48315828 tensor-analysis

2
MSC 97U20
PACS 01.30.Pp
R. A. Sharipov. Quick Introduction to Tensor Analysis: lecture notes.
Freely distributed on-line. Is free for individual use and educational purposes.
Any commercial use without written consent from the author is prohibited.
This book was written as lecture notes for classes that I taught to undergraduate
students majoring in physics in February 2004 during my time as a guest instructor
at The University of Akron, which was supported by Dr. Sergei F. Lyuksyutov’s
grant from the National Research Council under the COBASE program. These 4
classes have been taught in the frame of a regular Electromagnetism course as an
introduction to tensorial methods.
I wrote this book in a ”do-it-yourself” style so that I give only a draft of tensor
theory, which includes formulating definitions and theorems and giving basic ideas
and formulas. All other work such as proving consistence of definitions, deriving
formulas, proving theorems or completing details to proofs is left to the reader in
the form of numerous exercises. I hope that this style makes learning the subject
really quick and more effective for understanding and memorizing.
I am grateful to Department Chair Prof. Robert R. Mallik for the opportunity
to teach classes and thus to be involved fully in the atmosphere of an American
university. I am also grateful to
Mr. M. Boiwka (mboiwka@hotmail.com)
Mr. A. Calabrese (ajc10@uakron.edu)
Mr. J. Comer (funnybef@lycos.com)
Mr. A. Mozinski (arm5@uakron.edu)
Mr. M. J. Shepard (sheppp2000@yahoo.com)
for attending my classes and reading the manuscript of this book. I would like to
especially acknowledge and thank Mr. Jeff Comer for correcting the grammar and
wording in it.
Contacts to author.
Office: Mathematics Department, Bashkir State University,
32 Frunze street, 450074 Ufa, Russia
Phone: 7-(3472)-23-67-18
Fax: 7-(3472)-23-67-74
Home: 5 Rabochaya street, 450003 Ufa, Russia
Phone: 7-(917)-75-55-786
E-mails: R Sharipov@ic.bashedu.ru,
r-sharipov@mail.ru,
ra sharipov@hotmail.com,
URL: http://www.geocities.com/r-sharipov
CopyRight c Sharipov R.A., 2004

CONTENTS.
CONTENTS. ............................................................................................... 3.
CHAPTER I. PRELIMINARY INFORMATION. .......................................... 4.
§ 1. Geometrical and physical vectors. ........................................................... 4.
§ 2. Bound vectors and free vectors. .............................................................. 5.
§ 3. Euclidean space. .................................................................................... 8.
§ 4. Bases and Cartesian coordinates. ........................................................... 8.
§ 5. What if we need to change a basis ? ...................................................... 12.
§ 6. What happens to vectors when we change the basis ? ............................ 15.
§ 7. What is the novelty about vectors that we learned knowing
transformation formula for their coordinates? ....................................... 17.
CHAPTER II. TENSORS IN CARTESIAN COORDINATES. ..................... 18.
§ 8. Covectors. ........................................................................................... 18.
§ 9. Scalar product of vector and covector. .................................................. 19.
§ 10. Linear operators. ............................................................................... 20.
§ 11. Bilinear and quadratic forms. ............................................................. 23.
§ 12. General definition of tensors. .............................................................. 25.
§ 13. Dot product and metric tensor. .......................................................... 26.
§ 14. Multiplication by numbers and addition. ............................................. 27.
§ 15. Tensor product. ................................................................................. 28.
§ 16. Contraction. ...................................................................................... 28.
§ 17. Raising and lowering indices. .............................................................. 29.
§ 18. Some special tensors and some useful formulas. ................................... 29.
CHAPTER III. TENSOR FIELDS. DIFFERENTIATION OF TENSORS. ... 31.
§ 19. Tensor fields in Cartesian coordinates. ................................................ 31.
§ 20. Change of Cartesian coordinate system. .............................................. 32.
§ 21. Differentiation of tensor fields. ............................................................ 34.
§ 22. Gradient, divergency, and rotor. Laplace and d’Alambert operators. ..... 35.
CHAPTER IV. TENSOR FIELDS IN CURVILINEAR COORDINATES. ..... 38.
§ 23. General idea of curvilinear coordinates. ............................................... 38.
§ 24. Auxiliary Cartesian coordinate system. ............................................... 38.
§ 25. Coordinate lines and the coordinate grid. ............................................ 39.
§ 26. Moving frame of curvilinear coordinates. ............................................. 41.
§ 27. Dynamics of moving frame. ................................................................ 42.
§ 28. Formula for Christoffel symbols. ......................................................... 42.
§ 29. Tensor fields in curvilinear coordinates. ............................................... 43.
§ 30. Differentiation of tensor fields in curvilinear coordinates. ...................... 44.
§ 31. Concordance of metric and connection. ............................................... 46.
REFERENCES. ......................................................................................... 47.

CHAPTER I
PRELIMINARY INFORMATION.
§ 1. Geometrical and physical vectors.
Vector is usually understood as a segment of straight line equipped with an
arrow. Simplest example is displacement vector a. Say its length is 4 cm, i. e.
|a| = 4 cm.
You can draw it on the paper as shown on Fig. 1a. Then it means that point B is
4 cm apart from the point A in the direction pointed to by vector a. However, if
you take velocity vector v for a stream in a brook, you cannot draw it on the paper
immediately. You should ﬁrst adopt a scaling convention, for example, saying that
1 cm on paper represents 1 cm/sec (see Fig. 1b).
Conclusion 1.1. Vectors with physical meaning other than displacement vec-
tors have no unconditional geometric presentation. Their geometric presentation is
conventional; it depends on the scaling convention we choose.
Conclusion 1.2. There are plenty of physical vectors, which are not geomet-
rically visible, but can be measured and then drawn as geometric vectors.
One can consider unit vector m. Its length is equal to unity not 1 cm, not 1 km,
not 1 inch, and not 1 mile, but simply number 1:
|m| = 1.
Like physical vectors, unit vector m cannot be drawn without adopting some
scaling convention. The concept of a unit vector is a very convenient one. By

§ 2. BOUND VECTORS AND FREE VECTORS. 5
multiplying m to various scalar quantities, we can produce vectorial quantities of
various physical nature: velocity, acceleration, force, torque, etc.
Conclusion 1.3. Along with geometrical and physical vectors one can imagine
vectors whose length is a number with no unit of measure.
§ 2. Bound vectors and free vectors.
All displacement vectors are bound ones. They are bound to those points whose
displacement they represent. Free vectors are usually those representing global
physical parameters, e. g. vector of angular velocity ω for Earth rotation about
its axis. This vector produces the Coriolis force aﬀecting water streams in small
rivers and in oceans around the world. Though it is usually drawn attached to the
North pole, we can translate this vector to any point along any path provided we
keep its length and direction unchanged.

6 CHAPTER I. PRELIMINARY INFORMATION.
The next example illustrates the concept of a vector field. Consider the water
flow in a river at some fixed instant of time t. For each point P in the water the
velocity of the water jet passing through
this point is defined. Thus we have a
function
v = v(t, P). (2.1)
Its first argument is time variable t. The
second argument of function (2.1) is not
numeric. It is geometric object — a
point. Values of a function (2.1) are also
not numeric: they are vectors.
Definition2.1. A vector-valued func-
tion with point argument is called vector
field. If it has an additional argument t,
it is called a time-dependent vector field.
Let v be the value of function (2.1) at
the point A in a river. Then vector v is a
bound vector. It represents the velocity
of the water jet at the point A. Hence, it is bound to point A. Certainly, one can
translate it to the point B on the bank of the river (see Fig. 3). But there it loses
its original purpose, which is to mark the water velocity at the point A.
Conclusion 2.1. There exist functions with non-numeric arguments and non-
numeric values.
Exercise 2.1. What is a scalar field ? Suggest an appropriate definition by
analogy with definition 2.1.
Exercise 2.2 (for deep thinking). Let y = f(x) be a function with a non-
numeric argument. Can it be continuous ? Can it be differentiable ? In general,
answer is negative. However, in some cases one can extend the definition of conti-
nuity and the definition of derivatives in a way applicable to some functions with
non-numeric arguments. Suggest your version of such a generalization. If no ver-
sions, remember this problem and return to it later when you gain more experience.
Let A be some fixed point (on the ground, under the ground, in the sky, or in
outer space, wherever). Consider all vectors of some physical nature bound to this
point (say all force vectors). They constitute an infinite set. Let’s denote it VA.
We can perform certain algebraic operations over the vectors from VA:
(1) we can add any two of them;
(2) we can multiply any one of them by any real number α ∈ R;
These operations are called linear operations and VA is called a linear vector space.
Exercise 2.3. Remember the parallelogram method for adding two vectors
(draw picture). Remember how vectors are multiplied by a real number α. Consider
three cases: α > 0, α < 0, and α = 0. Remember what the zero vector is. How it
is represented geometrically ?

§ 2. BOUND VECTORS AND FREE VECTORS. 7
Exercise 2.4. Do you remember the exact mathematical definition of a linear
vector space ? If yes, write it. If no, visit Web page of Jim Hefferon
http://joshua.smcvt.edu/linearalgebra/
and download his book [1]. Keep this book for further references. If you find it use-
ful, you can acknowledge the author by sending him e-mail: jim@joshua.smcvt.edu
Conclusion 2.2. Thus, each point A of our geometric space is not so simple,
even if it is a point in a vacuum. It can be equipped with linear vector spaces of
various natures (such as a space of force vectors in the above example). This idea,
where each point of vacuum space is treated as a container for various physical
fields, is popular in modern physics. Mathematically it is realized in the concept of
bundles: vector bundles, tensor bundles, etc.
Free vectors, taken as they are, do not form a linear vector space. Let’s denote
by V the set of all free vectors. Then V is union of vector spaces VA associated
with all points A in space:
V =
A∈E
VA. (2.2)
The free vectors forming this set (2.2)
are too numerous: we should work to
make them confine the definition of a
linear vector space. Indeed, if we have
a vector a and if it is a free vector, we
can replicate it by parallel translations
that produce infinitely many copies of it
(see Fig. 4). All these clones of vector
a form a class, the class of vector a.
Let’s denote it as Cl(a). Vector a is a
representative of its class. However, we
can choose any other vector of this class
as a representative, say it can be vector
ã. Then we have
Cl(a) = Cl(ã).
Let’s treat Cl(a) as a whole unit, as one indivisible object. Then consider the set
of all such objects. This set is called a factor-set, or quotient set. It is denoted as
V/ ∼ . This quotient set V/ ∼ satisfies the definition of linear vector space. For
the sake of simplicity further we shall denote it by the same letter V as original
set (2.2), from which it is produced by the operation of factorization.
Exercise 2.5. Have you heard about binary relations, quotient sets, quotient
groups, quotient rings and so on ? If yes, try to remember strict mathematical
definitions for them. If not, then have a look to the references [2], [3], [4]. Cer-
tainly, you shouldn’t read all of these references, but remember that they are freely
available on demand.
CopyRight c Sharipov R.A., 2004.

3. Euclidean space.
What is our geometric space ? Is it a linear vector space ? By no means. It
is formed by points, not by vectors. Properties of our space were first system-
atically described by Euclid, the Greek
mathematician of antiquity. Therefore, it
is called Euclidean space and denoted by
E. Euclid suggested 5 axioms (5 postu-
lates) to describe E. However, his state-
ments were not satisfactorily strict from
a modern point of view. Currently E is
described by 20 axioms. In memory of
Euclid they are subdivided into 5 groups:
(1) axioms of incidence;
(2) axioms of order;
(3) axioms of congruence;
(4) axioms of continuity;
(5) axiom of parallels.
20-th axiom, which is also known as 5-th
postulate, is most famous.
Exercise 3.1. Visit the following Non-
Euclidean Geometry web-site and read a
few words about non-Euclidean geometry
and the role of Euclid’s 5-th postulate in
its discovery.
Usually nobody remembers all 20 of these axioms by heart, even me, though I
wrote a textbook on the foundations of Euclidean geometry in 1998. Furthermore,
dealing with the Euclidean space E, we shall rely only on common sense and on
our geometric intuition.
§ 4. Bases and Cartesian coordinates.
Thus, E is composed by points. Let’s choose one of them, denote it by O and
consider the vector space VO composed by displacement vectors. Then each point
B ∈ E can be uniquely identified with the displacement vector rB =
−→
OB. It is
called the radius-vector of the point B, while O is called origin. Passing from
points to their radius-vectors we identify E with the linear vector space VO . Then,
passing from the vectors to their classes, we can identify V with the space of free
vectors. This identification is a convenient tool in studying E without referring to
Euclidean axioms. However, we should remember that such identification is not
unique: it depends on our choice of the point O for the origin.
Definition 4.1. We say that three vectors e1, e2, e3 form a non-coplanar
triple of vectors if they cannot be laid onto the plane by parallel translations.
These three vectors can be bound to some point O common to all of them, or
they can be bound to different points in the space; it makes no difference. They
also can be treated as free vectors without any definite binding point.

