# Introducing the notion of tensors through a variation of a Feynman didactic
approach

Lucas Queiroz<EMAIL_ADDRESS>Faculdade de Física, Universidade
Federal do Pará, 66075-110, Belém, Pará, Brazil  Edson C. M. Nogueira
<EMAIL_ADDRESS>Faculdade de Física, Universidade Federal
do Pará, 66075-110, Belém, Pará, Brazil  Danilo T. Alves<EMAIL_ADDRESS>Faculdade de Física, Universidade Federal do Pará, 66075-110, Belém, Pará,
Brazil  Centro de Física, Universidade do Minho, P-4710-057, Braga, Portugal

(August 27, 2024)

###### Abstract

In one of his books [The Feynmann Lectures on Physics, vol. 2], Feynman
presents a didactic approach to introduce basic ideas about tensors, using, as
a first example, the dependence of the induced polarization of a crystal on
the direction of the applied electric field, and also presenting the energy
ellipsoid as a way of visualizing the polarization tensor. In the present
paper, we propose some variations on Feynman’s didactic approach, considering
as our basic models a single ground-state atom and a carbon dioxide
($\text{CO}_{2}$) molecule, instead of crystals, and introducing a visual
representation of tensors based on the ideas of the Lamé stress ellipsoid,
instead of the energy ellipsoid. With these changes, the resulting didactic
proposal presents a reduction in the prerequisites of physical and
mathematical concepts if compared to Feynman’s original approach, requiring,
for example, no differential calculus and only introductory vector algebra.
The text is written so that it can be used directly as a learning tool for
students (even those in the beginning of the undergraduate course), as well as
for teachers interested in preparing their own materials.

##  I Introduction

A physical system whose properties are the same in all directions is called
isotropic, whereas when they are different in distinct directions, it is
called anisotropic. Feynmam, in Chapter 31 of Ref. [1], points that students
in undergraduate courses have to deal, at some moment of their studies or
future careers, with real situations involving anisotropic properties, as the
electric conductivity, moment of inertia, stress caused by a force acting on a
body, among others. According to him, undergraduate students, in courses of
basic physics, should have some idea about tensors, which are mathematical
objects used to describe the anisotropic properties of systems [1]. Taking
into account that in Chapter 30 of Ref. [1] Feynman discusses the different
properties of crystalline substances in distinct directions, in Chapter 31 [1]
he presents a didactic approach to introduce the description of tensors, using
the dependence of the induced polarization of a crystal on the direction of
the applied electric field, as the basic example of an object with anisotropic
property. He also presents, in Chapter 31 [1], the energy ellipsoid as a way
of visualizing the polarization tensor.

In the present paper, we propose an introduction to the notion of tensors,
inspired by the aforementioned Feynman didactic approach, found in Sections
31-1 to 31-3 of Ref. [1], but considering a single ground state atom and a
$\text{CO}_{2}$ molecule, instead of a crystal, as the examples that guide the
initial discussions. Moreover, we propose a preliminary visual representation
of tensors, based on the ideas of the Lamé stress ellipsoid [2], instead of
the energy ellipsoid (as done in Ref. [1]). These proposed changes aim to
reduce the prerequisites of physical and mathematical concepts required to
follow the discussion, as well as to facilitate visual representations. For
example, to deal with crystals, in Ref. [1] the polarization P is used, which
is dipole moment per unit volume [3, 1]. Here, to deal with a single atom or
molecule, just the dipole moment p is necessary. As another example, the
perception of the isotropic polarizability of an atom with a spherically
symmetric electron cloud is more direct than that of a cubic crystal.
Therefore, the consideration of a single atom (to illustrate an isotropic
situation), or of a $\text{CO}_{2}$ molecule (to illustrate an anisotropic
one), can simplify the visualization of the polarization properties, if
compared to crystals. About the visualization of a tensor, to deal with an
energy ellipsoid requires, as discussed in Ref. [1], some notion of
differential and integral calculus, and also ideas on the energy per unit
volume required to polarize a crystal. On the other hand, only vector algebra
is required to deal with the Lamé ellipsoid. In this way, the modifications
proposed here are presented as a didactic proposal requiring less
prerequisites if compared to Feynman’s original approach, and can be used
directly as a learning tool for students, as well as for teachers interested
in preparing their own materials.

The paper is organized as follows. We start reviewing some basic concepts: in
Sec. II, we discuss coordinate transformation; in Sec. III, scalars; in Sec.
IV, vectors. In Sec. V, we discuss the polarizability of an isotropic
particle, taking a ground-state atom as example. In Sec. VI, we discuss the
polarizability of an anisotropic particle, considering a $\text{CO}_{2}$
molecule as our basic model: in Sec. VI.1, we discuss the polarizability
tensor; in Sec. VI.2, a diagonal matrix representation of this tensor; in Sec.
VI.3, a visual representation of this tensor, based on the on the ideas of the
Lamé stress ellipsoid; in Sec. VI.4, a visual representation of the
polarizability tensor for a rotated molecule; in Sec. VI.5, a non-diagonal
matrix representation of the tensor for a rotated molecule; in Sec. VI.6, we
return to the oriented molecule as discussed in Secs. VI.2 and VI.3, and
discuss the non-diagonal representation of the polarization tensor, now in a
rotated coordinate system. In Sec. VII we make a brief summary of the main
ideas discussed in this paper. Finally, in Sec. VIII, we present our final
comments.

##  II Basic ideas: coordinate transformation

Let us consider two points, $O$ and $S$, in space. Considering a Cartesian
coordinate system $xyz$, whose origin coincides, for convenience, with $O$,
the points $O$ and $S$ are located by the coordinates $(0,0,0)$ and $(x,y,z)$,
respectively [see Fig. 1].

Figure 1:  The points $O$ and $S$, and the Cartesian coordinate system $xyz$,
whose origin coincides with $O$.

Through the present text, we are discussing how certain quantities described
in a given coordinate system become described in another rotated one
(specifically, we use the correlation between these descriptions to
differentiate scalar, vectors and tensors). Then, let us consider another
Cartesian coordinate system $x^{\prime}y^{\prime}z^{\prime}$, rotated with
respect to $xyz$ as illustrated in Fig. 2, describing the same points $O$ and
$S$ by the coordinates $(0,0,0)$ and $(x^{\prime},y^{\prime},z^{\prime})$,
respectively.

Figure 2:  (a) Two points, $O$ and $S$, and two Cartesian coordinate systems:
$xyz$, and $x^{\prime}y^{\prime}z^{\prime}$, being the latter rotated with
respect to the former by an angle $\phi$, keeping the axis $z^{\prime}$
coinciding with $z$. (b) Another visualization of the system
$x^{\prime}y^{\prime}z^{\prime}$, rotated with respect to $xyz$, illustrated
in Fig. (a). Here the axis $z^{\prime}=z$ appears perpendicular to the paper.
Figure 3:  Illustration of the coordinate systems
$x^{\prime}y^{\prime}z^{\prime}$ and $xyz$, which are related by Eq. (4).

Naturally, the point $S$ itself has not been changed, but its description in
$xyz$ is different from that in $x^{\prime}y^{\prime}z^{\prime}$. The relation
between the coordinates $(x^{\prime},y^{\prime},z^{\prime}$) and $(x,y,z)$ are
given by (see Fig. 3):

$\displaystyle x^{\prime}$ $\displaystyle=$
$\displaystyle\cos\phi\;x+\sin\phi\;y,$ (1) $\displaystyle y^{\prime}$
$\displaystyle=$ $\displaystyle-\sin\phi\;x+\cos\phi\;y,$ (2) $\displaystyle
z^{\prime}$ $\displaystyle=$ $\displaystyle z.$ (3)

These relations can be written in matrix notation as

$\left[\begin{array}[]{c}x^{\prime}\\\ y^{\prime}\\\
z^{\prime}\end{array}\right]=\left[\begin{array}[]{ccc}\cos\phi&\sin\phi&0\\\
-\sin\phi&\cos\phi&0\\\ 0&0&1\end{array}\right]\left[\begin{array}[]{c}{x}\\\
{y}\\\ {z}\end{array}\right].$ (4)

Defining the square matrix in this equation as

$\mathbf{R}^{(0)}=\left[\begin{array}[]{ccc}\cos\phi&\sin\phi&0\\\
-\sin\phi&\cos\phi&0\\\ 0&0&1\end{array}\right],$ (5)

one can note that

$\mathbf{R}^{(0)}{\mathbf{R}}^{(0)T}=\mathbf{1},$ (6)

where the superscript $T$ represents the transpose of the matrix, and
$\mathbf{1}$ is the $3\times 3$ identity matrix. Eq. (4) can also be written
in a more compact manner, using the index notation, as

$x^{\prime}_{i}=\sum_{r=1}^{3}{R}^{(0)}_{ir}x_{r},$ (7)

where $(x_{1},x_{2},x_{3})=(x,y,z)$,
$(x^{\prime}_{1},x^{\prime}_{2},x^{\prime}_{3})=(x^{\prime},y^{\prime},z^{\prime})$,
and ${R}^{(0)}_{ir}$ are the elements of the matrix $\mathbf{R}^{(0)}$. In
index notation, Eq. (6) can be rewritten as

$\sum_{k=1}^{3}R^{(0)}_{ik}R^{(0)}_{jk}=\delta_{ij},$ (8)

where $\delta_{ij}$ is the Kronecker delta symbol, defined by [4]

$\delta_{ij}=\begin{cases}0&(\text{if $i\neq j$}),\\\ 1&(\text{if
$i=j$}).\end{cases}$ (9)

For a general rotated Cartesian coordinate system
$x^{\prime}y^{\prime}z^{\prime}$ (as illustrated in Fig. 4), the relation
between the coordinates $(x^{\prime},y^{\prime},z^{\prime}$) and $(x,y,z)$
takes the form [3]

$\left[\begin{array}[]{c}x^{\prime}\\\ y^{\prime}\\\
z^{\prime}\end{array}\right]=\left[\begin{array}[]{ccc}R_{11}&R_{12}&R_{13}\\\
R_{21}&R_{22}&R_{23}\\\
R_{31}&R_{32}&R_{33}\end{array}\right]\left[\begin{array}[]{c}{x}\\\ {y}\\\
{z}\end{array}\right],$ (10)

where $R_{ij}$ are the elements of a matrix R, which can be written, in terms
of the Euler angles, as shown in Appendix A. The explicit form of this matrix
is just for informational purposes, since it is not necessary to follow the
reasoning through this article. However, an important feature of R, relevant
to the present discussion, is that it is orthogonal, which means that

$\mathbf{R}\mathbf{R}^{T}=\mathbf{1}.$ (11)

Note that R is a generalization of the matrix $\textbf{R}^{(0)}$, and Eq. (11)
is a generalization of Eq. (6). Eqs. (10) and (11) can be written in index
notation, respectively, by

$x^{\prime}_{i}=\sum_{j=1}^{3}R_{ij}x_{j},$ (12)

and

$\sum_{k=1}^{3}R_{ik}R_{jk}=\delta_{ij}.$ (13)

Figure 4:  General rotation of the system $x^{\prime}y^{\prime}z^{\prime}$
with respect to $xyz$.

