1 Tensor Analysis

1.1 Notation and Conventions

Let uα denote the α-component of a vector u, e.g., if u = (1, 4, 2) then u2 = 4. How can we write avector in tensor notation? Let’s show it by using our example:

uα = 1δ1α + 4δ2α + 2δ3α,

where δαβ is the Kronecker symbol.

Representation of matrices follow the same rule: Mαβ is the αβ component of a matrix M . One couldsay that α denotes the row and β the column of a matrix, but one could also do it the other wayaround, where α represents the column and β the row. It is only important to be consistent. We willstick to the first convention. So what would the matrix

Mαβ = 1.5 δα1δβ2

look like? Well, all its components are zero except the component in the first row and the secondcolumn, which is equal to 1.5. Thus,

M =

0 1.5 00 0 00 0 0

.

In what follows we will call objects with one index a tensor of rank 1, an object with two indices atensor of rank 2, and so on.

The scalar product of two vectors u and v is commonly defined as:

s =d

∑

α=1

uαvα, (1)

where d is the dimensionality of the vector space. Real space has d = 3.

Note that the index in equation (1) occurs twice. Also note that there are no vector multiplications,where an index occurs three times. There simply are no (physically) meaningful constructs of theform

∑

α uα vα wα. Moreover, when an index occurs twice, such as in calculating the scalar productor in a matrix multiplication

uα =∑

β

Mαβvβ,

one sums over the index that occurs twice. This leads us to the Einstein summation convention, whichsays that whenever an index occurs twice in a product, then the summation over that index is implied.The equation

s = uαvα, (2)

is thus equivalent with equation (1). However, having the same index occur twice in a sum, does notimply summation convention, i.e.,

uα + vα 6=∑

α

uα + vα.

The dyadic product of two vectors u and v, see also http://en.wikipedia.org/wiki/Dyadic product,can be written as, uαvβ . The result can be interpreted as the αβ element of a matrix, specifically

M = u⊗ v (3a)

(M)αβ = uα vβ . (3b)

Thus, when we multiply two tensors of rank one, we can obtain either a tensor of rank zero, that is,a number, or a tensor of rank two. So what would be the meaning of the expression Mαα? Well, theindex α occurs twice as if we had a scalar product of two vectors uαvα, so we need to sum over theindex α, as if we had an expression uαvα. Thus, the trace of matrix M is given by

Tr(M) = Mαα. (4)

Likewise, when multiplying a “vector” (in fact, a tensor of rank 1) with a “matrix” (in fact, a tensor ofrank 2), we can either form an expression of the type uαMβγ , which would be a tensor of rank three,or one in which two indices are the same , e.g., uαMβα. This would then result in a tensor of rank 1.

Problems:

1. Let the matrix A and the vector r be defined by

Aαβ = 1δα1δβ1 + 2δα1δβ2 + 3δα2δβ1 and rα = 1δα1 + 2δα2.

Calculate the expressions:

(a) rαrβ. [1 mark]

(b) rαrα. [1 mark]

(c) Aαβrβ [1 mark]

(d) AαβAβγ [1 mark]

(e) AαβAβα [1 mark]

Hint: Try to do all the calculations in regular vector notation, i.e., first represent the vector andthe matrix in a form familiar to you. There is no need to state the results in tensor notation.

2. If the product Ar of a matrix A with a vector r is written as Aαβrβ , then what would theexpression Aαβrα correspond to? [1 mark]

3. Calculate Mαα for the matrix [1 mark]

M =

1.3 1.7 2.00.5 10.0 3.37.6 7.7 −7.6

4. What is the rank of the tensor formed by the product of the tensor M , which is of rank 6, andtensor N of rank 3: MαµνγδνNκµα? [1 mark]

5. Would an expression of the sort uα + vβ make sense? Give a reason for your answer. [1 mark]

1.2 The cross-product and the Levi-Civita tensor

Let us define a tensor of rank 3 with the following property:

εαβγ =

+1 if (α, β, γ) is an even permutation of (1,2,3)−1 if (α, β, γ) is an odd permutation of (1,2,3)0 else.

(5)

This tensor is called Levi-Civita tensor or also the antisymmetric symbol. It greatly facilitates anytensor algebra in which cross products or rotations occur. Before making use of the Levi-Civita tensor,

let us examine some of its properties. First, the tensor remains invariant under an even permutationof its indices, i.e.,

εαβγ = εβγα, (6)

while it changes sign upon reversing two indices, that is,

εαβγ = −εβαγ . (7)

The latter equation (as well as its definition) implies that εαββ = 0.

