of 10 ## Content

Gerardo Urrutia S´anchez
∗1
1
04510, M´exico
Noviembre 2013
Abstract
The paper ”Tensors: A guide for undergraduate students ”, AIP, 81, 498 (2013); doi: 10.1119/1.4802811
is an interesting publication aimed for students of Physics. But some treatments seem inecesarios and
generalize a all metric space. Also are considered the most important operators in physics for make the
transformation and derivates.
1 Introduction
The topic of tensor calculus is dark for underdgrad-
uate students. I am aware that there is much lit-
erature on the subject. But sometimes the concepts
are not clear enough. Perhaps it’s easier to take con-
cepts that underdgraduate students relate to then go
including more sophisticated concepts.
A ﬁrst idea is to change the notation to denote a
vector with multiple entries.

A −→
_
A
0
, A
1
, A
2
, . . . , A
N
_
= {A
i
} (1)
This tiny change in notation might seem extravagant
and unnatural but, as we shall see, it greatly simpli-
ﬁes the manipulations involved. A common notation
that we adopt is the Einstein summation rule. As
we will see, we need two types of indices that we
distinguish from each other by locating them either
in a lower position, as in A
i
or in an upper position,
as in A
a
. The Einstein summation rule is:
Rule 0: Whenever an index is repeated twice in a
product-once in an upper position and once in a
lower position-the index is called a dummy index
and summation over it is implied.
N

i=0
A
i
ˆ e
i
= A
i
ˆ e
i
(2)
A
ij
B
j
k
= A
i0
B
0
k
+ A
i1
B
1
k
+ A
i2
B
2
k
+· · · + A
iN
B
N
k
(3)
and
C
i
i
= C
0
0
+ C
1
1
+ C
2
2
+· · · + C
N
N
(4)
An index that is not a dummy index is called a free

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1
index. Dummy and free indices follow three rules
that, as trivial as they might sound, are so crucial
and used continuously that we better state them ex-
plicitly:
Rule 1: In the case of a multiple sum, diﬀerent let-
ters must denote the dummy indices A
ij
B
ij
. Its im-
plique a double sum over the two indices.
A
ij
B
ij
= A
0j
B
0j
+ A
1j
B
1j
+· · · + A
Nj
B
Nj
= A
00
B
00
+ A
01
B
01
+· · · + A
0N
B
0N
+A
10
B
10
+ A
11
B
11
+· · · + A
1N
B
1N
.
.
.
+A
N0
B
N0
+ A
N1
B
N1
+· · · + A
NN
B
NN
(5)
Rule 2 A dummy index may be renamed, at will and
as convenience requires, within any single factor (as
long there is no conﬂict with Rule 1).
A
ij
B
i
= A
kj
B
k
(6)
Rule 3 Any free index must appear with the same
name and position on each side of an equation; there-
after, if one wishes to rename it, it must be done
throughout the entire equation.
2 Review Orthonormal Bases
Let {ˆ e
i
; i = 1, 2, 3} be an orthonormal basis span-
ning the vectors of the ordinary Euclidean three-
dimensional (3D) space. The vectors may be associa-
tively and commutatively summed, there is a unique
zero-vector, and each vector has its unique opposite.
The vectors may be associatively multiplied by real
numbers, an operation that is distributive with re-
spect to both the sum of real numbers and the sum of
vectors. Between any two vectors is deﬁned a dot (or
scalar) product, a commutative rule that associates
a real number to each pair of vectors. This dot-
product rule is distributive with respect to a linear
combination of vectors and provides a non-negative
number whenever a vector is dotted with itself (only
the zero-vector dotted with itself gives the number
zero). A set of vectors {A
i
; i = 1, 2, . . . , N} is said
to be linearly independent if (note that a sum over
n is implied).
C
i
A
i
= 0 ⇒ C
i
= 0 ∀i (7)
In general, if N is the maximum allowed number of
linearly independent vectors, then the space is said
to be N-dimensional and, in this case, N linearly in-
dependent vectors are said to form a basisany vector
may be written as a linear combination of the basis.

