Decomposition of Lorentz transformations using orthogonal matrices

In the present note I want to explore the decomposition of an arbitrary Lorentz transformation $L$ in the form $L = R_1 \overline{L} R_2$ , where $R_1$ and $R_2$ are orthogonal Lorentz matrices and $\overline{L}$ is a simple Lorentz matrix (to be defined below). We will use throughout a metric tensor of the form $G = \text{diag}(1, -1, -1, -1)$ .

Any $4 \times 4$ matrix that preserves the quadratic form $x^T G x$ is called Lorentz. We have here

$x = \begin{bmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}$

where $x^0 \equiv ct$ is the temporal coordinate and $x^1, x^2, x^3$ are spatial coordinates. So, what this means is that if $A$ is a $4 \times 4$ matrix, it is Lorentz if $y = Ax$ is such that

$y^T G y = x^T A^T G A x = x^T G x$

Therefore $A$ is Lorentz iff $A^T G A = G$ .

The set of all Lorentz matrices thus defined form a group, the Lorentz group, under matrix multiplication. The identity element of the group is obviously the $4 \times 4$ identity matrix $I_4$ . To prove closure under matrix multiplication, suppose $A$ and $B$ are two Lorentz matrices. Then

$(AB)^T G (AB) = B^T (A^T G A) B = B^T G B = G$

so the product is also Lorentz. To prove that the inverse of a Lorentz matrix is also Lorentz, suppose $A$ is Lorentz so that $A^T G A = G$ . Then since $G^2 = I_4$ , left-multiplying both sides by $G$ gives

$(G A^T G) A = I_4$

so the inverse of $A$ is $G A^T G$ (and the inverse of $A^T$ is $G A G$ , by right-multiplying). But then we have

$(G A^T G)^T G (G A^T G) = G A G G G A^T G = G A G A^T G = I_4 G = G$

so the inverse of $A$ is also Lorentz. Therefore the Lorentz matrices form a group as claimed.

For two inertial reference frames in standard configuration, the Lorentz transformation will be a $4 \times 4$ matrix of the form

$\begin{bmatrix} a & b & 0 & 0 \\ b & a & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

such that $a^2 - b^2 = 1$ . Any $4 \times 4$ matrix of this form is said to be simple Lorentz. The relative velocity in the physical situation modelled by $A$ is recovered as

$\beta = \frac{v}{c} = -\frac{b}{a}$

Notice that since $A^{-1} = G A^T G$ , we have

$A^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} a & b & 0 & 0 \\ b & a & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} a & -b & 0 & 0 \\ b & -a & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$

$= \begin{bmatrix} a & -b & 0 & 0 \\ -b & a & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

which is itself a simple Lorentz matrix corresponding to a reversal in the sign of $\beta$ .

Notice also that the transpose of a Lorentz matrix is Lorentz. To see this, if $A$ is Lorentz then $A^T G A = G$ . Pre-multiplying this by $AG$ and post-multiplying by $A^{-1} G$ we get

$(AG)(A^T G A)(A^{-1} G) = A G G A^{-1} G$

$\iff (A G A^T)(G A A^{-1} G) = G$

$\iff A G A^T = G$

as required.

We now look in detail at the decomposition result mentioned at the start of this note. This expresses in mathematical terms the possibility of simplifying an arbitrary Lorentz matrix by a suitable rotation of axes. The result says that an arbitrary Lorentz matrix $L = (a^i_j)$ has the representation

$L = R_1 \overline{L} R_2$

where $\overline{L}$ is a simple Lorentz matrix with parameters $a = |a^0_0| = \epsilon a^0_0$ (with $\epsilon = \pm 1$ ) and $b = - \sqrt{(a^0_0)^2 - 1}$ , and $R_1$ and $R_2$ are orthogonal Lorentz matrices defined by

$R_1 = L R_2^T \overline{L}^{-1}$

and

$R_2 = [e_1 \ r^{\prime} \ s^{\prime} \ t^{\prime}]^T$

where $e_1 = (1, 0, 0, 0)$ , $r^{\prime} = (\epsilon/b)(0, a^0_1, a^0_2, a^0_3) \equiv (0, r)$ , $s^{\prime} = (0, s)$ , $t^{\prime} = (0, t)$ , with $s$ and $t$ chosen so that $[r \ s \ t]$ is $3 \times 3$ orthogonal.