§ 4. BASES AND CARTESIAN COORDINATES. 9
Definition 4.2. Any non-coplanar ordered triple of vectors e1, e2, e3 is called
a basis in our geometric space E.
Exercise 4.1. Formulate the definitions of bases on a plane and on a straight
line by analogy with definition 4.2.
Below we distinguish three types of bases: orthonormal basis (ONB), orthogonal
basis (OB), and skew-angular basis (SAB). Orthonormal basis is formed by three
mutually perpendicular unit vectors:
e1 ⊥ e2,
e2 ⊥ e3,
e3 ⊥ e1,
(4.1)
|e1| = 1,
|e2| = 1,
|e3| = 1.
(4.2)
For orthogonal basis, the three con-
ditions (4.1) are fulfilled, but lengths of
basis vectors are not specified.
And skew-angular basis is the most
general case. For this basis neither angles
nor lengths are specified. As we shall see
below, due to its asymmetry SAB can
reveal a lot of features that are hidden in
symmetric ONB.
Let’s choose some basis e1, e2, e3 in E. In the general case this is a skew-
angular basis. Assume that vectors e1, e2, e3 are bound to a common point O as
shown on Fig. 6 below. Otherwise they can be brought to this position by means
of parallel translations. Let a be some arbitrary vector. This vector also can be
translated to the point O. As a result we have four vectors e1, e2, e3, and a
beginning at the same point O. Drawing additional lines and vectors as shown on
Fig. 6, we get
a =
−→
OD =
−→
OA +
−→
OB +
−→
OC. (4.3)
Then from the following obvious relationships
e1 =
−−→
OE1, e2 =
−−→
OE2, e3 =
−−→
OE3,
−−→
OE1
−→
OA,
−−→
OE2
−→
OB,
−−→
OE3
−→
OC
we derive
−→
OA = α e1,
−→
OB = β e2,
−→
OC = γ e3, (4.4)
where α, β, γ are scalars. Now from (4.3) and (4.4) we obtain
a = α e1 + β e2 + γ e3. (4.5)

Exercise 4.2. Explain how, for what reasons, and in what order additional
lines on Fig. 6 are drawn.
Formula (4.5) is known as the expansion of vector a in the basis e1, e2, e3,
while α, β, γ are coordinates of vector a in this basis.
Exercise 4.3. Explain why α, β, and γ are uniquely determined by vector a.
Hint: remember what linear dependence and linear independence are. Give exact
mathematical statements for these concepts. Apply them to exercise 4.3.
Further we shall write formula (4.5) as follows
a = a1
e1 + a2
e2 + a3
e3 =
3
i=1
ai
ei, (4.6)
denoting α = a1
, β = a2
, and γ = a3
. Don’t confuse upper indices in (4.6)
with power exponentials, a1
here is not a, a2
is not a squared, and a3
is not a
cubed. Usage of upper indices and the implicit summation rule were suggested by
Einstein. They are known as Einstein’s tensorial notations.
Once we have chosen the basis e1, e2, e3 (no matter ONB, OB, or SAB), we
can associate vectors with columns of numbers:
a ←→
a1
a2
a3
, b ←→
b1
b2
b3
. (4.7)

§ 4. BASES AND CARTESIAN COORDINATES. 11
We can then produce algebraic operations with vectors, reducing them to arith-
metic operations with numbers:
a + b ←→
a1
a2
a3
+
b1
b2
b3
=
a1
+ b1
a2
+ b2
a3
+ b3
,
α a ←→ α
a1
a2
a3
=
α a1
α a2
α a3
.
Columns of numbers framed by matrix delimiters like those in (4.7) are called
vector-columns. They form linear vector spaces.
Exercise 4.4. Remember the exact mathematical definition for the real arith-
metic vector space Rn
, where n is a positive integer.
Definition 4.1. The Cartesian coordinate system is a basis complemented
with some fixed point that is called the origin.
Indeed, if we have an origin O, then we can associate each point A of our space
with its radius-vector rA =
−→
OA. Then, having expanded this vector in a basis, we
get three numbers that are called the Cartesian coordinates of A. Coordinates of
a point are also specified by upper indices since they are coordinates of the radius-
vector for that point. However, unlike coordinates of vectors, they are usually
not written in a column. The reason will be clear when we consider curvilinear
coordinates. So, writing A(a1
, a2
, a3
) is quite an acceptable notation for the point
A with coordinates a1
, a2
, and a3
.
The idea of cpecification of geometric objects by means of coordinates was
first raised by French mathematician and philosopher René Descartes (1596-1650).
Cartesian coordinates are named in memory of him.

§ 5. What if we need to change a basis ?
Why could we need to change a basis ? There may be various reasons: we may
dislike initial basis because it is too symmetric like ONB, or too asymmetric like
SAB. Maybe we are completely satisfied; but the wisdom is that looking on how
something changes we can learn more about this thing than if we observe it in a
static position. Suppose we have a basis e1, e2, e3, let’s call it the old basis, and
suppose we want to change it to a new one ˜e1, ˜e2, ˜e3. Let’s take the first vector
of the new basis e1. Being isolated from the other two vectors ˜e2 and ˜e3, it is
nothing, but quite an ordinary vector of space. In this capacity, vector ˜e1 can be
expanded in the old basis e1, e2, e3:
˜e1 = S1
e1 + S2
e2 + S3
e3 =
3
j=1
Sj
ej. (5.1)
Compare (5.1) and (4.6). Then we can take another vector ˜e2 and also expand it
in the old basis. But what letter should we choose for denoting the coefficients of
this expansion ? We can choose another letter; say the letter “R”:
˜e2 = R1
e1 + R2
e2 + R3
e3 =
3
j=1
Rj
ej. (5.2)
However, this is not the best decision. Indeed, vectors ˜e1 and ˜e2 differ only in
number, while for their coordinates we use different letters. A better way is to add
an extra index to S in (5.1). This is the lower index coinciding with the number
of the vector:
˜e1 = S1
1 e1 + S2
1 e2 + S3
1 e3 =
3
j=1
Sj
1 ej (5.3)
Color is of no importance; it is used for highlighting only. Instead of (5.2), for the
second vector e2 we write a formula similar to (5.3):
˜e2 = S1
2 e1 + S2
2 e2 + S3
2 e3 =
3
j=1
Sj
2 ej. (5.4)
And for third vector as well:
˜e3 = S1
3 e1 + S2
3 e2 + S3
3 e3 =
3
j=1
Sj
3 ej. (5.5)
When considered jointly, formulas (5.3), (5.4), and (5.5) are called transition
formulas. We use a left curly bracket to denote their union:



˜e1 = S1
1 e1 + S2
1 e2 + S3
1 e3,
˜e2 = S1
2 e1 + S2
2 e2 + S3
2 e3,
˜e3 = S1
3 e1 + S2
3 e2 + S3
3 e3.
(5.6)

§ 5. WHAT IF WE NEED TO CHANGE A BASIS ? 13
We also can write transition formulas (5.6) in a more symbolic form
˜ei =
3
j=1
Sj
i ej. (5.7)
Here index i runs over the range of integers from 1 to 3.
Look at index i in formula (5.7). It is a free index, it can freely take any
numeric value from its range: 1, 2, or 3. Note that i is the lower index in both
sides of formula (5.7). This is a general rule.
Rule 5.1. In correctly written tensorial formulas free indices are written on the
same level (upper or lower) in both sides of the equality. Each free index has only
one entry in each side of the equality.
Now look at index j. It is summation index. It is present only in right hand
side of formula (5.7), and it has exactly two entries (apart from that j = 1 under
the sum symbol): one in the upper level and one in the lower level. This is also
general rule for tensorial formulas.
Rule 5.2. In correctly written tensorial formulas each summation index should
have exactly two entries: one upper entry and one lower entry.
Proposing this rule 5.2, Einstein also suggested not to write the summation
symbols at all. Formula (5.7) then would look like ˜ei = Sj
i ej with implicit
summation with respect to the double index j. Many physicists (especially those
in astrophysics) prefer writing tensorial formulas in this way. However, I don’t like
omitting sums. It breaks the integrity of notations in science. Newcomers from
other branches of science would have difficulties in understanding formulas with
implicit summation.
Exercise 5.1. What happens if ˜e1 = e1 ? What are the numeric values of
coefficients S1
1 , S2
1 , and S3
1 in formula (5.3) for this case ?
Returning to transition formulas (5.6) and (5.7) note that coefficients in them
are parameterized by two indices running independently over the range of integer
numbers from 1 to 3. In other words, they form a two-dimensional array that
usually is represented as a table or as a matrix:
S =
S1
1 S1
2 S1
3
S2
1 S2
2 S2
3
S3
1 S3
2 S3
3
(5.8)
Matrix S is called a transition matrix or direct transition matrix since we
use it in passing from old basis to new one. In writing such matrices like S the
following rule applies.
Rule 5.3. For any double indexed array with indices on the same level (both
upper or both lower) the first index is a row number, while the second index is a
column number. If indices are on different levels (one upper and one lower), then
the upper index is a row number, while lower one is a column number.
Note that according to this rule 5.3, coefficients of formula (5.3), which are
written in line, constitute first column in matrix (5.8). So lines of formula (5.6)
turn into columns in matrix (5.8). It would be worthwhile to remember this fact.

If we represent each vector of the new basis ˜e1, ˜e2, ˜e3 as a column of its
coordinates in the old basis just like it was done for a and b in formula (4.7) above
e1 ←→
S1
1
S2
1
S3
1
, e2 ←→
S1
2
S2
2
S3
2
, e3 ←→
S1
3
S2
3
S3
3
, (5.9)
then these columns (5.9) are exactly the first, the second, and the third columns
in matrix (5.8). This is the easiest way to remember the structure of matrix S.
Exercise 5.2. What happens if ˜e1 = e1, ˜e2 = e2, and ˜e3 = e3 ? Find the
transition matrix for this case. Consider also the following two cases and write the
transition matrices for each of them:
(1) ˜e1 = e1, ˜e2 = e3, ˜e3 = e2;
(2) ˜e1 = e3, ˜e2 = e1, ˜e3 = e2.
Explain why the next case is impossible:
(3) ˜e1 = e1 − e2, ˜e2 = e2 − e3, ˜e3 = e3 − e1.
Now let’s swap bases. This means that we are going to consider ˜e1, ˜e2, ˜e3 as
the old basis, e1, e2, e3 as the new basis, and study the inverse transition. All
of the above stuff applies to this situation. However, in writing the transition
formulas (5.6), let’s use another letter for the coefficients. By tradition here the
letter “T” is used: 


e1 = T1
1 ˜e1 + T2
1 ˜e2 + T3
1 ˜e3,
e2 = T1
2 ˜e1 + T2
2 ˜e2 + T3
2 ˜e3,
e3 = T1
3 ˜e1 + T2
3 ˜e2 + T3
3 ˜e3.
(5.10)
Here is the short symbolic version of transition formulas (5.10):
ei =
3
j=1
Tj
i ˜ej. (5.11)
Denote by T the transition matrix constructed on the base of (5.10) and (5.11). It
is called the inverse transition matrix when compared to the direct transition
matrix S:
(e1, e2, e3)
S
−−−−→
←−−−−
T
(˜e1, ˜e2, ˜e3). (5.12)
Theorem 5.1. The inverse transition matrix T in (5.12) is the inverse matrix
for the direct transition matrix S, i. e. T = S−1
.
Exercise 5.3. What is the inverse matrix ? Remember the definition. How
is the inverse matrix A−1
calculated if A is known ? (Don’t say that you use a
computer package like Maple, MathCad, or any other; remember the algorithm for
calculating A−1
).