With the definition in equation (5), one can show that the cross-product, u = v ×w can be writtenas:

uα = εαβγvβwγ (8)

It is important that you convince yourself why equation (8) works. You can do this by setting α = 1(so you’d be looking at the x-component) and by applying the summation convention to the indicesβ and γ. Then you insert the definition of εαβγ and as result you will see that

u1 = v2w3 − v3w2.

Similar equations follow for the y and z components of u by permuting the indices.

A useful relation of the Levi-Civita tensor is

εαβγεαµν = δβµδγν − δβνδγµ. (9)

You will be asked to prove this equation in a problem. More importantly, however, you shouldmemorize this equation, which can be done as follows: The result of the product can only be afunction of the indices β, γ, µ and ν. It has to be positive, i.e., equal to (+1), if both εαβγ and εαµν

have the same sign. This is the case when β = µ and γ = ν (and µ 6= γ). It has to be negative, i.e.,equal to (-1), if εαβγ and εαµν have opposite sign, i.e., when µ = γ and β = ν and µ 6= β.

Let us discuss an example. Consider the product, s = (a×b) · (c×d). Using equation (9) along withthe summation convention, s can be expressed as:

s = (εαβγaβbγ) · (εαµνcµdν)

= εαβγεαµν aβbγcµdν

using equation (9) → = (δβµδγν − δβνδγµ) aβbγcµdν

summation convention → = aβcβbνdν − aβdβbγcγ

= (a · c)(b · d)− (a · d)(b · c) (10)

The advantage of using Levi-Civita tensors is fully borne out when some of the vectors in algebraicexpressions are replaced with differential operators and/or with vector functions, as it frequentlyhappens in electromagnetism. Most vector identities, which you needed to memorize in the past, canbe derived in a heart beat once you master the Levi Civita tensor.

Problems:

1. Show the validity of equation (9). [3 marks]

2. Show that εαβγMαβ = 0 if M is a symmetric matrix (Mαβ = Mβα). [1 mark]

3. Using the Levi-Civita tensor, prove the following vector identities:

(a) a · (b× c) = c · (a× b)[1 mark]

(b) a× b× c = b(a · c)− c(a · b)[2 marks]

1.3 Gradient, divergence, rotation, and Laplace operator in tensor notation

It is common to write partial derivatives as ∂/∂x, ∂/∂y, etc. However, in order to incorporate theminto tensor algebra, we will replace them with the symbols ∂1, ∂2, .... Thus, the derivative with respectto the α’th coordinate will be written as ∂α. So we can associate:

gradient: (∇f(r))α = ∂αf(r) (11a)

divergence: ∇ · u(r) = ∂αuα(r) (11b)

rotation: (∇× u(r))α = εαβγ∂βuγ(r) (11c)

Laplace operator: ∆ = ∂α∂α (11d)

All the rules for summation convention, differentiation, Levi-Civita tensor, etc. are maintained andapplied to expressions involving vector functions. One important consequence is that

∂αrβ = δαβ . (12)

Often, one deals with centro-symmetric vector functions f(r), where r = |r|, which is why it is worthmemorizing the expression for ∂αf(r).

∂αf(r) =∂f

∂r

∂r

∂rα(chain rule) (13a)

∂r

∂rα= ∂α

√rβrβ (13b)

chain rule again → =1

2

1√

rβrβ

∂αrγrγ (do not recycle the index β!!!) (13c)

product rule → =1

2

1√

rβrβ(2δαγrγ) (13d)

=rα

r(13e)

and thus finally: ∂αf = f ′(r)rα

r. (13f)

The expression (13f) is not only extremely useful in a variety of contexts in electromagnetism, suchas multipole expansion and dipole radiation, but in many other branches of physics and engineeringas well.

Problems:

1. Use tensor notation to show that ∆ = div grad but ∆ 6= grad div. [1 mark]

2. Let a vector field E be the negative gradient of a scalar field Φ(r), i.e., Eα = −∂αΦ(r). Usingwhat you have learned in this section show that [1 mark]

∇×E = 0.

3. Using the Levi Civita tensor, simplify the following expression for the case in which the compo-nents in u are constants: [1 mark]

(∇× u) · r.