A corollary of this result is that if $L = (a^i_j)$ connects two inertial frames, then the relative velocity between the frames is

$v = c \sqrt{1 - (a^0_0)^{-2}}$

To prove this decomposition result and its corollary, we begin by observing that

$||r||^2 = \big(\frac{\epsilon}{b}\big)^2 ((a^0_1)^2 + (a^0_2)^2 + (a^0_3)^2)$

But from the first element of the product $L G L^T = G$ , which is

$= \begin{bmatrix} a^0_0 & a^0_1 & a^0_2 & a^0_3 \\ a^1_0 & a^1_1 & a^1_2 & a^1_3 \\ a^2_0 & a^2_1 & a^2_2 & a^2_3 \\ a^3_0 & a^3_1 & a^3_2 & a^3_3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} a^0_0 & a^1_0 & a^2_0 & a^3_0 \\ a^0_1 & a^1_1 & a^2_1 & a^3_1 \\ a^0_2 & a^1_2 & a^2_2 & a^3_2 \\ a^0_3 & a^1_3 & a^2_3 & a^3_3 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$

we get

$[a^0_0 \ -a^0_1 \ -a^0_2 \ -a^0_3] \begin{bmatrix} a^0_0 \\ -a^0_1 \\ -a^0_2 \\ -a^0_3 \end{bmatrix} = 1$

$\iff (a^0_0)^2 - (a^0_1)^2 - (a^0_2)^2 - (a^0_3)^2 = 1$

Using this result in the expression for $||r||^2$ above we therefore have

$||r||^2 = \big(\frac{1}{b}\big)^2 ((a^0_0)^2 - 1) = \frac{b^2}{b^2} = 1$

Therefore the matrix

$[r \ s \ t] = \begin{bmatrix} r_1 & s_1 & t_1 \\ r_2 & s_2 & t_2 \\ r_3 & s_3 & t_3 \end{bmatrix}$

is orthogonal (i.e., its columns and rows are orthogonal unit vectors – recall that $s$ and $t$ are chosen so that this is true). Therefore the matrix $R_2$ has the form

$R_2 = [e_1 \ r^{\prime} \ s^{\prime} \ t^{\prime}]^T = \begin{bmatrix} e_1 \\ r^{\prime} \\ s^{\prime} \\ t^{\prime} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & r_1 & r_2 & r_3 \\ 0 & s_1 & s_2 & s_3 \\ 0 & t_1 & t_2 & t_3 \end{bmatrix}$

Therefore

$R_2^T = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & r_1 & s_1 & t_1 \\ 0 & r_2 & s_2 & t_2 \\ 0 & r_3 & s_3 & t_3 \end{bmatrix}$

This is a $4 \times 4$ orthogonal matrix which is also Lorentz. It is clearly orthogonal since $R_2^{-1} = R_2^T$ . To confirm that $R_2^T$ is Lorentz we compute

$R_2 G R_2^T = \begin{bmatrix} 1 & 0 & 0 & 0 \\0 & r_1 & r_2 & r_3 \\0 & s_1 & s_2 & s_3 \\0 & t_1 & t_2 & t_3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0\\0 & r_1 & s_1 & t_1 \\0 & r_2 & s_2 & t_2 \\0 & r_3 & s_3 & t_3 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 & 0 & 0 \\0 & -r_1 & -r_2 & -r_3 \\0 & -s_1 & -s_2 & -s_3 \\0 & -t_1 & -t_2 & -t_3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0\\0 & r_1 & s_1 & t_1 \\0 & r_2 & s_2 & t_2 \\0 & r_3 & s_3 & t_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{bmatrix}$

using the fact that $r$ , $s$ and $t$ are unit vectors. Therefore $R_2 G R_2^T = G$ , so $R_2^T$ is both orthogonal and Lorentz as claimed. Therefore we have

$R_1 \overline{L} R_2 = (L R_2^T \overline{L}^{-1})(\overline{L})(R_2) = L$

as claimed in the decomposition result above.

Now, since $R_1 = L R_2^T \overline{L}^{-1}$ is a product of Lorentz matrices, $R_1$ itself must be a Lorentz matrix. To show that it is orthogonal, we can write out $L R_2^T \overline{L}^{-1}$ explicitly as

$\begin{bmatrix} a_0 & b_0 & c_0 & d_0 \\a_1 & b_1 & c_1 & d_1 \\a_2 & b_2 & c_2 & d_2 \\a_3 & b_3 & c_3 & d_3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0\\0 & r_1 & s_1 & t_1 \\0 & r_2 & s_2 & t_2 \\0 & r_3 & s_3 & t_3 \end{bmatrix} \begin{bmatrix} \epsilon a_0 & \sqrt{a_0^2 - 1} & 0 & 0 \\\sqrt{a_0^2 - 1} & \epsilon a_0 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} (a_0) & (b_0 r_1 + c_0 r_2 + d_0 r_3) & (b_0 s_1 + c_0 s_2 + d_0 s_3) & (b_0 t_1 + c_0 t_2 + d_0 t_3) \\ \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot \end{bmatrix} \begin{bmatrix} \epsilon a_0 & \sqrt{a_0^2 - 1} & 0 & 0 \\\sqrt{a_0^2 - 1} & \epsilon a_0 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{bmatrix}$

where the omitted rows are of the same form as the first row, but with $a_i$ , $b_i$ , $c_i$ , $d_i$ instead of $a_0$ , $b_0$ , $c_0$ , $d_0$ , for $i = 1, 2, 3$ .

