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\myskip
The \emph{4-momentum} of a massive particle with
rest mass $m > 0$ is
\begin{align*}
    P^\mu
    &= mU^\mu \\
    &= m\gamma_{\bf{u}} (c, \bf{u}) \\
    &= \left( \frac{E}{c}, \bf{p} \right)
\end{align*}
where $E$ is the energy and $\bf{p}$ is the
3-momentum. Its square is
\[ P \cdot P = -\left( \frac{E}{c} \right)^2 +
|\bf{p}|^2 = -m^2 c^2 ,\]
consistent with the relation
\[ E = \sqrt{m^2 c^4 + |\bf{p}|^2 c^2} \]
between energy and momentum. In the Newtonian
limit $|\bf{u}|^2 \ll c^2$, we have $\bf{p} =
m\gamma_{\bf{u}} \bf{u} \approx m\bf{u}$ and
\begin{align*}
    E = m\gamma_{\bf{u}} c^2
    &\approx m \left( 1 +
    \half \frac{|\bf{u}|^2}{c^2} \right) c^2 \\
    E
    &\approx mc^2 + \half m|\bf{u}|^2
\end{align*}
In the rest frame of the particle, $\bf{u} =
\bf{0}$ and $E = mc^2$ which is the \emph{rest
energy}.

\myskip
Massless particles such as photons travel at the
speed of light (in any inertial frame) and have
null world lines. They have $E = |\bf{p}| c$.

\subsubsection*{Covectors and tensors}

Let
\[ Y_\mu = \pfrac{f}{X^\mu} \]
be the \emph{derivative} of a scalar field $f(X)$
on spacetime. How does this transform under a LT?
By the chain rule,
\begin{align*}
    \pfrac{f}{X'^\mu}
    &= \pfrac{X^\nu}{X'^\mu}
    \pfrac{f}{X^\nu} \\
    Y'_\mu
    &= (\Lambda^{-1})^\nu{}_\mu Y_\nu \tag{2}
\end{align*}
where
\[ (\Lambda^{-1})^\nu{}_\mu =
\pfrac{X^\nu}{X'^\mu} \]
is the inverse LT matrix, satisfying
\[ \Lambda^\mu{}_\nu (\Lambda^{-1})^\nu{}_\mu =
\pfrac{X'^\mu}{X^\nu} \pfrac{X^\nu}{X'^\rho} =
\pfrac{X'^\mu}{X'^\rho} = \delta^\mu{}_\rho \]
The derivative is an example of a \emph{covector}.
Generally a covector transforms as (2):
\[ Y' = \Lambda^{-1} Y .\]
A (4-)tensor of type $(r, s)$ is an object whose
$4^{r + s}$ components transform as
\[ T'^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s} =
\Lambda^{\mu_1}{}_{\rho_1} \cdots
\Lambda^{\mu_r}{}_{\rho_r}
(\Lambda^{-1})^{\sigma_1}{}_{\nu_1} \cdots
(\Lambda^{-1})^{\sigma_3}_{\nu_3} T^{\rho_1 \cdots
\rho_r}{}_{\sigma_1 \cdots \sigma_s} \]
under a LT. (Any ordering of the up and down
indices is possible, for example $A_\mu{}^\nu$ or
$B^\mu{}_{\nu\rho}^\sigma$).
\myskip
Each ``up'' index transforms like a vector. Each
``down'' index transforms like a co-vector.
\begin{itemize}
    \item A $(1, 0)$ tensor is a (4-)vector.
    \item A $(0, 1)$ tensor is a covector.
    \item A $(0, 0)$ tensor is a \emph{scalar} (a
        Lorentz invariant).
\end{itemize}
It is straightdorward to show the following:
\begin{itemize}
    \item A linear combination of $(r, s)$ tensors
        is also an $(r, s)$ tensor, for example
        \[ A\indices{^\mu^\nu_\rho_\sigma} =
        aB\indices{^\mu^\nu_\rho_\sigma} +
        bC\indices{^\mu^\nu_\rho_\sigma} \]
        where $a$ and $b$ are scalars. The free
        indices must match.
    \item The \emph{outer product} of tensors of
        type $(r_1, s_1)$ and $(r_2, s_2)$ is a
        tensor of type $(r_1 + r_2, s_1 + s_2)$,
        for example
        \[ A\indices{^\mu^\nu_\rho_\sigma} =
        B\indices{^\mu^\nu}
        C\indices{_\rho_\sigma} .\]
    \item The \emph{derivative} of a tensor field
        of type $(r, s)$ is a tensor of type $(r,
        s + 1)$, for example
        \[ A\indices{_\mu^\nu^\rho_\sigma} =
        \partial_\mu B\indices{^\nu^\rho_\sigma} \]
        where
        \[ \partial_\mu = \pfrac{}{X^\mu} \]
    \item The \emph{contraction} of an $(r, s)$
        tensor (with $r, s > 0$) on a pair of its
        indices produces an $(r - 1, s - 1)$
        tensor, for example
        \[ A\indices{^\mu_\rho} =
        B\indices{^\mu^\nu_\rho_\nu} \]
        This works only if the pair consists of
        one ``up'' and one ``down'' index.
\end{itemize}
The summation convention on spacetime requires the
repeated Greek index to appear once up and once
down.