§ 6. WHAT HAPPENS TO VECTORS WHEN WE CHANGE THE BASIS ? 15
Exercise 5.4. Remember what is the determinant of a matrix. How is it usually
calculated ? Can you calculate det(A−1
) if det A is already known ?
Exercise 5.5. What is matrix multiplication ? Remember how it is deﬁned.
Suppose you have a rectangular 5 × 3 matrix A and another rectangular matrix B
which is 4 × 5. Which of these two products A B or B A you can calculate ?
Exercise 5.6. Suppose that A and B are two rectangular matrices, and suppose
that C = A B. Remember the formula for the components in matrix C if the
components of A and B are known (they are denoted by Aij and Bpq). Rewrite
this formula for the case when the components of B are denoted by Bpq
. Which
indices (upper, or lower, or mixed) you would use for components of C in the last
case (see rules 5.1 and 5.2 of Einstein’s tensorial notation).
Exercise 5.7. Give some examples of matrix multiplication that are consis-
tent with Einstein’s tensorial notation and those that are not (please, do not use
examples that are already considered in exercise 5.6).
Let’s consider three bases: basis one e1, e2, e3, basis two ˜e1, ˜e2, ˜e3, and basis
three ˜˜e1, ˜˜e2, ˜˜e3. And let’s consider the transition matrices relating them:
(e1, e2, e3)
S
−−−−→
←−−−−
T
(˜e1, ˜e2, ˜e3)
˜S
−−−−→
←−−−−
˜T
(˜˜e1, ˜˜e2, ˜˜e3). (5.13)
Denote by ˜˜S and ˜˜T transition matrices relating basis one with basis three in
(5.13):
(e1, e2, e3)
˜˜S
−−−−→
←−−−−
˜˜T
(˜˜e1, ˜˜e2, ˜˜e3). (5.14)
Exercise 5.8. For matrices ˜˜S and ˜˜T in (5.14) prove that ˜˜S = S ˜S and ˜˜T = ˜T T.
Apply this result for proving theorem 5.1.
§ 6. What happens to vectors when we change the basis ?
The answer to this question is very simple. Really nothing ! Vectors do not
need a basis for their being. But their coordinates, they depend on our choice
of basis. And they change if we change the basis. Let’s study how they change.
Suppose we have some vector x expanded in the basis e1, e2, e3:
x = x1
e1 + x2
e2 + x3
e3 =
3
i=1
xi
ei. (6.1)
Then we keep vector x and change the basis e1, e2, e3 to another basis ˜e1, ˜e2, ˜e3.
As we already learned, this process is described by transition formula (5.11):
ei =
3
j=1
Tj
i ˜ej.

Let’s substitute this formula into (6.1) for ei:
x =
3
i=1
xi
3
j=1
Tj
i ej =
3
i=1
3
j=1
xi
Tj
i ˜ej =
3
j=1
3
i=1
xi
Tj
i ˜ej =
=
3
j=1
3
i=1
Tj
i xi
˜ej =
3
j=1
˜xj
˜ej, where ˜xj
=
3
i=1
Tj
i xi
.
Thus we have calculated the expansion of vector x in the new basis and have
derived the formula relating its new coordinates to its initial ones:
˜xj
=
3
i=1
Tj
i xi
. (6.2)
This formula is called a transformation formula, or direct transformation
formula. Like (5.7), it can be written in expanded form:



˜x1
= T1
1 x1
+ T1
2 x2
+ T1
3 x3
,
˜x2
= T2
1 x1
+ T2
2 x2
+ T2
3 x3
,
˜x3
= T3
1 x1
+ T3
2 x2
+ T3
3 x3
.
(6.3)
And the transformation formula (6.2) can be written in matrix form as well:
˜x1
˜x2
˜x3
=
T1
1 T1
2 T1
3
T2
1 T2
2 T2
3
T3
1 T3
2 T3
3
x1
x2
x3
. (6.4)
Like (5.7), formula (6.2) can be inverted. Here is the inverse transformation
formula expressing the initial coordinates of vector x through its new coordinates:
xj
=
3
i=1
Sj
i ˜xi
. (6.5)
Exercise 6.1. By analogy with the above calculations derive the inverse trans-
formation formula (6.5) using formula (5.7).
Exercise 6.2. By analogy with (6.3) and (6.4) write (6.5) in expanded form
and in matrix form.
Exercise 6.3. Derive formula (6.5) directly from (6.2) using the concept of the
inverse matrix S = T −1
.
Note that the direct transformation formula (6.2) uses the inverse transition
matrix T, and the inverse transformation formula (6.5) uses direct transition
matrix S. It’s funny, but it’s really so.

§ 7. WHAT IS THE NOVELTY ABOUT THE VECTORS . . . 17
§ 7. What is the novelty about the vectors that we learned
knowing transformation formula for their coordinates ?
Vectors are too common, too well-known things for one to expect that there
are some novelties about them. However, the novelty is that the method of
their treatment can be generalized and then applied to less customary objects.
Suppose, we cannot visually observe vectors (this is really so for some kinds of
them, see section 1), but suppose we can measure their coordinates in any basis
we choose for this purpose. What then do we know about vectors ? And how
can we tell them from other (non-vectorial) objects ? The answer is in formulas
(6.2) and (6.5). Coordinates of vectors (and only coordinates of vectors) will obey
transformation rules (6.2) and (6.5) under a change of basis. Other objects usually
have a diﬀerent number of numeric parameters related to the basis, and even if
they have exactly three coordinates (like vectors have), their coordinates behave
diﬀerently under a change of basis. So transformation formulas (6.2) and (6.5)
work like detectors, like a sieve for separating vectors from non-vectors. What
are here non-vectors, and what kind of geometrical and/or physical objects of a
non-vectorial nature could exist — these are questions for a separate discussion.
Furthermore, we shall consider only a part of the set of such objects, which are
called tensors.

CHAPTER II
TENSORS IN CARTESIAN COORDINATES.
§ 8. Covectors.
In previous 7 sections we learned the following important fact about vectors:
a vector is a physical object in each basis of our three-dimensional Euclidean
space E represented by three numbers such that these numbers obey certain
transformation rules when we change the basis. These certain transformation rules
are represented by formulas (6.2) and (6.5).
Now suppose that we have some other physical object that is represented by
three numbers in each basis, and suppose that these numbers obey some certain
transformation rules when we change the basis, but these rules are different from
(6.2) and (6.5). Is it possible ? One can try to find such an object in nature.
However, in mathematics we have another option. We can construct such an
object mentally, then study its properties, and finally look if it is represented
somehow in nature.
Let’s denote our hypothetical object by a, and denote by a1, a2, a3 that
three numbers which represent this object in the basis e1, e2, e3. By analogy
with vectors we shall call them coordinates. But in contrast to vectors, we
intentionally used lower indices when denoting them by a1, a2, a3. Let’s prescribe
the following transformation rules to a1, a2, a3 when we change e1, e2, e3 to
˜e1, ˜e2, ˜e3:
ãj =
3
i=1
Si
j ai, (8.1)
aj =
3
i=1
Ti
j ãi. (8.2)
Here S and T are the same transition matrices as in case of the vectors in (6.2)
and (6.5). Note that (8.1) is sufficient, formula (8.2) is derived from (8.1).
Exercise 8.1. Using the concept of the inverse matrix T = S−1
derive formula
(8.2) from formula (8.1). Compare exercise 8.1 and exercise 6.3.
Definition 8.1. A geometric object a in each basis represented by a triple
of coordinates a1, a2, a3 and such that its coordinates obey transformation rules
(8.1) and (8.2) under a change of basis is called a covector.
Looking at the above considerations one can think that we arbitrarily chose
the transformation formula (8.1). However, this is not so. The choice of the
transformation formula should be self-consistent in the following sense. Let
e1, e2, e3 and ˜e1, ˜e2, ˜e3 be two bases and let ˜˜e1, ˜˜e2, ˜˜e3 be the third basis in the

§ 9. SCALAR PRODUCT OF VECTOR AND COVECTOR. 19
space. Let’s call them basis one, basis two and basis three for short. We can pass
from basis one to basis three directly, see the right arrow in (5.14). Or we can use
basis two as an intermediate basis, see the right arrows in (5.13). In both cases the
ultimate result for the coordinates of a covector in basis three should be the same:
this is the self-consistence requirement. It means that coordinates of a geometric
object should depend on the basis, but not on the way that they were calculated.
Exercise 8.2. Using (5.13) and (5.14), and relying on the results of exer-
cise 5.8 prove that formulas (8.1) and (8.2) yield a self-consistent way of defining the
covector.
Exercise 8.3. Replace S by T in (8.1) and T by S in (8.2). Show that the
resulting formulas are not self-consistent.
What about the physical reality of covectors? Later on we shall see that
covectors do exist in nature. They are the nearest relatives of vectors. And
moreover, we shall see that some well-known physical objects we thought to be
vectors are of covectorial nature rather than vectorial.
§ 9. Scalar product of vector and covector.
Suppose we have a vector x and a covector a. Upon choosing some basis
e1, e2, e3, both of them have three coordinates: x1
, x2
, x3
for vector x, and
a1, a2, a3 for covector a. Let’s denote by a, x the following sum:
a, x =
3
i=1
ai xi
. (9.1)
The sum (9.1) is written in agreement with Einstein’s tensorial notation, see
rule 5.2 in section 5 above. It is a number depending on the vector x and on
the covector a. This number is called the scalar product of the vector x and the
covector a. We use angular brackets for this scalar product in order to distinguish
it from the scalar product of two vectors in E, which is also known as the dot
product.
Defining the scalar product a, x by means of sum (9.1) we used the coordi-
nates of vector x and of covector a, which are basis-dependent. However, the value
of sum (9.1) does not depend on any basis. Such numeric quantities that do not
depend on the choice of basis are called scalars or true scalars.
Exercise 9.1. Consider two bases e1, e2, e3 and ˜e1, ˜e2, ˜e3, and consider the
coordinates of vector x and covector a in both of them. Relying on transformation
rules (6.2), (6.5), (8.1), and (8.2) prove the equality
3
i=1
ai xi
=
3
i=1
ãi ˜xi
. (9.2)
Thus, you are proving the self-consistence of formula (9.1) and showing that the
scalar product a, x given by this formula is a true scalar quantity.

20 CHAPTER II. TENSORS IN CARTESIAN COORDINATES.
Exercise 9.2. Let α be a real number, let a and b be two covectors, and let
x and y be two vectors. Prove the following properties of the scalar product (9.1):
(1) a + b, x = a, x + b, x ;
(2) α a, x = α a, x ;
(3) a, x + y = a, x + a, y ;
(4) a, α x = α a, x .
Exercise 9.3. Explain why the scalar product a, x is sometimes called the
bilinear function of vectorial argument x and covectorial argument a. In this ca-
pacity, it can be denoted as f(a, x). Remember our discussion about functions with
non-numeric arguments in section 2.
Important note. The scalar product a, x is not symmetric. Moreover, the
formula
a, x = x, a
is incorrect in its right hand side since the ﬁrst argument of scalar product (9.1)
by deﬁnition should be a covector. In a similar way, the second argument should
be a vector. Therefore, we never can swap them.
§ 10. Linear operators.
In this section we consider more complicated geometric objects. For the sake
of certainty, let’s denote one of such objects by F. In each basis e1, e2, e3, it is
represented by a square 3 × 3 matrix F i
j of real numbers. Components of this
matrix play the same role as coordinates in the case of vectors or covectors. Let’s
prescribe the following transformation rules to F i
j :
˜Fi
j =
3
p=1
3
q=1
Ti
p Sq
j Fp
q , (10.1)
Fi
j =
3
p=1
3
q=1
Si
p Tq
j
˜Fp
q . (10.2)
Exercise 10.1. Using the concept of the inverse matrix T = S−1
prove that
formula (10.2) is derived from formula (10.1).
If we write matrices F i
j and ˜Fp
q according to the rule 5.3 (see in section 5), then
(10.1) and (10.2) can be written as two matrix equalities:
˜F = T F S, F = S ˜F T. (10.3)
Exercise 10.2. Remember matrix multiplication (we already considered it in
exercises 5.5 and 5.6) and derive (10.3) from (10.1) and (10.2).
Definition 10.1. A geometric object F in each basis represented by some
square matrix F i
j and such that components of its matrix F i
j obey transformation
rules (10.1) and (10.2) under a change of basis is called a linear operator.