##  III Basic ideas: scalars

From the point of view of the system $xyz$, the distance from $O$ to $S$ is
visually represented by a line segment, and numerically represented by a
number $d$, which is given, in terms of the coordinates and in index notation,
by

$d^{2}=\sum_{i=1}^{3}x_{i}x_{i}.$ (14)

Considering the system $x^{\prime}y^{\prime}z^{\prime}$, the distance from $O$
to $S$, represented by the number $d^{\prime}$ (everything in this system we
indicate by the superscript "$\prime$"), is given by

$d^{\prime 2}=\sum_{i=1}^{3}x^{\prime}_{i}x^{\prime}_{i}.$ (15)

Using Eqs. (7), (8), and (14) in Eq. (15), we obtain

$d^{\prime 2}=d^{2}.$ (16)

Then, the distance from $O$ to $S$ is represented by the same number in both
coordinate systems $xyz$ and $x^{\prime}y^{\prime}z^{\prime}$, as naturally
expected. The distance is an example of a scalar quantity, in the sense that
it is represented by a number invariant under a coordinate transformation, as
that given in Eq. (4). In other words, a scalar is a quantity characterized by
just one number, which is independent of the coordinate system we are using to
describe this quantity [5, 6]; or, according to Feynman [1],

> “… a number independent of the choice of axes”.

For a general rotated Cartesian coordinate system
$x^{\prime}y^{\prime}z^{\prime}$ (as illustrated in Fig. 4), using Eqs. (12)
and (13) in Eq. (15), we obtain again that $d=d^{\prime}$, so that the
distance from $O$ to $S$ is represented in the same manner in the system
$xyz$, or in any rotated system $x^{\prime}y^{\prime}z^{\prime}$. In other
words, the distance, or the value that represents it, does not change under
the operation of a rotation of the coordinate system, or is a symmetry or
symmetrical under this operation. Symmetry is a key concept in understanding
the laws of physics [7]. According to Feynman in Sec. 11-1 in Ref. [7],

> “Professor Hermann Weyl has given this definition of symmetry: a thing is
> symmetrical if one can subject it to a certain operation and it appears
> exactly the same after the operation.”

In this section, we considered the distance between two points as our base
example of a scalar, but other examples of scalar quantities include
temperature, energy, mass, charge, among others.

##  IV Basic ideas: vectors

Let us consider again two points, $O$ and $S$, in space [Fig. 1]. Let us
imagine a point charge $q>0$ at $S$, and another one $-q$ at $O$ [see Fig. 5],
so that, they form an electric dipole.

Figure 5:  (a) A dipole formed by a point charge $-q$ (with $q>0$) at the
point $O$, and $+q$ at $S$. (b) The dipole moment vector p, which
characterizes the dipole formed by these two point charges.

A visual representation of this dipole can be done by an arrow p (or also
$\vec{p}$), called dipole moment vector, pointing from the negative to the
positive charge (this is a convention) [3], whose length is directly
proportional to the product $qd$ ($d$ is the distance between these charges),
as illustrated in Fig. 5. When we consider a Cartesian coordinate system
$xyz$, p is described by the components $(p_{x},p_{y},p_{z})$ (see Fig. 6).

Figure 6:  The Cartesian coordinate system $xyz$, whose origin coincides with
$O$, and the dipole moment vector p, which characterizes the dipole formed by
the charges $-q$ (with $q>0$) and $+q$.

Another Cartesian coordinate system $x^{\prime}y^{\prime}z^{\prime}$, rotated
with respect to $xyz$, as illustrated in Fig. 7, describes p by
$(p_{x}^{\prime},p_{y}^{\prime},p_{z}^{\prime})$.

Figure 7:  The dipole moment vector p, and two Cartesian coordinate systems:
$xyz$, and $x^{\prime}y^{\prime}z^{\prime}$, being the latter rotated with
respect to the former by an angle $\phi$, keeping the axis $z^{\prime}$
coinciding with $z$.

Note that the vector p itself has not been changed, but its description in
$xyz$ is different from that in $x^{\prime}y^{\prime}z^{\prime}$. The relation
between the components $(p_{x}^{\prime},p_{y}^{\prime},p_{z}^{\prime})$ and
$(p_{x},p_{y},p_{z})$ are given by (see Fig. 8):

$\displaystyle p_{x}^{\prime}$ $\displaystyle=$
$\displaystyle\cos\phi\;p_{x}+\sin\phi\;p_{y},$ (17) $\displaystyle
p_{y}^{\prime}$ $\displaystyle=$
$\displaystyle-\sin\phi\;p_{x}+\cos\phi\;p_{y},$ (18) $\displaystyle
p_{z}^{\prime}$ $\displaystyle=$ $\displaystyle p_{z}.$ (19)

Figure 8:  Another visualization of the system
$x^{\prime}y^{\prime}z^{\prime}$ rotated with respect to $xyz$, illustrated in
Fig. 7. Here the axis $z^{\prime}=z$ appears perpendicular to the paper.

These relations can be written in matrix notation as

$\left[\begin{array}[]{c}p_{x}^{\prime}\\\ p_{y}^{\prime}\\\
p_{z}^{\prime}\end{array}\right]=\left[\begin{array}[]{ccc}\cos\phi&\sin\phi&0\\\
-\sin\phi&\cos\phi&0\\\
0&0&1\end{array}\right]\left[\begin{array}[]{c}p_{x}\\\ p_{y}\\\
p_{z}\end{array}\right],$ (20)

and in index notation as

$p^{\prime}_{i}=\sum_{r=1}^{3}{R}^{(0)}_{ir}p_{r},$ (21)

where $(p_{1},p_{2},p_{3})=(p_{x},p_{y},p_{z})$ and
$(p^{\prime}_{1},p^{\prime}_{2},p^{\prime}_{3})=(p_{x}^{\prime},p_{y}^{\prime},p_{z}^{\prime})$.
Note that the square matrix in this equation is the same found in Eq. (4).
This means that the components of p transform in the same way that the
coordinates of a point in space. We can say that the components of a general
vector v, under the rotation of the coordinate system shown in Fig. 3 and
described by Eq. (4), transform like the coordinates $(x,y,z)$, so that

$v_{i}^{\prime}=\sum_{r=1}^{3}R^{(0)}_{ir}v_{r}.$ (22)

In this context, a three-dimensional vector is a set of three quantities which
transform, under a rotation of the coordinate system, in the same manner that
the three-dimensional coordinates of a point in space [8].

A vector is a mathematical object characterized by a magnitude and a direction
associated with it [5, 6]. The coordinates of a point in space, can be
considered, themselves, as components of a vector r named as position vector
(see Fig. 9) [8].

Figure 9:  The position vector r, whose componets are the coordinates
$(x,y,z)$.

For a general rotated Cartesian coordinate system
$x^{\prime}y^{\prime}z^{\prime}$ (as illustrated in Fig. 4), the relation
between the components $(v_{x}^{\prime},v_{y}^{\prime},v_{z}^{\prime}$) and
$(v_{x},v_{y},v_{z})$, of a general vector v, takes the form

$v_{i}^{\prime}=\sum_{r=1}^{3}R_{ir}v_{r}.$ (23)

This means that the components of v transform in the same way that the
coordinates in Eq. (12).

Figure 10:  General rotation of the system $x^{\prime}y^{\prime}z^{\prime}$,
with respect to $xyz$. It is also shown the dipole moment vector p.

From the point of view of the system $xyz$, the magnitude of the vector v,
here called $v$, is such that

$v^{2}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}=\sum_{i=1}^{3}v_{i}v_{i}.$ (24)

From the point of view of the system $x^{\prime}y^{\prime}z^{\prime}$, the
magnitude of v is such that

$v^{\prime 2}=\sum_{i=1}^{3}v^{\prime}_{i}v^{\prime}_{i}.$ (25)

Using (23) and (13) in (25), we obtain that

$v^{\prime}=v.$ (26)

In other words, the magnitude of a vector v is a scalar.

According to Feynman (Chap. 11 in Ref. [7]), a vector is

> “A “directed quantity” (which is really 3 quantities; components $a_{x}$,
> $a_{y}$, $a_{z}$ on three axes)… represented by a single symbol $\vec{a}$”.

(Here, we are using, as a choice, the notation a instead of $\vec{a}$.) There
are some basic operations which involve vectors. First, two vectors u and v
can be added [7, 3],

$\textbf{u}+\textbf{v}=\textbf{v}+\textbf{u}=\textbf{w},$ (27)

so that, the components of w are

${w_{i}}={u_{i}}+{v_{i}}.$ (28)

Another basic operation with vectors is the multiplication by a scalar $\mu$,

$\mu\textbf{v}=\textbf{w},$ (29)

which means

${w_{i}}=\mu{v_{i}}.$ (30)

From two vectors, we can also build a scalar by means of an operation called
scalar product, defined by

$\textbf{v}\cdot\textbf{w}=\sum_{i=1}^{3}v_{i}w_{i},$ (31)

which, from a geometrical point of view is given by

$\textbf{v}\cdot\textbf{w}=vw\cos(\theta),$ (32)

where $\theta$ is the angle between the vectors. Note that, using (23) and
(13), we obtain that
$\sum_{i=1}^{3}v_{i}^{\prime}w_{i}^{\prime}=\sum_{i=1}^{3}v_{i}w_{i}$, so that
the result of the product $\textbf{v}\cdot\textbf{w}$ is an invariant under
rotations of the coordinate system, which characterizes it as a scalar [8].

The unit vectors $\hat{\textbf{x}}$, $\hat{\textbf{y}}$ and
$\hat{\textbf{z}}$, pointing to the $x$, $y$, and $z$ directions,
respectively, form a basis, so that any vector y can be written as

$\textbf{v}=v_{1}\hat{\textbf{x}}+v_{2}\hat{\textbf{y}}+v_{3}\hat{\textbf{z}},$
(33)

where $v_{1}=\textbf{v}\cdot\hat{\textbf{x}}$,
$v_{2}=\textbf{v}\cdot\hat{\textbf{y}}$, and
$v_{3}=\textbf{v}\cdot\hat{\textbf{z}}$. Using the notation
$\hat{\mathbf{e}}_{1}=\hat{\textbf{x}}$,
$\hat{\mathbf{e}}_{2}=\hat{\textbf{y}}$ and
$\hat{\mathbf{e}}_{3}=\hat{\textbf{z}}$, one can write

$\textbf{v}=\sum_{i=1}^{3}v_{i}\hat{\mathbf{e}}_{i}.$ (34)

In this section, we discussed vectors, focusing, as base examples, on the
electric dipole moment and position vectors. Other examples of vector
quantities include force, velocity, acceleration, among others [5].

