In the previous chapter, we used 2-component vectors to describe spacetime, with one component for time and the other for space. In the case of 3 spatial dimensions, we use 4-component vectors, namely \begin{equation} x^\nu = \pmatrix{x^0\cr x^1\cr x^2\cr x^3\cr} = \pmatrix{\cc t\cr x\cr y\cr z\cr} \end{equation} These are called contravariant vectors, and their indices are written “upstairs”, that is, as superscripts.

Just as before, Lorentz transformations are hyperbolic rotations, which must now be written as $4\times4$ matrices. For instance, a “boost” in the $x$ direction now takes the form \begin{equation} \pmatrix{\cc t'\cr x'\cr y'\cr z'\cr} = \pmatrix{  \cosh\beta& -\sinh\beta& 0& 0\cr -\sinh\beta&   \cosh\beta& 0& 0\cr 0& 0& 1& 0\cr 0& 0& 0& 1\cr} \pmatrix{\cc t\cr x\cr y\cr z\cr} \end{equation} A general Lorentz transformation can be written in the form \begin{equation} x'^\mu = \Lambda^\mu{}_\nu \, x^\nu \end{equation} where $\Lambda^\mu{}_\nu$ are (the components of) the appropriate $4\times4$ matrix, and where we have adopted the Einstein summation convention that repeated indices, in this case $\nu$, are to be summed from $0$ to $3$. In matrix notation, this can be written as \begin{equation} \bx' = \bL\bx \end{equation}

Why are some indices up and others down? In relativity, both special and general, it is essential to distinguish between 2 types of vectors. In addition to contravariant vectors, there are also covariant vectors, often referred to as dual vectors. The dual vector associated with $x^\mu$ is 1) \begin{equation} x_\mu = \pmatrix{-x_0& x_1& x_2& x_3\cr} = \pmatrix{-\cc t& x& y& z\cr} \end{equation} We won't have much need for covariant vectors, but note that the invariance of the interval can be nicely written as \begin{eqnarray} x_\mu x^\mu &=& -\csq t^2 + x^2 + y^2 + z^2 \nonumber\\ &=& x'_\mu x'^\mu \end{eqnarray} (Don't forget the summation convention!) In fact, this property can be taken as the definition of Lorentz transformations, and it is straightforward to determine which matrices $\Lambda^\mu{}_\nu$ are allowed.

Taking the derivative with respect to proper time leads to the 4-velocity \begin{equation} u^\mu = \frac{dx^\mu}{d\tau} = \frac{dx^\mu}{dt} \frac{dt}{d\tau} \end{equation} It is often useful to divide the components of the 4-velocity into space and time in the form \begin{equation} u = \pmatrix{\cc\gamma\cr \vv\gamma} = \pmatrix{\cc\cosh\beta\cr \vhat\,\cc\sinh\beta} \end{equation} where $\vhat$ is the unit vector in the direction of $\vv$. Note that the 4-velocity is a unit vector in the sense that \begin{equation} \cmsq\,{u_\mu u^\mu} = -1 \end{equation} The 4-momentum is is simply the 4-velocity times the rest mass, that is \begin{equation} p^\mu = m u^\mu = \pmatrix{\cm\,{E}\cr \PP} = \pmatrix{m\cc\gamma\cr m\vv\gamma} = \pmatrix{m\cc\cosh\beta\cr \vhat\,m\cc\sinh\beta} \end{equation} and note that \begin{equation} p_\mu p^\mu = -m^2 \csq \end{equation} which is equivalent to our earlier result \begin{equation} E^2 - p^2\csq = m^2 \cfour \end{equation}