§ 10. LINEAR OPERATORS. 21
Exercise 10.3. By analogy with exercise 8.2 prove the self-consistence of the
above definition of a linear operator.
Let’s take a linear operator F represented by matrix F i
j in some basis e1, e2, e3
and take some vector x with coordinates x1
, x2
, x3
in the same basis. Using F i
j
and xj
we can construct the following sum:
yi
=
3
j=1
Fi
j xj
. (10.4)
Index i in the sum (10.4) is a free index; it can deliberately take any one of three
values: i = 1, i = 2, or i = 3. For each specific value of i we get some specific
value of the sum (10.4). They are denoted by y1
, y2
, y3
according to (10.4). Now
suppose that we pass to another basis ˜e1, ˜e2, ˜e3 and do the same things. As a
result we get other three values ˜y1
, ˜y2
, ˜y3
given by formula
˜yp
=
3
q=1
˜Fp
q ˜xq
. (10.5)
Exercise 10.4. Relying upon (10.1) and (10.2) prove that the three numbers
y1
, y2
, y3
and the other three numbers ˜y1
, ˜y2
, ˜y3
are related as follows:
˜yj
=
3
i=1
Tj
i yi
, yj
=
3
i=1
Sj
i ˜yi
. (10.6)
Exercise 10.5. Looking at (10.6) and comparing it with (6.2) and (6.5) find
that the y1
, y2
, y3
and ˜y1
, ˜y2
, ˜y3
calculated by formulas (10.4) and (10.5) represent
the same vector, but in different bases.
Thus formula (10.4) defines the vectorial object y, while exercise 10.5 assures
the correctness of this definition. As a result we have vector y determined by a
linear operator F and by vector x. Therefore, we write
y = F(x) (10.7)
and say that y is obtained by applying linear operator F to vector x. Some people
like to write (10.7) without parentheses:
y = F x. (10.8)
Formula (10.8) is a more algebraistic form of formula (10.7). Here the action of
operator F upon vector x is designated like a kind of multiplication. There is also
a matrix representation of formula (10.8), in which x and y are represented as
columns:
y1
y2
y3
=
F1
1 F1
2 F1
3
F2
1 F2
2 F2
3
F3
1 F3
2 F3
3
x1
x2
x3
. (10.9)

Exercise 10.6. Derive (10.9) from (10.4).
Exercise 10.7. Let α be some real number and let x and y be two vectors.
Prove the following properties of a linear operator (10.7):
(1) F(x + y) = F(x) + F(y),
(2) F(α x) = α F(x).
Write these equalities in the more algebraistic style introduced by (10.8). Are they
really similar to the properties of multiplication ?
Exercise 10.8. Remember that for the product of two matrices
det(A B) = det A det B. (10.10)
Also remember the formula for det(A−1
). Apply these two formulas to (10.3) and
derive
det F = det ˜F. (10.11)
Formula (10.10) means that despite the fact that in various bases linear operator
F is represented by various matrices, the determinants of all these matrices are
equal to each other. Then we can define the determinant of linear operator F as
the number equal to the determinant of its matrix in any one arbitrarily chosen
basis e1, e2, e3:
det F = det F. (10.12)
Exercise 10.9 (for deep thinking). Square matrices have various attributes:
eigenvalues, eigenvectors, a characteristic polynomial, a rank (maybe you remember
some others). If we study these attributes for the matrix of a linear operator, which
of them can be raised one level up and considered as basis-independent attributes
of the linear operator itself ? Determinant (10.12) is an example of such attribute.
Exercise 10.10. Substitute the unit matrix for F i
j into (10.1) and verify that
˜Fi
j is also a unit matrix in this case. Interpret this fact.
Exercise 10.11. Let x = ei for some basis e1, e2, e3 in the space. Substitute
this vector x into (10.7) and by means of (10.4) derive the following formula:
F(ei) =
3
j=1
Fj
i ej. (10.13)
Compare (10.13) and (5.7). Discuss the similarities and differences of these two
formulas. The fact is that in some books the linear operator is determined first,
then its matrix is introduced by formula (10.13). Explain why if we know three
vectors F(e1), F(e2), and F(e3), then we can reconstruct the whole matrix of
operator F by means of formula (10.13).
Suppose we have two linear operators F and H. We can apply H to vector x
and then we can apply F to vector H(x). As a result we get
F ◦ H(x) = F(H(x)). (10.14)
Here F ◦ H is new linear operator introduced by formula (10.14). It is called a
composite operator, and the small circle sign denotes composition.

§ 11. BILINEAR AND QUADRATIC FORMS. 23
Exercise 10.12. Find the matrix of composite operator F ◦ H if the matrices
for F and H in the basis e1, e2, e3 are known.
Exercise 10.13. Remember the definition of the identity map in mathematics
(see on-line Math. Encyclopedia) and define the identity operator id. Find the
matrix of this operator.
Exercise 10.14. Remember the definition of the inverse map in mathematics
and define inverse operator F−1
for linear operator F. Find the matrix of this
operator if the matrix of F is known.
§ 11. Bilinear and quadratic forms.
Vectors, covectors, and linear operators are all examples of tensors (though we
have no definition of tensors yet). Now we consider another one class of tensorial
objects. For the sake of clarity, let’s denote by a one of such objects. In each
basis e1, e2, e3 this object is represented by some square 3 × 3 matrix aij of real
numbers. Under a change of basis these numbers are transformed as follows:
ãij =
3
p=1
3
q=1
Sp
i Sq
j apq, (11.1)
aij =
3
p=1
3
q=1
Tp
i Tq
j ãpq. (11.2)
Transformation rules (11.1) and (11.2) can be written in matrix form:
ã = S a S, a = T ã T. (11.3)
Here by S and T we denote the transposed matrices for S and T respectively.
Exercise 11.1. Derive (11.2) from (11.1), then (11.3) from (11.1) and (11.2).
Definition 11.1. A geometric object a in each basis represented by some
square matrix aij and such that components of its matrix aij obey transformation
rules (11.1) and (11.2) under a change of basis is called a bilinear form.
Let’s consider two arbitrary vectors x and y. We use their coordinates and the
components of bilinear form a in order to write the following sum:
a(x, y) =
3
i=1
3
j=1
aij xi
yj
. (11.4)
Exercise 11.2. Prove that the sum in the right hand side of formula (11.4)
does not depend on the basis, i. e. prove the equality
3
i=1
3
j=1
aij xi
yj
=
3
p=1
3
q=1
ãpq ˜xp
˜yq
.
This equality means that a(x, y) is a number determined by vectors x and y
irrespective of the choice of basis. Hence we can treat (11.4) as a scalar function
of two vectorial arguments.

Exercise 11.3. Let α be some real number, and let x, y, and z be three vectors.
Prove the following properties of function (11.4):
(1) a(x+y, z) = a(x, z)+a(y, z);
(2) a(α x, y) = α a(x, y);
(3) a(x, y+z) = a(x, y)+a(x, z);
(4) a(x, α y) = α a(x, y).
Due to these properties function (10.4) is called a bilinear function or a bilinear
form. It is linear with respect to each of its two arguments.
Note that scalar product (9.1) is also a bilinear function of its arguments.
However, there is a crucial difference between (9.1) and (11.4). The arguments of
scalar product (9.1) are of a different nature: the first argument is a covector, the
second is a vector. Therefore, we cannot swap them. In bilinear form (11.4) we
can swap arguments. As a result we get another bilinear function
b(x, y) = a(y, x). (11.5)
The matrices of a and b are related to each other as follows:
bij = aji, b = a . (11.6)
Definition 11.2. A bilinear form is called symmetric if a(x, y) = a(y, x).
Exercise 11.4. Prove the following identity for a symmetric bilinear form:
a(x, y) =
a(x + y, x + y) − a(x, x) − a(y, y)
2
. (11.7)
Definition 11.3. A quadratic form is a scalar function of one vectorial argu-
ment f(x) produced from some bilinear function a(x, y) by substituting y = x:
f(x) = a(x, x). (11.8)
Without a loss of generality a bilinear function a in (11.8) can be assumed
symmetric. Indeed, if a is not symmetric, we can produce symmetric bilinear
function
c(x, y) =
a(x, y) + a(y, x)
2
, (11.9)
and then from (11.8) due to (11.9) we derive
f(x) = a(x, x) =
a(x, x) + a(x, x)
2
= c(x, x).
This equality is the same as (11.8) with a replaced by c. Thus, each quadratic
function f is produced by some symmetric bilinear function a. And conversely,
comparing (11.8) and (11.7) we get that a is produced by f:
a(x, y) =
f(x + y) − f(x) − f(y)
2
. (11.10)
Formula (11.10) is called the recovery formula. It recovers bilinear function a
from quadratic function f produced in (11.8). Due to this formula, in referring to
a quadratic form we always imply some symmetric bilinear form like the geometric
tensorial object introduced by definition 11.1.

§ 12. GENERAL DEFINITION OF TENSORS. 25
§ 12. General definition of tensors.
Vectors, covectors, linear operators, and bilinear forms are examples of tensors.
They are geometric objects that are represented numerically when some basis in
the space is chosen. This numeric representation is specific to each of them:
vectors and covectors are represented by one-dimensional arrays, linear operators
and quadratic forms are represented by two-dimensional arrays. Apart from the
number of indices, their position does matter. The coordinates of a vector are
numerated by one upper index, which is called the contravariant index. The
coordinates of a covector are numerated by one lower index, which is called the
covariant index. In a matrix of bilinear form we use two lower indices; therefore
bilinear forms are called twice-covariant tensors. Linear operators are tensors
of mixed type; their components are numerated by one upper and one lower
index. The number of indices and their positions determine the transformation
rules, i. e. the way the components of each particular tensor behave under a change
of basis. In the general case, any tensor is represented by a multidimensional array
with a definite number of upper indices and a definite number of lower indices.
Let’s denote these numbers by r and s. Then we have a tensor of the type (r, s),
or sometimes the term valency is used. A tensor of type (r, s), or of valency (r, s)
is called an r-times contravariant and an s-times covariant tensor. This is
terminology; now let’s proceed to the exact definition. It is based on the following
general transformation formulas:
Xi1... ir
j1... js
=
3
...
3
h1, ... , hr
k1, ... , ks
Si1
h1
. . . Sir
hr
Tk1
j1
. . . Tks
js
˜Xh1... hr
k1... ks
, (12.1)
˜Xi1... ir
j1... js
=
3
...
3
h1, ... , hr
k1, ... , ks
Ti1
h1
. . . Tir
hr
Sk1
j1
. . . Sks
js
Xh1... hr
k1... ks
. (12.2)
Definition 12.1. A geometric object X in each basis represented by (r + s)-
dimensional array Xi1... ir
j1... js
of real numbers and such that the components of this
array obey the transformation rules (12.1) and (12.2) under a change of basis is
called tensor of type (r, s), or of valency (r, s).
Formula (12.2) is derived from (12.1), so it is sufficient to remember only one
of them. Let it be the formula (12.1). Though huge, formula (12.1) is easy to
remember. One should strictly follow the rules 5.1 and 5.2 from section 5.
Indices i1, . . . , ir and j1, . . . , js are free indices. In right hand side of the
equality (12.1) they are distributed in S-s and T-s, each having only one entry
and each keeping its position, i. e. upper indices i1, . . . , ir remain upper and lower
indices j1, . . . , js remain lower in right hand side of the equality (12.1).
Other indices h1, . . . , hr and k1, . . . , ks are summation indices; they enter the
right hand side of (12.1) pairwise: once as an upper index and once as a lower
index, once in S-s or T-s and once in components of array ˜Xh1... hr
k1... ks
.
When expressing Xi1... ir
j1... js
through ˜Xh1... hr
k1... ks
each upper index is served by direct
transition matrix S and produces one summation in (12.1):
X... iα ...
... ... ... = . . .
3
hα=1
. . . . . . S iα
hα
. . . ˜X... hα ...
... ... ... . (12.3)