##  V Polarizability of an isotropic particle

According to Feynman [1],

> “… in physics we usually start out by talking about the special case in
> which the polarizability is the same in all directions, to make life
> easier.”

Following this comment, let us start considering the case where the
polarizability is the same in all directions, in other words, the case where
this property is called isotropic. In this way, we consider a neutral atom in
its ground state. For any atom, no matter how many electrons it contains, in
its ground state the electron distribution around its nucleus has spherical
symmetry [9], as illustrated in Fig. 11. Let us also consider this neutral
atom in the presence of an external uniform electric field E, as illustrated
in Fig. 11. When E is applied, the positive nucleus of the atom is pushed in
the direction of the field, whereas, its electrons are pulled in the opposite
direction, so that the atom becomes polarized [see Fig. 11] [3]. [Note that
the electron cloud becomes deformed (see, for instance, Ref. [9]).] Thus the
electric field E induces in the atom a dipole moment p. Note in Fig. 11 that
the structure of the atom (with the deformed electron cloud) in region $A$, is
identical to that in region $B$, so that, no matter what is the direction of
E, by symmetry arguments it is not expected that p points to other direction
than parallel to E. In fact, the dipole moment p, in this case, is given by

$\text{{p}}=\alpha{\textbf{E}},$ (35)

where $\alpha$ is the atomic polarizability [3], which establishes the
connection between the induced dipole moment p and an incident field E. Note
that $\alpha$ is a scalar, which means that Eq. (35) remains valid if a
coordinate system is rotated with respect to the atom [see Fig. 12], or vice-
versa [see Fig. 12].

Figure 11:  (a) Representation of the electron distribution around the nucleus
of an atom in the ground state. Note that the electron distribution has
spherical symmetry. (b) Representation of the atom in the presence of an
external uniform electric field E. Note that the structure of the atom in
region $A$ (dark), is identical to that in region $B$ (lighter). (c)
Representation of the electron distribution around the nucleus of an atom in
the ground state, in the presence of E. Note that the symmetry between the
regions $A$ and $B$ remains, so that the induced dipole moment p has the same
direction than E.

Figure 12:  (a) General rotation of the system
$x^{\prime}y^{\prime}z^{\prime}$, with respect to $xyz$. It is also shown an
atom in the ground state, so that, its electronic distribution has spherical
symmetry. (b) The atom rotated with respect to a coordinate system $xyz$.

##  VI Polarizability of an anisotropic particle

###  VI.1 Polarizability tensor for a $\text{CO}_{2}$ molecule

When we have different polarizability in different directions, this property
is called anisotropic. In order to discuss this case, we consider a
$\text{CO}_{2}$ molecule, as illustrated in Fig. 13. Note that, different from
the atom considered before, this molecule does not have a spherical symmetry.
Despite this, it has a symmetry axis, named the molecule axis [3], which is
the one that crosses the nuclei of the atoms that form the molecule.

Let us consider the $\text{CO}_{2}$ molecule in the presence of an external
uniform electric field E. In Fig. 13, one can see the case where E is applied
parallel to the molecule axis. In this figure, one can note that the structure
of the $\text{CO}_{2}$ molecule in region $A$, is identical to that in region
$B$, so that, by symmetry arguments, it is not expected that the dipole moment
vector points to other direction than parallel to the molecule axis and in the
same direction of E. In fact, the dipole moment p in this case, is given by

$\textbf{p}=\alpha_{\parallel}\textbf{E},$ (36)

where $\alpha_{\parallel}>0$ is the polarizability of the molecule in the
direction of its axis (the subscript $\parallel$ refers to this direction).
When the electric field E is applied in a direction perpendicular to the
molecule axis, one can see in Fig. 13 that, the structure of the
$\text{CO}_{2}$ molecule in region $A$, is identical to that in region $B$, so
that, by symmetry arguments, it is not expected that the dipole moment vector
points to other direction than perpendicular to the molecule axis and in the
same direction of E. In fact, the dipole moment in this case, is given by

$\textbf{p}=\alpha_{\bot}\textbf{E},$ (37)

where $\alpha_{\bot}$ is the polarizability of the molecule in the directions
perpendicular to its axis (the subscript $\perp$ refers to these directions).
The set of two directions perpendicular to the molecule axis, together with
that parallel to this axis, as illustrated in Fig. 13, are known as the
principal axes of the $\text{CO}_{2}$ molecule.

Figure 13:  (a) Representation of the electron distribution of a
$\text{CO}_{2}$ molecule in the ground state. Note that, different from a
single atom, the electron distribution of this molecule does not have a
spherical symmetry. In (b) and (c), an uniform electric field E (represented
by the arrow) is applied parallel (b), and perpendicular (c) to the molecule
axis. Note that the structure of the $\text{CO}_{2}$ molecule in region A
(dark), is identical to that in region B (lighter), so that, by the symmetry
between the regions A and B, it is not expected that the dipole moment vector
points to other direction than the direction of E, which is expressed in Eq.
(36) [for (b)], and (37) [for (c)]. (d) A set of two directions perpendicular
(indicated by $\bot_{1}$ and $\bot_{2}$), and the one parallel (indicated by
$\parallel$) to the molecule axis.

Moreover, when the electric field E is applied in a direction not coinciding
with one of these principal axes, we have that the polarization p is no longer
in the same direction as the electric field E (this is discussed next). Thus,
the connection between the induced dipole moment p and the applied electric
field E is more complex than that of an isotropic atom. This connection
between p and E in this case is given by the polarizability tensor
$\overleftrightarrow{\alpha}$ of the $\text{CO}_{2}$ molecule. A physical
quantity that is characterized by magnitudes, which are associated to multiple
directions, is called a tensor [5]. In the next section, we discuss a
representation of $\overleftrightarrow{\alpha}$.

###  VI.2 Matrix representation of the polarizability tensor for a
$\text{CO}_{2}$ molecule

Let us start considering a laboratory Cartesian system $xyz$, and put the
$\text{CO}_{2}$ molecule oriented in space in a such way that its principal
axes are parallel to the $xyz$ axes, as shown in Fig. 14 (this is just a
convenient choice to start our discussion).

Figure 14:  (a) A $\text{CO}_{2}$ molecule, oriented in space in a such way
that its principal axes (dotted lines) are parallel to those of the laboratory
$xyz$-system. (b) Another point of view, in which the $z$-axis appears
perpendicular to the paper.

When we apply an electric field $\textbf{E}_{x}$ in the $x$-direction, this
produces, according to Eq. (37), an induced dipole moment $\textbf{p}_{x}$
only in the $x$-direction [see Fig. 15], given by

$\textbf{p}_{x}=\alpha_{\bot}\textbf{E}_{x}.$ (38)

When we apply an electric field $\textbf{E}_{y}$ in the $y$-direction, this
produces, according to Eq. (36), an induced dipole moment $\textbf{p}_{y}$
only in the $y$-direction [see Fig. 15], given by

$\textbf{p}_{y}=\alpha_{\parallel}\textbf{E}_{y}.$ (39)

Lastly, when an electric field $\textbf{E}_{z}$ is applied in the
$z$-direction, this produces an induced dipole moment $\textbf{p}_{z}$ only in
the $z$-direction [see Fig. 15], given by

$\textbf{p}_{z}=\alpha_{\bot}\textbf{E}_{z}.$ (40)

To analyze a superposition of the fields $\textbf{E}_{x}$ and
$\textbf{E}_{y}$, let us consider Feynman’s forwarding:

> “Suppose, in a particular crystal, we find that an electric field
> $\textbf{E}_{1}$ in the $x$-direction produces the polarization
> $\textbf{P}_{1}$ in the $x$-direction. Then we find that an electric field
> $\textbf{E}_{2}$ in the $y$-direction, with the same strength as
> $\textbf{E}_{1}$, produces a different polarization $\textbf{P}_{2}$ in the
> $y$-direction. What would happen if we put an electric field at
> $45\degree$?”

Figure 15:  A $\text{CO}_{2}$ molecule, oriented in space in a such way that
its principal axes (dotted lines) are parallel to those of the laboratory
$xyz$-system. (a) An electric field $\textbf{E}_{x}$, in the $x$-direction,
induces a dipole moment $\textbf{p}_{x}$, only in the $x$-direction. (b) An
electric field $\textbf{E}_{y}$, in the $y$-direction, induces a dipole moment
$\textbf{p}_{y}$, only in $y$-direction. (c) An electric field
$\textbf{E}_{z}$, in the $z$-direction, induces a dipole moment
$\textbf{p}_{z}$, only in $z$-direction.

This Feynman sentence refers to a crystal and a vector polarization
$\textbf{P}_{1}$ (dipole moment per unit volume of the crystal). Here, we use
his sentence, but adapting it to our case, by considering a $\text{CO}_{2}$
molecule, instead of a crystal, and replacing
$\textbf{P}_{1}\to\textbf{p}_{1}$. In this way, the induced dipole moments
$\textbf{p}_{x}$ and $\textbf{p}_{y}$, produced by the electric fields
$\textbf{E}_{x}=E_{x}\hat{\textbf{x}}$ and
$\textbf{E}_{y}=E_{y}\hat{\textbf{y}}$, respectively, are given by Eqs. (36)
and (37). Then, if we apply an electric field
$\textbf{E}=\textbf{E}_{x}+\textbf{E}_{y}$, the induced dipole moment p will
be the vector sum of $\textbf{p}_{x}$ and $\textbf{p}_{y}$, which is given by

$\textbf{p}=\textbf{p}_{x}+\textbf{p}_{y}=\alpha_{\perp}E_{x}\hat{\textbf{x}}+\alpha_{\parallel}E_{y}\hat{\textbf{y}}.$
(41)

Following [1], let us consider $E_{x}=E_{y}$ (it is $E_{1}=E_{2}$, in
Feynman’s notation), which means that E is applied at $45\degree$. Note that,
in this case, the induced dipole moment p is not in the same direction as the
electric field E, as shown in Fig. 16 (this figure corresponds to Fig. 31-1(a)
of Ref. [1]). This occurs because $\alpha_{\parallel}\neq\alpha_{\bot}$, which
results in $p_{y}\neq p_{x}$, even with $E_{x}=E_{y}$. The explanation given
by Feynman [1], in the context of a crystal, can be directly applied to this
case of a $\text{CO}_{2}$ molecule:

> “The polarization is no longer in the same direction as the electric field.
> You can see how that might come about. There may be charges which can move
> easily up and down, but which are rather stiff for sidewise motions. When a
> force is applied at $45\degree$, the charges move farther up than they do
> toward the side. The displacements are not in the direction of the external
> force, because there are asymmetric internal elastic forces.”

