% vim: tw=50 % 02/03/2023 09AM \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 $c$.