tensors

Introduction to tensors

Joakim Strandberg

July 2005

1 Contravariant and covariant vectors

1.1 The definitions of contravariant and covariant components

The purpose of this text is to be an introductory text to what tensors are and wherethey come from. To achieve this goal understanding of what contravariant and covariantvectors are is essential. Therefore this text begins with the definitions of these concepts.

Definition of contravariant components of a vector The contravariant compo-nents (a1, a2, . . . , an) of a n-dimensional vector v with respect to the linearly independentvectors x1,x2, . . . ,xn (called coordinate axes) are defined by:

v = a1x1 + a2x2 + · · ·+ anxn (1)

Definition of covariant components of a vector The covariant components (b1, b2,. . . , bn) of a n-dimensional vector v with respect to the linearly independent vectorsx1,x2, . . . ,xn (called coordinate axes) are defined by:

b1 = v · x1

b2 = v · x2

......

bn = v · xn

(2)

Why the components (a1, a2, . . . , an) are called contravariant and (b1, b2, . . . , bn) arecalled covariant is explained on page 2. Problem 1 illustrates how to compute the con-travariant and covariant components of a given vector.

Problem 1 Given the vectors x1 = (1, 0), x2 = (1, 2) and v = (1, 2).

a) Determine the contravariant components (a1, a2) of v with respect to the vectorsx1 and x2.

b) Determine the covariant components (b1, b2) of v with respect to the vectors x1

and x2.

Solution 1a The vectors x1 and x2 are clearly independent. Using the definition ofcontravariant components of a vector

v = a1x1 + a2x2 (3)

1

(12

)=(

10

)a1 +

(12

)a2 (4)

This is an equation system (1 10 2

)(a1

a2

)=(

12

)(5)

Using Cramer’s rule

a1 =

∣∣∣∣1 12 2

∣∣∣∣∣∣∣∣1 10 2

∣∣∣∣ =2− 2

2= 0 (6)

a2 =

∣∣∣∣1 10 2

∣∣∣∣2

=22

= 1 (7)

Thus (a1, a2) = (0, 1).

Solution 1b Using the definition of covariant components

b1 = v · x1 = (1, 2) · (1, 0) = 1 (8)

b2 = v · x2 = (1, 2) · (1, 2) = 5 (9)

Thus (b1, b2) = (1, 5).Notice that in this example (a1, a2) 6= (b1, b2).

Important! Generally the contravariant and covariant components of a vector are dif-ferent. But when the “coordinate axes” x1, . . . , xn are orthonormal (they all have unitlength and and are orthogonal) there is no difference between contravariant and covariantcomponents!

1.2 Why the components are called contravariant and covariant

Problem 2 Why are the components of a vector v defined by (1) and (2) called con-travariant and covariant components?

Solution 2 If the coordinate axes are scaled by a factor of c, then the contravariantcomponents become c times smaller but the covariant components become c times greater.Thus the contravariant components of a vector scale against the coordinate axes and thecovariant components scale with. In Latin contra means against and co means with.Thereof the names contravariant and covariant components. Problem 3 illustrates this.

Problem 3 Given the vectors x1 = (c, 0), x2 = (c, 2c) and v = (1, 2).

a) Determine the contravariant components (a1, a2) of v with respect to the vectorsx1 and x2.

b) Determine the covariant components (b1, b2) of v with respect to the vectors x1

and x2.

2

Solution 3a The equation system now becomes(c c0 2c

)(a1

a2

)=(

12

)(10)

Using Cramer’s rule

a1 =

∣∣∣∣1 c2 2c

∣∣∣∣∣∣∣∣c c0 2c

∣∣∣∣ =2c− 2c

2c2= 0 (11)

a2 =

∣∣∣∣c 10 2

∣∣∣∣2c2

=2c2c2

=1c

(12)

Thus (a1, a2) = 1c (0, 1).

Solution 3b Using the definition of covariant components

b1 = v · x1 = (1, 2) · (c, 0) = c (13)

b2 = v · x2 = (1, 2) · (c, 2c) = 5c (14)

Thus (b1, b2) = c(1, 5).

1.3 Why the contravariant and covariant components are inter-esting

Problem 4 Why are only the definitions (1) and (2) of the components of a vectorinteresting? For example, it’s possible to define components

c1 = v · x1/|x1|c2 = v · x2/|x2|...

...cn = v · xn/|xn|

(15)

Why not also give the components (c1, c2, . . . , cn) a name? Why are the contravariantand covariant components so “important” that they have been given names?

Solution 4 When parametrizations of the Euclidean plane and tensors of first orderhave been explained, it is shown on page 7 how the definition of work in physics give riseto contravariant and covariant components.

Problem 5 What is the definition of the Euclidean plane?

Solution 5 Intuitively the Euclidean plane can be imagined as a flat piece of paper.In this paper one can introduce two orthogonal coordinate axes usually denoted x- andy-axes. Two numbers x and y can then be measured to describe the position of a point.How is a point defined? Instead of saying that the two numbers describe the position ofa point, the two numbers are defined to be a point.

3

Definition of the Euclidean plane The Euclidean plane is defined as the set of allordered 2-tuples r = (x, y) called points where x and y are real numbers.

That the 2-tuples are ordered means that (x, y) 6= (y, x)⇔ x 6= y.

Definition of a parametrization of the Euclidean plane A parametrization ofthe Euclidean plane is a specification of unique coordinates (u1, u2) (where u1 and u2 arereal) to every point in the plane.

r(u1, u2) = (x(u1, u2), y(u1, u2)) (16)

This text does not consider all possible parametrizations but restricts itself to allowableparametrizations.

Important! Notice that coordinates (in a parametrization) are NOT written with sub-scripts (u1, u2) but with superscripts (u1, u2). The 2 reasons for this will be explainedon page ???. It’s not explained here in the text because the explanation requires theEinstein summation convention and contravariant tensors of first order.

Definition of an allowable parametrization of the Euclidean plane An allowableparametrization of the Euclidean plane satisfies:

i) Every point in the Euclidean plane is given unique coordinates (u1, u2) where(u1, u2 ∈ R)

r(u1, u2) = (x(u1, u2), y(u1, u2)) (17)

ii) The vectors ∂r∂u1 and ∂r

∂u2 are defined everywhere, linearly independent and 6= 0.