Moreover, if we replace “polarization of a crystal” by “dipole moment of a
$\text{CO}_{2}$ molecule”, in the text below [1], we have a comment also valid
for a $\text{CO}_{2}$ molecule:

> “There is, of course, nothing special about $45\degree$. It is generally
> true that the induced polarization of a crystal is not in the direction of
> the electric field.”

This can be seen, for instance, in Fig. 16 where we consider E applied at
$60\degree$ (this results in $E_{x}\neq E_{y}$).

Figure 16:  A $\text{CO}_{2}$ molecule, oriented in space in a such way that
its principal axes (dotted lines) are parallel to those of the laboratory
$xyz$-system. Note that the $z$-axis appears perpendicular to the paper. (a) A
field E, applied at $45\degree$. (b) A field E, applied at $60\degree$. Note
that, in (a) and (b), the induced dipole moment p is not in the same direction
of E.

In the most general case, when a field
$\textbf{E}=\textbf{E}_{x}+\textbf{E}_{y}+\textbf{E}_{z}$ is applied, we have

$\textbf{p}=\alpha_{\bot}E_{x}\hat{\textbf{x}}+\alpha_{\parallel}E_{y}\hat{\textbf{y}}+\alpha_{\bot}E_{z}\hat{\textbf{z}}.$
(42)

Note that, differently from the isotropic particle, for which we can always
write Eq. (35), for an anisotropic particle we cannot simply write p as the
field E multiplied by a constant, which means that the induced dipole moment p
may not be in the same direction as the field E. In this case, we write an
equation with a similar structure of Eq. (35) (something that characterizes
the polarizability, multiplied by the electric field) expressing Eq. (42) as a
matrix equation, i.e.

$\left[\begin{array}[]{c}p_{x}\\\ p_{y}\\\
p_{z}\end{array}\right]=\left[\begin{array}[]{ccc}\alpha_{\bot}&0&0\\\
0&\alpha_{\parallel}&0\\\
0&0&\alpha_{\perp}\end{array}\right]\left[\begin{array}[]{c}E_{x}\\\ E_{y}\\\
E_{z}\end{array}\right].$ (43)

Writing in a more compact form, one has

$\textbf{p}=\boldsymbol{\alpha}\textbf{E},$ (44)

where, here, the vectors p and E are represented by the column matrices in
left and right hand side of Eq. (43), respectively, whereas

$\boldsymbol{\alpha}=\left[\begin{array}[]{ccc}\alpha_{\bot}&0&0\\\
0&\alpha_{\parallel}&0\\\ 0&0&\alpha_{\perp}\end{array}\right].$ (45)

Note that, unlike Eq. (35), in Eq. (44) we are using the bold symbol
$\boldsymbol{\alpha}$. Comparing Eq. (35) with (44), one can see that the
latter one is more complicated, because it shows that the vector p is related
with the vector E by a second-rank Cartesian tensor, represented by the matrix
$\boldsymbol{\alpha}$, whereas the former [Eq. (35)] shows that for a
spherically symmetric charge distribution [illustrated in Fig. 11] the vector
p is related to the vector E by a single number (a scalar) $\alpha$ [10].

We can also write Eq. (44) in index notation (since this notation is commonly
used to deal with tensors, one can also call it as tensor notation), which is
given by

$p_{i}=\sum_{j=1}^{3}\alpha_{ij}E_{j},$ (46)

where $\alpha_{ij}$ are the elements of the matrix $\boldsymbol{\alpha}$,
which is the representation in the system $xyz$ of the polarizability tensor
$\overleftrightarrow{\alpha}$. According to Feynman [1],

> “The tensor $\alpha_{ij}$ should really be called a “tensor of second rank,”
> because it has two indexes. A vector - with one index - is a tensor of the
> first rank, and a scalar - with no index - is a tensor of zero rank.”

It is important to remark that $\overleftrightarrow{\alpha}$ is represented in
$xyz$ system by a diagonal matrix in Eq. (43), which occurs because the
$\text{CO}_{2}$ molecule was chosen having its principal axes parallel to the
$xyz$ axes. However, $\overleftrightarrow{\alpha}$ can have a non-diagonal
representation, as discussed in Secs. VI.5 and VI.6.

###  VI.3 A visual representation of the $\text{CO}_{2}$ polarizability tensor

The polarizability tensor $\overleftrightarrow{\alpha}$ establishes the
connection between the induced dipole moment p and the incident field E. Thus,
a way to have a certain visual representation of
$\overleftrightarrow{\alpha}$, is by means of a visual representation of the
behavior of p in terms of E.

Let us consider the $\text{CO}_{2}$ molecule illustrated in Fig. 14, and apply
E in different directions, but with a same magnitude. In a first moment, we
also consider, for simplicity, E having only $x$\- or $y$-component. In other
words, we consider the situations as illustrated in Figs. 15 and 15, but now
considering that all applied fields have the same magnitude.

When E points in the $x$-direction, one has the dipole moment, renamed
$\textbf{p}_{\text{(min)}}$ (this nomenclature is explained later), given by
$\textbf{p}_{\text{(min)}}=\alpha_{\bot}\textbf{E}$ [see Eq. (37)], as
illustrated in Fig. 17.

Figure 17:  A $\text{CO}_{2}$ molecule, oriented in space in a such way that
its principal axes (dotted lines) are parallel to those of the laboratory
$xyz$-system. Note that the $z$-axis appears perpendicular to the paper. Here
it is illustrated the two possible induced dipole moments p, in the presence
of fields E with a same magnitude, pointing along $y$-direction. (a) The
induced dipole moment $\textbf{p}_{\text{(min)}}$, when E points in the
positive $x$-direction. (b) The induced dipole moment
$\textbf{p}_{\text{(min)}}$, when E points in the negative $x$-direction.

If E points in the $y$-direction, one has the dipole moment, renamed
$\textbf{p}_{\text{(max)}}$, given by
$\textbf{p}_{\text{(max)}}=\alpha_{\parallel}\textbf{E}$ [see Eq. (36)], as
illustrated in Fig. 18.

Figure 18:  A $\text{CO}_{2}$ molecule, oriented in space in a such way that
its principal axes (dotted lines) are parallel to those of the laboratory
$xyz$-system. Note that the $z$-axis appears perpendicular to the paper. Here
it is illustrated the two possible induced dipole moments p, in the presence
of fields E with a same magnitude, pointing along the $y$-direction. (a) The
induced dipole moment $\textbf{p}_{\text{(max)}}$, when E points in the
positive $y$-direction. (b) The induced dipole moment
$\textbf{p}_{\text{(max)}}$, when E points in the negative $y$-direction.

When making a superposition of the images taken from Figs. 17 and 18, we have
Fig. 19, which illustrates the behavior of p in terms of E (with this field
having the same magnitude in all the cases, and pointing to $x$\- or
$y$-direction). We can say that Fig. 19 is a germinal visual representation of
$\overleftrightarrow{\alpha}$.

Figure 19:  A superposition of the images taken from Figs. 17 and 18. Here it
is illustrated the four possible induced dipole moments p for the
$\text{CO}_{2}$ molecule shown in Fig. 14, in the presence of fields E with a
same magnitude, pointing along $x$\- or $y$-direction.

The visual representation of $\overleftrightarrow{\alpha}$ in Fig. 19 does not
take into account the situation in which
$\textbf{E}=\textbf{E}_{x}+\textbf{E}_{y}$. For this more general case, one
has p given by Eq. (42) (with $E_{z}=0$), and two particular cases illustrated
in Figs. 16 and 16. Then, let us consider again the application, on a
$\text{CO}_{2}$ molecule as illustrated in Fig. 14, but with
$\textbf{E}=\textbf{E}_{x}+\textbf{E}_{y}$ applied in different directions,
under the condition that all the applied fields have a same magnitude

$E^{2}=E_{x}^{2}+E_{y}^{2}.$ (47)

Using Eqs. (38) and (39), we get

$\frac{p_{x}^{2}}{E^{2}\alpha_{\perp}^{2}}+\frac{p_{y}^{2}}{E^{2}\alpha_{\parallel}^{2}}=1.$
(48)

We define:

$\displaystyle p_{\text{(min)}}^{2}=E^{2}\alpha_{\perp}^{2},$ (49)
$\displaystyle p_{\text{(max)}}^{2}=E^{2}\alpha_{\parallel}^{2}.$ (50)

Then, we can write (48) as

$\frac{p_{x}^{2}}{p_{\text{(\text{min)}}}^{2}}+\frac{p_{y}^{2}}{p_{\text{(\text{max)}}}^{2}}=1.$
(51)

Note that this is the equation of an ellipse centered at the point $S$ (since
we are considering the vectors E and p with their origin in this point), as
shown in Fig. 20.

Figure 20:  Ellipse corresponding to Eq. (51), which defines all possible
induced dipole moments $\textbf{p}=\textbf{p}_{x}+\textbf{p}_{y}$ (which go
from the the point $S$ to the curve of the ellipse) for the $\text{CO}_{2}$
molecule shown in Fig. 14, in the presence of fields
$\textbf{E}=\textbf{E}_{x}+\textbf{E}_{y}$ with a same magnitude. This figure
is an extension of Fig. 19.

The values of $p_{x}$ and $p_{y}$ satisfying Eq. (51) define all possible
induced dipole moments p for the $\text{CO}_{2}$ molecule shown in Fig. 14, in
the presence of fields $\textbf{E}=\textbf{E}_{x}+\textbf{E}_{y}$ with a same
magnitude $E$. Note that the minimum magnitude of p is $p_{\text{(min)}}$,
along the minor axis of the ellipse, and the maximum magnitude is
$p_{\text{(max)}}$, along the larger axis. This justifies the nomenclature
introduced above.

