a comma to denote ordinary partial differentia-
tion. Keep in mind: the Christoffel terms in Eq. (5.79) account for the variation of the base vec-
tors in space. Using covariant differentiation, the gradient of a vector is then written
compactly as
.
(5.80)
Study Question 5.3
Using the spherical covariant base vectors in Eq. (5.31), prove that
(5.75a)
(5.75b)
.
(5.75c)
Therefore show that the Christoffel symbols for spherical
coordinates are
, ,
(5.76a)
, (5.76b)
,
(all
other
).
(5.76c)
Γ
˜
11
0
˜
=
Γ
˜
12
1
r
---
g
˜
2
=
Γ
˜
13
1
r
---
g
˜
3
=
Γ
˜
21
Γ
˜
12
=
Γ
˜
22
r
g
˜
1
–
=
Γ
˜
23
φ
cot
g
˜
3
=
Γ
˜
31
Γ
˜
13
=
Γ
˜
32
Γ
˜
23
=
Γ
˜
33
r
sin
2
θ
g
˜
1
–
θ
sin
θ
cos
g
˜
2
–
=
r
θ φ
, ,
{
}
Γ
r
θ
θ
Γ
θ
r
θ
1
r
---
=
=
Γ
r
φ
φ
Γ
φ
r
φ
1
r
---
=
=
Γ
θθ
r
r
–
=
Γ
θφ
φ
Γ
φθ
φ
φ
cot
=
=
Γ
φφ
r
r
sin
2
θ
–
=
Γ
φφ
θ
θ
sin
θ
cos
–
=
Γ
ij
k
0
=
Γ
˜
12
g
˜
1
η
2
Γ
˜
12
e
˜
θ
=
Γ
12
2
Γ
˜
12
g
˜
2
Γ
12
2
Γ
r
θ
θ
Γ
˜
12
g
˜
2
•
e
˜
θ
e
˜
θ
r
⁄
(
)
•
1
r
⁄
=
=
=
=
d
v
˜
d
x
˜
------
∂
v
i
∂η
j
--------
g
˜
i
g
˜
j
v
i
Γ
ij
k
g
˜
k
g
˜
j
+
=
d
v
˜
d
x
˜
------
∂
v
i
∂η
j
--------
g
˜
i
g
˜
j
v
k
Γ
kj
i
g
˜
i
g
˜
j
+
=
v
i
/
j
∂
v
i
∂η
j
--------
v
k
Γ
kj
i
+
≡
v
i
/
j
v
i
,
j
d
v
˜
d
x
˜
------
v
/
j
i
g
˜
i
g
˜
j
=
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DRAFT June 17, 2004
80
Example: Gradient of a vector in cylindrical coordinates
=
(5.81rr)
=
(5.81rt)
=
(5.81rz)
=
(5.81tr)
=
(5.81tt)
=
(5.81tz)
=
(5.81zr)
=
(5.81zt)
=
(5.81zz)
Hence, the gradient of a vector is given by
=
+
+
+ +
+
+ +
+
.
(5.82)
Substituting Eqs. (5.19), (5.26), and (5.29) into the above formula gives
=
+
+
+ +
+
+ +
+
.
(5.83)
v
1
/
1
∂
v
1
∂η
1
---------
v
1
Γ
11
1
v
2
Γ
21
1
v
3
Γ
31
1
+
+
+
=
∂
v
1
∂
r
--------
v
1
/
2
∂
v
1
∂η
2
---------
v
1
Γ
12
1
v
2
Γ
22
1
v
3
Γ
32
1
+
+
+
=
∂
v
1
∂θ
--------
v
2
r
–
v
1
/
3
∂
v
1
∂η
3
---------
v
1
Γ
13
1
v
2
Γ
23
1
v
3
Γ
33
1
+
+
+
=
∂
v
1
∂
z
--------
v
2
/
1
∂
v
2
∂η
1
---------
v
1
Γ
11
2
v
2
Γ
21
2
v
3
Γ
31
2
+
+
+
=
∂
v
2
∂
r
--------
v
2
/
2
∂
v
2
∂η
2
---------
v
1
Γ
12
2
v
2
Γ
22
2
v
3
Γ
32
2
+
+
+
=
∂
v
2
∂θ
--------
v
1
r
-----
+
v
2
/
3
∂
v
2
∂η
3
---------
v
1
Γ
13
2
v
2
Γ
23
2
v
3
Γ
33
2
+
+
+
=
∂
v
2
∂
z
--------
v
3
/
1
∂
v
3
∂η
1
---------
v
1
Γ
11
3
v
2
Γ
21
3
v
3
Γ
31
3
+
+
+
=
∂
v
3
∂
r
--------
v
3
/
2
∂
v
3
∂η
2
---------
v
1
Γ
12
3
v
2
Γ
22
3
v
3
Γ
32
3
+
+
+
=
∂
v
3
∂θ
--------
v
3
/
3
∂
v
3
∂η
3
---------
v
1
Γ
13
3
v
2
Γ
23
3
v
3
Γ
33
3
+
+
+
=
∂
v
3
∂
z
--------
d
v
˜
d
x
˜
------
∂
v
1
∂
r
--------
g
˜
1
g
˜
1
∂
v
1
∂θ
--------
v
2
r
–
g
˜
1
g
˜
2
∂
v
1
∂
z
--------
g
˜
1
g
˜
3
∂
v
2
∂
r
--------
g
˜
2
g
˜
1
∂
v
2
∂θ
--------
v
1
r
-----
+
g
˜
2
g
˜
2
∂
v
2
∂
z
--------
g
˜
2
g
˜
3
∂
v
3
∂
r
--------
g
˜
3
g
˜
1
∂
v
3
∂θ
--------
g
˜
3
g
˜
2
∂
v
3
∂
z
--------
g
˜
3
g
˜
3
d
v
˜
d
x
˜
------
∂
v
r
∂
r
--------
e
˜
r
e
˜
r
∂
v
r
∂θ
--------
v
θ
–
e
˜
r
e
˜
θ
r
-----
∂
v
r
∂
z
--------
e
˜
r
e
˜
z
1
r
---
∂
v
θ
∂
r
--------
r
e
˜
θ
(
)
e
˜
r
1
r
---
∂
v
θ
∂θ
--------
v
r
r
----
+
r
e
˜
θ
(
)
e
˜
θ
r
-----
1
r
---
∂
v
θ
∂
z
--------
r
e
˜
θ
(
)
e
˜
z
∂
v
z
∂
r
--------
e
˜
z
e
˜
r
∂
v
z
∂θ
--------
e
˜
z
e
˜
θ
r
-----
∂
v
z
∂
z
--------
e
˜
z
e
˜
z
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81
Upon simplification, the matrix of
with respect to the usual ortho
normalized
basis
is
w.r.t
unit basis.
