Lorentz Transformations

We normally think of a vector in Minkowski spacetime in the form \begin{equation} \xx = \begin{pmatrix}t\cr x\cr y\cr z\cr\end{pmatrix} \end{equation} But there is another natural way to package these four degrees of freedom. Consider the complex matrix \begin{equation} \XX = \begin{pmatrix}t+z& x-iy\cr x+iy& t-z\cr\end{pmatrix} \label{spacetime} \end{equation} This matrix is Hermitian, that is \begin{equation} \XX^\dagger = \XX \end{equation} where the Hermitian conjugate of a matrix, denoted by a dagger, is the matrix obtained by taking both the transpose and the (complex) conjugate of the original matrix. Note that $2\times2$ complex Hermitian matrices have precisely 4 (real) independent components; the diagonal elements must be real, and the off-diagonal elements must be (complex) conjugates of each other.

Lorentz transformations are transformations which preserve the “squared length” of $\xx$; this is the (squared) interval of special relativity, defined by \begin{equation} |\xx|^2 = x^2 + y^2 + z^2 - t^2 \end{equation} where the speed of light has been set to $1$. We will only consider those transformations which also preserve the relative orientation of the axes. These transformations form a group, the Lorentz group, denoted $SO(3,1)$. (The $1$ keeps track of the minus sign in the norm.)

How can we express this in terms of the matrix $\XX$? The beauty of this description lies in the fact that the norm of $\xx$ is just (minus) the determinant of $\XX$, that is \begin{equation} -\det\XX = |\xx|^2 = x^2 + y^2 + z^2 - t^2 \end{equation} We therefore seek linear transformations which preserve the determinant.

At first sight, this is easy. Since (complex) determinants satisfy \begin{equation} \det(\XX\YY) = (\det\XX) (\det\YY) \label{det} \end{equation} it would seem that all we need to do is multiply $\XX$ (on either side) by a matrix with determinant $1$. There are two problems with this. First of all, there is no simple condition to ensure that the product $\XX\YY$ is Hermitian! Furthermore, this approach will fail over the other divisions algebras; ($\ref{det}$) is not true for quaternionic matrices.

It is not that hard to solve the first problem. We consider conjugation of $\XX$ by a matrix $\MM$, that is, we consider the transformation \begin{equation} \XX \longmapsto \MM\XX\MM^\dagger \label{Mtrans} \end{equation} Using the property \begin{equation} (\XX\YY)^\dagger = \YY^\dagger \XX^\dagger \end{equation} it is easy to check that $\MM\XX\MM^\dagger$ is Hermitian if $\XX$ is — with no such restriction on $\MM$. We therefore seek complex matrices $\MM$ such that \begin{equation} \det(\MM\XX\MM^\dagger) = \det(\MM) \det(\XX) \det(\MM^\dagger) = \det\XX \end{equation} or equivalently \begin{equation} \det(\MM\MM^\dagger) = \det(\MM) \det(\MM^\dagger) = 1 \nonumber\\ \end{equation} Since \begin{equation} \det\MM^\dagger=\bar{\det\MM} \end{equation} we must have \begin{equation} |\det\MM|=1 \end{equation} However, over $\CC$, we can assume without loss of generality that \hbox{$\det\MM=1$}, since we can always multiply $\MM$ by a complex phase without affecting anything else. We have in fact shown that the group of complex matrices with determinant $1$, written as $SL(2,\CC)$, is essentially 1) the same as the Lorentz group, that is $SL(2,\CC) \approx SO(3,1)$, at least locally.