\myskip
Note that, for example
\[ \sum_{\mu = 0}^3 X\indices{^\mu}
X\indices{^\mu} = c^2 t^2 + |\bf{x}|^2 \]
is \emph{not} a scalar (Lorentz invariant), unlike
\[ X \cdot X = -c^2 t^2 + |\bf{x}|^2 .\]
A tensor is \emph{isotropic} if its components are
the same in every inertial frame. The metric
tensor
\[ \eta\indices{_\mu_\nu} = \diag(-1, 1, 1, 1) \]
is an isotropic $(0, 2)$ tensor, and its inverse
\[ \eta\indices{^\mu^\nu} = \diag(-1, 1, 1, 1) \]
is an isotropic $(2, 0)$ tensor (\emph{exercise}).
The Kronecker delta symbol
\[ \delta\indices{^\mu_\nu} = \diag(1, 1, 1, 1) \]
is an isotropic $(1, 1)$ tensor (\emph{exercise}).
The Levi-Civita epsilon symbol
\[ \eps\indices{_\mu_\nu_\rho_\sigma} = \begin{cases}
    +1 & \text{if $(\mu, \nu, \rho, \sigma)$ is an
    even permutation} \\
    -1 & \text{if $(\mu, \nu, \rho, \sigma)$ is an
    odd permutation} \\
        0 & \text{otherwise}
\end{cases} \]
is an isotropic \emph{pseudo-tensor} of type
$(0, 4)$ (see Example Sheet 3 Question 5). A
pseudo-tensor of type $(r, s)$ satisfies the same
transformation law as a tensor but with an extra
factor of $\det(\Lambda) = \pm 1$.

\myskip
The LT matrix $\Lambda$ is \emph{not} a tensor. A
\emph{tensor equation} is one in which every term
is a tensor of the same type and transforms
identically under a LT. Suh an equation is said to
be \emph{covariant} and is valid in any inertial
frame.

\subsubsection*{Vector calculus in spactime}

The spacetime derivative of a scalar field $f(X)$
is a covector field
\[ \partial_\mu f = (\partial_0 f, \partial_i f) =
\left( \frac{1}{c} \pfrac{f}{t}, \nabla f \right) \]
The spacetime divergence of a (4-)vector field
$V(X) = (V^0, \bf{v})$ is a scalar field:
\begin{align*}
    \partial_\mu V^\mu
    &= \partial \partial_0 V^0 + \partial_i V^i \\
    &= \frac{1}{c} \pfrac{V^0}{t} + \nabla \cdot
    \bf{v}
\end{align*}
In $\RR^3$ the curl involves the antisymmetric
part of the derivative of a vector field. The
spacetime equivalent is the antisymmetric
derivative of a covector field $Y(X)$, a tensor of
type $(0, 2)$:
\[ \partial_\mu Y\indices{_\nu} - \partial_\nu
Y\indices{_\nu} .\]
The generalisation of the Laplacian $\nabla^2$ is
the \emph{d'Alembertian}, a scalar differential
operator
\begin{align*}
    \Box
    &= \eta\indices{^\mu^\nu}
    \partial_\mu\partial_\nu \\
    &= -\partial_0\partial_0 + \partial_i
    \partial_i \\
    &= -\frac{1}{c^2} \pfrac[2]{}{t} + \nabla^2
\end{align*}
The wave equation for a  scalar field $f(X)$ is
$\Box f = 0$. This is a scalar equation only for
waves that travel at $c$.