Some examples of allowable parametrizations:

Problem 6 Determine which of the following parametrizations which are allowable para-metrizations.

a) r = (u1, u2)

b) r = (u1 + u2 + (u2)3, 2u2)

Solution 6a The parametrization r = (u1, u2) leads to

∂r∂u1

= (1, 0)

∂r∂u2

= (0, 1)(18)

It is clear that the vectors ∂r∂u1 and ∂r

∂u2 are defined everywhere, linearly independent and6= 0.

4

Solution 6b The parametrization r = (u1 + u2 + (u2)3, 2u2) leads to

∂r∂u1

= (1, 0)

∂r∂u2

= (1 + 3(u2)2, 2)(19)

Here the vectors ∂r∂u1 and ∂r

∂u2 are defined everywhere. They satisfy 6= 0 because (u2)2 ≥ 0and are linearly independent. Notice that the vector ∂r

∂u2 is dependent on the point underconsideration. This example shows that the vectors ∂r

∂u1 and ∂r∂u2 may be different from

point to point.

Problem 7 Why is the second requirement (ii) important in the definition of allowableparametrizations?

Solution 7 The second requirement (ii) guarantees that a vector defined at a point inthe Euclidean plane always can be written as a linear combination of ∂r

∂u1 and ∂r∂u2 defined

at the point.

Important! From now on when coordinates are introduced in the Euclidean plane itis assumed they are allowed parametrizations.

Problem 8 Why is the letter r used to define a point? Why not use the notationp = (x, y) instead, where the letter p has been chosen since it’s the first letter in theword point as is done in the book Elementary differential geometry by Barret O’Neill?

Solution 8 This text follows the notational convention used in the book Vektoranalysby Anders Ramgard. Ramgard does not explain why he chose r to denote a point orvector in Euclidean n-space. It may be because of the following three reasons:

(i) In Euclidean 2-space where r = (x, y), one can introduce a parametrization (r, φ)called polar coordinates r = (r cosφ, r sinφ). Notice that the length of the vectorr is |r| = r, where r originates from the word radius (here radius of a circle).

(ii) In Euclidean 3-space where r = (x, y, z), one can introduce a parametrization(r, φ, θ) called spherical coordinates r = (r sin θ cosφ, r sin θ sinφ, cos θ). The lengthof the vector r is |r| = r, where r originates from the radius of the sphere.

(iii) Euclidean n-space is denoted Rn, where the letter R here comes from the word real.

2 First order tensors defined on the Euclidean plane

The following problem leads to the definition of contravariant tensors of 1:st order.

Problem 9 Let a vector v be given by the contravariant components (a1, a2) at a fixedpoint P in the Euclidean plane using the coordinates u1, u2:

v = a1 ∂r∂u1

+ a2 ∂r∂u2

(20)

5

Using another coordinate system u1, u2, the contravariant components of the vector vare (a1, a2) and v can be written:

v = a1 ∂r∂u1 + a2 ∂r

∂u2 (21)

What is the relation between (a1, a2) and (a1, a2)? Express (a1, a2) as a function of(a1, a2).

Solution 9 Using the chain-rule (here α = 1, 2):

∂r∂uα

=∂r∂u1

∂u1

∂uα+

∂r∂u2

∂u2

∂uα(22)

It is allowed to use the chain-rule here since ∂r∂u1 , ∂r

∂u2 , ∂u1

∂uα and ∂u2

∂uα are continuousfunctions. That ∂u1

∂uα and ∂u2

∂uα are continuous functions is shown in problem 11. Nowusing equations (20), (21) and (22) leads to expressing (a1, a2) as a function of (a1, a2)

v = a1 ∂r∂u1

+ a2 ∂r∂u2

= a1

(∂r∂u1

∂u1

∂u1+

∂r∂u2

∂u2

∂u1

)+ a2

(∂r∂u1

∂u1

∂u2+

∂r∂u2

∂u2

∂u2

)=(a1 ∂u

1

∂u1+ a2 ∂u

1

∂u2

)∂r∂u1 +

(a1 ∂u

2

∂u1+ a2 ∂u

2

∂u2

)∂r∂u2

= a1 ∂r∂u1 + a2 ∂r

∂u2

(23)

Identifying the coefficients in front of ∂r∂u1 and ∂r

∂u2 in the last two steps

a1 = a1 ∂u1

∂u1+ a2 ∂u

1

∂u2(24)

a2 = a1 ∂u2

∂u1+ a2 ∂u

2

∂u2(25)

Summarizing (here β = 1, 2):

aβ = a1 ∂uβ

∂u1+ a2 ∂u

β

∂u2(26)

Equation (26) is the source of inspiration for tensors (1:st order contravariant tensors).There are two kinds of tensors of 1:st order:

Definition of a contravariant tensor of 1:st order (or a contravariant vector)on the Euclidean plane Let a 2-tuple of real numbers (a1, a2) be associated with apoint P in the Euclidean plane with coordinates u1, u2. Associate also with the point Pa 2-tuple of real numbers a1, a2 with respect to the coordinates u1, u2. If these numberssatisfy:

aβ = a1 ∂uβ

∂u1+ a2 ∂u

β

∂u2(27)

we say that a contravariant vector at P is given. The numbers a1, a2 and a1, a2 are calledthe components of the contravariant vector in the respective coordinate systems u1, u2

or u1, u2. Contravariant vectors are indicated by a superscript index.

6

Definition of a covariant tensor of 1:st order (or a covariant vector) on theEuclidean plane Let a 2-tuple of real numbers (b1, b2) be associated with a point Pin the Euclidean plane with coordinates u1, u2. Associate also with the point P a 2-tupleof real numbers b1, b2 with respect to the coordinates u1, u2. If these numbers satisfy:

bβ = b1∂u1

∂uβ+ b2

∂u2

∂uβ(28)

we say that a covariant vector at P is given. The numbers b1, b2 and b1, b2 are called thecomponents of the covariant vector in the respective coordinate systems u1, u2 or u1, u2.Covariant vectors are indicated by a subscript index.

Problem 10 Determine if the vector (du1, du2) is contravariant, covariant or neither.