(5.84)
This is the formula usually cited in handbooks.
d
v
˜
d
x
˜
⁄
e
˜
r
e
˜
θ
e
˜
z
,
,
{
}
d
v
˜
d
x
˜
------
∂
v
r
∂
r
--------
1
r
---
∂
v
r
∂θ
--------
v
θ
–
∂
v
r
∂
z
--------
∂
v
θ
∂
r
--------
1
r
---
∂
v
θ
∂θ
--------
v
r
r
----
+
∂
v
θ
∂
z
--------
∂
v
z
∂
r
--------
1
r
---
∂
v
z
∂θ
--------
∂
v
z
∂
z
--------
=
e
˜
r
e
˜
θ
e
˜
z
,
,
{
}
Study Question 5.4
Again we will duplicate the methods of the preceding example for
spherical coordinates. However, rather than duplicating the entire hideous analyses, we
here compute only the
component of
. In particular, we seek
.
(5.85)
(a) Noting from Eqs. (5.31) and (5.32) how and are related to
, show that
.
(5.86)
(b) Explain why Eq. (5.80) therefore implies that
.
(5.87)
(c) With respect to the orthonormal basis
recall from Eq. (5.33) that
and
. Use the Christoffel symbols of Eq. (5.76) in the formula (5.79) to show that
.
(5.88)
(d) The final step is to substitute this result into Eq. (5.87) to deduce that
.
(5.89)
Cite a textbook (or other reference) that tabulates formulas for gradients in spherical coordi-
nates. Does your Eq. (5.89) agree with the formula for the
component of
pro-
vided in the textbook?
r
φ
d
v
˜
d
x
˜
⁄
e
˜
r
d
v
˜
d
x
˜
------
e
˜
φ
•
•
e
˜
r
e
˜
φ
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
e
˜
r
d
v
˜
d
x
˜
------
e
˜
φ
•
•
1
r
θ
sin
--------------
g
˜
1
d
v
˜
d
x
˜
------
g
˜
3
•
•
=
e
˜
r
d
v
˜
d
x
˜
------
e
˜
φ
•
•
v
3
;
1
r
θ
sin
--------------
=
e
˜
r
e
˜
θ
e
˜
φ
,
,
{
}
v
1
=
v
r
v
3
=
v
φ
r
θ
sin
⁄
v
1
/
3
∂
v
r
∂φ
--------
v
φ
sin
θ
–
(
)
+
=
e
˜
r
d
v
˜
d
x
˜
------
e
˜
φ
•
•
1
r
θ
sin
--------------
∂
v
r
∂φ
--------
v
φ
r
-----
–
=
r
φ
d
v
˜
d
x
˜
⁄
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DRAFT June 17, 2004
82
Covariant differentiation of covariant components
Recalling, Eq. (5.63), we now consider
a similarly-defined
basis-dependent
vector:
,
(5.90)
and analogously to Eq. (5.65) we define
.
(5.91)
Given a vector for which the covariant components are known, an analysis similar to
that of the previous sections eventually reveals that the gradient of the vector can be written:
,
(5.92)
where
.
(5.93)
The question naturally arises: what connection, if any, does
have with
? To answer this
question, differentiate both sides of Eq. (2.12) with respect to
:
,
(5.94)
or
and therefore
.
(5.95)
In other words,
is just the negative of
. Consequently, equation (5.92) becomes
where
.
(5.96)
We now have an equation for increments of the
contravariant
base vectors
(5.97)
which should be compared with Eq. (5.69)
Product rules for covariant differentiation
Most of
the usual rules of differential calculus
apply
. For example,
and
, etc.
(5.98)
P
˜
i
k
∂
g
˜
k
∂η
i
---------
≡
P
ij
k
P
˜
i
k
g
˜
j
•
≡
v
˜
v
k
d
v
˜
d
x
˜
------
v
i
/
j
g
˜
i
g
˜
j
=
v
i
/
j
∂
v
i
∂η
j
--------
v
k
P
ij
k
+
≡
P
ij
k
Γ
ij
k
η
k
∂
g
˜
i
∂η
k
---------
g
˜
j
•
g
˜
i
∂
g
˜
j
∂η
k
---------
•
+
0
=
P
˜
k
i
g
˜
j
•
g
˜
i
Γ
˜
jk
•
+
0
=
P
kj
i
Γ
kj
i
+
0
=
P
ij
k
Γ
ij
k
d
v
˜
d
x
˜
------
v
i
/
j
g
˜
i
g
˜
j
=
v
i
/
j
∂
v
i
∂η
j
--------
v
k
Γ
ij
k
–
≡
d
g
˜
k
Γ
ij
k
g
˜
j
–
(
)
d
η
i
=
v
i
w
j
(
)
/
k
v
i
w
j
/
k
v
i
/
k
w
j
+
=
v
i
w
j
(
)
/
k
v
i
w
j
/
k
v
i
/
k
w
j
+
=
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DRAFT June 17, 2004
83
5.7 Backward and forward gradient operators
The gradient
that we defined in the preceding section is really a
backward
operating
gradient. By this we mean that the component follows an index ordering that is analogous
to the index ordering on the dyad
. This dyad has components
, which is comparable
to the index ordering in the
Cartesian ij
component formula
.