In a similar way, each lower index is served by inverse transition matrix T and
also produces one summation in formula (12.1):
X... ... ...
... jα ... = . . .
3
kα=1
. . . . . . Tkα
jα
. . . ˜X... ... ...
... kα .... (12.4)
Formulas (12.3) and (12.4) are the same as (12.1) and used to highlight how
(12.1) is written. So tensors are defined. Further we shall consider more examples
showing that many well-known objects undergo the definition 12.1.
Exercise 12.1. Verify that formulas (6.5), (8.2), (10.2), and (11.2) are spe-
cial cases of formula (12.1). What are the valencies of vectors, covectors, linear
operators, and bilinear forms when they are considered as tensors.
Exercise 12.2. Let aij be the matrix of some bilinear form a. Let’s denote
by bij
components of inverse matrix for aij . Prove that matrix bij
under a change
of basis transforms like matrix of twice-contravariant tensor. Hence it determines
tensor b of valency (2, 0). Tensor b is called a dual bilinear form for a.
§ 13. Dot product and metric tensor.
The covectors, linear operators, and bilinear forms that we considered above
were artificially constructed tensors. However there are some tensors of natural
origin. Let’s remember that we live in a space with measure. We can measure
distance between points (hence we can measure length of vectors) and we can
measure angles between two directions in our space. Therefore for any two vectors
x and y we can define their natural scalar product (or dot product):
(x, y) = |x| |y| cos(ϕ), (13.1)
where ϕ is the angle between vectors x and y.
Exercise 13.1. Remember the following properties of the scalar product (13.1):
(1) (x + y, z) = (x, z) + (y, z);
(2) (α x, y) = α (x, y);
(3) (x, y + z) = (x, y) + (x, z);
(4) (x, α y) = α (x, y);
(5) (x, y) = (y, x);
(6) (x, x) 0 and (x, x) = 0 implies x = 0.
These properties are usually considered in courses on analytic geometry or vector
algebra, see Vector Lessons on the Web.
Note that the first four properties of the scalar product (13.1) are quite
similar to those for quadratic forms, see exercise 11.3. This is not an occasional
coincidence.
Exercise 13.2. Let’s consider two arbitrary vectors x and y expanded in some
basis e1, e2, e3. This means that we have the following expressions for them:
x =
3
i=1
xi
ei, y =
3
j=1
xj
ej. (13.2)

§ 14. MULTIPLICATION BY NUMBERS AND ADDITION. 27
Substitute (13.2) into (13.1) and using properties (1)– (4) listed in exercise 13.1
derive the following formula for the scalar product of x and y:
(x, y) =
3
i=1
3
j=1
(ei, ej) xi
yj
. (13.3)
Exercise 13.3. Denote gij = (ei, ej) and rewrite formula (13.3) as
(x, y) =
3
i=1
3
j=1
gij xi
yj
. (13.4)
Compare (13.4) with formula (11.4). Consider some other basis ˜e1, ˜e2, ˜e3, denote
˜gpq = (˜ep, ˜eq) and by means of transition formulas (5.7) and (5.11) prove that
matrices gij and ˜gpq are components of a geometric object obeying transformation
rules (11.1) and (11.2) under a change of base. Thus you prove that the Gram
matrix
gij = (ei, ej) (13.5)
deﬁnes tensor of type (0, 2). This is very important tensor; it is called the metric
tensor. It describes not only the scalar product in form of (13.4), but the whole
geometry of our space. Evidences for this fact are below.
Matrix (13.5) is symmetric due to property (5) in exercise 13.1. Now, comparing
(13.4) and (11.4) and keeping in mind the tensorial nature of matrix (13.5), we
conclude that the scalar product is a symmetric bilinear form:
(x, y) = g(x, y). (13.6)
The quadratic form corresponding to (13.6) is very simple: f(x) = g(x, x) = |x|2
.
The inverse matrix for (13.5) is denoted by the same symbol g but with upper
indices: gij
. It determines a tensor of type (2, 0), this tensor is called dual metric
tensor (see exercise 12.2 for more details).
§ 14. Multiplication by numbers and addition.
Tensor operations are used to produce new tensors from those we already have.
The most simple of them are multiplication by number and addition. If
we have some tensor X of type (r, s) and a real number α, then in some base
e1, e2, e3 we have the array of components of tensor X; let’s denote it Xi1... ir
j1... js
.
Then by multiplying all the components of this array by α we get another array
Y i1... ir
j1... js
= α Xi1... ir
j1... js
. (14.1)
Choosing another base ˜e1, ˜e2, ˜e3 and repeating this operation we get
˜Y i1... ir
j1... js
= α ˜Xi1... ir
j1... js
. (14.2)

Exercise 14.1. Prove that arrays ˜Y i1... ir
j1... js
and Y i1... ir
j1... js
are related to each other
in the same way as arrays ˜Xi1... ir
j1... js
and Xi1... ir
j1... js
, i. e. according to transformation
formulas (12.1) and (12.2). In doing this you prove that formula (14.1) applied in
all bases produces new tensor Y = α X from initial tensor X.
Formula (14.1) defines the multiplication of tensors by numbers. In
exercise 14.1 you prove its consistence. The next formula defines the addition of
tensors:
Xi1... ir
j1... js
+ Y i1... ir
j1... js
= Zi1... ir
j1... js
. (14.3)
Having two tensors X and Y both of type (r, s) we produce a third tensor Z of the
same type (r, s) by means of formula (14.3). It’s natural to denote Z = X + Y.
Exercise 14.2. By analogy with exercise 14.1 prove the consistence of formula
(14.3).
Exercise 14.3. What happens if we multiply tensor X by the number zero and
by the number minus one ? What would you call the resulting tensors?
§ 15. Tensor product.
The tensor product is defined by a more tricky formula. Suppose we have tensor
X of type (r, s) and tensor Y of type (p, q), then we can write:
Z
i1... ir+p
j1... js+q
= Xi1... ir
j1... js
Y
ir+1... ir+p
js+1... js+q
. (15.1)
Formula (15.1) produces new tensor Z of the type (r + p, s + q). It is called the
tensor product of X and Y and denoted Z = X ⊗ Y. Don’t mix the tensor
product and the cross product. They are different.
Exercise 15.1. By analogy with exercise 14.1 prove the consistence of formula
(15.1).
Exercise 15.2. Give an example of two tensors such that X ⊗ Y = Y ⊗ X.
§ 16. Contraction.
As we have seen above, the tensor product increases the number of indices.
Usually the tensor Z = X ⊗ Y has more indices than X and Y. Contraction is an
operation that decreases the number of indices. Suppose we have tensor X of the
type (r + 1, s + 1). Then we can produce tensor Z of type (r, s) by means of the
following formula:
Zi1... ir
j1... js
=
n
ρ=1
X
i1... im−1 ρ im... ir
j1... jk−1 ρ jk... js
. (16.1)
What we do ? Tensor X has at least one upper index and at least one lower index.
We choose the m-th upper index and replace it by the summation index ρ. In the
same way we replace the k-th lower index by ρ. Other r upper indices and s lower
indices are free. They are numerated in some convenient way, say as in formula
(16.1). Then we perform summation with respect to index ρ. The contraction is
over. This operation is called a contraction with respect to m-th upper and
k-th lower indices. Thus, if we have many upper an many lower indices in tensor
X, we can perform various types of contractions to this tensor.

§ 18. SOME SPECIAL TENSORS AND SOME USEFUL FORMULAS. 29
Exercise 16.1. Prove the consistence of formula (16.1).
Exercise 16.2. Look at formula (9.1) and interpret this formula as the con-
traction of the tensor product a ⊗ x. Find similar interpretations for (10.4), (11.4),
and (13.4).
§ 17. Raising and lowering indices.
Suppose that X is some tensor of type (r, s). Let’s choose its α-th lower index:
X... ... ...
... k ... . The symbols used for the other indices are of no importance. Therefore,
we denoted them by dots. Then let’s consider the tensor product Y = g ⊗ X:
Y ... p q ...
... k ... = gpq
X... ... ...
... k ... . (17.1)
Here g is the dual metric tensor with the components gpq
(see section 13 above).
In the next step let’s contract (17.1) with respect to the pair of indices k and q.
For this purpose we replace them both by s and perform the summation:
X... p ...
... ... ... =
3
s=1
gps
X... ... ...
... s ... . (17.2)
This operation (17.2) is called the index raising procedure. It is invertible.
The inverse operation is called the index lowering procedure:
X... ... ...
... p ... =
3
s=1
gps X... s ...
... ... .... (17.3)
Like (17.2), the index lowering procedure (17.3) comprises two tensorial operations:
the tensor product and contraction.
§ 18. Some special tensors and some useful formulas.
Kronecker symbol is a well known object. This is a two-dimensional array
representing the unit matrix. It is determined as follows:
δi
j =
1 for i = j,
0 for i = j.
(18.1)
We can determine two other versions of Kronecker symbol:
δij
= δij =
1 for i = j,
0 for i = j.
(18.2)
Exercise 18.1. Prove that deﬁnition (18.1) is invariant under a change of basis,
if we interpret the Kronecker symbol as a tensor. Show that both deﬁnitions in
(18.2) are not invariant under a change of basis.
Exercise 18.2. Lower index i of tensor (18.1) by means of (17.3). What ten-
sorial object do you get as a result of this operation?

Exercise 18.3. Likewise, raise index J in (18.1).
Another well known object is the Levi-Civita symbol. This is a three-
dimensional array determined by the following formula:
jkq = jkq
=



0, if among j, k, q, there are
at least two equal num-
bers;
1, if (j k q) is even permuta-
tion of numbers (1 2 3);
−1, if (j k q) is odd permuta-
tion of numbers (1 2 3).
(18.3)
The Levi-Civita symbol (18.3) is not a tensor. However, we can produce two
tensors by means of Levi-Civita symbol. The ﬁrst of them
ωijk = det(gij) ijk (18.4)
is known as the volume tensor. Another one is the dual volume tensor:
ωijk
= det(gij) ijk
. (18.5)
Let’s take two vectors x and y. Then using (18.4) we can produce covector a:
ai =
3
j=1
3
k=1
ωijk xj
yk
. (18.6)
Then we can apply index raising procedure (17.2) and produce vector a:
ar
=
3
i=1
3
j=1
3
k=1
gri
ωijk xj
yk
. (18.7)
Formula (18.7) is known as formula for the vectorial product (cross product) in
skew-angular basis.
Exercise 18.4. Prove that the vector a with components (18.7) coincides with
cross product of vectors x and y, i. e. a = [x, y].