Solution 10 The vector (du1, du2) is a contravariant vector since the components du1

and du2 transform contravariantly under change of coordinate system from (u1, u2) to(u1, u2) accordning to the chain-rule

du1 = du1 ∂u1

∂u1 + du2 ∂u1

∂u2 (29)

du2 = du1 ∂u2

∂u1 + du2 ∂u2

∂u2 (30)

Summarizing:

duγ = du1 ∂uγ

∂u1 + du2 ∂uγ

∂u2 (γ = 1, 2) (31)

That the vector (du1, du2) transforms contravariantly is part of the reason why coordi-nates are written with superscripts.

Returning to problem 4 posed on page 3 Why are the contravariant and covariantcomponents so “important” that they have been given names?

Solution When calculating the work performed by a force on a particle under an in-finitesimal displacement in the Euclidean plane, its natural to assume that the workperformed is independent of the coordinate system used. If the work performed is inde-pendent of the coordinate system then it turns out that if the displacement is expressedby its contravariant components then the force must be expressed by its covariant compo-nents. To see this, assume that a force F is exerted on a particle which moves (du1, du2)in an ordinary Cartesian coordinate system r = (u1, u2). The work done dW is

dW = F1du1 + F2du

2 (32)

With new coordinates (u1, u2), the force is F = (F1, F2). We would like to write

dW = F1du1 + F2du

2 (33)

What then is the connection between (F1, F2) and (F1, F2)? Using the fact that du1 anddu2 transform contravariantly

duγ = du1 ∂uγ

∂u1 + du2 ∂uγ

∂u2(34)

7

dW = F1du1 + F2du

2

= F1

(du1 ∂u

1

∂u1 + du2 ∂u1

∂u2

)+ F2

(du1 ∂u

2

∂u1 + du2 ∂u2

∂u2

)=(F1∂u1

∂u1 + F2∂u2

∂u1

)du1 +

(F1∂u1

∂u2 + F2∂u2

∂u2

)du2

= F1du1 + F2du

2

(35)

Identifying:

F1 = F1∂u1

∂u1 + F2∂u2

∂u1 (36)

F2 = F1∂u1

∂u2 + F2∂u2

∂u2 (37)

Equations (36) and (37) can be summarized by:

Fβ = F1∂u1

∂uβ+ F2

∂u2

∂uβ(38)

Comparing (38) and (28) shows that the components (F1, F2) must transform covariantlyif the work performed by the force is independent of the choice of coordinates used.

Problem 11 A contravariant 1:st order tensor is a 2-tuple of real numbers (a1, a2) thatis transformed under change from coordinates (u1, u2) to (u1, u2) as:

aβ = a1 ∂uβ

∂u1+ a2 ∂u

β

∂u2(39)

But is it obvious that ∂uβ

∂u1 and ∂uβ

∂u2 always exists? It is sufficient in this problem toinvestigate if ∂u1

∂u1 exists under change of allowable parametrizations (it is not difficultto generalize). If ∂u1

∂u1 doesn’t always exists, then a first order tensor is defined by anexpression that is not always defined. How can one define something by something thatis not defined???

Solution 11 The expression ∂u1

∂u1 is always well defined under coordinate change betweenallowable parametrizations. To see this, use u1 = u1(x, y) and the chain rule:

∂u1

∂u1=∂u1

∂x

∂x

∂u1+∂u1

∂y

∂y

∂u1(40)

The expression ∂u1

∂u1 is defined if ∂u1

∂x , ∂x∂u1 , ∂u

1

∂y and ∂y∂u1 are defined (it is then also allowed

to use the chain-rule). Since ∂r∂u1 = (∂x(u

1,u2)∂u1 , ∂y(u

1,u2)∂u1 ) is well defined according to the

definition of allowable parametrizations ⇒ ∂x∂u1 and ∂y

∂u1 are defined. But is ∂u1

∂x and ∂u1

∂y

defined? Since ∂r∂u1 and ∂r

∂u2 are defined, the differentials dx and dy can be written:

dx =∂x

∂u1 du1 +

∂x

∂u2 du2 (41)

dy =∂y

∂u1 du1 +

∂y

∂u2 du2 (42)(

dxdy

)=(

∂x∂u1

∂x∂u2

∂y∂u1

∂y∂u2

)(du1

du2

)(43)

8

Using Cramer’s rule:

du1 =

∣∣∣∣dx ∂x∂u2

dy ∂y∂u2

∣∣∣∣∣∣∣∣ ∂x∂u1∂x∂u2

∂y∂u1

∂y∂u2

∣∣∣∣ =∂y∂u2

∂x∂u1

∂y∂u2 − ∂y

∂u1∂x∂u2

dx−∂x∂u2

∂x∂u1

∂y∂u2 − ∂y

∂u1∂x∂u2

dy (44)

If ∂u1

∂x , ∂u1

∂y are defined then du1 can be written:

du1 =∂u1

∂xdx+

∂u1

∂ydy (45)

Identifying:∂u1

∂x=

∂y∂u2

∂x∂u1

∂y∂u2 − ∂y

∂u1∂x∂u2

(46)

∂u1

∂y= −

∂x∂u2

∂x∂u1

∂y∂u2 − ∂y

∂u1∂x∂u2

(47)

The denominator ∂x∂u1

∂y∂u2 − ∂y

∂u1∂x∂u2 is equal to the determinant∣∣∣∣ ∂x∂u1

∂x∂u2

∂y∂u1

∂y∂u2

∣∣∣∣ (48)

which is equal to ±the area spanned by the vectors ∂r∂u1 and ∂r

∂u2 . And this area is 6= 0since ∂r

∂u1 and ∂r∂u2 are two linearly independent vectors 6= 0. The nominators ∂y

∂u2 and ∂x∂u2

are defined since ∂r∂u2 is defined. Thus the right-hand sides of equations (46) and (47) are

defined, and hence the left-hand sides ∂u1

∂x and ∂u1

∂y are also defined. Thus the expression∂u1

∂u1 is always defined under coordinate change between allowed parametrizations.

Problem 12 Determine if the vector ∂r∂u1 is contravariant, covariant or neither.