In Cartesian coordinates, the gradient
is a second order tensor whose
Cartesian ij
components are given by
. But couldn’t we just have well have defined a second order
tensor whose
Cartesian ij
components would be
. The only difference between these
two choices is that the index placement is swapped. Thus, the transpose of one choice gives
the other choice.
Following definitions and notation used by Malvern [11], we will denote the backward
operating gradient by
and the forward operating gradient by
. The “ ” operates on
arguments to its right, while the operates on arguments to its left. As mentioned earlier,
our previously defined vector gradient is actually a backward del definition:
means the same thing as
.
Thus,
(5.99)
means the same thing as
Thus,
(5.100)
The component ordering for the forward gradient operator is defined in a manner that is
analogous to the dyad
, whereas the component ordering for the backward
gradient is analogous to that on the dyad
. This leads naturally to the ques-
tion of whether or not it is possible to define right and left gradient operators in a manner that
permits some “heuristic assistance.”
Recall that
(5.101)
Suppose that we wish to identify a left-acting operator such that
(5.102)
Let’s suppose we desire to define this operator such that it follows a product rule so that
(5.103)
This suggests that we should define
and (5.104)
d
v
˜
d
x
˜
⁄
ij
v
˜
d
˜
v
i
d
j
∂
v
i
∂
x
j
⁄
d
v
˜
d
x
˜
⁄
∂
v
i
∂
x
j
⁄
∂
v
j
∂
x
i
⁄
v
˜
∇
∇
v
˜
∇
∇
v
˜
∇
d
v
˜
d
x
˜
------
v
˜
∇
v
/
j
i
g
˜
i
g
˜
j
=
∇
v
˜
d
v
˜
d
x
˜
------
T
∇
v
˜
v
/
i
j
g
˜
i
g
˜
j
=
d
˜
v
˜
d
i
v
j
(
)
g
˜
i
g
˜
j
=
v
˜
d
˜
v
j
d
i
(
)
g
˜
i
g
˜
j
=
d
v
˜
d
x
˜
------
∂
v
i
∂η
j
--------
g
˜
i
g
˜
j
v
k
Γ
kj
i
g
˜
i
g
˜
j
+
=
∇
d
v
˜
d
x
˜
------
v
i
g
˜
i
(
)∇
=
v
i
g
˜
i
(
)∇
g
˜
i
v
i
( )∇
[
]
v
i
g
˜
i
( )∇
[
]
+
=
v
i
( )∇
∂
v
i
∂η
j
--------
g
˜
j
=
g
˜
k
( )∇
Γ
kj
i
g
˜
i
g
˜
j
=
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DRAFT June 17, 2004
84
Similarly, for the forward operation gradient, we define
(5.105)
from which it follows
and
(5.106)
These definitions are cute, but we caution against using them too cavalierly. A careful applica-
tion of the original definition of the gradient is probably safer.
5.8 Divergence of a vector
The divergence of a vector is denoted
and it is defined by
(5.107)
The simplicity of the component expression for the trace depends on the expression chosen
for the gradient operation. For example, taking the trace of Eq. (5.96) gives a valid but some-
what ugly expression:
(5.108)
A much simpler expression for the divergence can be obtained by taking the trace of Eq. (5.80)
to give
(5.109)
Written out explicitly, this result is
(5.110)
It will later be shown [in Eq. (5.126)] that
(5.111)
Therefore, Eq. (5.110) gives the very useful formula [7]:
(5.112)
∇
v
i
g
˜
i
(
)
∇
v
i
( )
g
˜
i
v
i
∇
g
˜
i
+
=
∇
v
i
( )
∂
v
i
∂η
j
--------
g
˜
j
=
∇
g
˜
k
( )
Γ
kj
i
g
˜
j
g
˜
i
=
v
˜
∇
v
˜
•
∇
v
˜
•
tr
d
v
˜
d
x
˜
------
=
∇
v
˜
•
v
i
/
j
g
˜
i
g
˜
j
•
(
)
v
i
/
j
g
ij
( )
∂
v
i
∂η
j
--------
v
k
Γ
ij
k
–
g
ij
=
=
=
∇
v
˜
•
v
/
j
i
g
˜
i
g
˜
j
•
v
/
j
i
δ
i
j
v
/
i
i
=
=
=
∇
v
˜
•
∂
v
i
∂η
i
--------
v
k
Γ
ki
i
+
=
Γ
ki
i
1
J
---
∂
J
∂η
k
---------
=
∇
v
˜
•
1
J
---
∂
Jv
k
(
)
∂η
k
----------------
=
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DRAFT June 17, 2004
85
5.9 Curl of a vector
The curl of a vector is denoted
and it is defined to equal the axial vector
1
associ-
ated with
. Thus, recalling Eq. (5.100),
(5.113)
In the last step, we have used the skew-symmetry of the permutation tensor. Substituting Eq.
(5.77) gives
(5.114)
This result can be written as
(5.115)
We can alternatively start with the covariant expression for the gradient given in Eq. (5.96).