CHAPTER III
TENSOR FIELDS. DIFFERENTIATION OF TENSORS.
§ 19. Tensor fields in Cartesian coordinates.
The tensors that we defined in section 12 are free tensors. Indeed, their
components are arrays related to bases, while any basis is a triple of free vectors
(not bound to any point). Hence, the tensors previously considered are also not
bound to any point.
Now suppose we want to bind our tensor to some point in space, then another
tensor to another point and so on. Doing so we can fill our space with tensors, one
per each point. In this case we say that we have a tensor field. In order to mark a
point P to which our particular tensor is bound we shall write P as an argument:
X = X(P). (19.1)
Usually the valencies of all tensors composing the tensor field are the same. Let
them all be of type (r, s). Then if we choose some basis e1, e2, e3, we can represent
any tensor of our tensor field as an array Xi1... ir
j1... js
with r + s indices:
Xi1... ir
j1... js
= Xi1... ir
j1... js
(P). (19.2)
Thus, the tensor field (19.1) is a tensor-valued function with argument P being a
point of three-dimensional Euclidean space E, and (19.2) is the basis representation
for (19.1). For each fixed set of numeric values of indices i1, . . . , ir, j1, . . . , js in
(19.2), we have a numeric function with a point-valued argument. Dealing with
point-valued arguments is not so convenient, for example, if we want to calculate
derivatives. Therefore, we need to replace P by something numeric. Remember
that we have already chosen a basis. If, in addition, we fix some point O as an
origin, then we get Cartesian coordinate system in space and hence can represent
P by its radius-vector rP =
−→
OP and by its coordinates x1
, x2
, x3
:
Xi1... ir
j1... js
= Xi1... ir
j1... js
(x1
, x2
, x3
). (19.3)
Conclusion 19.1. In contrast to free tensors, tensor fields are related not to
bases, but to whole coordinate systems (including the origin). In each coordinate
system they are represented by functional arrays, i. e. by arrays of functions (see
(19.3)).
A functional array (19.3) is a coordinate representation of a tensor field (19.1).
What happens when we change the coordinate system ? Dealing with (19.2), we

32 CHAPTER III. TENSOR FIELDS. DIFFERENTIATION OF TENSORS.
need only to recalculate the components of the array Xi1... ir
j1... js
in the basis by means
of (12.2):
˜Xi1... ir
j1... js
(P) =
3
...
3
h1, ... , hr
k1, ... , ks
Ti1
h1
. . . Tir
hr
Sk1
j1
. . . Sks
js
Xh1... hr
k1... ks
(P). (19.4)
In the case of (19.3), we need to recalculate the components of the array Xi1... ir
j1... js
in the new basis
˜Xi1... ir
j1... js
(˜x1
, ˜x2
, ˜x3
) =
3
...
3
h1, ... , hr
k1, ... , ks
Ti1
h1
. . . Tir
hr
Sk1
j1
. . . Sks
js
Xh1... hr
k1... ks
(x1
, x2
, x3
), (19.5)
using (12.2), and we also need to express the old coordinates x1
, x2
, x3
of the
point P in right hand side of (19.5) through new coordinates of the same point:



x1
= x1
(˜x1
, ˜x2
, ˜x3
),
x2
= x2
(˜x1
, ˜x2
, ˜x3
),
x3
= x3
(˜x1
, ˜x2
, ˜x3
).
(19.6)
Like (12.2), formula (19.5) can be inverted by means of (12.1):
Xi1... ir
j1... js
(x1, x2, x3) =
3
...
3
h1, ... , hr
k1, ... , ks
Si1
h1
. . . Sir
hr
Tk1
j1
. . . Tks
js
˜Xh1... hr
k1... ks
(˜x1
, ˜x2
, ˜x3
). (19.7)
But now, apart from (19.7), we should have inverse formulas for (19.6) as well:



˜x1
= x1
(x1
, x2
, x3
),
˜x2
= x2
(x1
, x2
, x3
),
˜x3
= x3
(x1
, x2
, x3
).
(19.8)
THe couple of formulas (19.5) and (19.6), and another couple of formulas (19.7)
and (19.8), in the case of tensor fields play the same role as transformation
formulas (12.1) and (12.2) in the case of free tensors.
§ 20. Change of Cartesian coordinate system.
Note that formulas (19.6) and (19.8) are written in abstract form. They only
indicate the functional dependence of new coordinates of the point P from old
ones and vice versa. Now we shall specify them for the case when one Cartesian
coordinate system is changed to another Cartesian coordinate system. Remember
that each Cartesian coordinate system is determined by some basis and some fixed
point (the origin). We consider two Cartesian coordinate systems. Let the origins
of the first and second systems be at the points O and Õ, respectively. Denote by
e1, e2, e3 the basis of the first coordinate system, and by ˜e1, ˜e2, ˜e3 the basis of
the second coordinate system (see Fig. 7 below).
Let P be some point in the space for whose coordinates we are going to
derive the specializations of formulas (19.6) and (19.8). Denote by rP and ˜rP the

§ 20. CHANGE OF CARTESIAN COORDINATE SYSTEM. 33
radius-vectors of this point in our two
coordinate systems. Then rP =
−→
OP and
˜rP =
−→
ÕP. Hence
rP =
−→
O Õ + ˜rP . (20.1)
Vector
−→
O Õ determines the origin shift
from the old to the new coordinate sys-
tem. We expand this vector in the basis
e1, e2, e3:
a =
−→
O Õ =
3
i=1
ai
ei. (20.2)
Radius-vectors rP and ˜rP are expanded
in the bases of their own coordinate sys-
tems:
rP =
3
i=1
xi
ei,
˜rP =
3
i=1
˜xi
˜ei,
(20.3)
Exercise 20.1. Using (20.1), (20.2), (20.3), and (5.7) derive the following for-
mula relating the coordinates of the point P in the two coordinate systems in Fig. 7:
xi
= ai
+
3
j=1
Si
j ˜xj
. (20.4)
Compare (20.4) with (6.5). Explain the differences in these two formulas.
Exercise 20.2. Derive the following inverse formula for (20.4):
˜xi
= ãi
+
3
j=1
Ti
j xj
. (20.5)
Prove that ai
in (20.4) and ãi
in (20.5) are related to each other as follows:
ãi
= −
3
j=1
Ti
j aj
, ai
= −
3
j=1
Si
j ãj
. (20.6)
Compare (20.6) with (6.2) and (6.5). Explain the minus signs in these formulas.
Formula (20.4) can be written in the following expanded form:



x1
= S1
1 ˜x1
+ S1
2 ˜x2
+ S1
3 ˜x3
+ a1
,
x2
= S2
1 ˜x1
+ S2
2 ˜x2
+ S2
3 ˜x3
+ a2
,
x3
= S3
1 ˜x1
+ S3
2 ˜x2
+ S3
3 ˜x3
+ a3
.
(20.7)

This is the required specialization for (19.6). In a similar way we can expand
(20.5):



˜x1
= T1
1 x1
+ T1
2 x2
+ T1
3 x3
+ ã1
,
˜x2
= T2
1 x1
+ T2
2 x2
+ T2
3 x3
+ ã2
,
˜x3
= T3
1 x1
+ T3
2 x2
+ T3
3 x3
+ ã3
.
(20.8)
This is the required specialization for (19.8). Formulas (20.7) and (20.8) are used
to accompany the main transformation formulas (19.5) and (19.7).
§ 21. Differentiation of tensor fields.
In this section we consider two different types of derivatives that are usually
applied to tensor fields: differentiation with respect to spacial variables x1
, x2
, x3
and differentiation with respect to external parameters other than x1
, x2
, x3
, if
they are present. The second type of derivatives are simpler to understand. Let’s
consider them to start. Suppose we have tensor field X of type (r, s) and depending
on the additional parameter t (for instance, this could be a time variable). Then,
upon choosing some Cartesian coordinate system, we can write
∂Xi1... ir
j1... js
∂t
= lim
h→0
Xi1... ir
j1... js
(t + h, x1
, x2
, x3
) − Xi1... ir
j1... js
(t, x1
, x2
, x3
)
h
. (21.1)
The left hand side of (21.1) is a tensor since the fraction in right hand side is
constructed by means of tensorial operations (14.1) and (14.3). Passing to the limit
h → 0 does not destroy the tensorial nature of this fraction since the transition
matrices S and T in (19.5), (19.7), (20.7), (20.8) are all time-independent.
Conclusion 21.1. Differentiation with respect to external parameters (like t
in (21.1)) is a tensorial operation producing new tensors from existing ones.
Exercise 21.1. Give a more detailed explanation of why the time derivative
(21.1) represents a tensor of type (r, s).
Now let’s consider the spacial derivative of tensor field X, i. e. its derivative
with respect to a spacial variable, e. g. with respect to x1
. Here we also can write
∂Xi1... ir
j1... js
∂x1
= lim
h→0
Xi1... ir
j1... js
(x1
+ h, x2
, x3
) − Xi1... ir
j1... js
(x1
, x2
, x3
)
h
, (21.2)
but in numerator of the fraction in the right hand side of (21.2) we get the
difference of two tensors bound to different points of space: to the point P with
coordinates x1
, x2
, x3
and to the point P with coordinates x1
+ h, x2
, x3
. To
which point should be attributed the difference of two such tensors? This is not
clear. Therefore, we should treat partial derivatives like (21.2) in a different way.
Let’s choose some additional symbol, say it can be q, and consider the partial
derivative of Xi1... ir
j1... js
with respect to the spacial variable xq
:
Y i1... ir
q j1... js
=
∂Xi1... ir
j1... js
∂xq
. (21.3)
Partial derivatives (21.2), taken as a whole, form an (r + s + 1)-dimensional array

§ 22. GRADIENT, DIVERGENCY, AND ROTOR . . . 35
with one extra dimension due to index q. We write it as a lower index in Y i1... ir
q j1... js
due to the following theorem concerning (21.3).
Theorem 21.1. For any tensor field X of type (r, s) partial derivatives (21.3)
with respect to spacial variables x1
, x2
, x3
in any Cartesian coordinate system
represent another tensor field Y of the type (r, s + 1).
Thus differentiation with respect to x1
, x2
, x3
produces new tensors from
already existing ones. For the sake of beauty and convenience this operation is
denoted by the nabla sign: Y = X. In index form this looks like
Y i1... ir
q j1... js
= qXi1... ir
j1... js
. (21.4)
Simplifying the notations we also write:
q =
∂
∂xq
. (21.5)
Warning 21.1. Theorem 21.1 and the equality (21.5) are valid only for Carte-
sian coordinate systems. In curvilinear coordinates things are different.
Exercise 21.2. Prove theorem 21.1. For this purpose consider another Carte-
sian coordinate system ˜x1
, ˜x2
, ˜x3
related to x1
, x2
, x3
via (20.7) and (20.8). Then
in the new coordinate system consider the partial derivatives
˜Y i1... ir
q j1... js
=
∂ ˜Xi1... ir
j1... js
∂˜xq
(21.6)
and derive relationships binding (21.6) and (21.3).
§ 22. Gradient, divergency, and rotor.
Laplace and d’Alambert operators.
The tensorial nature of partial derivatives established by theorem 21.1 is a very
useful feature. We can apply it to extend the scope of classical operations of
vector analysis. Let’s consider the gradient, grad = . Usually the gradient
operator is applied to scalar fields, i. e. to functions ϕ = ϕ(P) or ϕ = ϕ(x1
, x2
, x3
)
in coordinate form:
aq = qϕ =
∂ϕ
∂xq
. (22.1)
Note that in (22.1) we used a lower index q for aq. This means that a = gradϕ
is a covector. Indeed, according to theorem 21.1, the nabla operator applied to
a scalar field, which is tensor field of type (0, 0), produces a tensor field of type
(0, 1). In order to get the vector form of the gradient one should raise index q:
aq
=
3
i=1
gqi
ai =
3
i=1
gqi
iϕ. (22.2)
Let’s write (22.2) in the form of a differential operator (without applying to ϕ):
q
=
3
i=1
gqi
i. (22.3)

In this form the gradient operator (22.3) can be applied not only to scalar fields,
but also to vector fields, covector fields and to any other tensor fields.
Usually in physics we do not distinguish between the vectorial gradient q
and
the covectorial gradient q because we use orthonormal coordinates with ONB as
a basis. In this case dual metric tensor is given by unit matrix (gij
= δij
) and
components of q
and q coincide by value.
Divergency is the second differential operation of vector analysis. Usually it
is applied to a vector field and is given by formula:
div X =
3
i=1
iXi
. (22.4)
As we see, (22.4) is produced by contraction (see section 16) from tensor qXi
.
Therefore we can generalize formula (22.4) and apply divergency operator to
arbitrary tensor field with at least one upper index:
(div X)... ... ...
... ... ... =
3
s=1
sX... .s. ...
... ... ... . (22.5)
The Laplace operator is defined as the divergency applied to a vectorial gradient
of something, it is denoted by the triangle sign: = div grad. From (22.3) and
(22.5) for Laplace operator we derive the following formula:
=
3
i=1
3
j=1
gij
i j. (22.6)
Denote by the following differential operator:
=
1
c2
∂2
∂t2
− . (22.7)
Operator (22.7) is called the d’Alambert operator or wave operator. In
general relativity upon introducing the additional coordinate x0
= c t one usually
rewrites the d’Alambert operator in a form quite similar to (22.6) (see my book
[5], it is free for download from http://samizdat.mines.edu/).
And finally, let’s consider the rotor operator or curl operator (the term
“rotor” is derived from “rotation” so that “rotor” and “curl” have approximately
the same meaning). The rotor operator is usually applied to a vector field and
produces another vector field: Y = rot X. Here is the formula for the r-th
coordinate of rot X:
(rot X)r
=
3
i=1
3
j=1
3
k=1
gri
ωijk
j
Xk
. (22.8)
The volume tensor ω in (22.8) is given by formula (18.4), while the vectorial
gradient operator j
is defined in (22.3).