For a more general visual representation of $\overleftrightarrow{\alpha}$, let
us consider the application, on a $\text{CO}_{2}$ molecule as illustrated in
Fig. 14, of a field $\textbf{E}=\textbf{E}_{x}+\textbf{E}_{y}+\textbf{E}_{z}$
applied in different directions, under the condition that all the applied
fields have a same magnitude

$E^{2}=E_{x}^{2}+E_{y}^{2}+E_{z}^{2}.$ (52)

Using Eqs. (38) - (40), we have

$\frac{p_{x}^{2}}{E^{2}\alpha_{\perp}^{2}}+\frac{p_{y}^{2}}{E^{2}\alpha_{\parallel}^{2}}+\frac{p_{z}^{2}}{E^{2}\alpha_{\perp}^{2}}=1.$
(53)

Using Eqs. (49) and (50), we have

$\frac{p_{x}^{2}}{p_{\text{(\text{min)}}}^{2}}+\frac{p_{y}^{2}}{p_{\text{(\text{max)}}}^{2}}+\frac{p_{z}^{2}}{p_{\text{(\text{min)}}}^{2}}=1,$
(54)

which is the equation of an ellipsoid (called in this work Lamé’s ellipsoid),
centered in the point $S$, as shown in Fig. 21.

Figure 21:  Lamé’s ellipsoid, corresponding to Eq. (54), which defines all
possible induced dipole moments
$\textbf{p}=\textbf{p}_{x}+\textbf{p}_{y}+\textbf{p}_{z}$ for the
$\text{CO}_{2}$ molecule shown in Fig. 14, in the presence of fields
$\textbf{E}=\textbf{E}_{x}+\textbf{E}_{y}+\textbf{E}_{z}$ with a same
magnitude. In this sense, this Lamé ellipsoid is a way to get some
visualization of the polarizability tensor for the $\text{CO}_{2}$ molecule
shown in Fig. 14. This figure is an extension of Fig. 20.

This Lamé ellipsoid defines all possible induced dipole moments p for the
$\text{CO}_{2}$ molecule shown in Fig. 14, in the presence of fields E with a
same magnitude. Since the tensor $\overleftrightarrow{\alpha}$ establishes the
behavior of p in terms of E, the Lamé ellipsoid is a way to get some
visualization of this tensor. It is important to remark that the Lamé
ellipsoid was considered originally for a visual representation of the stress
tensor [2]. Here, we constructed a correspondent Lamé ellipsoid for the
polarizability tensor, and simply called it as the Lamé ellipsoid.

###  VI.4 A visual representation of the polarizability tensor for a rotated
$\text{CO}_{2}$ molecule

In this section, we study what happens if the $\text{CO}_{2}$ molecule,
instead of being oriented in space as shown in Fig. 14, is rotated of an angle
$\phi$ with the $y$-direction, as illustrated in Fig. 22. First, it is
important to consider a preliminary example of a rotating object characterized
by a vector. Thus, note that the dipole moment vector p of a
$\text{H}_{2}\text{O}$ molecule changes to a different vector
$\tilde{\textbf{p}}$ when this molecule is rotated. Here, in a similar way,
the polarizability tensor $\overleftrightarrow{\alpha}$, for the
$\text{CO}_{2}$ molecule in Fig. 14, changes to a different tensor
$\widetilde{\overleftrightarrow{\alpha}}$ when this molecule is rotated as in
Fig. 22. This difference occurs because the polarizability tensor establishes
the spatial connection between the induced dipole moment and an incident
electric field, thus, when the molecule rotates, this spatial connection
changes, which means a change in its polarizability tensor.

As we can see in Fig. 22, when the molecule is rotated in the $xy$-plane, and
makes an angle $\phi$ with the $y$-direction, its principal axes
$\tilde{\parallel}$ and $\tilde{\bot}_{1}$ are no longer parallel to the $y$
and $x$, respectively. As has been done before, let us apply electric fields E
on this rotated $\text{CO}_{2}$ molecule, with E applied in different
directions, but under the condition that they have a same magnitude $E$. We
could write $\textbf{E}=\textbf{E}_{x}+\textbf{E}_{y}+\textbf{E}_{z}$, as done
before, but for convenience we write

$\textbf{E}=\textbf{E}_{\tilde{\bot}_{1}}+\textbf{E}_{\tilde{\parallel}}+\textbf{E}_{\tilde{\bot}_{2}},$
(55)

where now we are decomposing the field in the directions ${\tilde{\bot}_{1}}$,
${\tilde{\parallel}}$, and ${\tilde{\bot}_{2}}$. When E points in the
${\tilde{\bot}_{1}}$-direction ($\textbf{E}=\textbf{E}_{\tilde{\bot}_{1}}$),
one has the dipole moment, renamed $\tilde{\textbf{p}}_{\bot_{1}}$, given by
[see Eq. (37)]

$\tilde{\textbf{p}}_{\bot_{1}}=\alpha_{\bot}\textbf{E}_{\tilde{\bot}_{1}},$
(56)

and illustrated in Fig. 17. If E points in the ${\tilde{\parallel}}$-direction
($\textbf{E}=\textbf{E}_{\tilde{\parallel}}$), one has the dipole moment,
renamed $\tilde{\textbf{p}}_{{\tilde{\parallel}}}$, given by [see Eq. (36)]

$\tilde{\textbf{p}}_{\tilde{\parallel}}=\alpha_{\parallel}\textbf{E}_{\tilde{\parallel}},$
(57)

as illustrated in Fig. 18. When E points in the ${\tilde{\bot}_{2}}$-direction
($\textbf{E}=\textbf{E}_{\tilde{\bot}_{2}}$), one has the dipole moment,
renamed $\tilde{\textbf{p}}_{\tilde{\bot}_{2}}$, given by [see Eq. (37)]

$\tilde{\textbf{p}}_{\tilde{\bot}_{2}}=\alpha_{\bot}\textbf{E}_{\tilde{\bot}_{2}}.$
(58)

Note that the constants $\alpha_{\bot}$ and $\alpha_{\parallel}$ appearing in
Eqs. (56) - (58) are the same as those in Eqs. (36) and (37). One can write

$E^{2}={E}_{\tilde{\bot}_{1}}^{2}+{E}_{\tilde{\parallel}}^{2}+{E}_{\tilde{\bot}_{2}}^{2}.$
(59)

Using Eqs. (56)-(58) in Eq. (59), we have

$\frac{\tilde{{p}}_{\tilde{\bot}_{1}}}{E^{2}\alpha_{\perp}^{2}}+\frac{\tilde{{p}}_{\tilde{\parallel}}}{E^{2}\alpha_{\parallel}^{2}}+\frac{\tilde{{p}}_{\tilde{\bot}_{2}}}{E^{2}\alpha_{\perp}^{2}}=1.$
(60)

In this way, from Eqs. (49) and (50), we have

$\frac{\tilde{{p}}_{\tilde{\bot}_{1}}}{p_{\text{\text{(min)}}}^{2}}+\frac{\tilde{{p}}_{\tilde{\parallel}}}{p_{\text{\text{(max)}}}^{2}}+\frac{\tilde{{p}}_{\tilde{\bot}_{2}}}{p_{\text{\text{(min)}}}^{2}}=1.$
(61)

This is the equation of the Lamé ellipsoid (centered in the point $S$), which
is a visual representation of the polarizability tensor
$\widetilde{\overleftrightarrow{\alpha}}$, for the rotated $\text{CO}_{2}$
molecule in Fig. 22. As we can see, as the $\text{CO}_{2}$ molecule rotates,
this Lamé ellipsoide rotates together.

###  VI.5 Matrix representation of the polarizability tensor for a rotated
$\text{CO}_{2}$ molecule

In this section, we discuss the representation of
$\widetilde{\overleftrightarrow{\alpha}}$ (corresponding to the
$\text{CO}_{2}$ molecule illustrated in Fig. 22) in the $xyz$-system. If we
apply an electric field $\textbf{E}_{x}$ in the $x$-direction, one can obtain
that this field can be decomposed into two fields, namely
$\textbf{E}_{\parallel}^{\left(x\right)}$ and
$\textbf{E}_{\perp}^{\left(x\right)}$ [the superscript $(x)$ indicates that
these fields result from the application of $\textbf{E}_{x}$], which are
applied in the directions parallel and perpendicular to the molecule axis,
respectively [see Fig. 23]. We can write their magnitudes in terms of
$|\textbf{E}_{x}|=E_{x}$ as

$\displaystyle E_{\parallel}^{\left(x\right)}$ $\displaystyle=E_{x}\sin\phi,$
(62) $\displaystyle E_{\perp}^{\left(x\right)}$ $\displaystyle=E_{x}\cos\phi,$
(63)

so that we can write $\textbf{E}_{\parallel}^{\left(x\right)}$ as

$\displaystyle\textbf{E}_{\parallel}^{\left(x\right)}$
$\displaystyle=E_{\parallel}^{\left(x\right)}\sin\phi\hat{\textbf{x}}+E_{\parallel}^{\left(x\right)}\cos\phi\hat{\textbf{y}}$
$\displaystyle=E_{x}\sin^{2}\phi\hat{\textbf{x}}+E_{x}\cos\phi\sin\phi\hat{\textbf{y}},$
(64)

and $\textbf{E}_{\perp}^{\left(x\right)}$ as

$\displaystyle\textbf{E}_{\perp}^{\left(x\right)}$
$\displaystyle=E_{\perp}^{\left(x\right)}\cos\phi\hat{\textbf{x}}-E_{\perp}^{\left(x\right)}\sin\phi\hat{\textbf{y}}$
$\displaystyle=E_{x}\cos^{2}\phi\hat{\textbf{x}}-E_{x}\cos\phi\sin\phi\hat{\textbf{y}}.$
(65)

Figure 22:  Illustration of the rotation by an angle $\phi$ (with the
$y$-direction) of the $\text{CO}_{2}$ molecule illustrated in Fig. 14. In (a),
we show a 3D, and in (b), a 2D visualization.

Thus, according to Eqs. (36) and (64), the field
$\textbf{E}_{\parallel}^{\left(x\right)}$ produces a dipole moment
$\textbf{p}_{\parallel}^{\left(x\right)}$, given by

$\displaystyle\textbf{p}_{\parallel}^{\left(x\right)}$
$\displaystyle=\alpha_{\parallel}\textbf{E}_{\parallel}^{\left(x\right)}$
$\displaystyle=\alpha_{\parallel}E_{x}\sin^{2}\phi\hat{\textbf{x}}+\alpha_{\parallel}E_{x}\cos\phi\sin\phi\hat{\textbf{y}},$
(66)

whereas, according to Eqs. (37) and (65), the field
$\textbf{E}_{\perp}^{\left(x\right)}$ produces a dipole moment
$\textbf{p}_{\perp}^{\left(x\right)}$, given by

$\displaystyle\textbf{p}_{\perp}^{\left(x\right)}$
$\displaystyle=\alpha_{\perp}\textbf{E}_{\perp}^{\left(x\right)}$
$\displaystyle=\alpha_{\perp}E_{x}\cos^{2}\phi\hat{\textbf{x}}-\alpha_{\perp}E_{x}\cos\phi\sin\phi\hat{\textbf{y}}.$
(67)

The dipole moment vectors $\textbf{p}_{\parallel}^{\left(x\right)}$ and
$\textbf{p}_{\perp}^{\left(x\right)}$ produce a resultant dipole moment vector
given by the sum
$\textbf{p}^{\left(x\right)}=\textbf{p}_{\parallel}^{\left(x\right)}+\textbf{p}_{\perp}^{\left(x\right)}$
[see Fig. 23], so that, from Eqs. (66) and (67), it can be written as

$\displaystyle\textbf{p}^{\left(x\right)}$
$\displaystyle=(\alpha_{\parallel}\sin^{2}\phi+\alpha_{\perp}\cos^{2}\phi)E_{x}\hat{\textbf{x}}$
$\displaystyle\quad+(\alpha_{\parallel}-\alpha_{\perp})\sin\phi\cos\phi
E_{x}\hat{\textbf{y}}.$ (68)

Figure 23:  (a) $\textbf{E}_{x}$ and its components in $\parallel$ and $\perp$
directions with respect to the molecule axis. (b) The correspondent
polarization $\textbf{p}^{(x)}$ and its components in $\parallel$ and $\perp$
directions.