V ˆ e
i
(8)
In our ordinary 3D space a basis of vectors may al-
ways be chosen to be orthonormal. ie.
ˆ e
i
· ˆ e
j
= δ
j
i
(9)
where δ
j
i
is the is the Kronecker delta, a quantity
whose 9 components are equal to 0 if i = j or equal
to 1 if i = j. The properties of the dot product is
A
i
= ˆ e
i

A (10)
and

B = (A
i
ˆ e
i
) · (B
j
ˆ e
j
) = (A
i
B
j
) (ˆ e
i
· ˆ e
j
)
= (A
i
B
j
) δ
j
i
= A
i
B
i
(11)
As done here, and in several calculations that fol-
low, we shall redundantly collect some factors within
parentheses with the aim of clarifying the algebra in-
volved without the need for any other explanation.
In the 3D space, a cross (or vector) product is de-
ﬁned as

B ≡ det
_
_
ˆ e
1
ˆ e
2
ˆ e
3
A
1
A
2
A
3
B
1
B
2
B
3
_
_
=
ijk
A
i
B
j
ˆ e
c
(12)
2
where
ijk
is the Levi-Civita symbol a quantity
whose 27 components are equal to +1 or −1 accord-
ing to whether (i, j, k) forms an even or odd permu-
tation of the sequence (1, 2, 3) and equal to 0 other-
wise (if two or more indices take on the same value).
A ﬁeld is a function of the space coordinates x ≡
(x
1
, x
2
, x
3
), and possibly of time as well, and may
be a scalar or a vector ﬁeld depending on whether
the function is a scalar or a vector. In this section,
we denote the diﬀerential operator by ∂
i
≡ ∂/∂x
i
.
Given a vector ﬁeld A = A(x) = A
i
(x) ˆ e
i
its diver-
gence is the scalar ﬁeld given by
∇· A = ∂
i
A
i
(13)
and its curl is the vector ﬁeld given by
∇×A ≡ det
_
_
ˆ e
1
ˆ e
2
ˆ e
3

1

2

3
A
1
A
2
A
3
_
_
=
ijk

i
A
j
ˆ e
c
(14)
Given a scalar ﬁeld φ(x) its gradient is the vector
ﬁeld deﬁned by
∇φ(x) = ˆ e
i

i
φ (15)
and its Laplacian is deﬁned as the divergence of its
φ ≡ ∇
2
φ ≡= ∇· ∇φ = ∂
i

i
φ (16)
Why do we need scalars, vectors, and, in general,
tensors? What we need in deﬁning a physical quan-
tity is for it to have some character of objectivity
in the sense that it does not have to depend on
the coordinate system used. For instance, a one-
component function of position, such as a tempera-
ture ﬁeld specifying the temperature T at each point
P of space, must provide a unique value T when the
point P is considered, regardless of the coordinate
system used.
Not every quantity speciﬁed by a single number is a
scalar. For example, in our 3D space whose points
are parameterized by a Cartesian coordinate system,
all displacements from a given point, such as the ori-
gin, will be speciﬁed by three quantities (x
1
, x
2
, x
3
).
Now, the displacement that in one coordinate sys-
tem is speciﬁed, for instance, as (1, 0, 0), would be
speciﬁed as (0, 1, 0) in a coordinate system rotated
clockwise by π/2 around the third axis. Thus, each
component of the displacement is indeed a single-
component quantity, yet these components depend
on the coordinate system used (they are not scalars)
and do not have that objectivity character we desire.
An objectivity-preserving, single-component quan-
tity must transform in a speciﬁc way under a coor-
dinate transformationit must be an invariant.
Likewise, multi-component quantities must have an
objectivity character (and are thereby called vectors
and tensors), a circumstance that translates into spe-
ciﬁc rules about how their components must trans-
form in order to preserve that objectivity. As not
all single-component quantities are scalars, similarly,
not all multi-component quantities are vectors or
tensors.
3 Arbitrary Bases: Duality
We consider a set of basis vectors {ˆ e
i
; i = 1, 2, 3} for
which
g
ij
≡ ˆ e
i
· ˆ e
j
= δ
j
i
(17)
It is appropriate to remark at this point that the
(symmetric) matrix [g
ij
], whose elements are g
ij
, has
a determinant G ≡ det [g
ij
] = 0. In fact, given {ˆ e
i
}
as a basis its vectors are linearly independent so that
C
i
ˆ e
j
= 0 ⇒C
i
= 0 ∀i (18)
But this equation implies that
_
C
i
ˆ e
i
_
· ˆ e
j
= C
i
(ˆ e
1
· ˆ e
j
) = C
i
g
ij
= 0 ⇒C
1
= 0 ∀i
(19)
However, a homogeneous linear equation such as
C
i
g
ij
= 0 possesses the trivial solution
_
C
i
= 0
_
3
if and only if det [g
ij
] = 0. (Incidentally, this im-
plies that det [g
ij
] is a non-singular matrix; in other
words, it admits an inverse.) We shall denote the
elements of the inverse matrix by g
ij
, so that
[g
ij
]
−1