Namely,
where
.
(5.116)
which gives
(5.117)
Recalling from Eq. (3.47) that
, the expanded component form of this result is
(5.118)
where (recall)
(5.119)
1. The axial vector associated with any tensor is defined by
, where is the permutation tensor.
v
˜
∇
v
˜
×
T
˜˜
1
2
---
ξ
˜˜˜
:T
˜˜
–
ξ
˜˜˜
∇
v
˜
∇
v
˜
×
1
2
---
ξ
˜˜˜
:
d
v
˜
d
x
˜
------
T
–
1
2
---
ξ
˜˜˜
:
d
v
˜
d
x
˜
------
=
=
∇
v
˜
×
1
2
---
ξ
˜˜˜
:
∂
v
i
∂η
j
--------
v
k
Γ
kj
i
+
g
˜
i
g
˜
j
=
∇
v
˜
×
1
2
---
∂
v
i
∂η
j
--------
v
k
Γ
kj
i
+
g
˜
i
g
˜
j
×
=
d
v
˜
d
x
˜
------
v
i
/
j
g
˜
i
g
˜
j
=
v
i
/
j
∂
v
i
∂η
j
--------
v
k
Γ
ij
k
–
≡
∇
v
˜
×
1
2
---
ξ
˜˜˜
:
v
i
/
j
g
˜
i
g
˜
j
[
]
1
2
---
ξ
kij
v
i
/
j
g
˜
k
=
=
ξ
kij
1
J
---
ε
ijk
=
∇
v
˜
×
1
2
J
------
v
2/3
v
3/2
–
(
)
g
˜
1
v
3/1
v
1/3
–
(
)
g
˜
2
v
1/2
v
2/1
–
(
)
g
˜
1
+
+
[
]
=
J
g
˜
1
g
˜
2
×
(
)
g
˜
3
•
=
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DRAFT June 17, 2004
86
5.10 Gradient of a tensor
Using methods similar to the preceding sections, one can apply the direct notation defini-
tion of the gradient
to eventually prove that
, where
,
(5.120)
or, if the tensor is known in its covariant components,
, where
.
(5.121)
For mixed tensor components, we have
, where
.
(5.122)
Note that
there is a Christoffel term for each subscript on the tensor. The term is negative if the
summed index is a subscript on
T
and positive if the summed index is a superscript on
T
.
Ricci’s lemma
Recall from Eq. (3.18) that the metric coefficients are components of the iden-
tity tensor. Knowing that the gradient of the identity tensor must be zero, it follows that
and
.
(5.123)
Corollary
Recall that the Jacobian equals the volume of the parallelepiped formed by the
three base vectors
. Since the base vectors vary with position, it follows that var-
ies with position. In other words, the Jacobian may be regarded as a function of the coordi-
nates. Taking the derivative of the Jacobian with respect to the coordinate
and applying
the chain rule gives
, where we have used Eq. (2.30).
(5.124)
Recalling from Eq. (5.123) that the metric components have vanishing
covariant derivatives
, Eq.
(5.125)
=
d
T
˜˜
d
x
˜
⁄
d
T
˜˜
d
x
˜
-------
T
ij
/
k
g
˜
i
g
˜
j
g
˜
k
=
T
ij
/
k
∂
T
ij
∂η
k
----------
T
mj
Γ
mk
i
T
im
Γ
mk
j
+
+
≡
d
T
˜˜
d
x
˜
-------
T
ij
/
k
g
˜
i
g
˜
j
g
˜
k
=
T
ij
/
k
∂
T
ij
∂η
k
---------
T
mj
Γ
ik
m
–
T
im
Γ
jk
m
–
≡
d
T
˜˜
d
x
˜
-------
T
•
j
/
k
i
g
˜
i
g
˜
j
g
˜
k
=
T
•
j
/
k
i
∂
T
•
j
i
∂η
k
----------
T
•
j
m
Γ
mk
i
T
•
m
i
Γ
jk
m
–
+
≡
g
ij
/
k
0
=
g
ij
/
k
0
=
J
g
˜
1
g
˜
2
g
˜
3
,
,
[
]
J
J
η
k
∂
J
∂η
k
---------
∂
J
∂
g
ij
---------
∂
g
ij
∂η
k
---------
1
2
---
Jg
ij
∂
g
ij
∂η
k
---------
=
=
∂
g
ij
∂η
k
---------
g
mj
Γ
ik
m
g
mi
Γ
jk
m
+
=
∂
J
∂η
k
---------
1
2
---
Jg
ij
g
mj
Γ
ik
m
g
mi
Γ
jk
m
+
[
]
=
1
2
---
J
δ
m
i
Γ
ik
m
δ
m
j
Γ
jk
m
+
[
]
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DRAFT June 17, 2004
87
=
=
(5.126)
or
(5.127)
5.11 Christoffel symbols of the first kind
Section 5.6 showed how to compute the Christoffel symbols of the second kind by directly
applying the formula:
.
(5.128)
This method was tractable because the space was Euclidean and we could therefore utilize the
existence of an underlying rectangular Cartesian basis to determine the change in the base
vectors with respect to the coordinates. However, for non-Euclidean spaces, you have only
the metric coefficients
, and the simplest way to compute the Christoffel symbols of the sec-
ond kind then begins by first computing the
Christoffel symbols of the first kind
,
which
are related to the Christoffel symbols of the second kind by
.
(5.129)
These
are frequently denoted in the literature as
, presumably to emphasize that
they are not components of any third-order-tensor, in the sense discussed on page __. Substi-
tuting Eq. (5.128) into Eq. (5.129) gives
,
(5.130)
where the final term results from lowering the index on
.