If we apply an electric field $\textbf{E}_{y}$ in the $y$-direction, one can
obtain that this field can be decomposed into two fields, namely
$\textbf{E}_{\parallel}^{\left(y\right)}$ and
$\textbf{E}_{\perp}^{\left(y\right)}$ [the superscript $(y)$ indicates that
these fields result from the application of $\textbf{E}_{y}$], which are
applied in the directions parallel and perpendicular to the molecule axis,
respectively [see Fig. 24]. We can write their magnitudes in terms of
$|\textbf{E}_{y}|=E_{y}$ as

$\displaystyle E_{\parallel}^{\left(y\right)}$ $\displaystyle=E_{y}\cos\phi,$
(69) $\displaystyle E_{\perp}^{\left(y\right)}$ $\displaystyle=E_{y}\sin\phi,$
(70)

so that we can write $\textbf{E}_{\parallel}^{\left(y\right)}$ as

$\displaystyle\textbf{E}_{\parallel}^{\left(y\right)}$
$\displaystyle=E_{\parallel}^{\left(y\right)}\sin\phi\hat{\textbf{x}}+E_{\parallel}^{\left(y\right)}\cos\phi\hat{\textbf{y}}$
$\displaystyle=E_{y}\sin\phi\cos\phi\hat{\textbf{x}}+E_{y}\cos^{2}\phi\hat{\textbf{y}},$
(71)

and $\textbf{E}_{\perp}^{\left(y\right)}$ as

$\displaystyle\textbf{E}_{\perp}^{\left(y\right)}$
$\displaystyle=-E_{\perp}^{\left(y\right)}\cos\phi\hat{\textbf{x}}+E_{\perp}^{\left(y\right)}\sin\phi\hat{\textbf{y}}$
$\displaystyle=-E_{y}\sin\phi\cos\phi\hat{\textbf{x}}+E_{y}\sin^{2}\phi\hat{\textbf{y}}.$
(72)

Thus, according to Eqs. (36) and (71), the field
$\textbf{E}_{\parallel}^{\left(y\right)}$ produces a dipole moment
$\textbf{p}_{\parallel}^{\left(y\right)}$, given by

$\displaystyle\textbf{p}_{\parallel}^{\left(y\right)}$
$\displaystyle=\alpha_{\parallel}\textbf{E}_{\parallel}^{\left(y\right)}$
$\displaystyle=\alpha_{\parallel}E_{y}\sin\phi\cos\phi\hat{\textbf{x}}+\alpha_{\parallel}E_{y}\cos^{2}\phi\hat{\textbf{y}},$
(73)

whereas, according to Eqs. (37) and (72), the field
$\textbf{E}_{\perp}^{\left(y\right)}$ produces a dipole moment
$\textbf{p}_{\perp}^{\left(y\right)}$, given by

$\displaystyle\textbf{p}_{\perp}^{\left(y\right)}$
$\displaystyle=\alpha_{\perp}\textbf{E}_{\perp}^{\left(y\right)}$
$\displaystyle=-\alpha_{\perp}E_{y}\sin\phi\cos\phi\hat{\textbf{x}}+\alpha_{\perp}E_{y}\sin^{2}\phi\hat{\textbf{y}}.$
(74)

The dipole moment vectors $\textbf{p}_{\parallel}^{\left(y\right)}$ and
$\textbf{p}_{\perp}^{\left(y\right)}$ produce a resultant dipole moment vector
given by the sum
$\textbf{p}^{\left(y\right)}=\textbf{p}_{\parallel}^{\left(y\right)}+\textbf{p}_{\perp}^{\left(y\right)}$
[see Fig. 24], so that, from Eqs. (73) and (74), it can be written as

$\displaystyle\textbf{p}^{\left(y\right)}$
$\displaystyle=\left(\alpha_{\parallel}-\alpha_{\perp}\right)\sin\phi\cos\phi
E_{y}\hat{\textbf{x}}$
$\displaystyle\quad+\left(\alpha_{\parallel}\cos^{2}\phi+\alpha_{\perp}\sin^{2}\phi\right)E_{y}\hat{\textbf{y}}.$
(75)

Figure 24:  (a) $\textbf{E}_{y}$ and its components in $\parallel$ and $\perp$
directions with respect to the molecule axis. (b) The correspondent
polarization $\textbf{p}^{(y)}$ and its components in $\parallel$ and $\perp$
directions.

We can also apply an electric field $\textbf{E}_{z}$ in the $z$-direction, but
in this case, we simply have that this field, according to Eq. (37), produces
a dipole moment $\textbf{p}^{\left(z\right)}$, given by

$\textbf{p}^{\left(z\right)}=\alpha_{\perp}\textbf{E}_{z}=\alpha_{\perp}E_{z}\hat{\textbf{z}},$
(76)

since the molecule still has a principal axis (perpendicular to the molecule
axis) parallel to the $z$-direction.

As we did in the previous section, let us investigate a superposition of the
fields $\textbf{E}_{x}$ and $\textbf{E}_{y}$. Thus, if we apply an electric
field $\textbf{E}=E_{x}\hat{\textbf{x}}+E_{y}\hat{\textbf{y}}$, it produces a
dipole moment p that can be written as a sum of the dipole moments produced by
the fields $\textbf{E}_{x}$ and $\textbf{E}_{y}$ separately. Thus, we can
write p as

$\textbf{p}=\textbf{p}^{\left(x\right)}+\textbf{p}^{\left(y\right)},$ (77)

which, from Eqs. (68) and (75), can be written as

$\displaystyle\textbf{p}=$
$\displaystyle[(\alpha_{\parallel}\sin^{2}\phi+\alpha_{\perp}\cos^{2}\phi)E_{x}+(\alpha_{\parallel}-\alpha_{\perp})\sin\phi$
$\displaystyle\times\cos\phi
E_{y}]\hat{\textbf{x}}+[(\alpha_{\parallel}-\alpha_{\perp})\sin\phi\cos\phi
E_{x}$
$\displaystyle+(\alpha_{\parallel}\cos^{2}\phi+\alpha_{\perp}\sin^{2}\phi)E_{y}]\hat{\textbf{y}}.$
(78)

An illustration of the relation between p and E, given in Eq. (78), is shown
in Fig. 25, for the case $E_{x}=E_{y}$, which means that E is applied at
$45\degree$. Figure 25 corresponds to Fig. 31-1(b) of Ref. [1].

Figure 25:  An illustration of the relation between p and E, given in Eq.
(78), for the case $E_{x}=E_{y}$, which means that E is applied at
$45\degree$.

When a general field
$\textbf{E}=E_{x}\hat{\textbf{x}}+E_{y}\hat{\textbf{y}}+E_{z}\hat{\textbf{z}}$
is applied, we have

$\textbf{p}=\textbf{p}^{\left(x\right)}+\textbf{p}^{\left(y\right)}+\textbf{p}^{\left(z\right)},$
(79)

which, from Eqs. (68), (75) and (76), can be written as

$\displaystyle\textbf{p}=$
$\displaystyle[(\alpha_{\parallel}\sin^{2}\phi+\alpha_{\perp}\cos^{2}\phi)E_{x}+(\alpha_{\parallel}-\alpha_{\perp})\sin\phi$
$\displaystyle\times\cos\phi
E_{y}]\hat{\textbf{x}}+[(\alpha_{\parallel}-\alpha_{\perp})\sin\phi\cos\phi
E_{x}$
$\displaystyle+(\alpha_{\parallel}\cos^{2}\phi+\alpha_{\perp}\sin^{2}\phi)E_{y}]\hat{\textbf{y}}+\alpha_{\perp}E_{z}\hat{\textbf{z}}.$
(80)

We can express this equation as a matrix equation as

$\textbf{p}=\widetilde{\boldsymbol{\alpha}}\textbf{E}$ (81)

where, here, p and E are represented by the column matrices, and
$\tilde{\boldsymbol{\alpha}}$ is given by

$\widetilde{\boldsymbol{\alpha}}=\left[\begin{array}[]{ccc}\alpha_{\parallel}\sin^{2}\phi+\alpha_{\perp}\cos^{2}\phi&(\alpha_{\parallel}-\alpha_{\perp})\sin\phi\cos\phi&0\\\
(\alpha_{\parallel}-\alpha_{\perp})\sin\phi\cos\phi&\alpha_{\parallel}\cos^{2}\phi+\alpha_{\perp}\sin^{2}\phi&0\\\
0&0&\alpha_{\perp}\end{array}\right],$ (82)

which is the representation in the system $xyz$ of the polarizability tensor
$\widetilde{\overleftrightarrow{\alpha}}$ of the rotated $\text{CO}_{2}$
molecule illustrated in Fig. 22. Comparing the matrix $\boldsymbol{\alpha}$ in
Eq. (43), with the matrix $\widetilde{\boldsymbol{\alpha}}$ in Eq. (82), one
can see that in the latter appears off-diagonal elements. Furthermore, note
that
$\widetilde{\alpha}_{xy}=\widetilde{\alpha}_{yx}=(\alpha_{\parallel}-\alpha_{\perp})\sin\phi\cos\phi$,
$\widetilde{\alpha}_{xz}=\widetilde{\alpha}_{zx}=0$ and
$\widetilde{\alpha}_{yz}=\widetilde{\alpha}_{zy}=0$. In a compact form,

$\tilde{\alpha}_{ij}=\tilde{\alpha}_{ji},$ (83)

which means that the matrix $\widetilde{\boldsymbol{\alpha}}$ is symmetric.

Following Feynman [1],

> “We want now to treat the general case of an arbitrary orientation of a
> crystal with respect to the coordinate axes.”