_
g
ij
¸
(20)
which is equivalent to writing
[g
ij
]
_
g
ij
¸
= I (21)
or
g
ij
g
ij
= δ
j
i
(22)
where I is the identity matrix and the indices in the
Kronecker delta have been written according to our
Rule 3. Clearly
_
g
ij
¸
also symmetric and
det
_
g
ij
¸
= 1/ det [g
ij
] = 1/G (23)
a result that follows from eq (23) and the fact that
the determinant of a product of matrices is equal to
the product of their determinants. Any vector can
be expressed as a linear combination of the basis {ˆ e
i
}
so that

A = A
i
ˆ e
i
. Now
ˆ e
i
·

A = e
i
·
_
A
j
ˆ e
j
_
= (ˆ e
i
· ˆ e
j
) A
j
= g
ij
A
j
= δ
i
j
A
j
= A
i
(24)
and

B =
_
A
i
ˆ e
i
_
·
_
B
j
ˆ e
j
_
= (ˆ e
i
· ˆ e
j
) A
i
B
j
= g
ij
A
i
B
j
= δ
j
i
A
i
B
j
= A
j
B
j
(25)
In conclusion, A
i
= ˆ e
i
·

A and A · B = A
j
B
j
when
we are not in Euclidean space. A natural question to
ask is whether there exists some other basis, which
we provisionally denote by {e
i
}, such that
A
i
= ˆ e
i
·

A (26)
Although this provisional notion is motivated by our
desire to follow Rule 3, we will see that such a basis
does exist and is unique. Moreover, if we call {e
i
}
the dual basis of the original basis {e
i
}, it turns out
that the dual of {e
i
} is simply {e
i
}. Now
A
i
= ˆ e
i
·

A = ˆ e
i
·
_
A
j
ˆ e
j
_
=
_
ˆ e
i
· ˆ e
j
_
A
j
(27)
which holds provided the vectors {e
i
} are solutions
of the equation
_
ˆ e
i
· ˆ e
j
_
= δ
i
j
(28)
If the vectors {e
i
} exist they must be expressible
as a linear combination of the basis {ˆ e
i
}. Therefore
ˆ e
i
= C
ik
ˆ e
k
and the left-hand side of eq (28) becomes
_
C
ik
ˆ e
k
_
· ˆ e
j
= C
ik
(ˆ e
k
· ˆ e
j
) = C
ik
g
ij
, which means
c
ik
g
ij
= δ
i
j
(29)
Because the matrix [g
ij
] is non-singular. The eq(29)
has the solution
_
c
iK
¸
= [g
ij
]
−1
=
_
g
ij
¸
, which
means that the unique solution of eq. (28) is given
by
ˆ e
i
= g
ij
ˆ e
j
(30)
The basis {e
i
} is called the dual basis of {e
i
} and
is obtained from the latter by transforming it via
the matrix inverse of [ˆ e
i
· ˆ e
j
]. Similarly, the dual ba-
sis of the basis {e
i
} is obtained from the latter by
transforming it via the matrix inverse of
_
ˆ e
i
· ˆ e
j
¸
To
understand this, we note that Eqs. (29) and (27)
ˆ e
i
· ˆ e
j
=
_
g
ik
ˆ e
k
_
· ˆ e
j
=
_
ˆ e
k
· ˆ e
j
_
= g
ik
δ
j
k
= g
ij
(31)
so that
g
ij
= ˆ e
i
· ˆ e
j
(32)
so that
4
g
ij
= ˆ e
i
· ˆ e
j
(33)
whereby
_
ˆ e
i
· ˆ e
j
¸
−1

