Generators of the Lorentz Group

Let

$$x = \left( \begin{array}{c} x^0 x^1 x^2 x^3 \end{array} \right)$$ (16.30)

be a column vector. Note that we no longer indicate a vector by using a vector arrow and/or boldface - those are reserved for the spatial part of the four-vector only. Then a ``matrix'' scalar product is formed in the usual way by

$$(a,b) = \tilde{a}b$$ (16.31)

where $\tilde{a}$ is the (row vector) transpose of $a$. The metrix tensor is just a matrix:

$$g = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 &... ... \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right)$$ (16.32)

and $g^2 = \stackrel{\Leftrightarrow}{\bf I}$. Finally,

$$gx = \left( \begin{array}{c} x^0 -x^1 -x^2 -x^3 ... ...in{array}{c} x_0 x_1 x_2 x_3 \end{array} \right) .$$ (16.33)

In this compact notation we define the scalar product in this metric to be

$$a \cdot b = (a,gb) = (ga,b) = \tilde{a}gb = a^\alpha g_{\alpha \beta} b^\beta = a^\alpha b_\alpha .$$ (16.34)

We seek the set (group, we hope) of linear transformations that leaves $(x,gx) = x \cdot x$ invariant. Since this is the ``norm'' (squared) of a four vector, these are ``length preserving'' transformations in this four dimensional metric. That is, we want all matrices $A$ such that

$$x' = Ax$$ (16.35)

leaves the norm of $x$ invariant,

$$x' \cdot x' = \tilde{x}' g x' = \tilde{x} g x = x \cdot x$$ (16.36)

or

$$\tilde{x} \tilde{A} g A x = \tilde{x} g x$$ (16.37)

or

$$\tilde{A} g A = g .$$ (16.38)

Clearly this last condition is sufficient to ensure this property in $A$.

Now,

$$\det \left\vert \tilde{A} g A \right\vert = \det \left\vert... ... \left\vert A \right\vert )^2 = \det \left\vert g \right\vert$$ (16.39)

where the last equality is required. But $\det \left\vert g \right\vert = -1 \ne 0$, so

$$\det \left\vert A \right\vert = \pm 1$$ (16.40)

is a constraint on the allowed matrices (transformations) $A$. There are thus two classes of transformations we can consider. The

proper Lorentz transformations

with $\det \left\vert A \right\vert = +1$; and

improper Lorentz transformations

with $\det \left\vert A \right\vert = \pm1$.

Proper L. T.'s contain the identity (and thus can form a group by themselves), but improper L. T.'s can have either sign of the determinant. This is a signal that the metric we are using is ``indefinite''. Two examples of improper transformations that illustrate this point are spatial inversions (with $\det \left\vert A \right\vert = -1$) and $A = -I$ (space and time inversion, with $\det \left\vert A \right\vert = +1$).

In very general terms, the proper transformations are the continuously connected ones that form a Lie group, the improper ones include one or more inversions and are not equal to the product of any two proper transformations. The proper transformations are a subgroup of the full group -- this is not true of the improper ones, which, among other things, lack the identity. With this in mind, let us review the properties of infinitesimal linear transformations, preparatory to deducing the particular ones that form the homogeneous Lorentz group.