§ 22. GRADIENT, DIVERGENCY, AND ROTOR . . . 37
Exercise 22.1. Formula (22.8) can be generalized for the case when X is an
arbitrary tensor ﬁeld with at least one upper index. By analogy with (22.5) suggest
your version of such a generalization.
Note that formulas (22.6) and (22.8) for the Laplace operator and for the rotor
are diﬀerent from those that are commonly used. Here are standard formulas:
=
∂
∂x1
2
+
∂
∂x2
2
+
∂
∂x3
2
, (22.9)
rot X = det
e1 e2 e3
∂
∂x1
∂
∂x2
∂
∂x3
X1
X2
X3
. (22.10)
The truth is that formulas (22.6) and (22.8) are written for a general skew-angular
coordinate system with a SAB as a basis. The standard formulas (22.10) are valid
only for orthonormal coordinates with ONB as a basis.
Exercise 22.2. Show that in case of orthonormal coordinates, when gij
= δij
,
formula (22.6) for the Laplace operator reduces to the standard formula (22.9).
The coordinates of the vector rot X in a skew-angular coordinate system are
given by formula (22.8). Then for vector rot X itself we have the expansion:
rot X =
3
r=1
(rot X)r
er. (22.11)
Exercise 22.3. Substitute (22.8) into (22.11) and show that in the case of a
orthonormal coordinate system the resulting formula (22.11) reduces to (22.10).

CHAPTER IV
TENSOR FIELDS IN CURVILINEAR COORDINATES.
§ 23. General idea of curvilinear coordinates.
What are coordinates, if we forget for a moment about radius-vectors, bases
and axes? What is the pure idea of coordinates? The pure idea is in representing
points of space by triples of numbers. This means that we should have one to
one map P (y1
, y2
, y3
) in the whole space or at least in some domain, where
we are going to use our coordinates y1
, y2
, y3
. In Cartesian coordinates this
map P (y1
, y2
, y3
) is constructed by means of vectors and bases. Arranging
other coordinate systems one can use other methods. For example, in spherical
coordinates y1
= r is a distance from the point P to the center of sphere, y2
= θ
and y3
= ϕ are two angles. By the way, spherical coordinates are one of the
simplest examples of curvilinear coordinates. Furthermore, let’s keep in mind
spherical coordinates when thinking about more general and hence more abstract
curvilinear coordinate systems.
§ 24. Auxiliary Cartesian coordinate system.
Now we know almost everything about Cartesian coordinates and almost noth-
ing about the abstract curvilinear coordinate system y1
, y2
, y3
that we are going
to study. Therefore, the best idea is to represent each point P by its radius-
vector rP in some auxiliary Cartesian coordinate system and then consider a map
rP (y1
, y2
, y3
). The radius-vector itself is represented by three coordinates in
the basis e1, e2, e3 of the auxiliary Cartesian coordinate system:
rP =
3
i=1
xi
ei. (24.1)
Therefore, we have a one-to-one map (x1
, x2
, x3
) (y1
, y2
, y3
). Hurrah! This is a
numeric map. We can treat it numerically. In the left direction it is represented
by three functions of three variables:



x1
= x1
(y1
, y2
, y3
),
x2
= x2
(y1
, y2
, y3
),
x3
= x3
(y1
, y2
, y3
).
(24.2)
In the right direction we again have three functions of three variables:



y1
= y1
(x1
, x2
, x3
),
y2
= y2
(x1
, x2
, x3
),
y3
= y3
(x1
, x2
, x3
).
(24.3)

§ 25. COORDINATE LINES AND THE COORDINATE GRID. 39
Further we shall assume all functions in (24.2) and (24.3) to be differentiable and
consider their partial derivatives. Let’s denote
Si
j =
∂xi
∂yj
, Ti
j =
∂yi
∂xj
. (24.4)
Partial derivatives (24.4) can be arranged into two square matrices S and T
respectively. In mathematics such matrices are called Jacobi matrices. The
components of matrix S in that form, as they are defined in (24.4), are functions
of y1
, y2
, y3
. The components of T are functions of x1
, x2
, x3
:
Si
j = Si
j(y1
, y2
, y3
), Ti
j = Ti
j (x1
, x2
, x3
). (24.5)
However, by substituting (24.3) into the arguments of Si
j, or by substituting (24.2)
into the arguments of T i
j , we can make them have a common set of arguments:
Si
j = Si
j(x1
, x2
, x3
), Ti
j = Ti
j (x1
, x2
, x3
), (24.6)
Si
j = Si
j(y1
, y2
, y3
), Ti
j = Ti
j (y1
, y2
, y3
), (24.7)
When brought to the form (24.6), or when brought to the form (24.7) (but not in
form of (24.5)), matrices S and T are inverse of each other:
T = S−1
. (24.8)
This relationship (24.8) is due to the fact that numeric maps (24.2) and (24.3) are
inverse of each other.
Exercise 24.1. You certainly know the following formula:
df(x1
(y), x2
(y), x3
(y))
dy
=
3
i=1
fi (x1
(y), x2
(y), x3
(y))
dxi
(y)
dy
, where fi =
∂f
∂xi
.
It’s for the differentiation of composite function. Apply this formula to functions
(24.2) and derive the relationship (24.8).
§ 25. Coordinate lines and the coordinate grid.
Let’s substitute (24.2) into (24.1) and take into account that (24.2) now assumed
to contain differentiable functions. Then the vector-function
R(y1
, y2
, y3
) = rP =
3
i=1
xi
(y1
, y2
, y3
) ei (25.1)
is a differentiable function of three variables y1
, y2
, y3
. The vector-function
R(y1
, y2
, y3
) determined by (25.1) is called a basic vector-function of a curvi-
linear coordinate system. Let P0 be some fixed point of space given by its
curvilinear coordinates y1
0, y2
0, y3
0. Here zero is not the tensorial index, we use it
in order to emphasize that P0 is fixed point, and that its coordinates y1
0, y2
0, y3
0

40 CHAPTER IV. TENSOR FIELDS IN CURVILINEAR COORDINATES.
are three fixed numbers. In the next step let’s undo one of them, say first one, by
setting
y1
= y1
0 + t, y2
= y2
0, y3
= y3
0. (25.2)
Substituting (25.2) into (25.1) we get a vector-function of one variable t:
R1(t) = R(y1
0 + t, y2
0, y3
0), (25.3)
If we treat t as time variable (though it
may have a unit other than time), then
(25.3) describes a curve (the trajectory
of a particle). At time instant t = 0
this curve passes through the fixed point
P0. Same is true for curves given by two
other vector-functions similar to (25.4):
R2(t) = R(y1
0, y2
0 + t, y3
0), (25.4)
R3(t) = R(y1
0, y2
0, y3
0 + t). (25.5)
This means that all three curves given by
vector-functions (25.3), (25.4), and (25.5)
are intersected at the point P0 as shown
on Fig. 8. Arrowheads on these lines
indicate the directions in which parameter t increases. Curves (25.3), (25.4),
and (25.5) are called coordinate lines. They are subdivided into three families.
Curves within one family do not intersect each other. Curves from different
families intersect so that any regular point of space is an intersection of exactly
three coordinate curves (one per family).
Coordinate lines taken in whole form
a coordinate grid. This is an infinitely
dense grid. But usually, when drawing,
it is represented as a grid with finite
density. On Fig. 9 the coordinate grid
of curvilinear coordinates is compared to
that of the Cartesian coordinate system.
Another example of coordinate grid is
on Fig. 2. Indeed, meridians and par-
allels are coordinate lines of a spherical
coordinate system. The parallels do not
intersect, but the meridians forming one
family of coordinate lines do intersect at the North and at South Poles. This
means that North and South Poles are singular points for spherical coordinates.
Exercise 25.1. Remember the exact definition of spherical coordinates and
find all singular points for them.

§ 26. MOVING FRAME OF CURVILINEAR COORDINATES. 41
§ 26. Moving frame of curvilinear coordinates.
Let’s consider the three coordinate lines shown on Fig. 8 again. And let’s find
tangent vectors to them at the point P0. For this purpose we should differentiate
vector-functions (25.3), (25.4), and (25.5) with respect to the time variable t and
then substitute t = 0 into the derivatives:
Ei =
dRi
dt t=0
=
∂R
∂yi
at the point P0
. (26.1)
Now let’s substitute the expansion (25.1) into (26.1) and remember (24.4):
Ei =
∂R
∂yi
=
3
j=1
∂xj
∂yi
ej =
3
j=1
Sj
i ej. (26.2)
All calculations in (26.2) are still in reference to the point P0. Though P0 is a
fixed point, it is an arbitrary fixed point. Therefore, the equality (26.2) is valid at
any point. Now let’s omit the intermediate calculations and write (26.2) as
Ei =
3
i=1
Sj
i ej. (26.3)
And then compare (26.3) with (5.7). They are strikingly similar, and det S = 0
due to (24.8). Formula (26.3) means that tangent vectors to coordinate lines
E1, E2, E3 form a basis (see Fig. 10),
matrices (24.4) are transition matrices
to this basis and back to the Cartesian
basis.
Despite obvious similarity of the for-
mulas (26.3) and (5.7), there is some
crucial difference of basis E1, E2, E3 as
compared to e1, e2, e3. Vectors E1, E2,
E3 are not free. They are bound to that
point where derivatives (24.4) are calcu-
lated. And they move when we move this
point. For this reason basis E1, E2, E3
is called moving frame of the curvi-
linear coordinate system. During their
motion the vectors of the moving frame
E1, E2, E3 are not simply translated
from point to point, they can change
their lengths and the angles they form with each other. Therefore, in general
the moving frame E1, E2, E3 is a skew-angular basis. In some cases vectors
E1, E2, E3 can be orthogonal to each other at all points of space. In that case we
say that we have an orthogonal curvilinear coordinate system. Most of the well
known curvilinear coordinate systems are orthogonal, e. g. spherical, cylindrical,
elliptic, parabolic, toroidal, and others. However, there is no curvilinear coordinate
system with the moving frame being ONB ! We shall not prove this fact since it
leads deep into differential geometry.

