# Special Relativity/Mathematical Appendix#Mathematics of the Lorentz Transformation Equations

## Mathematics of the Lorentz Transformation Equations

Consider two observers $O$ and $O^'$, moving at velocity $v \,$ relative
to each other who synchronise their clocks so that $t=t^'=0$ as they
pass each other. They both observe the same event as a flash of light.
How will the coordinates recorded by the observers of the event that
produced the light be interrelated?

The relationship between the coordinates can be derived using linear
algebra on the basis of the postulates of relativity and an extra
homogeneity and isotropy assumption.

**The homogeneity and isotropy assumption**: space is uniform and
homogeneous in all directions. If this were not the case then when
comparing lengths between coordinate systems the lengths would depend
upon the position of the measurement. For instance, if $x^' = a x^2 \,$
the distance between two points would depend upon position.

The linear equations relating coordinates in the primed and unprimed
frames are:

:   $x^' = a_{11} x + a_{12} y + a_{13} z  + a_{14} t \,$
:   $y^' = a_{21} x + a_{22} y + a_{23} z  + a_{24} t \,$
:   $z^' = a_{31} x + a_{32} y + a_{33} z  + a_{34} t \,$
:   $t^' = a_{41} x + a_{42} y + a_{43} z  + a_{44} t \,$

There is no relative motion in the $y$ or $z$ directions so, according
to the \'relativity\' postulate:

:   $z^' = z \,$
:   $y^' = y \,$

Hence:

:   $a_{22} = 1 \,$ and $a_{21} = a_{23} = a_{24} = 0 \,$
:   $a_{33} = 1 \,$ and $a_{31} = a_{32} = a_{34} = 0 \,$

So the following equations remain to be solved:

:   $x^' = a_{11} x + a_{12} y + a_{13} z  + a_{14} t \,$
:   $t^' = a_{41} x + a_{42} y + a_{43} z  + a_{44} t \,$

If space is isotropic (the same in all directions) then the motion of
clocks should be independent of the y and z axes (otherwise clocks
placed symmetrically around the x-axis would appear to disagree). Hence:

:   $a_{42} = a_{43} = 0 \,$

so:

:   $t^' = a_{41} x + a_{44} t \,$

Events satisfying $x^' = 0 \,$ must also satisfy $x = vt \,$. So:

:   $0 = a_{11} vt + a_{12} y + a_{13} z  + a_{14} t \,$

and

:   $-a_{11} vt = a_{12} y + a_{13} z  + a_{14} t \,$

Given that the equations are linear then $a_{12} y + a_{13} z = 0 \,$
and:

:   $-a_{11} vt = a_{14} t \,$

and

:   $-a_{11} v = a_{14} \,$

Therefore the correct transformation equation for $x^' \,$ is:

:   $x^' = a_{11} (x - vt) \,$

The analysis to date gives the following equations:

:   $x^' = a_{11} (x - vt) \,$
:   $y^' = y \,$
:   $z^' = z \,$
:   $t^' = a_{41} x + a_{44} t \,$

Assuming that the speed of light is constant, the coordinates of a flash
of light that expands as a sphere will satisfy the following equations
in each coordinate system:

:   $x^2 + y^2 + z^2 = c^2t^2 \,$
:   $x^{'2} + y^{'2} + z^{'2} = c^2t^{'2} \,$

Substituting the coordinate transformation equations into the second
equation gives:

:   $a_{11}^2(x - vt)^2 + y^2 + z^2 = c^2(a_{41}x + a_{44}t)^2 \,$

rearranging:

:   $(a_{11}^2 - c^2 a_{41}^2)x^2 + y^2 + z^2 - 2(va_{11}^2 + c^2a_{41} a_{44}) xt = (c^2 a_{44}^2 - v^2 a_{11}^2)t^2 \,$

We demand that this is equivalent with

:   $x^2 + y^2 + z^2 = c^2t^2 \,$

So we get:

:   $c^2 a_{44}^2 - v^2 a_{11}^2 = c^2 \,$
:   $a_{11}^2 - c^2 a_{41}^2 = 1 \,$
:   $va_{11}^2 + c^2a_{41} a_{44} = 0 \,$

Solving these 3 simultaneous equations gives:

:   $a_{44} = \frac{1}{\sqrt{(1 - v^2/c^2)}} \,$
:   $a_{11} = \frac{1}{\sqrt{(1 - v^2/c^2)}} \,$
:   $a_{41} = -\frac{v/c^2}{\sqrt{(1 - v^2/c^2)}} \,$

Substituting these values into:

:   $x^' = a_{11} (x - vt) \,$
:   $y^' = y \,$
:   $z^' = z \,$
:   $t^' = a_{41} x + a_{44} t \,$

gives:

:   $x^' = \frac{x - vt}{\sqrt{(1 - v^2/c^2)}} \,$
:   $y^' = y \,$
:   $z^' = z \,$
:   $t^' = \frac{t - (v/c^2)x}{\sqrt{(1 - v^2/c^2)}} \,$

The inverse transformation is:

:   $x = \frac{x^' + vt^'}{\sqrt{(1 - v^2/c^2)}} \,$
:   $y = y^' \,$
:   $z = z^' \,$
:   $t = \frac{t^' + (v/c^2)x^'}{\sqrt{(1 - v^2/c^2)}} \,$

### Einstein\'s original approach

How would two observers measure the position and timing of an event by
using light rays if the speed of light were constant? The modern
analysis of this problem, exposing the assumptions involved, is given
above but Einstein\'s original reasoning (Einstein 1905,1920) is as
follows.

Light is transmitted along the positive x axis according to the equation
$x = ct$ where $c$ is the velocity of light. This can be rewritten as:

$$x - ct = 0$$

Another observer, moving relatively to the first may find different
values for x and t but the same equation will apply:

$$x^' - ct^' = 0$$

A simple relationship between these formulae, which apply to the same
event is:

$$(x^' - ct^') = \lambda (x - ct)$$

Light is transmitted along the negative x axis according to the equation
$x = -ct$ where $c$ is the velocity of light. This can be rewritten as:

$$x + ct = 0$$

$$x^' + ct^' = 0$$

And:

$$(x^' + ct^') = \mu (x + ct)$$

Adding the equations and substituting $a = \frac{\lambda + \mu}{2}$ and
$b = \frac{\lambda - \mu}{2}$:

:   \(1\) $x' = ax - bct$
:   \(2\) $ct' = act - bx$

The origin of one set of coordinates can be set so that $x^' = 0$ hence:

$$x = \frac{bc}{a} t$$

If $v$ is the velocity of one observer relative to the other then
$v = \frac{x}{t}$ and:

:   \(3\) $v = \frac{bc}{a}$

At $t = 0$:

:   \(4\) $x^' = ax$

Therefore two points separated by unit distance in the primed frame of
reference ie: when $x^' = 1$ have the following separation in the
unprimed frame:

:   \(5\) $\Delta x = \frac{1}{a}$

Now $t$ can be eliminated from equations (1) and (2) and combined with
$v = \frac{bc}{a}$ and (4) to give in the case where $x=1$ and
$t^' = 0$:

:   \(6\) $x^' = a (1 -\frac{v^2}{c^2}) x$

And, if $\Delta x = 1$:

:   \(7\) $\Delta x^' = a (1 -\frac{v^2}{c^2})$

Now if the two moving systems are identical and the situation is
symmetrical a measurement in the unprimed system of a division showing
one metre on a measuring rod in the primed system is going to be
identical to a measurement in the primed system of a division showing
one metre on a measuring rod in the unprimed system. Thus (5) and (7)
can be combined so that:

$$\frac {1}{a} = a (1 -\frac{v^2}{c^2})$$

So:

$$a^2 = \frac {1}{(1 -\frac{v^2}{c^2})}$$

Inserting this value for $a$ into equations (1) and (2) and solving for
$b$ gives:

$$x^' = \frac {x - vt}{\sqrt {1 -\frac{v^2}{c^2}}}$$

$$t^' = \frac {t - (v/c^2)x}{\sqrt {1 -\frac{v^2}{c^2}}}$$

These are the **Lorentz Transformation Equations** for events on the x
axis.

Einstein, A. (1920). Relativity. The Special and General Theory. Methuen
& Co Ltd 1920. Written December, 1916. Robert W. Lawson (Authorised
translation). <http://www.bartleby.com/173/>