(5.131)
Recalling that
, this result also reveals that
is symmetric in its first two indices:
.
(5.132)
Now note that
.
(5.133)
1
2
---
J
Γ
mk
m
Γ
mk
m
+
[
]
J
Γ
mk
m
Γ
mk
m
1
J
---
∂
J
∂η
k
---------
=
Γ
ij
k
g
˜
k
∂
g
˜
i
∂η
j
--------
•
=
g
ij
Γ
ijk
Γ
ijk
Γ
ij
m
g
mk
=
Γ
ijk
ij k
,
[
]
Γ
ijk
g
˜
m
∂
g
˜
i
∂η
j
--------
•
g
mk
∂
g
˜
i
∂η
j
--------
g
˜
m
•
g
mk
∂
g
˜
i
∂η
j
--------
g
˜
k
•
=
=
=
g
˜
m
Γ
ijk
Γ
˜
ij
g
˜
k
•
=
Γ
˜
ij
Γ
˜
ji
=
Γ
ijk
Γ
ijk
Γ
jik
=
∂
g
ik
(
)
∂η
j
---------------
∂
g
˜
i
g
˜
k
•
(
)
∂η
j
-------------------------
∂
g
˜
i
∂η
j
--------
g
˜
k
•
g
˜
i
∂
g
˜
k
∂η
j
---------
•
+
Γ
˜
ij
g
˜
k
•
g
˜
i
Γ
˜
kj
•
+
=
=
=
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DRAFT June 17, 2004
88
1
.
(5.134)
This relationship can be easily remembered by noting the structure of the indices.
Note, for
example, that the middle subscript on both ‘s is the same as that on the coordinate .
Direct substitution of Eq. (5.134) into (5.135) validates the following expression that can be
used to directly
obtain the Christoffel symbols of the first kind given only the metric coeffi-
cients
:
.
(5.135)
This formula (which can be verified by substituting Eq. (5.134) into (5.135) represents the
easiest way to obtain the Christoffel symbols of the first kind when only the metric coeffi-
cients are known. This formula has easily remembered index structure: for
, the index
symbols on each of the coordinates are ordered
.
Once the
are known, the Christ-
offel symbols of the second kind are obtained by solving Eq. (5.129) to give
.
(5.136)
5.12 The fourth-order Riemann-Christoffel curvature tensor
Recall that this document has limited its scope to curvilinear systems embedded in a
Euclidean space
of the same dimension. When the Euclidean space is three-dimensional, then
this document has focused on alternative coordinate systems that still define points in this
space and therefore require three coordinates. An example of a two-dimensional curvilinear
space embedded in Euclidean space is the surface of a sphere. Only two coordinates are
required to specify a point on a sphere and this two dimensional space is “
Riemannian”
(i.e.,
non-Euclidean) because it is not possible for us to construct a rectangular coordinate grid on a
sphere. In ordinary engineering mechanics applications, the mathematics of reduced dimen-
sion spaces within ordinary physical space is needed to study plate and beam theory.
We have focused on the three-dimensional Euclidean space that we
think
we live it. In this
modern (post-Einstein) era, this notion of a Euclidean physical world is now recognized to be
only an approximation (referred to as Newtonian space). Einstein and colleagues threw a
wrench in our thinking by introducing the notion that space and time are
curved
.
We have now mentioned two situations (prosaic shell/beam theory and exciting relativity
theory) where understanding some basics of non-Euclidean spaces is needed. We have men-
1. An alternative way to obtain this result is to simply apply Eq. (5.129) to Eq. (5.125).
∂
g
ik
∂η
j
----------
Γ
ijk
Γ
kji
+
=
Γ
η
j
Γ
ijk
1
2
---
∂
g
jk
∂η
i
----------
∂
g
ki
∂η
j
----------
∂
g
ij
∂η
k
---------
–
+
=
Γ
ijk
η
i j k
, ,
Γ
ijk
Γ
ij
n
Γ
ijk
g
kn
=
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DRAFT June 17, 2004
89
tioned that a Riemannian space is one that does not permit construction of a rectangular coor-
dinate grid. How is this statement cast in mathematical terms? In other words, what process is
needed to decide if a space is Euclidean or Riemannian in the first place. The answer is tied to
a special fourth-order tensor called the Riemann-Christoffel tensor soon to be defined. If this
tensor turns out to be zero, then your space is Euclidean. Otherwise, it is Riemannian.
The Riemann-Christoffel tensor is defined
(5.137)
or
(5.138)
This tensor is skew-symmetric in the first two indices and in the last two indices. It is major
symmetric:
(5.139)
In a two-dimensional space, only the
component is independent, the others being either
zero or related to this component by
.