_
g
ij
¸
−1
= [g
ij
]. Thus, the vec-
tor dual to ˆ e
i
is
g
ij
ˆ e
j
= g
ij
_
g
jk
ˆ e
k
_
=
_
g
ij
g
jk
_
ˆ e
k
= δ
k
i
ˆ e
k
= ˆ e
i
(34)
In other words, if the basis dual to {e
i
} is {e
i
}, then
the casis dual to the latter is {e
i
}:
{ˆ e
i
= g
ij
ˆ e
j
} ⇐⇒{ˆ e
i
= g
ij
ˆ e
j
} (35)
We therefore see that once a non-orthonormal ba-
sis is considered, another basisits dualnaturally
emerges. Accordingly, any vector A may then be
written as

A = A
j
ˆ e
j
= A
j
ˆ e
j
(36)
where the components Aa can be found by dotting
ea with eq. (36) and using Eq. (28) to get
A
i
=

A· ˆ e
i
(37)
Similarly, the components A
i
can be found by dot-
ting e
i
with Eq. (36) to get
A
i
=

A· ˆ e
i
(38)
the components labeled by an upper index are called
the contravariant components of

A, and those la-
beled by a lower index are called the covariant com-
ponents of

A. A relation between the contravariant
and covariant components may be readily obtained
from the equality A
i
ˆ e
j
= A
i
ˆ e
j
dotting it with either
e
i
or e
j
and using Eq. (28) either Eq. (17) or Eq.
(32) gives
A
i
= g
ij
A
j
(39)
A
i
= g
ij
A
j
(40)
Now

B = g
ij
A
i
B
j
= A
i
B
i
(41)
We notice here that in our 3D space, a ready way to
obtain the dual vectors ˆ e
i
of a given basis {ˆ e
i
} is by
using the relation
e
i
=
1
V
ˆ e
j
× ˆ e
k
(42)
With V = ˆ e
i
· (ˆ e
j
× ˆ e
k
) is the volume of the paral-
lelepiped spanned by the vectors of {e
i
}.
To ﬁnd the contravariant and covariant components
of the cross product of two vectors, let us ﬁrst show
that for any given six vectors, which shall here be
denoted as S
1
, S
2
, S
3
, T
1
, T
2
, T
3
one has
(S
1
· S
2
×S
3
) (T
1
· T
2
×T
3
) = det [S
i
· T
j
] (43)
Using Eqs. (11) and (12), we see that the left-hand
side becomes the product of determinants and ap-
plying Eq. (43) to the basis {ˆ e
i
} we get
V
2
= (ˆ e
1
· ˆ e
2
× ˆ e
3
)
2
= det [ˆ e
i
· ˆ e
j
] = det [g
ij
] ≡ G
(44)
We note that
_
ˆ e
1
· ˆ e
2
× ˆ e
3
_
2
= det
_
ˆ e
i
· ˆ e
j
¸
= det
_
g
ij
¸
≡ 1/G
(45)
due to the fact that det
_
g
ij
¸
= 1/ det [g
ij
] Armed
with Eqs.(44)(45) and (38)(40), we are now ready to
give an expression for the cross product of two vec-
tors

C =

A ×

B. For the covariant components, we
have
5
C
k
=

Cˆ e
k
=
_

B
_
· ˆ e
k
= A
i
B
j
(ˆ e
i
× ˆ e
j
) · ˆ e
k
= V
ijk
A
i
B
j
=

G
ijk
A
i
B
j
(46)
and similarly, for the covariant components, we have
C
k
=

Cˆ e
k
=
_

B
_
· ˆ e
k
= A
i
B
j
_
ˆ e
i
× ˆ e
j
_
· ˆ e
k
=
1

G

ijk
A
i
B
j
(47)
where
ijk
=
ijk
with the indices located to be con-
sistent with our Rule 0. In conclusion we have
_

B
_
=

G
ijk
A
i
B
j
ˆ e
k
=
1

G

ijk
A
i
B
j
ˆ e
k
(48)
4 Changing Bases for Tensors
We now wish to move from a given basis {ˆ e
α
} to
another basis denoted as {ˆ e
α
}. From the require-
ment that {ˆ e
α
} be a basis, it follows that each vector
ˆ e
α
of the unprimed basis can be written as a linear
combination of the primed basis vectors
ˆ e
α
= R
β

