04 Jul 2025

Lorentz Transformations from the Light Postulate

A first-principles derivation of the direct and inverse Lorentz transformations and the invariant spacetime interval.

bsc semester-i special-relativity lorentz-transformation

Use the standard configuration: $S’$ moves with constant velocity $v\hat{\mathbf x}$ relative to $S$, the axes are parallel, and the origins coincide at $t=t’=0$. We require $\lvert v\rvert<c$. Homogeneity of space and time requires a linear transformation. Since the origin $x’=0$ follows $x=vt$,

\[x'=A(x-vt).\]

Write the most general linear time relation as $t’=B(t-Dx)$. A light pulse emitted at the common origin satisfies $x=ct$ and $x’=ct’$. Substitution gives

\[A(c-v)t=cB(1-Dc)t. \tag{1}\]

For the pulse moving toward $-x$, $x=-ct$ and $x’=-ct’$, so

\[A(-c-v)t=-cB(1+Dc)t,\]

or

\[A(c+v)=cB(1+Dc). \tag{2}\]

Adding and subtracting (1) and (2),

\[2Ac=2cB\quad\Rightarrow\quad A=B,\] \[2Av=2c^2BD\quad\Rightarrow\quad D=\frac{v}{c^2}.\]

Thus

\[x'=A(x-vt), \qquad t'=A\left(t-\frac{vx}{c^2}\right).\]

The relativity principle and reciprocity make the inverse transformation identical in form after $v\to-v$ and interchange of primed and unprimed coordinates:

\[x=A(x'+vt'), \qquad t=A\left(t'+\frac{vx'}{c^2}\right).\]

Insert the direct expressions into the first inverse equation:

\[\begin{aligned} x &=A\left[A(x-vt)+vA\left(t-\frac{vx}{c^2}\right)\right]\\ &=A^2x\left(1-\frac{v^2}{c^2}\right). \end{aligned}\]

Therefore

\[A^2(1-\beta^2)=1, \qquad A=\gamma\equiv\frac{1}{\sqrt{1-\beta^2}},\]

where the positive root is selected by continuity with $A=1$ at $v=0$. The Lorentz transformations are

\[\boxed{x'=\gamma(x-vt)},\qquad \boxed{t'=\gamma\left(t-\frac{vx}{c^2}\right)},\] \[\boxed{y'=y},\qquad \boxed{z'=z}.\]

Their inverse is obtained by changing the sign of $v$:

\[\boxed{x=\gamma(x'+vt')},\qquad \boxed{t=\gamma\left(t'+\frac{vx'}{c^2}\right)}, \qquad y=y',\quad z=z'.\]

The $ct'$ axis is $x=vt$, while the $x'$ axis is $ct=(v/c)x$; each tilts toward, but never crosses, its neighboring light line $x=\pm ct$.

Interval check

Direct substitution gives

\[\begin{aligned} c^2t'^2-x'^2 &=\gamma^2\left[ c^2\left(t-\frac{vx}{c^2}\right)^2-(x-vt)^2 \right]\\ &=\gamma^2\left[ c^2t^2-2vtx+\frac{v^2x^2}{c^2} -x^2+2vtx-v^2t^2 \right]\\ &=\gamma^2(1-\beta^2)(c^2t^2-x^2)\\ &=c^2t^2-x^2. \end{aligned}\]

Including the unchanged transverse coordinates,

\[\boxed{c^2t'^2-x'^2-y'^2-z'^2 =c^2t^2-x^2-y^2-z^2}.\]

This invariant will be used to classify event order. For fixed event coordinates in the non-relativistic limit $\lvert v\rvert/c\to0$, $\gamma\to1$ and $vx/c^2\to0$, recovering $x’\simeq x-vt$ and $t’\simeq t$.

The inverse transformation, light-ray invariance, and interval identity are verified in the Maxima worksheet; every printed residual is zero.