Solution 12 Investigating how the vector ∂r∂u1 transforms when coordinates are changed

from uα to uβ .∂r∂u1 =

(∂x

∂u1 ,∂y

∂u1

)(49)

Using the chain-rule∂x

∂u1 =∂x

∂u1

∂u1

∂u1 +∂x

∂u2

∂u2

∂u1 (50)

∂y

∂u1 =∂y

∂u1

∂u1

∂u1 +∂y

∂u2

∂u2

∂u1 (51)

Thus

∂r∂u1 =

(∂x

∂u1

∂u1

∂u1 +∂x

∂u2

∂u2

∂u1 ,∂y

∂u1

∂u1

∂u1 +∂y

∂u2

∂u2

∂u1

)=(∂x

∂u1,∂y

∂u1

)∂u1

∂u1 +(∂x

∂u2,∂y

∂u2

)∂u2

∂u1

=∂r∂u1

∂u1

∂u1 +∂r∂u2

∂u2

∂u1

(52)

9

∂r∂uβ

=∂r∂u1

∂u1

∂uβ+

∂r∂u2

∂u2

∂uβ(53)

This means that ∂r∂uβ

does not define a contravariant nor covariant vector, but it isthe “vector” ( ∂r

∂u1 ,∂r∂u2 ) that transforms covariantly. Covarians is usually denoted by a

subscript, which inspires the following notation that is frequently used

rα ≡∂r∂uα

(54)

rα ≡∂r∂uα

(55)

Using this notation, it’s possible to write

rβ = r1∂u1

∂uβ+ r2

∂u2

∂uβ(56)

Problem 13 Do the covariant components bβ of a vector v (eq. (2) defined on page1) with respect to the coordinate axes r1 and r2 transform covariantly according to thedefinition of covariant 1:st order tensors?

bβ = v · rβ (57)

Solution 13 Yes, they do.

bβ = v · rβ

= v ·(

r1∂u1

∂uβ+ r2

∂u2

∂uβ

)= (v · r1)

∂u1

∂uβ+ (v · r2)

∂u2

∂uβ

= b1∂u1

∂uβ+ b2

∂u2

∂uβ

(58)

3 First order tensors defined on Euclidean n-space

3.1 The generalization

Mathematicians like to generalize things. How can 1:st order tensors defined on theEuclidean plane be generalized? Probably the first obvious generalization is to generalizefrom to the Euclidean plane to Euclidean n-space.

Definition of Euclidean n-space Let n be a positive integer. Euclidean n-space isdefined as the set of all ordered n-tuples r = (p1, . . . , pn) called points where p1, . . . , pnare real numbers (the index k of pk does not imply covarians, but is only an index).

Definition of an allowable parametrization of Euclidean n-space An allowableparametrization of Euclidean n-space satisfies:

i) Every point in Euclidean n-space is given unique coordinates (u1, . . . , un) where(u1, . . . , un ∈ R)

r(u1, . . . , un) = (p1(u1, . . . , un), . . . , pn(u1, . . . , un)) (59)

10

ii) The vectors ∂r∂u1 , . . . ,

∂r∂un are defined everywhere, linearly independent and 6= 0.

As before, when coordinates are introduced it is assumed they are allowed parametriza-tions. Generalizing problem 9:

Problem 14 Let a n-dimensional vector v be given by the contravariant components(a1, . . . , an) at a fixed point P in Euclidean n-space using the coordinates u1, . . . , un:

v = a1 ∂r∂u1

+ · · ·+ an∂r∂un

(60)

Using another coordinate system u1, . . . , un, the contravariant components of the vectorv are (a1, . . . , an) and v can be written:

v = a1 ∂r∂u1 + · · ·+ an

∂r∂un

(61)

What is the relation between (a1, a2) and (a1, a2)? Express (a1, a2) as a function of(a1, a2).

Solution 14 Using a similar argument as in problem 9

aβ = a1 ∂uβ

∂u1+ · · ·+ an

∂uβ

∂un(62)

Therefore the generalization of 1:st order tensors to be defined on Euclidean n-spacebecomes:

Definition of a contravariant tensor of 1:st order (or a contravariant vector)on Euclidean n-space A contravariant tensor of 1:st order is a quantity whose ncomponents aα are transformed according to

aβ = a1 ∂uβ

∂u1+ . . .+ an

∂uβ

∂un(63)

under change of coordinate system in Euclidean n-space.

Definition of a covariant tensor of 1:st order (or a covariant vector) on Eu-clidean n-space A covariant tensor of 1:st order is a quantity whose n components bβare transformed according to

bβ = b1∂u1

∂uβ+ . . .+ bn

∂un

∂uβ(64)

under change of coordinate system in Euclidean n-space.

3.2 The Einstein summation convention

Here it is convenient to introduce the Einstein summation convention that is used ex-tensively in tensor analysis1 and relativity theory. The point of the Einstein summation

1The term tensor analysis was coined by Einstein in 1916.

11

convention is to simplify notation. Consider the definition of 1:st order contravarianttensors. The notation may first be simplified by the introduction of a summation sign:

aβ = a1 ∂uβ

∂u1+ . . .+ an

∂uβ

∂un

=n∑α=1

aα∂uβ

∂uα

(65)

This sum is written in the Einstein notation as:n∑α=1

aα∂uβ

∂uα= aα

∂uβ

∂uα(66)

In the expression ∂uβ

∂uα the index β is considered superscript and the index α is consideredsubscript.

SUMMATION CONVENTION If in a product a letter figures twice, once as su-perscript and once as subscript, summation must be carried out from 1 to n with respectto this letter. The summation sign

∑will be omitted.

For example

aαbα =n∑α=1

aαbα = a1b1 + a2b2 + · · ·+ anbn (67)

Applying the Einstein summation to the definition of 1:st order covariant tensors:

bβ = b1∂u1

∂uβ+ . . .+ bn

∂un

∂uβ

=n∑γ=1

bγ∂uγ

∂uβ

= bγ∂uγ

∂uβ

(68)

Equation (20) on page 5 can be written

v = aβrβ (69)

Rewriting equation (56) with the Einstein summation notation (and of course generalizedto Euclidean n-space)

rβ = rα∂uα

∂uβ(70)

An index with respect to which summation must be carried out is called a summationindex or dummy index. The other indices are said to be free indices. Dummy indicesmay be changed during computations without warning, for example

aαbα = aibi = a1b1 + a2b2 + · · ·+ anbn (71)

aαbα + aγcγ = aα(bα + cα) (72)

Problem 15 Why are the coordinates (u1, u2, . . . , un) written with superscripts?