depends on the Christoffel symbols. For a Cartesian sys-
tem, all Christoffel symbols are zero. Hence, in a Cartesian system,
. Thus, if a space
is
capable
of supporting a Cartesian system (i.e., if the space is Euclidean), then the Riemann-
Christoffel tensor must be zero. This would be true even if you aren’t actually using a Carte-
sian system. For example, ordinary 3D Newtonian space is Euclidean and therefore its Rie-
mann-Christoffel tensor must be zero even if you happen to employ a different set of three
curvilinear coordinates such as spherical coordinates
. This would follow because the
transformation relations are linear. Hence, if there exists a Cartesian system (in which the Rie-
mann-Christoffel tensor is zero), a linear transformation to a different, possibly curvilinear,
system would result again in the zero tensor. For the Riemann-Christoffel tensor to
not
be
zero, you would have to be working in a reduced dimension space [such as the
surface
of a
sphere where the coordinates are
]. The Riemann-Christoffel tensor is, therefore, a mea-
sure of curvature of a Riemannian space. Because the Riemann-Christoffel tensor transforms
like a tensor, it is
not
a basis-intrinsic quantity despite the fact that it has been here defined in
terms of basis-intrinsic quantities. A similar situation was encountered with the metric coeffi-
cients
, which were defined
. The metrics
are components of the identity
tensor. Therefore the product of these components times base vectors
is basis-inde-
pendent (it equals the identity tensor). Similarly, the product of the Riemann-Christoffel com-
ponents times basis vectors is basis-independent:
(5.140)
R
ijkl
∂Γ
jli
∂η
k
-----------
∂Γ
jki
∂η
l
------------
–
Γ
ilp
Γ
jk
p
Γ
ikp
Γ
jl
p
–
+
=
R
ijkl
1
2
---
∂
2
g
il
∂η
j
∂η
k
------------------
∂
2
g
jl
∂η
i
∂η
k
------------------
–
∂
2
g
ik
∂η
j
∂η
l
-----------------
–
∂
2
g
jk
∂η
i
∂η
l
-----------------
+
g
mn
Γ
jkn
Γ
ilm
Γ
jlm
Γ
ikn
–
(
)
+
=
R
ijkl
R
jikl
–
R
ijlk
–
R
klij
=
=
=
R
1212
R
1212
R
2112
–
R
1221
–
R
2121
=
=
=
R
ijkl
R
ijkl
0
=
r
θ φ
, ,
(
)
θ φ
,
(
)
g
ij
g
ij
g
˜
i
g
˜
j
•
=
g
ij
g
ij
g
˜
i
g
˜
j
R
˜˜˜˜
R
ijkl
g
˜
i
g
˜
j
g
˜
k
g
˜
l
=
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DRAFT June 17, 2004
90
or
(5.141)
or
(5.142)
or
(5.143)
where
(5.144)
Noting that the indices on this tensor may be permuted in any manner, note that the first and
third terms in Eq. (5.143) may be canceled, giving
(5.145)
or, rearranging slightly,
(5.146)
or
(5.147)
where
(5.148)
6. Embedded bases and objective rates
We introduced the mapping tensor in Eq. (2.1) to serve as a mere “helper” tensor.
Namely, if the basis
exists in a Euclidean space, then we defined
(6.1)
Here, the basis,
is the same as the “laboratory” basis that we had originally used
R
˜˜˜˜
∂Γ
jli
∂η
k
-----------
∂Γ
jki
∂η
l
------------
–
Γ
ilp
Γ
jk
p
Γ
ikp
Γ
jl
p
–
+
g
˜
i
g
˜
j
g
˜
k
g
˜
l
=
R
˜˜˜˜
∂ Γ
˜
jl
g
˜
i
•
(
)
∂η
k
-------------------------
∂ Γ
˜
jk
g
˜
i
•
(
)
∂η
l
-------------------------
–
Γ
˜
il
g
˜
p
•
(
) Γ
˜
jk
g
˜
p
•
(
)
Γ
˜
ik
g
˜
p
•
(
) Γ
˜
jl
g
˜
p
•
(
)
–
+
g
˜
i
g
˜
j
g
˜
k
g
˜
=
R
˜˜˜˜
Γ
˜
jlk
g
˜
i
•
Γ
˜
jl
Γ
˜
ik
•
Γ
˜
jkl
g
˜
i
•
–
Γ
˜
jk
Γ
˜
il
•
–
Γ
˜
il
g
˜
p
•
(
) Γ
˜
jk
g
˜
p
•
(
)
Γ
˜
ik
g
˜
p
•
(
) Γ
˜
jl
g
˜
p
•
(
)
–
+
+
[
]
g
˜
i
g
˜
j
g
˜
k
g
˜
l
=
Γ
˜
ijk
∂Γ
˜
ij
∂η
k
----------
∂
2
g
˜
i
∂η
j
∂η
k
------------------
∂
3
x
˜
∂η
i
∂η
j
∂η
k
---------------------------
=
=
=
R
˜˜˜˜
Γ
˜
jl
Γ
˜
ik
•
Γ
˜
jk
Γ
˜
il
•
–
Γ
˜
il
Γ
˜
jk
•
Γ
˜
ik
Γ
˜
jl
•
–
+
[
]
g
˜
i
g
˜
j
g
˜
k
g
˜
l
=
R
˜˜˜˜
Γ
˜
ik
Γ
˜
jl
•
Γ
˜
il
Γ
˜
jk
•
–
Γ
˜
jk
Γ
˜
il
•
Γ
˜
jl
Γ
˜
ik
•
–
+
[
]
g
˜
i
g
˜
j
g
˜
k
g
˜
l
–
=
R
˜˜˜˜
g
ikjl
g
iljk
–
g
jkil
g
jlik
–
+
[
]
g
˜
i
g
˜
j
g
˜
k
g
˜
l
–
=
g
ijkl
Γ
˜
ij
Γ
˜
kl
•
≡
∂
2
x
˜
∂η
i
∂η
j
-----------------
∂
2
x
˜
∂η
k
∂η
l
------------------
•
=
F
˜˜
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
g
˜
i
F
˜˜
E
˜
i
•
=
E
˜
1
E
˜
2
E
˜
3
,
,
{
}
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http://me.unm.edu/~rmbrann/gobag.html
DRAFT June 17, 2004
91
in Section 2. In continuum mechanics (especially the older literature), a special “convected”
coordinate system is often adopted in which the coordinates
of a point currently located a
a position at time
t
are identified to equal the
initial
Cartesian coordinates
of the same
material particle at time . For this special case, the coordinate grid is identical to the defor-
mation of the initial coordinate grid when it is convected along with the material particles. For
this special case, the mapping tensor in Eq. (6.1) is identical to the deformation gradient ten-
sor from traditional continuum mechanics. In this physical context, the term “covariant” is
especially apropos because the covariant base vector varies coincidently with the underly-
ing material particles. That is, the unit cube is defined by three unit vectors
will,
upon deformation, deform to a parallelepiped whose sides move with the material and the
sides of this parallelepiped are given by the three convected base vectors
.