α
ˆ e
β
(49)
and, given {ˆ e
α
} as a basis, each vector ˆ e
α
of the
primed basis can be written as a linear combination
of the unprimed basis vectors
ˆ e
α
= R
β
α
ˆ e
β
(50)
Inserting Eq. (50) into (49), we get ˆ e
α
=
R
β

α
_
R
γ
β

ˆ e
γ
_
=
_
R
β

α
R
γ
β

_
ˆ e
γ
whereby
R
β

α
R
γ
β

= δ
γ
α
or
_
R
β

α
_ _
R
β
α

_
= I (51)
The matrix
_
R
β
α

_
is invertible and its inverse is the
matrix
_
R
β

α
_
. In particular, det
_
R
β
α

_
= det
_
R
β

α
_
and these nonzero determinants are reciprocal to
each other.
Any vector can now be written as

A ≡ A
α
ˆ e
α
≡ A
α
ˆ e
α

≡ A
α

ˆ e
α
= A
α
ˆ e
α
(52)
The metric of the primed system is
g
α

β
≡ ˆ e
α
ˆ e
β
= (R
γ
α
ˆ e
γ

_
R
ω
β
ˆ e
ω
_
= R
γ
α
R
ω
β
(ˆ e
γ
· ˆ e
ω
)
(53)
or
g
α

β
= R
γ
α
R
ω
β
g
γω
(54)
and
g
α

β

= R
α

γ
R
β

ω
g
γω
(55)
Now it is easy to guess that
A
α

= R
α

β
A
β
(56)
or
A
α
= R
β
α
A
β
(57)
For the general basis
ˆ e
α

= R
α

β
ˆ e
β
(58)
The dot product also must be scalar (invariant)
A
α
B
α

= (R
γ
α
A
γ
)
_
R
α

ω
B
ω
_
=
_
R
γ
α
R
α

ω
_
A
γ
B
ω
= δ
γ
ω
A
γ
B
ω
= A
γ
B
γ
= A
α
B
α
(59)
Thus, when going from one basis to another in our
3D space as dictated by the transformation (50),
scalar quantities are, by deﬁnition, invariant. Or,
6
if we like, quantities that are invariant under trans-
formation (50) are legitimate one-component physi-
cal quantities. On the other hand, the components
of a legitimate 3-component physical quantity (vec-
tor) must transform according to either Eq. (56)
(contravariant components) or Eq. (57) (covariant
components).
We are then led to deﬁne a second-rank tensor as a 9-
component quantity T whose covariant, contravari-
ant, and mixed components transform, respectively,
as
T
α

β
= R
γ
α
R
ω
β
T
γω
(60)
T
α

β

= R
α

γ
R
β

ω
T
γω
(61)
T
β

α
= R
γ
α
R
β

ω
T
ω
γ
(62)
Equations (60)-(62) may also be considered as the
deﬁning relations of, respectively, a (0, 2)-type, (2,
0)-type, and (1, 1)-type second-rank tensor. Con-
travariant and covariant vectors are (1, 0)-type and
(0, 1)-type ﬁrst-rank tensors, and scalars are (0, 0)-
type zero-rank tensors. Meanwhile, (p, q)-type ten-
sors of rank (p + q) may be easily deﬁned by gener-
alizing the relations (55)-(57).
Now we have δ
β

α
= R
γ
α
R
β

ω
δ
ω
γ
. Sure enough, we
ﬁnd that R
γ
α

_
R
β

ω
δ
ω
γ
_
= R
γ
α
R
β

γ
= δ
β

α
. This re-
sult also shows that the Kronecker delta δ
β
α
has the
same components in all coordinate systems, a prop-
erty not shared by either δ
αβ
orδ
αβ
. Also has the
property R
−1
= R
T
.
A tensor is symmetric if it is invariant under the ex-
change of two equal-variance (both upper or lower)
indices, such as T
αβ
= T
βα
, whereas a tensor is an-
tisymmetric if it changes sign under the exchange
of two equal-variance indices, such as T
αβ
= −T
βα
.
The importance of these properties is due to the
fact that they hold in any coordinate system. If
T
αβ
= ±T
βα
then
T
α