12

Solution 15 One could be led to believe that the vector (u1, u2, . . . , un) transformscontravariantly under coordinate changes because the indices of the coordinate variablesare written with superscripts. This is not true. It is the vector (du1, du2, . . . , dun)that transforms contravariantly. In expressions duα, ∂uα

∂uβ, ∂r∂uβ

the index α is consideredsuperscript and β subscript. This convention together with the Einstein summationconvention simplifies the tensor notation.

4 Higher order tensors

Now that first order tensors are generalized to be defined on Euclidean n-space, canthe tensor concept be generalized further? Yes, it can and the way first order tensorsare generalized is inspired by the definition of work in physics. The following problemsillustrate this.

Problem 16 Let aα be the contravariant components of an arbitrary vector v whichmeans aγ = aα ∂u

γ

∂uα . Consider the expression

I = bαaα (73)

Show that if I is invariant under coordinate change, then bα are the covariant componentsof a vector w which means bα transforms covariantly.

Solution 16 Let aγ , bγ be the components with respect to the coordinate system uα.Since I is invariant

I = bγaγ = bαa

α (74)

Inserting aγ = ∂uγ

∂uα aα

bγ∂uγ

∂uαaα = bαa

α (75)

This expression is valid for an arbitrary aα. Thus it’s possible to identify

bγ∂uγ

∂uα= bα (76)

Multiply both sides with ∂uα

∂uβand sum with respect to α from 1 to n

bγ∂uγ

∂uα∂uα

∂uβ= bα

∂uα

∂uβ(77)

Here is now used∂uγ

∂uα∂uα

∂uβ= δγβ (78)

where δγβ is the Kronecker delta which is = 1 when β = γ and = 0 otherwise. Equation(78) follows from the fact that the variables u1, u2, . . . , un are independent and the chain-rule

δγβ =∂uβ

∂uγ

=∂uβ

∂u1

∂u1

∂uγ+∂uβ

∂u2

∂u2

∂uγ+ · · ·+ ∂uβ

∂un∂un

∂uγ(no summation over n)

=∂uγ

∂uα∂uα

∂uβ

(79)

13

This meansbγδ

γβ = bα

∂uα

∂uβ(80)

bβ = bα∂uα

∂uβ(81)

Thus bα transforms covariantly. This problem can be generalized, for example to define2:nd order covariant tensors as in problem 17 and 2:nd order contravariant tensors as inproblem 18.

Problem 17 Let bα and cβ be two arbitrary contravariant 1:st order tensors. Consider

I = aαβbαcβ (82)

If I is invariant under coordinate change, determine the transformation behaviour of aαβ .

Solution 17 Since I is invariant

I = aωλbωcλ = aαβb

αcβ (83)

Inserting bω

= bα ∂uω

∂uα and cλ = cβ ∂uλ

∂uβ

I = aωλ

(bα∂uω

∂uα

)(cβ∂uλ

∂uβ

)= aαβb

αcβ (84)

Since bα and cβ are arbitrary it’s possible to identify

aαβ = aωλ∂uω

∂uα∂uλ

∂uβ(85)

Multiply both sides by ∂uα

∂uµ∂uβ

∂uν and sum with respect to both α and β from 1 to n

aαβ∂uα

∂uµ∂uβ

∂uν= aωλ

∂uω

∂uα∂uα

∂uµ∂uλ

∂uβ∂uβ

∂uν= aωλδ

ωµδ

λν (86)

aµν = aαβ∂uα

∂uµ∂uβ

∂uν(87)

If I is invariant under coordinate change then aαβ defines a 2:nd order covariant tensor.

Definition of a covariant tensor of 2:nd order defined on Euclidean n-spaceA covariant tensor of order 2 is a quantity whose n2 components aαβ are transformedaccording to

aµν = aαβ∂uα

∂uµ∂uβ

∂uν(88)

under change of coordinate system in Euclidean n-space.

Problem 18 Let bα and cβ be two arbitrary covariant 1:st order tensors. Consider

I = aαβbαcβ (89)

If I is invariant under coordinate change, determine the transformation behaviour of aαβ .

14

Solution 18 Since I is invariant

I = aωλbωcλ = aαβbαcβ (90)

Inserting bω = bα∂uα

∂uω and cλ = cβ∂uβ

∂uλ

I = aωλ(bα∂uα

∂uω

)(cβ∂uβ

∂uλ

)= aαβbαcβ (91)

Since bα and cβ are arbitrary it’s possible to identify

aαβ = aωλ∂uα

∂uω∂uβ

∂uλ(92)

Multiply both sides by ∂uµ

∂uα∂uν

∂uβand sum with respect to both α and β from 1 to n

aαβ∂uµ

∂uα∂uν

∂uβ= aωλ

∂uα

∂uω∂uµ

∂uα∂uβ

∂uλ∂uν

∂uβ= aωλδµωδ

νλ (93)

aµν = aαβ∂uµ

∂uα∂uν

∂uβ(94)

If I is invariant under coordinate change then aαβ defines a 2:nd order contravarianttensor.

Definition of a contravariant tensor of 2:nd order defined on Euclidean n-spaceA contravariant tensor of order 2 is a quantity whose n2 components aαβ are transformedaccording to

aµν = aαβ∂uµ

∂uα∂uν

∂uβ(95)

under change of coordinate system in Euclidean n-space.

Now, there is a third kind of tensor, the mixed 2:nd order tensor.

Problem 19 Let bα be an arbitrary covariant 1:st order tensor and cβ an arbitrarycontravariant 1:st order tensor. Consider

I = a βα bαcβ (96)

If I is invariant under coordinate change, determine the transformation behaviour of a βα .

Important! Notice that a βα is not written as aβα. This is because in tensor notation

it’s not only if the index is superscript or subscript that matters, but the position is alsoimportant. This has to do with the possibility of associating a new tensor with a givenone through the metric tensor by raising or lowering an index and is explained later inthis text.