We now focus on proving that the top row and first column of this product are $(\pm 1, 0, 0, 0)$ . If this is the case, then the remaining submatrix must be orthogonal since $R_1$ is Lorentz (the equation $R_1^T G R_1 = G$ can only be satisfied if the submatrix is orthogonal) and this will imply that the entire $R_1$ matrix is orthogonal.

The $00$ – element of the product is

$\epsilon a_0^2 + (\sqrt{a_0^2 - 1})(b_0 r_1 + c_0 r_2 + d_0 r_3)$

$= \epsilon a_0^2 + (\sqrt{a_0^2 - 1})(b_0^2 + c_0^2 + d_0^2) \cdot \frac{\epsilon}{-\sqrt{a_0^2 - 1}}$

(since $r_1$ , $r_2$ , $r_3$ are the last three elements in the first row of $L$ multiplied by $\frac{\epsilon}{-\sqrt{a_0^2 - 1}}$ )

$= \epsilon (a_0^2 - b_0^2 - c_0^2 - d_0^2) = \epsilon$

(since this is the same quadratic form as in the first element of $L G L^T = G$ derived above, which equals 1).

The next element in the top row is

$a_0 \sqrt{a_0^2 - 1} + \epsilon a_0 (b_0 r_1 + c_0 r_2 + d_0 r_3)$

$= a_0 \sqrt{a_0^2 - 1} + \epsilon a_0(b_0^2 + c_0^2 + d_0^2) \cdot \frac{\epsilon}{-\sqrt{a_0^2 - 1}}$

$= a_0 \sqrt{a_0^2 - 1} - a_0 \sqrt{a_0^2 - 1} = 0$

(since $b_0^2 + c_0^2 + d_0^2 = a_0^2 - 1$ ).

For the third element in the top row we have

$b_0 s_1 + c_0 s_2 + d_0 s_3 = -\frac{\sqrt{a_0^2 - 1}}{\epsilon}(r_1 s_1 + r_2 s_2 + r_3 s_3) = 0$

since $r$ and $s$ are orthogonal.

Finally, for the fourth element in the top row we have

$b_0 t_1 + c_0 t_2 + d_0 t_3 = -\frac{\sqrt{a_0^2 - 1}}{\epsilon}(r_1 t_1 + r_2 t_2 + r_3 t_3) = 0$

since $r$ and $t$ are orthogonal.

Next, we consider the first column of the product. For each $i = 1, 2, 3$ , we have

$a_i \epsilon a_0 + \sqrt{a_0^2 - 1} (b_i r_1 + c_i r_2 + d_i r_3)$

$= a_i \epsilon a_0 + \sqrt{a_0^2 - 1} (b_i b_0 + c_i c_0 + d_i d_0)\cdot \frac{\epsilon}{-\sqrt{a_0^2 - 1}}$

$= a_i \epsilon a_0 - \epsilon (b_i b_0 + c_i c_0 + d_i d_0)$

$= \epsilon(a_i a_0 - b_i b_0 - c_i c_0 - d_i d_0) = 0$

since the quadratic form inside the bracket is zero when $i \neq 0$ (it is an off-diagonal element of $L G L^T = G$ ).

Therefore the product matrix is of the form

$R_1 = \begin{bmatrix} \epsilon & 0 & 0 & 0 \\0 & \ & \ & \ \\0 & \ & R & \ \\0 & \ & \ & \ \end{bmatrix}$

where the $3 \times 3$ submatrix $R$ must be orthogonal, since $R_1$ is Lorentz. This proves the main decomposition result.

To prove the corollary, note that in a simple Lorentz matrix the relative velocity is given by

$\beta = \frac{v}{c} = -\frac{b}{a} = \frac{\sqrt{(a_0^0)^2 - 1}}{|a_0^0|} = \sqrt{1 - (a_0^0)^{-2}}$

and in $L = R_1 \overline{L} R_2$ this is unchanged.

Decomposition of Lorentz transformations using orthogonal matrices

Published by Dr Christian P. H. Salas

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