β
= R
γ
α
R
ω
β
T
γω
= ±R
γ
α
R
ω
β
T
ωγ
= ±R
ω
β
R
γ
α
T
ωγ
= ±T
β

α

(63)
and similarly for two contravariant indices. From
any tensor with two equal-variance indices, such
as T
αβ
, we may construct a symmetric and an
antisymmetric tensor, namely T
αβ
± T
βα
. Like-
wise, any such tensor may be written as a sum
of a symmetric and an antisymmetric tensor as
T
αβ
=
1
2
[(T
αβ
+ T
βα
) + (T
αβ
−T
βα
)]. Symme-
try/antisymmetry is not deﬁned for two indices of
diﬀerent variance because in this case the property
would not be coordinate-independent.
5 Fields
A ﬁeld is a function of the space coordinates
(and possibly of time as well). To represent an
objectivity-preserving physical quantity the ﬁeld
function must have a tensorial character. Specif-
ically, in our ordinary 3D space, it must be a
3
r
-component tensor ﬁeld of rank r, with r =
0, 1, 2, 3, . . . : a (0, 0)-type tensor (scalar ﬁeld), (1,
0)-type tensor (contravariant vector ﬁeld), (0, 1)-
type tensor (covariant vector ﬁeld) or, in general, a
(p, q)-type tensor ﬁeld of rank r = p + q. Let us
then consider our ordinary Euclidean 3D space pa-
rameterized by a set of arbitrary coordinates, which
we shall denote by x ≡
_
x
1
, x
2
, x
3
_
. We assume
that they are related to the Cartesian coordinates-
hereafter denoted by x’ ≡
_
x
1

, x
2

, x
3

_
by an in-
vertible relation, so that
x
α

= x
α
_
x
1
, x
2
, x
3
_
≡ x
α

(x) (64)
and
x
α
= x
α
_
x
1

, x
2

, x
3

_
≡ x
α
(x

) (65)
The Jacobian determinants are nonzero
7
J = det
_
∂x
α

∂x
β
_
= det
_
R
α

β
_
= 0 (66)
and
J
−1
= det
_
∂x
α
∂x
β

_
= det
_
R
α
β

¸
= 0 (67)
where we have set
∂x
α
∂x
β

= R
α
β
and
∂x
α

∂x
β
= R
α

β
(68)
These are elements of matrices inverse to each other
R
α
β
R
β

γ
=
∂x
α
∂x
β

∂x
β

∂x
γ
=
∂x
α
∂x
γ
(69)
Now we have
ˆ e
α
=
∂x
β

∂x
α
ˆ e
β
= R
β

α
ˆ e
β
(70)
and
dx
α
=
∂x
α
∂x
β

dx
β

≡ R
α
β
dx
β

(71)
For ease of notation you should consider ∂
α
=

∂x
α
because now we derivate a tensor

β
A
α
= ∂
β
(R
γ
α
A
γ
) = R
γ
α
(∂
β
A
γ
) + A
γ

β
R
γ
α

= R
γ
α

_
R
ω
β

ω
A
γ
_
+ A
γ
R
γ
β

α

= R
ω
β
R
γ
α

ω
A
γ
+ A
γ
R
γ
β

γ

(72)
For a vector in any basis

β

A = ∂
β
(A
γ
ˆ e
γ
) = (∂
β
A
γ
) ˆ e
γ
+ A
γ
(∂
β
ˆ e
γ
) (73)
In the last term, ∂
β
ˆ e
γ
, may be rewritten as a linear
combination of the basis vectors to give