Solution 19 Since I is invariant

I = a λω b

ωcλ = a β

α bαcβ (97)

Inserting bω

= bα ∂uω

∂uα and cλ = cβ∂uβ

∂uλ

I = a λω

(bα∂uω

∂uα

)(cβ∂uβ

∂uλ

)= a β

α bαcβ (98)

15

Since bα and cβ are arbitrary it’s possible to identify

a βα = a λ

ω

∂uω

∂uα∂uβ

∂uλ(99)

Multiply both sides by ∂uα

∂uµ∂uν

∂uβand sum with respect to both α and β from 1 to n

a βα

∂uα

∂uµ∂uν

∂uβ= a λ

ω

∂uω

∂uα∂uα

∂uµ∂uβ

∂uλ∂uν

∂uβ= a λ

ω δωµδνλ (100)

a νµ = a β

α

∂uα

∂uµ∂uν

∂uβ(101)

If I is invariant under coordinate change then a βα defines a 2:nd order mixed tensor. It

is covariant to 1:st order and contravariant to 1:st order.

Definition of a mixed tensor of 2:nd order defined on Euclidean n-space Amixed tensor of order 2, contravariant to order 1 and covariant to order 1, is a quantitywhose n2 components a β

α are transformed according to

a νµ = a β

α

∂uα

∂uµ∂uν

∂uβ(102)

under change of coordinate system in Euclidean n-space.

Continuing in the same way, to define a general order tensor, consider

I = c α1α2···αrν1ν2···νs aν1aν2 · · · aνnbα1bα2 · · · bαn (103)

If I is invariant under coordinate change, then c α1α2···αrν1ν2···νs defines a mixed tensor,

contravariant to order r and covariant to order s, and transforms as

c β1β2···βrµ1µ2···µs = c α1α2···αr

ν1ν2···νs∂uν1

∂uµ1

∂uν2

∂uµ2· · · ∂u

νs

∂uµs∂uβ1

∂uα1

∂uβ2

∂uα2· · · ∂u

βr

∂uαr(104)

Definition of a general order mixed tensor defined on Euclidean n-space Amixed tensor, contravariant to order r and covariant to order s, is a quantity whose nr+s

components c α1α2···αrν1ν2···νs are transformed according to

c β1β2···βrµ1µ2···µs = c α1α2···αr

ν1ν2···νs∂uν1

∂uµ1

∂uν2

∂uµ2· · · ∂u

νs

∂uµs∂uβ1

∂uα1

∂uβ2

∂uα2· · · ∂u

βr

∂uαr(105)

Problem 20 What’s the definition of a zero order tensor?

Solution 20 A zero order tensor is defined as a scalar. Why is this so? For example,a 2:nd order tensor has n2 components, a 1:st order tensor has n1 components, and thusa zero order tensor should have n0 = 1 components.

Now that the tensor concept has been introduced, the rest of this text is devoted toexplaining why tensors are written c α1α2···αr

ν1ν2···νs and not cα1α2···αrν1ν2···νs .

4.1 The connection with vectors and matrices

Here it is convenient to give a first motivation to why a 2:nd order mixed tensor writtenas a β

α , instead of aβα. When 1:st order tensors are viewed as vectors and 2:nd ordertensors are viewed as matrices, the convention is to let the first index α in a β

α denotethe row number and the second index β denote the column number in the matrix.

16

CONVENTION First order tensor: The index denotes the row in a column vector.Second order tensor: The first index denotes the row and the second index denotes thecolumn in a matrix.

For example if a βα is defined on the Euclidean plane and a 1

1 = 1, a 21 = 2, a 1

2 = 3,a 22 = 4, then this can be written in matrix notation

(a βα ) =

(1 23 4

)(106)

If bβ is defined on Euclidean 4-space and b1 = 1, b2 = 3, b3 = 5, b4 = 7 then this iswritten as a column matrix

(bβ) =

1357

(107)

If the tensor notation would be aβα where α and β are right on top of each other, thenwith the above convention it would not be possible to decide which index denotes therow and column in the matrix. Maybe one could choose the convention so that the lowerindex always denotes the row and the upper index always denotes the column. But then,which index denotes the row in the tensor aαβ?

4.2 The Kronecker delta and symmetric tensors

In equation (78) the Kronecker delta δνµ was only introduced as a symbol that takes thevalue 1 when µ = ν and 0 otherwise. But the Kronecker delta is more than a symbol. Ina given coordinate system uα it is defined as

δνµ =∂uν

∂uµ(108)

and is a mixed second order tensor, contravariant to order 1 and covariant to order 1. Ifthe lower index µ denotes the row and ν the column then δνµ can be written

(δνµ) =

1 0 · · · 00 1 · · · 0...

......

...0 0 · · · 1

(109)

Notice that the indices over the Kronecker delta tensor is written on top of each otherand not δ ν

µ nor δνµ . This is because one can change places of the indices δνµ = δµν .A tensor with this property is called a symmetric tensor and it does not matter whichindex denotes the row and which index denotes the column. Since it is a mixed tensor ittransforms as

δν

µ = δβα∂uα

∂uν∂uµ

∂uβ

=∂uα

∂uν∂uµ

∂uα

=∂uµ

∂uν

= δµν

(110)

17

Thus the Kronecker delta tensor has the same components in all coordinate systems.In Cartesian coordinate systems where there is no difference between contravariant andcovariant components it is usually written with both indices subscript δµν .

4.3 Matrix multiplication and contraction

When doing calculations with 1:st and 2:nd order tensors, it sometimes simplifies cal-culations to take advantage of matrix multiplication. The following problem illustratesthis.

Problem 21 Let Aαβ and Bαβ be two 2:nd order tensors given by

(Aαβ) =(

1 23 4

)(111)

(Bαβ) =(

4 32 1

)(112)

Calculate AαγBγβ !

