Special Relativity/Mathematical Appendix

Mathematics of the Lorentz Transformation Equations

Consider two observers $O$ and $O^{\text{'}}$, moving at velocity $v$ relative to each other, who observe the same event such as a flash of light. How will the coordinates recorded by the two observers be interrelated?

These 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 homogenous 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 $x^{\text{'}}=a{x}^2$ the distance between two points would depend upon position.

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

$x^{\text{'}}=a_{11}x+a_{12}y+a_{13}z+a_{14}t$

$y^{\text{'}}=a_{21}x+a_{22}y+a_{23}z+a_{24}t$

$z^{\text{'}}=a_{31}x+a_{32}y+a_{33}z+a_{34}t$

$t^{\text{'}}=a_{41}x+a_{42}y+a_{43}z+a_{44}t$

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

$z^{\text{'}}=z$

$y^{\text{'}}=y$

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:

$x^{\text{'}}=a_{11}x+a_{12}y+a_{13}z+a_{14}t$

$t^{\text{'}}=a_{41}x+a_{42}y+a_{43}z+a_{44}t$

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:

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

Events satisfying $x^{\text{'}}=0$ 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 $x^{\text{'}}$ is:

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

The analysis to date gives the following equations:

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

$y^{\text{'}}=y$

$z^{\text{'}}=z$

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

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:

$x^2+y^2+z^2=c^2{t}^2$

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

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:

$(a_{11}^2-c^2{a}_{41}^2)x^2+y^2+z^2-2(v{a}_{11}^2+c^2{a}_{41}{a}_{44})xt=(c^2{a}_{44}^2-v^2{a}_{11}^2)t^2$

We demand that this is equivalent with

$x^2+y^2+z^2=c^2{t}^2$

So we get:

$c^2{a}_{44}^2-v^2{a}_{11}^2=c^2$

$a_{11}^2-c^2{a}_{41}^2=1$

$v{a}_{11}^2+c^2{a}_{41}{a}_{44}=0$

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:

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

$y^{\text{'}}=y$

$z^{\text{'}}=z$

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

gives:

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

$y^{\text{'}}=y$

$z^{\text{'}}=z$

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

The inverse transformation is:

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

$y=y^{\text{'}}$

$z=z^{\text{'}}$

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