What do the Lorentz transformations look like? What matrices $\MM$ correspond to spatial rotations? First of all, a rotation by $2\theta$ in the $xy$-plane is given by \begin{equation} \RRR_z = \begin{pmatrix}e^{-i\theta}& 0\cr 0& e^{i\theta}\cr\end{pmatrix} \label{Rz} \end{equation} It is straightforward to check that if \begin{equation} \XX' = \RRR_z \XX \RRR_z^\dagger = \begin{pmatrix}t'+z'& x'-iy'\cr x'+iy'& t'-z'\cr\end{pmatrix} \end{equation} then \begin{align} t' &= t \nonumber\\ x' &= x\cos2\theta-y\sin2\theta \label{Lorentz4}\\ y' &= x\sin2\theta+y\cos2\theta \nonumber\\ z' &= z \nonumber \end{align} which corresponds to a counterclockwise rotation by $2\theta$ in the $xy$-plane as claimed. Of course, ($\ref{Lorentz4}$) can be put in matrix form, as \begin{equation} \xx' = {\LAMBDA} \xx \end{equation} from which the traditional representation of this Lorentz transformation as a $4\times4$ matrix $\LAMBDA$ can easily be determined. Similarly, rotations in the $yz$- and $zx$-planes are given, respectively, by \begin{equation} \RRR_x = \begin{pmatrix}   \cos\theta& -i\sin\theta\cr -i\sin\theta&   \cos\theta\cr \end{pmatrix} \qquad \RRR_y = \begin{pmatrix} \cos\theta& -\sin\theta\cr \sin\theta&   \cos\theta\cr \end{pmatrix} \end{equation} Any rotation can be built up out of these generators.

What about the other Lorentz transformations, namely the ones which “rotate” the time axis? These transformations, called boosts, are at the heart of special relativity; geometrically boosts are simply hyperbolic rotations. A boost in the $zt$-plane takes the form \begin{equation} \BB_z = \begin{pmatrix}e^\beta& 0\cr 0& e^{-\beta}\cr\end{pmatrix} \end{equation} It is straightforward to check that if \begin{equation} \XX' = \BB_z \XX \BB_z^\dagger \end{equation} then \begin{align} t' &= t\cosh2\beta + z\sinh2\beta \nonumber\\ x' &= x \\ y' &= y \nonumber\\ z' &= t\sinh2\beta + z\cosh2\beta \nonumber \end{align} corresponding to a boost in the $zt$-plane by $2\beta$, that is, to a relative speed of $c\tanh2\beta$. Similarly, boosts in the $xt$- and $yt$-planes are given, respectively, by \begin{equation} \BB_x = \begin{pmatrix} \cosh\beta& \sinh\beta\cr \sinh\beta& \cosh\beta\cr \end{pmatrix} \qquad \BB_y = \begin{pmatrix} \cosh\beta& -i\sinh\beta\cr i\sinh\beta&   \cosh\beta\cr \end{pmatrix} \end{equation}

What happens over the other division algebras? Simply replace the complex number $x+iy$ in $\XX$ by a division algebra element $a$, so that \begin{equation} \XX = \begin{pmatrix}t+z& \bar{a}\cr a& t-z\cr\end{pmatrix} \end{equation} Since $a$ has 1, 2, 4, or 8 components depending on whether $a$ is in $\RR$, $\CC$, $\HH$, or $\OO$, respectively, $\XX$ corresponds to a vector in a spacetime with 3, 4, 6, or 10 dimensions. As before, the determinant gives the Lorentzian norm \begin{equation} - \det\XX = |a|^2 + z^2 - t^2 \end{equation} Note that there is no problem defining the determinant here, since even in the octonionic case the components of $\XX$ lie in a complex subalgebra.

We are therefore led to seek transformations of the form ($\ref{Mtrans}$) which preserve the determinant. Even over the quaternions, however, we immediately have a problem: As already noted, ($\ref{det}$) is not true for quaternionic matrices! Even worse, it is not at all obvious how to define the determinant in the first place for non-Hermitian matrices.

Fortunately, there is another identity which comes to the rescue here. \begin{equation} \det(\MM\XX\MM^\dagger) = \det(\MM^\dagger\MM) \det\XX \label{det2} \end{equation} so we need to look for quaternionic matrices $\MM$ satisfying \begin{equation} \det(\MM^\dagger\MM) = 1 \label{detID} \end{equation} The order doesn't matter here, since \begin{equation} \det(\MM\MM^\dagger) \equiv \det(\MM^\dagger\MM) \end{equation} holds over $\HH$.