Solution 21 One way of solving this problem is to write out what the expressionAαγB

γβ is short hand notation for and calculate each case individually:

A11B11 +A12B

21 (α = 1, β = 1)

A11B12 +A12B

22 (α = 1, β = 2)

A21B11 +A22B

21 (α = 2, β = 1)

A21B12 +A22B

22 (α = 2, β = 2)

(113)

Here is an alternative solution is given. In this problem it’s advantegous to make use ofmatrix multiplication. Assume there is a tensor Cβα = AαγB

γβ , where the indices α andβ in Cβα are written right on top of each other because it’s not yet decided which indexdenotes the row and column. Writing out what the expression Cβα = AαγB

γβ is shorthand notation for

C11 = A11B

11 +A12B21

C21 = A11B

12 +A12B22

C12 = A21B

11 +A22B21

C22 = A21B

12 +A22B22

(114)

Rearranging

C11 = A11B

11 +A12B21

C12 = A21B

11 +A22B21

C21 = A11B

12 +A12B22

C22 = A21B

12 +A22B22

(115)

This can be written (C1

1

C12

)=(A11 A12

A21 A22

)(B11

B21

)(116)

18

(C2

1

C22

)=(A11 A12

A21 A22

)(B12

B22

)(117)

Or even more compact (C1

1 C21

C12 C2

2

)=(A11 A12

A21 A22

)(B11 B12

B21 B22

)(118)

Notice that in the matrix with the A:s and B:s, the α denotes the row and β the columnin Aαβ and Bαβ . Considering the matrix with the C:s, we let α denote the row and βthe column in Cβα and writes this from now on C β

α . Thus(C 1

1 C 21

C 12 C 2

2

)=(A11 A12

A21 A22

)(B11 B12

B21 B22

)(119)

Or in tensor notation C βα = AαγB

γβ . Returning now to the original problem, where thesolution is given by

(C βα ) =

(1 23 4

)(4 32 1

)=(

8 520 13

) (120)

Notice that the method used in this problem can be generalized. If instead

(Aαβ) =

A11 A12 · · · A1n

A21 A22 · · · A2n

......

......

An1 An2 · · · Ann

(121)

(Bαβ) =

B11 B12 · · · B1n

B21 B22 · · · B2n

......

......

Bn1 Bn2 · · · Bnn

(122)

then C βα = AαγB

γβ is given by

(C βα ) =

A11 A12 · · · A1n

A21 A22 · · · A2n

......

......

An1 An2 · · · Ann

B11 B12 · · · B1n

B21 B22 · · · B2n

......

......

Bn1 Bn2 · · · Bnn

(123)

If we put two indices, a contravariant and a covariant, equal to each other in a mixedtensor and sum with respect to this pair of indices we obtain a tensor of order one lesscontravariant and one less covariant. This process is called contraction of the giventensor. Contraction of a mixed tensor of second order for example a β

α gives a scalar

a αα = a 1

1 + a 22 + · · ·+ a n

n (124)

Rule to memorize If two 2:nd order tensors are multiplied with each other (for exam-ple AαγBγβ or AαγB

γβ) and the column index of the tensor with matrix representationA is contracted with the row index of the tensor with matrix representation B then theresulting tensor can be calculated by matrix multiplication AB.

19

4.4 The metric tensor

Problem 22 Given the contravariant components aβ and the vectors rβ , determine thecovariant components aα.

Solution 22 Using the definitions of contravariant and covariant components, equa-tions (2) and (56) on pages 1 and 10, aα = rα · v and v = aβrβ

aα = rα · v = rα · aβrβ = rα · rβaβ (125)

Here it is convenient to introduce the metric tensor gαβ which is defined as:

gαβ = rα · rβ (126)

Thusaα = gαβa

β (127)

Problem 23 Show that the metric tensor gαβ is a 2:nd order covariant tensor.

Solution 23 According to the definition of the metric tensor gαβ

gµν = rµ · rν

=(

rα∂uα

∂uµ

)·(

rβ∂uβ

∂uν

)= rα · rβ

∂uα

∂uµ∂uβ

∂uν

= gαβ∂uα

∂uµ∂uβ

∂uν

(128)

which shows that the metric tensor gαβ is indeed a 2:nd order covariant tensor.

Problem 24 In problem 22 the covariant and contravariant components are relatedthrough the metric tensor as aα = gαβa

β . When memorizing this formula, how shall oneremember that it is aα = gαβa

β and not aα = gβαaβ?

Solution 24 The metric tensor is by definition a symmetric tensor gαβ = gβα and thusboth expressions are correct.

Problem 25 Given the covariant components aβ and the vectors rβ , determine thecontravariant components aα.

Solution 25 According to problem 22 the connection between contravariant and co-variant components is given by aα = gαβa

β . Assuming we’re working on Euclidean2-space and since the column index β in gαβ is contracted with the row index β in aβ

then aα = gαβaβ can be written in matrix notation as(

a1

a2

)=(g11 g12g21 g22

)(a1

a2

)(129)

Problem 26 shows that the matrix (gαβ) is invertible, and thus it’s possible to multiplyboth sides with the inverse matrix (gαβ)−1 which gives(

a1

a2

)=(g11 g12g21 g22

)−1(a1

a2

)(130)

20

There is a tensor gαβ which is defined as the inverse of the metric tensor:

(gαβ) = (gαβ)−1 (131)

Here

(gαβ) = (gαβ)−1

=(g11 g12g21 g22

)−1

=1

g11g22 − g21g12

(g22 −g12−g21 g11

) (132)

In the Differential geometry litterature one often finds the notation g for the determinantof the matrix (gαβ). Here it means that g = g11g22 − g21g12 and one can write

(gαβ) =1g

(g22 −g12−g21 g11

)(133)

which means that g11 = g22g , g12 = − g12g , g21 = − g21g , g22 = g11

g . With this notation it’spossible to write (

a1

a2

)=(g11 g12

g21 g22

)(a1

a2

)(134)

Writing this in tensor notationaα = gαβaβ (135)

Equation (135) is not only valid in Euclidean 2-space but also in Euclidean n-space.

Problem 26 On Euclidean n-space the vectors r1, r2, . . . , rn are linearly independentvectors. Show that the n× n-matrix (gαβ) is invertible

(gαβ) =

g11 g12 · · · g1ng21 g22 · · · g2n...

......

...gn1 gn2 · · · gnn

(136)

where gαβ = rα · rβ .

Solution 26 The proof is divided into two steps. The first step proofs the assertionwhen gαβ is defined on Euclidean 2-space. The second step proofs the assertion whengαβ is defined on Euclidean n-space.