By contrast, consider the contravariant base vectors, which (recall) are related to the map-
ping tensor by
(6.2)
These base vectors do
not
move coincidently with the material particles. Instead, the contra-
variant base vectors move somewhat contrary to the material motion. In particular, if
are the outward unit normals to the unit cube, then material deformation will
generally move the cube to a parallelepiped whose faces now have outward unit normals par-
allel to
. In general, the motion of the normal to a plane moves somewhat con-
trary to the motion of material fibers that were originally parallel to the plane’s normal.
Of course, it is not really necessary for the initial coordinate system to be Cartesian itself.
When the initial coordinate system is permitted to be curvilinear, then we will denote its asso-
ciated set of base vectors by
. As before, these
covariant
base vectors deform to
new orientations given by
(6.3)
The associated dual basis is given by
(6.4)
Here is here retaining its meaning as the physical deformation gradient tensor, which
necessitates introducing new symbols to denote the mapping tensors for the individual curvi-
linear bases. Namely, we will presume the existence of tensors and such that
and
(6.5)
and
(6.6)
η
k
x
˜
X
k
0
F
˜˜
g
˜
i
E
˜
1
E
˜
2
E
˜
3
,
,
{
}
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
g
˜
k
F
˜˜
1
–
E
˜
k
•
=
E
˜
1
E
˜
2
E
˜
3
,
,
{
}
g
˜
1
g
˜
2
g
˜
3
,
,
{
}
G
˜
1
G
˜
2
G
˜
3
,
,
{
}
g
˜
i
F
˜˜
G
˜
i
•
=
g
˜
k
F
˜˜
T
–
G
˜
k
•
=
F
˜˜
h
˜˜
H
˜˜
g
˜
i
h
˜˜
E
˜
i
•
=
g
˜
k
h
˜˜
T
–
E
˜
k
•
=
G
˜
i
H
˜˜
E
˜
i
•
=
G
˜
k
H
˜˜
T
–
E
˜
k
•
=
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DRAFT June 17, 2004
92
from which it follows that
and
, where
(6.7)
and
, where
(6.8)
Furthermore, substituting Eqs. (6.5) and (6.6) into (6.3) implies that
(6.9)
In continuum mechanics, the Seth-Hill generalized strain [13] measure is defined in direct
notation as
(6.10)
Here, the Seth-Hill parameter is selected according to whichever strain measure the analyst
prefers. Namely,
-->
True/Swainger strain
-->
engineering/Cauchy strain
-->
Green/Lagrange strain
-->
Almansi
-->
logarithmic/Hencky
strain
Of course, the case of
must be applied in the limit.
The Lagrangian strain measure corresponds to choosing
to give
(6.11)
The “Euler” strain measure corresponds to choosing
to give
(6.12)
Strain measures in older literature look drastically different from this because they are
typically expressed in terms of initial and deformed metric tensors. What is the connection?
First note from Eq. (6.3) that
(6.13)
Furthermore, by definition,
(6.14)
Thus, dotting Eq. (6.11) from the left by
and from the right by
gives
(6.15)
This result shows that the difference between deformed and initial covariant metrics (which
g
ij
E
˜
i
y
˜˜
E
˜
j
•
•
=
g
ij
E
˜
i
y
˜˜
1
–
E
˜
j
•
•
=
y
˜˜
h
˜˜
T
h
˜˜
•
≡
G
ij
E
˜
i
Y
˜˜
E
˜
j
•
•
=
G
ij
E
˜
i
Y
˜˜
1
–
E
˜
j
•
•
=
Y
˜˜
H
˜˜
T
H
˜˜
•
≡
F
˜˜
h
˜˜
H
˜˜
1
–
•
=
e
˜˜
1
k
---
F
˜˜
T
F
˜˜
•
(
)
k
2
/
I
˜˜
–
[
]
=
k
k
1
–
=
k
=1
k
=2
k
= 2
–
k
=0
k
=0
k
=2
e
˜˜
Lagr
1
2
---
F
˜˜
T
F
˜˜
•
(
)
I
˜˜
–
[
]
=
k
= 2
–
e
˜˜
Euler
1
2
---
I
˜˜
F
˜˜
T
F
˜˜
•
(
)
1
–
–
[
]
=
g
ij
G
˜
i
F
˜˜
T
F
˜˜
•
G
˜
j
•
•
=
G
ij
G
˜
i
G
˜
j
•
≡
G
˜
i
I
˜˜
G
˜
j
•
•
=
G
˜
i
G
˜
j
G
˜
i
e
˜˜
Lagr
G
˜
j
•
•
1
2
---
g
ij
G
ij
–
[
]
=
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DRAFT June 17, 2004
93
appears frequently in older literature) is identically equal to the covariant components of the
Lagrangian strain tensor with respect to the initial basis.
Similarly, you can show that
(6.16)
In general, if you wish to convert a component equation from the older literature into modern
invariant direct notation form, you can use Eqs. (6.5), (6.6) and (6.9) as long as you can deduce
which basis applies to the component formulas. Converting an index equation to direct sym-
bolic form is harder than the reverse (which is a key argument used in favor of direct sym-
bolic formulations of governing equations). Consider, for example, the definition of the
second Piola Kirchhoff tensor:
, where
.