Over the octonions, the situation is even worse: because of the lack of associativity, ($\ref{Mtrans}$) is not well-defined — nor is it clear that the right-hand-side is Hermitian! We resolve this by restricting to those matrices $\MM$ for which $\MM\XX\MM^\dagger$ is well-defined for all Hermitian $\XX$. It turns out to be sufficient to assume that the components of $\MM$ lie in a complex subalgebra of $\OO$. 2) For such matrices, the transformation ($\ref{Mtrans}$) involves only two independent directions, and is therefore quaternionic. In particular, ($\ref{det2}$) will hold.

It appears to be straightforward to generalize the rotations and boosts given above. $\RRR_y$ still rotates the real direction in $a$ with $z$; $\BB_z$ still yields a boost in the $z$-direction; $\BB_x$ still yields a boost in the $x$-direction. The remaining transformations are nearly as easy. $\BB_y$ yields a boost in the $i$-direction, with obvious generalizations to the other spatial directions obtained by replacing $i$ by $j$, $k$, etc. Similarly, $\RRR_x$ and $\RRR_z$ yield rotations in the plane defined by either the real part of $a$ or $z$, respectively, and the $i$-direction — again generalizing to $j$, $k$, etc.

Counting up what we've got, we see that we have boosts in all directions, as well as all rotations involving either $z$ or the real part of $a$. That's more than enough to generate all the Lorentz transformations except those involving two imaginary directions in $a$. But we know how to do these!

In §2, we obtained a rotation in the $jk$-plane for single quaternions by conjugating with $e^{i\theta}$. That works here as well! Conjugating $\XX$ by the matrix \begin{equation} \RRR_i = \begin{pmatrix}e^{i\theta}& 0\cr 0& e^{i\theta}\cr\end{pmatrix} = e^{i\theta}\II \end{equation} won't touch the diagonal of $\XX$, but will precisely rotate $a$ (and $\bar{a}$!)\ by $2\theta$ in the $jk$-plane. Similarly, replacing $i$ with $j$ and $k$ yields rotations in the other imaginary planes. The phase freedom in the complex case has become an essential ingredient in performing these “internal” rotations over $\HH$!

Over the octonions, we must be a bit more careful. As noted in §3, conjugation by $e^{i\theta}$, and hence by $\RRR_i$, rotates three planes, not just one. But again, we know how to solve this problem: Use flips! For instance, \begin{equation} \XX \longmapsto (i\cos\theta+j\sin\theta) i \XX i (i\cos\theta+j\sin\theta) \end{equation} is a rotation by an angle $2\theta$ in the $ij$-plane. Yet again, we see that the lack of associativity has come to the rescue; the ability to nest transformations is crucial to this construction.

We have therefore obtained an explicit form for the generators of Lorentz transformations in 3, 4, 6, and 10 dimensions. But we have actually shown more, namely that \begin{align} SL(2,\RR) &\approx SO(2,1) \\ SL(2,\CC) &\approx SO(3,1) \\ SL(2,\HH) &\approx SO(5,1) \\ SL(2,\OO) &\approx SO(9,1) \end{align} where in each case the correspondence is local (due to the sign ambiguity). However, the groups $SL(2,\HH)$ and $SL(2,\OO)$ require some explanation. The notation “$SL$” normally means those matrices with determinant $1$, but the determinant of a quaternionic matrix is not well-defined. The generalization? To require precisely ($\ref{detID}$)! Over the octonions, all we must do is add the restriction that ($\ref{Mtrans}$) be well-defined. 3)

1) We say “essentially” because the correspondence is not one-to-one; changing the sign of $\MM$ does not affect the transformation ($\ref{Mtrans}$). This means that $SL(2,\CC)$ is really the double-covering of $SO(3,1)$.
2) The only other possibility is for the columns of the imaginary part of $\MM$ to be real multiples of each other [ 19 ].
3) Wait a minute, how can $SL(2,\OO)$ be a group when $\OO$ isn't associative? Simple; the multiplication in $SL(2,\OO)$ is not matrix multiplication, but composition, that is \begin{align} (\MM_1 \bullet \MM_2) [\XX] = \MM_1 \big[ \MM_2 [\XX] \big] = \MM_1 ( \MM_2\XX\MM_2^\dagger) \MM_1^\dagger \nonumber \end{align} where we have used square brackets to denote the action of an element of the group on a vector, and a bullet to indicate the group operation, which is associative.

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