# Difference between “Lorentz transformation” and “proper orthochronous”

I'm doing an assignment and I've been given a list of $4 \times 4$ matrices and asked:

Which of the following are Lorentz transformation matrices? Which are proper and orthochronous?

But, as far as I can tell from my notes and internet trawling, the two are synonymous. As in, a matrix $L$ is a proper orthochronous Lorentz transformation if it satisfies:

1. $L^{t}gL=g$, where $g={\rm diag}(-1,1,1,1)$

2. $L_{0}^{0}>0$

3. $det(L)=1$

Am I missing something?

• I have just completed my answer by writing out in full how to tell the difference between the $P$, $T$ and $P\,T$ cosets within $O(1,3)$ and $SO^+(1,3)$ itself. – Selene Routley May 20 '14 at 9:36

Your definitions are in fact those for proper, orthochronous Lorentz transformation, not for general Lorentz transformations, that's why you're having trouble telling the difference! (If it makes you feel any better, yesterday a collegue and I were trying to debug his test setup and two hours of complex testing passed before we two geniusses realised we hadn't switched the power to a key bit of kit on!)

A general Lorentz transformation is defined by criterion 1) alone - it is simply any linear transformation that preserves the quadratic form $t^2 - x^2 - y^2 - z^2$.

The proper, orthochronous transformations are those that belong to the identity connected component $SO^+(1,\,3)$ of the full Lorentz group $O(1,\,3)$. That is, the proper, orthochronous transformations are those that can be reached from the $4\times4$ identity matrix by following a continuous path through the Lorentz group. Equivalently, they are the matrices that are on paths through the Lorentz group defined by the differential equation:

$$\begin{array}{lcl}\mathrm{d}_s L &=& (a_x(s)\, J_x + a_2(s)\, J_y+a_z(s)\, J_z + b_x(s)\, K_x + b_y(s)\, K_y+b_z(s)\, K_z)\,L\\L(0) &=& \mathrm{id}\end{array}\tag{1}$$

where $\mathrm{id}$ is the $4\times 4$ identity, $a_j(s),\,b(s)$ are continuous functions of the parameter $s$ and the $J_j,\,K_J$ are six matrices $4\times 4$ that span the Lie algebra of the Lorentz group, i.e. the real vector space of all possible "tangents to the identity", i.e. all possible values of $\mathrm{d}_s L|_{s=0}$. One possible set is:

$$\begin{array}{lcllcllcl}J_x&=&\left(\begin{array}{cccc}0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&1&0\end{array}\right)&J_y&=&\left(\begin{array}{cccc}0&0&0&0\\0&0&0&1\\0&0&0&0\\0&-1&0&0\end{array}\right)&J_z&=&\left(\begin{array}{cccc}0&0&0&0\\0&0&-1&0\\0&1&0&0\\0&0&0&0\end{array}\right)\\K_x&=&\left(\begin{array}{cccc}0&10&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)&K_y&=&\left(\begin{array}{cccc}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0\end{array}\right)&K_z&=&\left(\begin{array}{cccc}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0\end{array}\right)\end{array}\tag{2}$$

(See how the $J_j$ are skew-Hermitian, thus have pure imaginary eigenvalues, so that $\exp(a_j\, J_j)$ has stuff like $\sin,\,\cos$ of an angle and is a rotation matrix, whereas the $K_j$ are Hermitian, with purely real eigenvalues, so that $\exp(b_j\, K_j)$ has stuff like $\sinh,\,\cosh$ of a rapidity and is a pure boost matrix).

An intuitive description: imagine you are sitting at the console of your spaceship's "hyperdrive": it has two track balls each with their own levers marked "spin" and "boost" and a set of accelerometers - linear and rotational. Your spaceship is initially moving inertially. You roll the trackballs around to set the axis of rotation and direction of boost respectively. When you pull on the levers, the spin lever accelerates the angular speed about the rotation axis, the boost lever accelerates the linear velocity in the boost direction. Otherwise put, the "rotate" trackball and its lever set the superposition weights $a_j(s)$ of the $J_j$ in (1) when we use the definitions in (2) and the "boost" trackball and its lever set the weights $b_j$ of the $K_k$. You go through a control sequence, ending so that your accelerometers read nought, so that now a set of $x,\,y,\,z$ axes attached to your spaceship is moving inertially relative to the beginning frame. The proper, orthochronous transformations are precisely every transformation between the beginning frame and an inertial frame that you can reach with your controls.

However, there are other transformations possible that preserve the quadratic form $t^2 - x^2 - y^2 - z^2$ that don't fulfill your criteria 2. and 3. but they follow only a "simple" pattern that makes them "not much different" from the identity connected component. A discrete subgroup of the full Lorentz group is $\{\mathrm{id},\,P,\,T,\,P\,T\}$ with

$$P=\text{"parity flipper"} = \mathrm{diag}[1,\,-1,\,-1,\,-1];\\T=\text{''time flipper''} = \mathrm{diag}[-1,\,1,\,1,\,1]$$

With the exception of $\mathrm{id}$, none of these can be reached from the identity by paths fulfilling (1). They belong to different connected components from the identity component $SO^+(1,\,3)$. Indeed, the identity connected component is a normal subgroup of the full Lorentz group $SO(1,\,3)$ and the quotient $O(1,\,3) / SO^+(1,\,3)$ is the little group $\{\mathrm{id},\,P,\,T,\,P\,T\}$. So any full Lorentz transformation can be represented as a proper orthochronous transformation followed by one of $P,\,T$ or $P\,T$. There are four separate connected components to the full Lorentz group. (an aside: $\{\mathrm{id},\,P,\,T,\,P\,T\}$ is the Klein "fourgroup": the only possible group of four elements aside from $\mathbb{Z}_4$).

To sniff out a non-proper or non-orthochronous transformation, you do one of two things:

1. Compute the matrix's determinant. If it is -1, then you know it has to include one of $P$ or $T$, so it's not proper or not orthochronous. You can further differentiate the $P$ and $T$ cosets by looking at the $L_0^0$ component of the transformation: the $T$ coset has $L_0^0<0$, since such a transformation swaps the roles of the "future" and "past" (actually reflects Minkowsky vector space in the $t=0$ plane).

2. If the determinant is $+1$, then it may belong to the $P\,T$ coset of $O(1,\,3)$. As in point 1, the $T$ coset and the $P\,T$ coset can be recognised as transformations with $L_0^0<0$

• "If it makes you feel any better, yesterday a collegue and I were trying to debug his test setup and two hours of complex testing passed before we two geniusses realised we hadn't switched the power to a key bit of kit on!" We've all been there, but at least the cup holder was working while you did the tests... – dmckee --- ex-moderator kitten May 20 '14 at 5:48

That is true.

However, all Lorentz transformations need not be proper ($det(L) = 1$) or orthochronous $L_0^0 \geq 1$ for example,

• parity transformation $diag(1,-1,-1,-1)$ is improper (since $det(L) = -1$) but orthochronous (since $L_0^0 = 1)$

• whereas, time reversal $diag(-1,1,1,1)$ is a non orthochronous transformation (since $L_0^0 = -1)$