(6.17)
Here,
and is the Cauchy stress. The tensor is called the “Kirchhoff” stress,
and it is identically equal to the Cauchy stress for incompressible materials. Dotting both
sides of the above equation by the initial contravariant base vectors gives
(6.18)
(6.19)
This shows that the contravariant components of the second Piola Kirchhoff tensor with
respect to the
initial
basis are equal to the contravariant components of the Kirchhoff tensor
with respect to the
current
basis. Results like this are worth noting and recording in your per-
sonal file so that you can quickly convert older curvilinear constitutive model equations into
modern direct notation form.
G
˜
i
e
˜˜
Euler
G
˜
j
•
•
1
2
---
G
ij
g
ij
–
[
]
=
s
˜˜
F
˜˜
1
–
τ
˜˜
F
˜˜
T
–
•
•
≡
τ
˜˜
J
σ
˜˜
=
J
det
F
˜˜
( )
=
σ
˜˜
τ
˜˜
G
˜
i
s
˜˜
G
˜
j
•
•
G
˜
i
F
˜˜
1
–
τ
˜˜
F
˜˜
T
–
G
˜
j
•
•
•
•
=
G
˜
i
s
˜˜
G
˜
j
•
•
g
˜
i
τ
˜˜
g
˜
j
•
•
=
rmbrann@me.unm.edu
http://me.unm.edu/~rmbrann/gobag.html
DRAFT June 17, 2004
94
7. Concluding remarks.
R.B. Bird et al. [1] astutely remark “...some authors feel it is stylish to use general tensor
notation even when it is not needed, and hence it is necessary to be familiar with the use of
base vectors and covariant and contravariant components in order to study the literature...”
Modern researchers realize that operations such as the dot product, cross product, and
gradient are proper tensor operations. Such operations commute with basis and/or coordi-
nate transformations as long as the computational procedure for evaluating them is defined
in a manner appropriate for the underlying coordinates or basis. In other words, one can
apply the applicable formula for such an operation in
any convenient system
and transform the
result to a second system, and the result will be the same as if the operation had been applied
directly in the second system at the outset.
Once an operation is known to be a proper tensor operation, it is endowed with a struc-
tured (direct) notation symbolic form. Derivations of new identities usually begin by casting a
direct notation formula in a conveniently simple system such as a principal basis or maybe
just ordinary rectangular Cartesian coordinates. From there, proper tensor operations are per-
formed within that system. In the final step, the result is re-cast in structured (direct) notation.
A structured result can then be justifiably expressed into
any other system
because all the oper-
ations used in the derivation had been proper operations. Consequently, one should always
perform derivations using only proper tensor operations using whatever coordinate system
makes the analysis accessible to the largest audience of readers. The last step is to cast the
final result back in direct notation, thereby allowing it to be recast in any other desired basis
or coordinate system.
rmbrann@me.unm.edu
http://me.unm.edu/~rmbrann/gobag.html
DRAFT June 17, 2004
95
8. REFERENCES
All of the following references are
books
, not journal articles. This choice has been made to
emphasize that the subject of curvilinear analysis has been around for centuries -- there exists
no new literature for us to quote that has not already been quoted in one or more of the fol-
lowing texts. So why, you might ask, is this new manuscript called for? The answer is that
the present manuscript is designed specifically for new students and researchers. It contains
many heuristically derived results. Any self-respecting student should follow up this manu-
script’s elementary introduction with careful reading of the following references.
Greenberg’s applied math book should be part of any engineer’s personal library. Readers
should make an effort to read all footnotes and student exercises in Greenberg’s book —
they’re both enlightening and entertaining! Simmond’s book is a good choice for continued
pursuit of curvilinear coordinates. Tables of gradient formulas for particular coordinate sys-
tems can be found in the appendix of Bird, Armstrong, and Hassager’s book.
1
Bird, R.B., Armstrong, R.C., Hassager, O.,
Dynamics of Polymeric Liquids
. Wiley. (1987).
2
Schreyer, H.L.,
Introduction to Continuum Mechanics
, University of New Mexico internal report.
(1976)
3
Arfken, G.B and Weber, H.J.,
Mathematical Methods for Physicists
. Academic Press (1995).
4
McConnel, A.J.,
Applications of Tensor Analysis
, Dover, NY, (1957)
5
Lovelock, David and Rund, Hanno.
Tensors, differential forms, and variational principles.
, Wiley,
NY (1975).
6
Ramkrishna, D. and Amundson, NR.
Linear Operator Methods in Chemical Engineering
, Pren-
tice-Hall, New Jersey, (1985).
7
Simmonds, James G.
A Brief on Tensor Analysis
, Springer-Verlag, NY (1994).
8
Buck, R.C.,
Advanced Calculus
, McGraw-Hill, NY (1978)
9
Rudin, W.,
Real and Complex Analysis
, 3rd Ed., McGraw-Hill, NY (1987).
10
Greenberg, M.D.,
Foundations of applied mathematics
, Prentice Hall, New Jersey (1978).
11
Malvern, L.E.,
Introduction to the mechanics of a continuous medium.
Prentice-Hall, New Jersey
(1969).
12
Papastravridis, John G.,
Tensor Calculus and Analytical Dynamics
, CRC Press, (1998).
13
Narasimhan, Mysore N.,
Principles of continuum mechanics.
, Wiley-Interscience, NY (1993.
rmbrann@me.unm.edu
http://me.unm.edu/~rmbrann/gobag.html
DRAFT June 17, 2004
96
14
Abraham, R., Marsden, J.E., and Ratiu, T.
Manifolds, Tensor Analysis, and Applications, 2nd
Ed.
(Vol. 75 in series entitled:
Applied Mathematical Sciences
). Springer, N.Y. (1988).