β
ˆ e
γ
= (∂
β
ˆ e
γ
)
α
ˆ e
α
= [ˆ e
α
· (∂
β
ˆ e
γ
)] ˆ e
α
≡ Γ
α
βγ
ˆ e
α
(74)
Then
Γ
α
βγ
ˆ e
α
≡ ˆ e
α
· (∂
β
ˆ e
γ
) (75)
When ∂
β
e
γ
= ∂
γ
e
β
then Γ
α
βγ
= Γ
α
γβ
ie in metric
spaces untwisted.
We deﬁne the covariant derivative
A
α
;
β
= ∂
β
A
α
+ Γ
α
βγ
A
γ
(76)
and
A
α
;
β
= ∂
β
A
α
−Γ
α
βγ
A
γ
(77)
The Kristoﬀel symbols have the following properties
that can be derived easily with the tools we have
learned (but I will to save sheets)
Γ
α
βγ
=
1
2
g
αω
[∂
β
g
ωγ
+ ∂
γ
g
ωβ
−∂
ω
g
βγ
] (78)
The following properties are also met
Γ
αβγ
≡ g
αβ
Γ
ω
βγ
=
1
2
g
αω
g
ωµ
[∂
ω
g
µγ
+ ∂
γ
g
µβ
−∂
µ
g
βγ
]
=
1
2
δ
µ
α
[∂
β
g
µγ
+ ∂
γ
g
µβ
−∂
µ
]
(79)
which gives
Γ
αβγ
= g
αω
Γ
ω
βγ
=
1
2
[∂
β
g
αγ
+ ∂
γ
g
αβ
−∂
α
g
βγ
] (80)
The covariant derivates of higher rank tensors is
A
αβ
;
γ
≡ ∂
γ
A
αβ
+ Γ
α
γω
A
ωβ
+ Γ
β
γω
A
αω
(81)
8
A
αβ
;
γ
≡ ∂
γ
A
αβ
−Γ
ω
αγ
A
ωβ
−Γ
ω
αω
A
αω
(82)
and
A
β
α
;
γ
≡ ∂
γ
A
β
α
+ Γ
β
γω
A
ω
α
−Γ
ω
αγ
A
β
ω
(83)
The covariant derivative of a product follows the
same rules as the ordinary derivative. For instance
_
A
α
B
β
_
;
γ
= A
α
;
γ
B
β
+ B
β
;
γ
A
α
(84)
Another important rule is to the metric
g
αβ
;
γ
= 0 (85)
We are then led to deﬁne the divergence of a vector
in an arbitrary coordinate system as
∇·

A ≡ A
α
;
α
≡ ∂
α
A
α
+ Γ
α
αγ
A
γ
(86)
Notice that the divergence of a vector is deﬁned in
terms of its contravariant components. If the covari-
ant components are available, then ∇·

A ≡ A
α
;
α
=
_
g
αβ
A
β
_
;
α
= g
αβ
A
β
;
α
we ;a deﬁne the divergence
of (the components of) a covariant vector to be the
divergence of the associated contravariant (compo-
nents of the) vector.
Remarkably, to evaluate the divergence of a vector,
it is not necessary to compute the Christoﬀel sym-
bols. In fact, the ones appearing in Eq. (86) are
Γ
α
αγ
=
1
2
g
αγ

γ
g
αω
(87)
where the cancellation between the ﬁrst and third
terms on the right-hand side of the ﬁrst equality
arises upon exchanging the dummy indices α and ω
and taking advantage of the symmetry of the metric
tensor. However, for the matrix whose elements are
g
αω
, we may write
[g
αω
] = [g
ωα
] = [g
ωα
]
−1
=
_
G
αβ
G
_
(88)
However, from the deﬁnition of a determinant we
also have G ≡ g
αβ
G
αβ
whereby ∂G/∂g
αω
= G
αω
and g
αβ
= G
αβ
/G = (1/G)(∂G/∂g
αω
) The Eq (87)
Γ
α
αγ

γ
(

G)

G
(89)
which can be inserted into Eq. (86) to give
∇·

A =

α

GA
α

G
(90)
and
φ =

α
_

Gg
αβ

β
φ
_

G
(91)
6 Conclusions
The tensor calculus helps us to perform all opera-
tions with vectors in space. Also helps us to fully
understand the operations. We uses a comfortable
and elegant notation. The tensor calculus is essential
to understand topics of the Modern Physics.
References
 Franco B, ”Tensors: A guide for undergraduate students ”, AIP, 81, 498 (2013); doi: 10.1119/1.4802811
 BF. Schutz, ” Geometrical Methods of Mathematical Physics ”,Cambridge University Press; First
Published edition (January 28, 1980).
9
 M Alcubierre; ” Introduction to 3+1 Numerical Relativity ”,Oxford University Press, USA (June 16,
2008)
10

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