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\subsubsection*{Tensor Fields}

A tensor field over $\RR^3$ assigns a tensor $T_{i
\cdots k}(\bf{x})$ to each point $\bf{x} \in
\RR^3$. This generalises the vector field
\[ \bf{F} : \RR^3 \to \RR^3 \]
to
\[ T : \RR^3 \to \RR^m \]
with $m = \text{ \# components of the tensor}$.
\myskip
Tensor fields have one further operation: we can
differentiate to build higher rank tensors.

\begin{example*}
    If $\phi$ is a scalar field then
    \[ \nabla \phi = \pfrac{\phi}{x^i} \bf{e}_i \]
    is a vector field, and so
    \[ \pfrac{\phi}{x^i} \]
    transforms as a 1-tensor.
\end{example*}

\noindent
More generally, if $T$ is a $p$-tensor field then
we can construct a $(p + q)$-tensor field
\[ X_{i_1 \cdots i_q j_1 \cdots j_p}(\bf{x}) =
\pfrac{}{x^{i_1}} \cdots \pfrac{}{x^{i_q}} T_{j_1
\cdots j_p}(\bf{x}) \]
To check that this is indeed a tensor, we use
\[ x_i' = R_{ij} x_j \implies x_j = R_{ij} x_i' \]
\[ \implies \pfrac{}{x_i'} = \pfrac{x_j}{x_i'}
\pfrac{}{x_j} = R_{ij} \pfrac{}{x^j} \]
($\pfrac{}{x}$ transforms as a tensor.)

\subsection{Physical Examples}

The simplest examples of tensors are just
matrices. \myskip
In a material, an applied electric field $\bf{E}$
will give a current $\bf{J}$ is given by
\[ J_i = \sigma_{ij} E_j \]
where $\sigma_{ij}$ is the \emph{conductivity
tensor}. This is the grown-up version of
\emph{Ohm's law}.

\begin{note*}
    In 3D, isotropic materials necessarily have
    \[ \sigma_{ij} = \sigma \delta_{ij} \]
    with $\sigma$ the conductivity. \\
    In 2D (i.e. thin materials) then isotropy
    means
    \begin{align*}
        \sigma_{ij}
        &= \delta_{xx} \delta_{ij} + \delta_{xy}
        \eps_{ij} \\
        &=
        \begin{pmatrix}
            \delta_{xx} & \delta_{xy} \\
            -\delta_{xy} & \delta_{xx}
        \end{pmatrix}
    \end{align*}
    ($\sigma_{xy}$ is the \emph{Hall
    conductivity}).
\end{note*}

\noindent
In Newtonian mechanics, a rigid body has
\[ \bf{L} = I \bf{\omega} \]
($\bf{L}$ is angular momentum, $\bf{\omega}$ is
angular velocity), where $I$ is the inertia
tensor. If the body is made of particles of mass
$m_a$, rotating as
\[ \dot{\bf{x}}_a = \bf{\omega} \times \bf{x}_a \]
then
\begin{align*}
    \bf{L}
    &= \sum_a m_a \bf{x}_a \times \dot{\bf{x}}_a
    \\
    &= \sum_a m_a \bf{x}_a \times (\bf{\omega}
    \times \bf{x}_a) \\
    &= \sum_a m_a (|\bf{x}_a|^2 \bf{\omega} -
    (\bf{x}_a \cdot \bf{\omega}) \bf{x}_a) \\
    \implies \bf{L}
    &= I_{ij} \omega_j
\end{align*}
with
\[ I_{ij} = \sum_a m_a (|\bf{x}_a|^2 \delta_{ij} -
(\bf{x}_a)_i (\bf{x}_a)_j) \]
For a continuous object,
\[ I_{ij} = \int_V \rho(\bf{x}) (|\bf{x}|^2
\delta_{ij} - x_i x_j) \dd V \]

\begin{example*}[A Sphere]
    A ball of radius $R$ and density $\rho(r)$ has
    \begin{align*}
        I_{ij}
        &= \int_V \rho(r) (r^2 \delta_{ij} - x_i
        x_j) \dd V \\
        &= \frac{8\pi}{3} \delta_{ij} \int_0^R \dd
        r \rho(r) r^4
    \end{align*}
\end{example*}

\begin{example*}[A Cylinder]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/d0d0728ea2c711ec.png}
    \end{center}
    \[ M = 2\pi a^2 L \rho \]
    In cylindrical polar,
    \[ x = r \cos \phi \qquad x = r \sin \phi \]
    \begin{align*}
        I_{33}
        &= \int_V \rho(x^2 + y^2) \dd V \\
        &= \rho \int_0^{2\pi} \dd \phi \int_0^a
        \dd r \int_{-L}^{+L} \dd z \cdot r \cdot
        r^2 \\
        &= \rho \pi L a^4 \\
        I_{33}
        &= \int_V \rho (y^2 + z^2) \dd V \\
        &= \rho \int_0^{2\pi} \dd \phi \int_0^a
        \dd r \int_{-L}{+L} r(r^2 \sin^2 \phi +
        z^2) \\
        &= \rho \pi a^2 L \left( \half a^2 +
        \frac{2}{3} L^2 \right) \\
        &= I_{22}
        &&\text{by symmetry} \\
        I_{13}
        &= -\rho \int_V xz \dd V \\
        &= -\rho \int_0^{2\pi} \dd \phi \int_0^a
        \dd r \int_{-L}^{+L} \dd z r^2 z \cos \phi
        \\
        &= -\rho \int_0^{2\pi} \dd \phi \cos\phi
        C \\
        &= 0
    \end{align*}
    All other off diagonal entries vanish
    similarly, so for a cylinder
    \[ I = \mathrm{diag} \left( M \left(
    \frac{a^2}{4} + \frac{L^2}{3} \right), M
    \left( \frac{a^2}{4} + \frac{L^2}{3} \right),
    \half Ma^2 \right) \]
\end{example*}

\noindent
For a general body, and a general choice of basis,
$I_{ij}$ will not be diagonal. However, $I_{ij} =
I_{ji}$ so there exist an $R \in \mathrm{SO}(3)$
such that
\[ I' = RIR^\top = \mathrm{diag}(I_1, I_2, I_3) \]
i.e. every body has a preferred set of axes such
that $I$ is diagonal.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/3926a064a2c911ec.png}
\end{center}
From $\bf{L} = I \bf{\omega}$, if the angular
velocity, $\bf{\omega}$ is aligned with one of
these axes then $\bf{L} \parallel \bf{\omega}$.
Otherwise $\bf{L}$ is not parallel to
$\bf{\omega}$ and this is the reason things
wobble! (see classical dynamics).