§ 27. Dynamics of moving frame.
Thus, we know that the moving frame moves. Let’s describe this motion
quantitatively. According to (24.5) the components of matrix S in (26.3) are
functions of the curvilinear coordinates y1
, y2
, y3
. Therefore, differentiating Ei
with respect to yj
we should expect to get some nonzero vector ∂Ei/∂yj
. This
vector can be expanded back in moving frame E1, E2, E3. This expansion is
written as
∂Ei
∂yj
=
3
k=1
Γk
ij Ek. (27.1)
Formula (27.1) is known as the derivational formula. Coefficients Γk
ij in (27.1)
are called Christoffel symbols or connection components.
Exercise 27.1. Relying upon formula (25.1) and (26.1) draw the vectors of the
moving frame for cylindrical coordinates.
Exercise 27.2. Do the same for spherical coordinates.
Exercise 27.3. Relying upon formula (27.1) and results of exercise 27.1 calcu-
late the Christoffel symbols for cylindrical coordinates.
Exercise 27.4. Do the same for spherical coordinates.
Exercise 27.5. Remember formula (26.2) from which you derive
Ei =
∂R
∂yi
. (27.2)
Substitute (27.2) into left hand side of the derivational formula (27.1) and relying on
the properties of mixed derivatives prove that the Christoffel symbols are symmetric
with respect to their lower indices: Γk
ij = Γk
ji.
Note that Christoffel symbols Γk
ij form a three-dimensional array with one
upper index and two lower indices. However, they do not represent a tensor. We
shall not prove this fact since it again leads deep into differential geometry.
§ 28. Formula for Christoffel symbols.
Let’s take formula (26.3) and substitute it into both sides of (27.1). As a result
we get the following equality for Christoffel symbols Γk
ij :
3
q=1
∂Sq
i
∂yj
eq =
3
k=1
3
q=1
Γk
ij Sq
k eq. (28.1)
Cartesian basis vectors eq do not depend on yj
; therefore, they are not differen-
tiated when we substitute (26.3) into (27.1). Both sides of (28.1) are expansions
in the base e1, e2, e3 of the auxiliary Cartesian coordinate system. Due to the
uniqueness of such expansions we have the following equality derived from (28.1):
∂Sq
i
∂yj
=
3
k=1
Γk
ij Sq
k. (28.2)

§ 29. TENSOR FIELDS IN CURVILINEAR COORDINATES. 43
Exercise 28.1. Using concept of the inverse matrix ( T = S−1
) derive the
following formula for the Christoffel symbols Γk
ij from (28.2):
Γk
ij =
3
q=1
Tk
q
∂Sq
i
∂yj
. (28.3)
Due to (24.4) this formula (28.3) can be transformed in the following way:
Γk
ij =
3
q=1
Tk
q
∂Sq
i
∂yj
=
3
q=1
Tk
q
∂2
xq
∂yi ∂yj
=
3
q=1
Tk
q
∂Sq
j
∂yi
. (28.4)
Formulas (28.4) are of no practical use because they express Γk
ij through an exter-
nal thing like transition matrices to and from the auxiliary Cartesian coordinate
system. However, they will help us below in understanding the differentiation of
tensors.
§ 29. Tensor fields in curvilinear coordinates.
As we remember, tensors are geometric objects related to bases and represented
by arrays if some basis is specified. Each curvilinear coordinate system provides us
a numeric representation for points, and in addition to this it provides the basis.
This is the moving frame. Therefore, we can refer tensorial objects to curvilinear
coordinate systems, where they are represented as arrays of functions:
Xi1... ir
j1... js
= Xi1... ir
j1... js
(y1
, y2
, y3
). (29.1)
We also can have two curvilinear coordinate systems and can pass from one to
another by means of transition functions:



˜y1
= ˜y1
(y1
, y2
, y3
),
˜y2
= ˜y2
(y1
, y2
, y3
),
˜y3
= ˜y3
(y1
, y2
, y3
),



y1
= y1
(˜y1
, ˜y2
, ˜y3
),
y2
= y2
(˜y1
, ˜y2
, ˜y3
),
y3
= y3
(˜y1
, ˜y2
, ˜y3
).
(29.2)
If we call ˜y1
, ˜y2
, ˜y3
the new coordinates, and y1
, y2
, y3
the old coordinates, then
transition matrices S and T are given by the following formulas:
Si
j =
∂yi
∂˜yj
, Ti
j =
∂˜yi
∂yj
. (29.3)
They relate moving frames of two curvilinear coordinate systems:
˜Ei =
3
j=1
S j
i Ej, Ej =
3
i=1
Ti
j
˜Ei. (29.4)

Exercise 29.1. Derive (29.3) from (29.4) and (29.2) using some auxiliary Carte-
sian coordinates with basis e1, e2, e3 as intermediate coordinate system:
(E1, E2, E3)
S
←−−−−
−−−−→
T
(e1, e2, e3)
˜S
−−−−→
←−−−−
˜T
(˜E1, ˜E2, ˜E3) (29.5)
Compare (29.5) with (5.13) and explain differences you have detected.
Transformation formulas for tensor fields for two curvilinear coordinate systems
are the same as in (19.4) and (19.5):
˜Xi1... ir
j1... js
(˜y1
, ˜y2
, ˜y3
) =
3
...
3
h1, ... , hr
k1, ... , ks
Ti1
h1
. . . Tir
hr
Sk1
j1
. . . Sks
js
Xh1... hr
k1... ks
(y1
, y2
, y3
), (29.6)
Xi1... ir
j1... js
(y1, y2, y3) =
3
...
3
h1, ... , hr
k1, ... , ks
Si1
h1
. . . Sir
hr
Tk1
j1
. . . Tks
js
˜Xh1... hr
k1... ks
(˜y1
, ˜y2
, ˜y3
). (29.7)
But formulas (19.6) and (19.8) should be replaced by (29.2).
§ 30. Differentiation of tensor fields
in curvilinear coordinates.
We already know how to differentiate tensor fields in Cartesian coordinates (see
section 21). We know that operator produces tensor field of type (r, s + 1)
when applied to a tensor field of type (r, s). The only thing we need now is to
transform to a curvilinear coordinate system. In order to calculate tensor X in
curvilinear coordinates, let’s first transform X into auxiliary Cartesian coordinates,
then apply , and then transform X back into curvilinear coordinates:
Xh1... hr
k1... ks
(y1
, y2
, y3
)
S,T
−−−−→ Xh1... hr
k1... ks
(x1
, x2
, x3
)


p


q=∂/∂xq
pXi1... ir
j1... js
(y1
, y2
, y3
)
T,S
←−−−− qXh1... hr
k1... ks
(x1
, x2
, x3
)
(30.1)
Matrices (24.4) are used in (30.1). From (12.3) and (12.4) we know that the
transformation of each index is a separate multiplicative procedure. When applied
to the α-th upper index, the whole chain of transformations (30.1) looks like
pX... iα ...
... ... ... =
3
q=1
Sq
p . . .
3
hα=1
T iα
hα
. . . q . . .
3
mα=1
S hα
mα
. . . X... mα ...
... ... ... . (30.2)
Note that q = ∂/∂xq
is a differential operator and due to (24.4) we have
3
q=1
Sq
p
∂
∂xq
=
∂
∂yp
. (30.3)

§ 30. DIFFERENTIATION OF TENSOR FIELDS IN CURVILINEAR COORDINATES. 45
Any diﬀerential operator when applied to a product produces a sum with as
many summands as there were multiplicand in the product. Here is the summand
produced by term S hα
mα
in formula (30.2):
pX... iα ...
... ... ... = . . . +
3
mα=1
3
hα=1
T iα
hα
S hα
mα
∂yp
X... mα ...
... ... ... + . . . . (30.4)
Comparing (30.4) with (28.3) or (28.4) we can transform it into the following
equality:
pX... iα ...
... ... ... = . . . +
3
mα=1
Γiα
pmα
X... mα ...
... ... ... + . . . . (30.5)
Now let’s consider the transformation of the α-th lower index in (30.1):
pX... ... ...
... jα ... =
3
q=1
Sq
p . . .
3
kα=1
Skα
jα
. . . q . . .
3
nα=1
Tnα
kα
. . . X... ... ...
... nα .... (30.6)
Applying (30.3) to (30.6) with the same logic as in deriving (30.4) we get
pX... ... ...
... jα ... = . . . +
3
nα=1
3
kα=1
Skα
jα
Tnα
kα
∂yp
X... ... ...
... nα ... + . . . . (30.7)
In order to simplify (30.7) we need the following formula derived from (28.3):
Γk
ij = −
3
q=1
Sq
i
∂Tk
q
∂yj
. (30.8)
Applying (30.8) to (30.7) we obtain
pX... ... ...
... jα ... = . . . −
3
nα=1
Γnα
pjα
X... ... ...
... nα ... + . . . . (30.9)
Now we should gather (30.5), (30.9), and add the term produced when q in (30.2)
(or equivalently in (30.4)) acts upon components of tensor X. As a result we get
the following general formula for pXi1... ir
j1... js
:
pXi1... ir
j1... js
=
∂Xi1... ir
j1... js
∂yp
+
r
α=1
3
mα=1
Γiα
pmα
Xi1... mα ... ir
j1... ... ... js
−
−
s
α=1
3
nα=1
Γnα
pjα
Xi1... ... ... ir
j1... nα ... js
.
(30.10)
The operator p determined by this formula is called the covariant derivative.
Exercise 30.1. Apply the general formula (30.10) to a vector ﬁeld and calculate
the covariant derivative pXq
.

Exercise 30.2. Apply the general formula (30.10) to a covector field and cal-
culate the covariant derivative pXq.
Exercise 30.3. Apply the general formula (30.10) to an operator field and find
pFq
m. Consider special case when p is applied to the Kronecker symbol δq
m.
Exercise 30.4. Apply the general formula (30.10) to a bilinear form and find
paqm.
Exercise 30.5. Apply the general formula (30.10) to a tensor product a ⊗ x
for the case when x is a vector and a is a covector. Verify formula (a ⊗ x) =
a ⊗ x + a ⊗ x.
Exercise 30.6. Apply the general formula (30.10) to the contraction C(F) for
the case when F is an operator field. Verify the formula C(F) = C( F).
Exercise 30.7. Derive (30.8) from (28.3).
§ 31. Concordance of metric and connection.
Let’s remember that we consider curvilinear coordinates in Euclidean space E.
In this space we have the scalar product (13.1) and the metric tensor (13.5).
Exercise 31.1. Transform the metric tensor (13.5) to curvilinear coordinates
using transition matrices (24.4) and show that here it is given by formula
gij = (Ei, Ej). (31.1)
In Cartesian coordinates all components of the metric tensor are constant since
the basis vectors e1, e2, e3 are constant. The covariant derivative (30.10) in
Cartesian coordinates reduces to differentiation p = ∂/∂xp
. Therefore,
pgij = 0. (31.2)
But g is a tensor. If all of its components in some coordinate system are
zero, then they are identically zero in any other coordinate system (explain why).
Therefore the identity (31.2) is valid in curvilinear coordinates as well.
Exercise 31.2. Prove (31.2) by direct calculations using formula (27.1).
The identity (31.2) is known as the concordance condition for the metric gij
and connection Γk
ij . It is very important for general relativity.
Remember that the metric tensor enters into many useful formulas for the
gradient, divergency, rotor, and Laplace operator in section 22. What is important
is that all of these formulas remain valid in curvilinear coordinates, with the only
difference being that you should understand that p is not the partial derivative
∂/∂xp
, but the covariant derivative in the sense of formula (30.10).
Exercise 31.3. Calculate rot A, div H, grad ϕ (vectorial gradient) in cylindrical
and spherical coordinates.
Exercise 31.4. Calculate the Laplace operator ϕ applied to the scalar field
ϕ in cylindrical and in spherical coordinates.

REFERENCES.
1. Heﬀeron J., Linear algebra, Electronic textbook, free for downloading from Web site of Saint
Michael’s College, Colchester, VM 05439, USA; Download [PDF] or [PS] ﬁle.
2. Lehnen A. P., An elementary introduction to logic and set theory, On-line resource, Madison
Area Technical College, Madison, WI 53704, USA.
3. Konstantopoulos T., Basic background for a course of information and cryptography, On-
line materials, Feb. 2000, Electrical & Computer Engineering Dept., University of Texas at
Austin, Austin, Texas 78712, USA.
4. Vaughan-Lee M., B2 ring theory preliminaries, On-line lecture materials, Sept. 2000, Univer-
sity of Oxford, Math. Institute, Oxford, OX1 3LB, UK.
5. Sharipov R. A., Classical electrodynamics and theory of relativity, Bashkir State University,
Ufa, Russia, 1997; English tr., 2003, physics/0311011 in Electronic Archive http://arXiv.org.

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