Here, we replace, in this Feynman sentence, a crystal by a $\text{CO}_{2}$
molecule, so that we are going to discuss the case of a $\text{CO}_{2}$
molecule with an arbitrary orientation with respect to the coordinate axes
$xyz$, as illustrated in Fig. 26. In this case, the components of p are
related with the components of E by [1, 3]

$\begin{array}[]{l}p_{x}=\alpha_{xx}E_{x}+\alpha_{xy}E_{y}+\alpha_{xz}E_{z},\\\
p_{y}=\alpha_{yx}E_{x}+\alpha_{yy}E_{y}+\alpha_{yz}E_{z},\\\
p_{z}=\alpha_{zx}E_{x}+\alpha_{zy}E_{y}+\alpha_{zz}E_{z},\end{array}$ (84)

which can be expressed in matrix notation as

$\left[\begin{array}[]{c}p_{x}\\\ p_{y}\\\
p_{z}\end{array}\right]=\left[\begin{array}[]{ccc}\alpha_{xx}&\alpha_{xy}&\alpha_{xz}\\\
\alpha_{yx}&\alpha_{yy}&\alpha_{yz}\\\
\alpha_{zx}&\alpha_{zy}&\alpha_{zz}\end{array}\right]\left[\begin{array}[]{c}E_{x}\\\
E_{y}\\\ E_{z}\end{array}\right].$ (85)

Note that, in this case, we have

$\boldsymbol{\alpha}=\left[\begin{array}[]{ccc}\alpha_{xx}&\alpha_{xy}&\alpha_{xz}\\\
\alpha_{yx}&\alpha_{yy}&\alpha_{yz}\\\
\alpha_{zx}&\alpha_{zy}&\alpha_{zz}\end{array}\right]$ (86)

where all the coefficients $\alpha_{ij}$ (with $i,j=x,y,z$) of this matrix,
can be non nulls. Despite this, the polarizability tensor has, in general, at
most six independent components. This is a consequence of the fact that the
polarizability tensor is a symmetric tensor, which means that its elements
$\alpha_{ij}$ have the property

$\alpha_{ij}=\alpha_{ji},$ (87)

as showed in Eqs. (82) and (83).

Figure 26:  Illustration of a $\text{CO}_{2}$ molecule with an arbitrary
orientation with respect to the coordinate axes $xyz$.

###  VI.6 Matrix representation of the polarizability tensor for a
$\text{CO}_{2}$ molecule in a rotated coordinate system

In Sec. III, we presented the distance $d$ as an example of a scalar, in the
sense that it is an invariant under the rotation of the coordinate system. In
Sec. IV, we defined a vector as an object whose components transform, under a
rotation of the coordinate system, in the same manner that the coordinates of
a point in space. In this section, we return in considering the molecule as
oriented in Sec. VI.2 and VI.3, and discuss how the representation of the
tensor $\overleftrightarrow{\alpha}$, given by the matrix
$\boldsymbol{\alpha}$ in Eq. (43), transforms under a rotation of the
coordinate system, as illustrated in Fig. 27. We remark that, in the present
case, the $\text{CO}_{2}$ molecule stays put in space, whereas the coordinate
system is rotated, so that the tensor $\overleftrightarrow{\alpha}$ itself has
not been changed, but its description in $xyz$ is different from that in
$x^{\prime}y^{\prime}z^{\prime}$.

Figure 27:  A visualization of the system $x^{\prime}y^{\prime}z^{\prime}$
rotated with respect to $xyz$. Here the axis $z^{\prime}=z$ appears
perpendicular to the paper.

Let us remember that, for the coordinate transformation in Fig. 27, the
relation between the coordinates $(x^{\prime},y^{\prime},z^{\prime}$) and
$(x,y,z)$ is given by the matrix $\mathbf{R}_{0}$ [see Eqs. (4) and (5)]. A
shortcut to discover how is the matrix representation
${\boldsymbol{\alpha}^{\prime}}$, of the tensor $\overleftrightarrow{\alpha}$
in the system $x^{\prime}y^{\prime}z^{\prime}$, requires to note that, from
the point of view of the system $x^{\prime}y^{\prime}z^{\prime}$, the
$\text{CO}_{2}$ molecule seems as illustrated in Fig. 28. Moreover, note that,
the non-rotated $\text{CO}_{2}$ molecule seems to the system
$x^{\prime}y^{\prime}z^{\prime}$ in the same way that a rotated
$\text{CO}_{2}$ molecule seems to the system $xyz$, as shown in Fig. 22. Thus,
in a similar reasoning used to find Eq. (82), we can also obtain that
${\boldsymbol{\alpha}^{\prime}}$, the matrix representation of the tensor
$\overleftrightarrow{\alpha}$ in the system $x^{\prime}y^{\prime}z^{\prime}$,
is given by

${\boldsymbol{\alpha}^{\prime}}=\left[\begin{array}[]{ccc}\alpha_{\parallel}\sin^{2}\phi+\alpha_{\perp}\cos^{2}\phi&(\alpha_{\parallel}-\alpha_{\perp})\sin\phi\cos\phi&0\\\
(\alpha_{\parallel}-\alpha_{\perp})\sin\phi\cos\phi&\alpha_{\parallel}\cos^{2}\phi+\alpha_{\perp}\sin^{2}\phi&0\\\
0&0&\alpha_{\perp}\end{array}\right].$ (88)

It is direct to verify that

${\boldsymbol{\alpha}^{\prime}}=\mathbf{R}^{(0)}\boldsymbol{\alpha}\mathbf{R}^{(0)T},$
(89)

which in tensor notation is written as
${\alpha}_{ij}^{\prime}=\sum_{r=1}^{3}\sum_{s=1}^{3}{R}^{(0)}_{ir}\alpha_{rs}{R}^{(0)T}_{sj}$,
which results

${\alpha}_{ij}^{\prime}=\sum_{r=1}^{3}\sum_{s=1}^{3}{R}^{(0)}_{ir}R^{(0)}_{js}\alpha_{rs}.$
(90)

Now, let us remember that the components of a vector transform as given in Eq.
(22). Thus, we can write, considering the components of two vectors v and w,

$v_{i}^{\prime}w_{j}^{\prime}=\sum_{r=1}^{3}\sum_{s=1}^{3}{R}^{(0)}_{ir}R^{(0)}_{js}v_{r}w_{s}.$
(91)

Comparing Eqs. (90) and (91), we see that the quantities ${\alpha}_{ij}$
transform into ${\alpha}_{ij}^{\prime}$, under the coordinate transformation
(7), like the product of components of two vectors [8].

For a general rotation, as illustrated in Fig. 4 and given by Eqs. (10) and
(12), one has that Eqs. (89) and (90), are generalized, respectively, to

$\displaystyle{\boldsymbol{\alpha}^{\prime}}$ $\displaystyle=$
$\displaystyle\mathbf{R}\boldsymbol{\alpha}\mathbf{R}^{T},$ (92)
$\displaystyle{\alpha}_{ij}^{\prime}$ $\displaystyle=$
$\displaystyle\sum_{r=1}^{3}\sum_{s=1}^{3}{R}_{ir}R_{js}\alpha_{rs},$ (93)

Thus, knowing the representation $\boldsymbol{\alpha}$ of the polarizability
tensor relative to an arbitrarily set of axes, we can know its representation
${\boldsymbol{\alpha}^{\prime}}$ in any other rotated system. Adapting to this
case the words of Feynman [1], the polarizability of the $\text{CO}_{2}$
molecule

> “… is described completely by giving the components of the polarization
> tensor $\alpha_{ij}$ with respect to any arbitrarily chosen set of axes.”

And, just as we can associate a position vector $\textbf{r}=(x,y,z)$, or
velocity $\textbf{v}=(v_{x},v_{y},v_{z})$, with the molecule, so that the
components of these vectors change in a certain definite way if we change the
coordinate system [see Eq. (23)], so to the molecule we can also

> “… associate its polarization tensor $\alpha_{ij}$, whose nine components
> will transform in a certain definite way if the coordinate system is
> changed.”

Considering again a general rotation, given in Eqs. (10) and (12), Eq. (91) is
generalized to

$v_{i}^{\prime}w_{j}^{\prime}=\sum_{r=1}^{3}\sum_{s=1}^{3}{R}_{ir}R_{js}v_{r}w_{s}.$
(94)

Comparing Eqs. (93) and (94), we see that the quantities ${\alpha}_{ij}$
transform into ${\alpha}_{ij}^{\prime}$, under the coordinate transformation
(12), like the product of components of two vectors [8]. In other words, a
general second rank tensor is a set of nine quantities (in the present three-
dimensional context) which, under rotations of the coordinate system,
transform like the products of the components of two vectors [8].

Figure 28:  Orientation of the $\text{CO}_{2}$ molecule, from the point of
view of the system $x^{\prime}y^{\prime}z^{\prime}$.

Considering two vectors v and w, a second rank tensor can be built by means of
the tensor product $\textbf{v}\otimes\textbf{w}$, defined so that their
components are “the products of the components of the two vectors of the
product” [11]. Thus,

$(\textbf{v}\otimes\textbf{w})_{ij}=v_{i}w_{j}$ (95)

To illustrate the use of tensor products, let us use them to rewrite the
polarizability tensor in Eq. (45), for a $\text{CO}_{2}$ molecule shown in
Fig. 14. From the definition in Eq. (95), we have:

$\hat{\textbf{x}}\otimes\hat{\textbf{x}}=\left[\begin{array}[]{ccc}1&0&0\\\
0&0&0\\\ 0&0&0\end{array}\right],$ (96)
$\hat{\textbf{y}}\otimes\hat{\textbf{y}}=\left[\begin{array}[]{ccc}0&0&0\\\
0&1&0\\\ 0&0&0\end{array}\right],$ (97)
$\hat{\textbf{z}}\otimes\hat{\textbf{z}}=\left[\begin{array}[]{ccc}0&0&0\\\
0&0&0\\\ 0&0&1\end{array}\right].$ (98)

Then, we can write

$\boldsymbol{\alpha}=\alpha_{\bot}\hat{\textbf{x}}\otimes\hat{\textbf{x}}+\alpha_{\parallel}\hat{\textbf{y}}\otimes\hat{\textbf{y}}+\alpha_{\bot}\hat{\textbf{z}}\otimes\hat{\textbf{z}}.$
(99)

###  VI.7 A general anisotropic particle

A general anisotropic object is one whose diagonal representation of its
polarizability tensor is given by