1.4 Taylor series expansion

Without providing the derivation, which proceeds very much like that in one dimension, I would like tostate the Taylor series expansion for more than one variable in tensor notation. If rα can be considered(sufficiently) small, then a function f(R + r) can be Taylor expanded according to:

f(R + r) = f(R) +1

1!∂αf(R)rα +

1

2!∂α∂βf(R)rαrβ +

1

3!∂α∂β∂γf(R)rαrβrγ + ... (14)

I hope you get the idea.

Problems:

1. Expand the function 1/|R + r| into a second-order Taylor series with respect to the “smallparameter”’ r. [2 marks]

2. Expand the function exp[iq · r] into a second-order Taylor series expansion with respect to r.Hint: Don’t make your life unnecessarily difficult. [1 mark]

1.5 Change of basis

Consider an orthonormal basis C = (e1, e2, e3). By definition, the basis vectors satisfy

eα · eβ = δαβ (15)

If we were to define a matrix U , whose components are constructed according to

Uαβ = (eα)β , (16)

then the matrix product of U with its transpose, UT, would satisfy:

(

UUT)

αβ= Uαγ(UT)γβ (17a)

= UαγUβγ (17b)

use definition of U and eq. (15) → = δαβ . (17c)

Matrices satisfying the properties defined in Eq. (17) are called orthonormal matrices. (If the com-ponents can be complex numbers and UT is replaced with the conjugate transpose U+, then U wouldbe called a unitary matrix.) Now consider the expression V defined by

V =1

2EαεαβEβ, (18)

where E is a vector, e.g., the electric field vector, and εαβ is a matrix, for instance the dielectric tensor.V (we may call it the electrostatic field energy) is the scalar product of the vector E with the productof the matrix (or tensor) εαβ and the vector E. V will remain unchanged if we reexpress it as

V =1

2Eα′δα′αεαβδββ′Eβ′ , (19)

We can now represent δα′α as the (α′α) component of the product of an orthogonal matrix with itstranspose, i.e.,

δα′α = Uα′α′′Uαα′′ . (20)

A similar comment applies to δβ′β . Thus,

V =1

2(Eα′Uα′α′′)

(

Uαα′′εαβUββ′′

) (

Uβ′β′′Eβ′

)

. (21)

The interpretation of the first term on the r.h.s. of equation (21),

Eα′′ := Eα′Uα′α′′ (22)

is the following: The Eα′ denote the vector (electric field) components in the coordinates systemspanned by the basis C, while the Eα′′ are the vector (electric field) components expressed in the“original” coordinate system. By association, one may argue that the tensor ε, whose components aredefined as

εα′′β′′ = εαβ Uαα′′Uββ′′ (23)

can be interpreted as the dielectric constant in the “original” coordinate system. Comparing equations(22) and (23) suggests that a tensor M of rank n can be expressed in the original coordinate systemas:

Mα1,α2,...,αn= Mβ1,β2,...,βn

Uα1β1Uα2β2

...Uαnβn. (24)

Problems:

1. An orthogonal matrix is given by: A =

sin ϑ cos ϕ cos ϑ cosϕ − sin ϕsin ϑ sin ϕ cos ϑ sin ϕ cos ϕ

cos ϑ − sin ϑ 0

Show that for each individual combination of (α, β), the following equation is satisfied:

AαγAβγ = δαβ [2 marks]

2. Find the inverse transform to that described in equation (22) and generalize to the case of tensorsof arbitrary rank, which is described in equation (24). [2 marks]

3. Let R be an orthogonal matrix. Which of the following expressions will certainly be identical toδβγ?

(a) RαβRαγ (b) RαβRγα (c) RβαRαγ (d) RβαRγα [2 marks]

4. Let rα = Uβαrβ be the transformation from one orthogonal coordinate system to another one.How do the derivatives ∂α transform? Hint: Require that equation (12) holds in the originaland in the new coordinate system.

Bonus problem

Graphite consists of loosely stacked layers, which are made up of tightly bonded carbon atoms. Thelayers look like a honeycomb lattice and have a 60◦ degree symmetry rotation about the z axis.Moreover, the xy, yz, and xz planes are mirror planes. How many independent components doesgraphite’s dielectric tensor have? [5 marks]

Hint: Whenever you apply a symmetry operation to a tensor that describes a material property ofthe system (such as the dielectric permettivity) you map the system onto itself, and hence you mapthe tensor onto itself. You can therefore require that in equation (24), M must be equal to M if thenew coordinate system is equivalent to the original coordinate system. The “non-trivial” part of thisbonus problem is whether εxx = εyy or εxx 6= εyy.