Step 1: When defined on the Euclidean plane the metric 4 components g11, g12, g21,g22 and this is written in matrix notation as:

(gαβ) =(g11 g12g21 g22

)(137)

The matrix (gαβ) is invertible if and only if it’s column vectors are linearly independent.Assuming that they are linearly dependent will lead to a contradiction. If they are linearlydependent, this means there is a real number k 6= 0 such that(

g11g21

)= k

(g12g22

)(138)

21

Notice that g12 = r1 · r2 = g21. Notice also that g12 <√g11g22. That this is so can be

seen from the definition of scalar product.

g12 = r1 · r2

= |r1||r2| cos θ(139)

where 0 ≤ θ < π is the angle between the two vectors r1 and r2. Since r1 and r2 arelinearly independent the angle θ 6= 0 which means that cos θ < 1 and

g12 < |r1||r2|

=√|r1|2|r2|2

=√g11g22

(140)

Now using equations

g12 <√g11g22

=√

(kg12)(g21k

)=√g12g21

= g12

(141)

Thus we arrive at the conclusion g12 < g12 which is obviously not true. Thus the twocolumn vectors in the matrix (gαβ) are linearly independent and the invertibilty of (gαβ)follows.

Step 2: Now to prove the general case when gαβ is defined on Euclidean n-space. Iftwo column vectors in the matrix (gαβ) are linearly dependent there is a real numberk 6= 0 such that (here γ is an integer satisfying 1 ≤ γ ≤ n)

Row 1...

Row γ...

Row n

g1α...gγα

...gnα

= k

g1β...gγβ

...gnβ

(142)

for arbitrary integers α and β 6= α such that 1 ≤ α, β ≤ n. Row α and row β means

gαα = kgαβ (143)

gβα = kgββ (144)

Using gαβ = gβα and gαβ <√gααgββ

gαβ <√gααgββ

=√

(kgαβ)(gβαk

)=√gαβgβα

= gαβ

(145)

Thus we arrive at the contradiction gαβ < gαβ , which means that the column vectors inthe matrix (gαβ) must be linearly independent and the matrix (gαβ) is therefore invertible.

22

Problem 27 Given the contravariant components aα of a vector v and the vectors rα,the vector v can be written

v = aαrα (146)

But if the covariant components aβ of the vector v were given instead of the contravariantcomponents aα, the vector v is also specified and therefore there should exist vectors rβ

such that it’s possible to writev = aβrβ (147)

Determine the vectors rβ !

Solution 27 Since the vectors rβ are given, the metric tensor gαβ = rα ·rβ is also given.This in turn implies that the tensor (gαβ) = (gαβ)−1 also is given. The relationshipbetween the contravariant components aα and the covariant components aβ of a vectoris given by

aα = gαβaβ (148)

Assume that the contravariant components of vector rγ is written aαγ and the covariantcomponents of this vector is written a γ

β . This means

aαγ = gαβa γβ (149)

Since aαγ are the contravariant components it’s possible to write

rγ = aαγrα (150)

Butrγ = δγβrβ (151)

Thus a γβ = δγβ

aαγ = gαβa γβ

= gαβδγβ

= gαγ

(152)

rγ = gαγrα (153)

Problem 28 Sincegαβ = rα · rβ (154)

one could suspect thatgαβ = rα · rβ (155)

Show that this is so!

Solution 28 Starting with the expression rα · rβ

rα · rβ = gαµrµ · gβνrν= gαµgβνgµν

(156)

23

Here we shall use gαµgµν = δαν , which is a relationship between gαβ and gαβ that is useda lot in tensor calculations. This follows from (gαβ) = (gαβ)−1, which means that

g11 g12 · · · g1n

g21 g22 · · · g2n

......

......

gn1 gn2 · · · gnn

g11 g12 · · · g1ng21 g22 · · · g2n...

......

...gn1 gn2 · · · gnn

=

1 0 · · · 00 1 · · · 0...

......

...0 0 · · · 1

(157)

Continuing

rα · rβ = gαµgµνgβν

= δαν gβν

= gβα

(158)

Since (gαβ) is a symmetric matrix, the inverse matrix (gαβ) is also a symmetric matrixwhich means gαβ = gβα.

Problem 29 The vectors rβ are linearly independent. A natural question thus arises,are the vectors rγ linearly independent?

Solution 29 Yes, they are linearly independent.

4.5 Raising and lowering an index

We have seen thataα = gαβaβ rα = gαβrβ (159)

aα = gαβaβ rα = gαβrβ (160)

Here we show that generally a νµ 6= aµν

a νµ = gµαa

αν

= gµαgνβaαβ

= (gµα)((aαβ)(gνβ)

) (161)

(gαβ) =(

1 11 2

)(162)

(gαβ) =(

2 −1−1 1

)(163)

(aαβ) =(

1 23 4

)(164)

(aαβ) =(

1 23 4

)(2 −1−1 1

)=(

0 12 1

) (165)

(a νµ ) =

(1 11 2

)(0 12 1

)=(

2 24 3

) (166)

24

4.6 Summary

Definition of a tensor of arbitrary order on Euclidean n-space

Let r, s and n be integers ≥ 1.

• A zero order tensor is defined as a scalar.

• A contravariant tensor of order r is a quantity whose nr components aα1α2···αr aretransformed according to

aβ1β2···βr = aα1α2···αr ∂uβ1

∂uα1

∂uβ2

∂uα2· · · ∂u

βr

∂uαr(167)

under change of coordinate system in Euclidean n-space.

• A covariant tensor of order s is a quantity whose ns components bν1ν2···νs aretransformed according to

bµ1µ2···µs = bν1ν2···νs∂uν1

∂uµ1

∂uν2

∂uµ2· · · ∂u

νs

∂uµs(168)

under change of coordinate system in Euclidean n-space.

• A mixed tensor, contravariant to order r and covariant to order s, is a quantitywhose nr+s components c α1α2···αr

ν1ν2···νs are transformed according to

c β1β2···βrµ1µ2···µs = c α1α2···αr

ν1ν2···νs∂uν1

∂uµ1

∂uν2

∂uµ2· · · ∂u

νs

∂uµs∂uβ1

∂uα1

∂uβ2

∂uα2· · · ∂u

βr

∂uαr(169)

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