$\boldsymbol{\alpha}=\left[\begin{array}[]{ccc}\alpha_{a}&0&0\\\
0&\alpha_{b}&0\\\ 0&0&\alpha_{c}\end{array}\right].$ (100)

where $\alpha_{a}\neq\alpha_{b}\neq\alpha_{c}$. Let us consider the principal
axis $a$, $b$, and $c$ (perpendicular to each other) of this anisotropic
object parallel to $x$, $y$, and $z$, respectively. Under the action of an
external electric field E, we have the induced dipole p given by
$\textbf{p}={\textbf{p}}_{x}+{\textbf{p}}_{y}+{\textbf{p}}_{z}$, where:

$\displaystyle{\textbf{p}}_{x}$ $\displaystyle=$
$\displaystyle\alpha_{a}\textbf{E}_{x},$ (101) $\displaystyle{\textbf{p}}_{y}$
$\displaystyle=$ $\displaystyle\alpha_{b}\textbf{E}_{y},$ (102)
$\displaystyle{\textbf{p}}_{z}$ $\displaystyle=$
$\displaystyle\alpha_{c}\textbf{E}_{z}.$ (103)

Using Eqs. (101), (102), and (103), in

$E^{2}={E}_{a}^{2}+{E}_{b}^{2}+{E}_{c}^{2},$ (104)

we have

$\frac{p_{x}^{2}}{E^{2}\alpha_{a}^{2}}+\frac{p_{y}^{2}}{E^{2}\alpha_{b}^{2}}+\frac{p_{z}^{2}}{E^{2}\alpha_{c}^{2}}=1.$
(105)

Choosing $\alpha_{a}<\alpha_{c}<\alpha_{b}$, and defining

$\displaystyle p_{a\text{(min)}}^{2}=E^{2}\alpha_{a}^{2},$ (106)
$\displaystyle p_{b{\text{(max)}}}^{2}=E^{2}\alpha_{b}^{2},$ (107)
$\displaystyle p_{c{\text{(int)}}}^{2}=E^{2}\alpha_{c}^{2},$ (108)

we can rewrite (105) as

$\frac{p_{x}^{2}}{p_{a{\text{(min)}}}^{2}}+\frac{p_{y}^{2}}{p_{b{\text{(max)}}}^{2}}+\frac{p_{z}^{2}}{p_{c{\text{(int)}}}^{2}}=1.$
(109)

Note that the minimum magnitude of p, $p_{a\text{(min)}}$, occurs when the
field E is parallel to the $a$-axis, whereas the maximum,
$p_{b{\text{(max)}}}$, when the field is parallel to the $b$-axis. A certain
intermediate value, $p_{c{\text{(int)}}}$, occurs when the field E is parallel
to the $c$-axis.

##  VII Summary

Inspired by Ref. [5], we organize, in the table shown in Fig. 29, some basic
ideas about a scalar, vector, and second-rank tensor, thus summarizing the
main ideas discussed in this paper.

Figure 29:  Summary of the main ideas discussed in this paper. Each column
exhibits information about a given mathematical object: a scalar (column 1), a
vector (column 2), and a second-rank tensor (column 3). In line 1, we show,
for each mathematical object, a certain visual representation. In line 2, it
is shown the numerical representations in a coordinate system $xyz$. In line
3, we show the numerical representations in a coordinate system
$x^{\prime}y^{\prime}z^{\prime}$, rotated in relation to $xyz$ according to
Eq. (12).

##  VIII Final remarks

We proposed an introduction to the notion of tensors, inspired by the Feynman
didactic approach found in Ref. [1], but with some variations. Instead of
crystals, as the base models, we considered a single ground-state atom (Sec.
V) and a $\text{CO}_{2}$ molecule (Sec. VI), and introduced a visual
representation of tensors based on the ideas of the Lamé stress ellipsoid
(Sec. VI.3), rather than the energy ellipsoid.

To deal with a single atom or molecule, just the dipole moment p is necessary
[as discussed in Eqs. (35) and (44)], whereas to deal with crystals, the
dipole moment per unit volume $(\textbf{P})$ is considered in Ref. [1]. The
perception of the isotropic polarizability of an atom with a spherically
symmetric electron cloud, as discussed using Fig. 11, is more direct than
that, for instance, of a cubic crystal. The visualization of the
polarizabilities along the principal axes of a single $\text{CO}_{2}$
molecule, exploring its symmetries as shown in Fig. 13, is more
straightforward than dealing with a crystal. The visual representation of the
$\text{CO}_{2}$ polarizability tensor was done here by the construction of the
correspondent Lamé ellipsoid, as shown in Eqs. (52)-(54), which did not
require differential calculus, just introductory vector algebra. In
counterpart, to deal with the energy ellipsoid, as discussed in Ref. [1], it
is required some notion of differential and integral calculus, and also ideas
on the energy per unit volume required to polarize a crystal.

In conclusion, the introduction to tensors presented here requires less
mathematical tools and physical concepts than the original Feynman approach.
Thus, it can be used by students still in earlier levels, and helping them to
follow the original Feynman approach [1].

##  Appendix A A general form for R

For a general rotated Cartesian coordinate system
$x^{\prime}y^{\prime}z^{\prime}$ (as illustrated in Fig. 4), the relation
between the coordinates $(x^{\prime},y^{\prime},z^{\prime}$) and $(x,y,z)$
takes the form in Eq. (10) where the square matrix $\mathbf{R}$ is orthogonal.
Although the explicit form of $\mathbf{R}$ is not necessary to follow the
reasoning through the main text of this article, we exhibit, for informational
purposes, the general aspect of this matrix in terms of Euler angles (for more
details about Euler angles, see, for instance, Refs. [12, 13, 14, 15]).
Denoting the Euler angles by $(\phi,\theta,\psi)$, according to the convention
usually adopted in quantum mechanics [14, 15], we have
$x_{i}^{\prime}=\sum_{j}R_{ij}x_{j}$, where $R_{ij}$ are the elements of the
Euler rotation matrix $R(\phi,\theta,\psi)$, given by

$\displaystyle R_{11}$ $\displaystyle=$
$\displaystyle\cos(\theta)\cos(\psi)\cos(\phi)-\sin(\psi)\sin(\phi),$
$\displaystyle R_{12}$ $\displaystyle=$
$\displaystyle\cos(\theta)\cos(\psi)\sin(\phi)+\cos(\phi)\sin(\psi),$
$\displaystyle R_{13}$ $\displaystyle=$
$\displaystyle-\sin(\theta)\cos(\psi),$ $\displaystyle R_{21}$
$\displaystyle=$
$\displaystyle-\cos(\psi)\sin(\phi)-\cos(\theta)\sin(\psi)\cos(\phi),$
$\displaystyle R_{22}$ $\displaystyle=$
$\displaystyle\cos(\psi)\cos(\phi)-\cos(\theta)\sin(\psi)\sin(\phi),$
$\displaystyle R_{23}$ $\displaystyle=$ $\displaystyle\sin(\theta)\sin(\psi),$
$\displaystyle R_{31}$ $\displaystyle=$ $\displaystyle\sin(\theta)\cos(\phi),$
$\displaystyle R_{32}$ $\displaystyle=$ $\displaystyle\sin(\theta)\sin(\phi),$
$\displaystyle R_{33}$ $\displaystyle=$ $\displaystyle\cos(\theta).$

Note that, when considering $\theta=0$ and $\psi=0$, we obtain

$\displaystyle R_{11}$ $\displaystyle=$ $\displaystyle\cos(\phi),$
$\displaystyle R_{12}$ $\displaystyle=$ $\displaystyle\sin(\phi),$
$\displaystyle R_{13}$ $\displaystyle=$ $\displaystyle 0,$ $\displaystyle
R_{21}$ $\displaystyle=$ $\displaystyle-\sin(\phi),$ $\displaystyle R_{22}$
$\displaystyle=$ $\displaystyle\cos(\phi),$ $\displaystyle R_{23}$
$\displaystyle=$ $\displaystyle 0,$ $\displaystyle R_{31}$ $\displaystyle=$
$\displaystyle 0,$ $\displaystyle R_{32}$ $\displaystyle=$ $\displaystyle 0,$
$\displaystyle R_{33}$ $\displaystyle=$ $\displaystyle 1,$

and we recover the matrix $\textbf{R}^{(0)}$ [given in Eq. (5)], showing that
R is a generalization of this matrix.

###### Acknowledgements.

L.Q. and E.C.M.N. were supported by the Coordenação de Aperfeiçoamento de
Pessoal de Nível Superior - Brasil (CAPES), Finance Code 001.

## References

  * Feynman  _et al._ [2006a] R. Feynman, R. Leighton, and M. Sands,  _The Feynman Lectures on Physics_, Vol. II (Pearson/Addison-Wesley, 2006).
  * Fung [1965] Y. C. Fung,  _Foundations of Solid Mechanics_ (Prentice-Hall, INC, 1965).
  * Griffiths [1999] D. J. Griffiths,  _Introduction to Electrodynamics_ (Prentice Hall, 1999).
  * Butkov [1968] E. Butkov,  _Mathematical physics_ (Addison-Wesley, 1968).
  * Fleisch [2011] D. A. Fleisch,  _A Student’s Guide to Vectors and Tensors_ (Cambridge University Press, 2011).
  * Arfken and Weber [2005] G. B. Arfken and H. J. Weber,  _Mathematical Methods for Physicists_, sixth ed. (Elsevier Science, 2005).
  * Feynman  _et al._ [2006b] R. Feynman, R. Leighton, and M. Sands,  _The Feynman Lectures on Physics_, Vol. I (Pearson/Addison-Wesley, 2006).
  * Landau and Lifshitz [2013] L. Landau and E. Lifshitz,  _The Classical Theory of Fields_, Course of Theoretical Physics, Vol. 2 (Elsevier Science, 2013).
  * Purcell [2011] E. Purcell,  _Electricity and Magnetism_, Vol. II (Cambridge University Press, 2011).
  * Keith D. Bonin [1956] V. V. K. Keith D. Bonin,  _Electric-dipole polarizabilities of atoms, molecules and clusters_ (World Scientific, 1956).
  * Cohen-Tannoudji  _et al._ [2019] C. Cohen-Tannoudji, B. Diu, and F. Laloë,  _Quantum Mechanics, Volume 1: Basic Concepts, Tools, and Applications_ (Wiley, 2019).
  * Landau and Lifshitz [1976] L. Landau and E. Lifshitz,  _Mechanics_, Course of Theoretical Physics, Vol. 1 (Elsevier Science, 1976).
  * Goldstein [1980] H. Goldstein,  _Classical Mechanics_, Addison-Wesley series in physics (Addison-Wesley Publishing Company, 1980).
  * Sakurai [1994] J. J. Sakurai,  _Modern Quantum Mechanics; rev. ed._ (Addison-Wesley, Reading, MA, 1994).
  * Ballentine [2014] L. E. Ballentine,  _Quantum mechanics: a modern development; 2nd ed._ (World Scientific, Hackensack, NJ, 2014).