Other tensors; the tensor of inertia
There are many other examples of tensors appearing in physics. For example, in a metal, or in any conductor, one often finds that the current density $\FLPj$ is approximately proportional to the electric field $\FLPE$; the proportionality constant is called the conductivity $\sigma$: \begin{equation*} \FLPj=\sigma\FLPE. \end{equation*}  For crystals, however, the relation between $\FLPj$ and $\FLPE$ is more complicated; the conductivity is not the same in all directions. The conductivity is a tensor, and we write \begin{equation*} j_i=\sum\sigma_{ij}E_j. \end{equation*}  Another example of a physical tensor is the moment of inertia. In Chapter 18 of Volume I we saw that a solid object rotating about a fixed axis has an angular momentum $L$ proportional to the angular velocity $\omega$, and we called the proportionality factor $I$, the moment of inertia: \begin{equation*} L=I\omega. \end{equation*}  For an arbitrarily shaped object, the moment of inertia depends on its orientation with respect to the axis of rotation. For instance, a rectangular block will have different moments about each of its three orthogonal axes. Now angular velocity $\FLPomega$ and angular momentum $\FLPL$ are both vectors. For rotations about one of the axes of symmetry, they are parallel. But if the moment of inertia is different for the three principal axes, then $\FLPomega$ and $\FLPL$ are, in general, not in the same direction (see Fig. 31–4). They are related in a way analogous to the relation between $\FLPE$ and $\FLPP$. In general, we must write \begin{equation} \begin{alignedat}{4} L_x&=I_{xx}&&\omega_x+I_{xy}&&\omega_y+I_{xz}&&\omega_z,\\[3pt] L_y&=I_{yx}&&\omega_x+I_{yy}&&\omega_y+I_{yz}&&\omega_z,\\[3pt] L_z&=I_{zx}&&\omega_x+I_{zy}&&\omega_y+I_{zz}&&\omega_z. \end{alignedat} \label{Eq:II:31:16} \end{equation}  The nine coefficients $I_{ij}$ are called the tensor of inertia. Following the analogy with the polarization, the kinetic energy for any angular momentum must be some quadratic form in the components $\omega_x$, $\omega_y$, and $\omega_z$: \begin{equation} \label{Eq:II:31:17} \text{KE}=\tfrac{1}{2}\sum_{ij}I_{ij}\omega_i\omega_j. \end{equation}  We can use the energy to define the ellipsoid of inertia. Also, energy arguments can be used to show that the tensor is symmetric—that $I_{ij}=I_{ji}$. The tensor of inertia for a rigid body can be worked out if the shape of the object is known. We need only to write down the total kinetic energy of all the particles in the body. A particle of mass $m$ and velocity $\FLPv$ has the kinetic energy $\tfrac{1}{2}mv^2$, and the total kinetic energy is just the sum \begin{equation*} \sum\tfrac{1}{2}mv^2 \end{equation*}  over all of the particles of the body. The velocity $\FLPv$ of each particle is related to the angular velocity $\FLPomega$ of the solid body. Let’s assume that the body is rotating about its center of mass, which we take to be at rest. Then if $\FLPr$ is the displacement of a particle from the center of mass, its velocity $\FLPv$ is given by $\FLPomega\times\FLPr$. So the total kinetic energy is \begin{equation} \label{Eq:II:31:18} \text{KE}=\sum\tfrac{1}{2}m(\FLPomega\times\FLPr)^2. \end{equation}  Now all we have to do is write $\FLPomega\times\FLPr$ out in terms of the components $\omega_x$, $\omega_y$, $\omega_z$, and $x$, $y$, $z$, and compare the result with Eq. (31.17); we find $I_{ij}$ by identifying terms. Carrying out the algebra, we write \begin{align*} (\FLPomega\times\FLPr)^2&= (\FLPomega\times\FLPr)_x^2+ (\FLPomega\times\FLPr)_y^2+ (\FLPomega\times\FLPr)_z^2\\[1ex] &=(\omega_yz-\omega_zy)^2+ (\omega_zx-\omega_xz)^2+ (\omega_xy-\omega_yx)^2\\[1ex] &=\begin{alignedat}[t]{7} &+\;\omega_y^2&&z^2&&-\;2\omega_y&&\omega_z&&zy&&\;+\;\omega_z^2&&y^2\\ &+\;\omega_z^2&&x^2&&-\;2\omega_z&&\omega_x&&xz&&\;+\;\omega_x^2&&z^2\\[.3ex] &+\;\omega_x^2&&y^2&&-\;2\omega_x&&\omega_y&&yx&&\;+\;\omega_y^2&&x^2. \end{alignedat} \end{align*}  Multiplying this equation by $m/2$, summing over all particles, and comparing with Eq. (31.17), we see that $I_{xx}$, for instance, is given by \begin{equation*} I_{xx}=\sum m(y^2+z^2). \end{equation*}  This is the formula we have had before (Chapter 19, Vol. I) for the moment of inertia of a body about the $x$-axis. Since $r^2=x^2+y^2+z^2$, we can also write this term as \begin{equation*} I_{xx}=\sum m(r^2-x^2). \end{equation*}  Working out all of the other terms, the tensor of inertia can be written as  \begin{equation} \label{Eq:II:31:19} I_{ij}= \begin{bmatrix} \sum m(r^2-x^2) & -\sum mxy & -\sum mxz\\ -\sum myx & \sum m(r^2-y^2) & -\sum myz\\ -\sum mzx & -\sum mzy & \sum m(r^2-z^2) \end{bmatrix}. \end{equation} 
 \begin{gather} \label{Eq:II:31:19} I_{ij}=\\[1ex] \left[\begin{alignedat}{4} \sum &m(r^2\!-x^2)& -\!&\sum\!mxy& -\!&\sum\!mxz&\\[.5ex] -\!&\sum\!myx& \sum &m(r^2\!-y^2)& -\!&\sum\!myz&\\[.5ex] -\!&\sum\!mzx& -\!&\sum\!mzy& \sum &m(r^2\!-z^2)& \end{alignedat}\right]\notag \end{gather} 
 If you wish, this may be written in “tensor notation” as \begin{equation} \label{Eq:II:31:20} I_{ij}=\sum m(r^2\delta_{ij}-r_ir_j), \end{equation}  where the $r_i$ are the components $(x,y,z)$ of the position vector of a particle and the $\sum$ means to sum over all the particles. The moment of inertia, then, is a tensor of the second rank whose terms are a property of the body and relate $\FLPL$ to $\FLPomega$ by \begin{equation} \label{Eq:II:31:21} L_i=\sum_jI_{ij}\omega_j. \end{equation}   For a body of any shape whatever, we can find the ellipsoid of inertia and, therefore, the three principal axes. Referred to these axes, the tensor will be diagonal, so for any object there are always three orthogonal axes for which the angular velocity and angular momentum are parallel. They are called the principal axes of inertia.