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: $S^{\prime}$ moves with constant velocity $v\hat{\mathbf x}$ relative to $S$, the axes are parallel, and the origins coincide at $t=t^{\prime}=0$. We require $\lvert v\rvert<c$. Homogeneity of space and time requires a linear transformation. Since the origin $x^{\prime}=0$ follows $x=vt$,

\[x^{\prime}=A(x-vt).\]

Write the most general linear time relation as $t^{\prime}=B(t-Dx)$. A light pulse emitted at the common origin satisfies $x=ct$ and $x^{\prime}=ct^{\prime}$. Substitution gives

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

For the pulse moving toward $-x$, $x=-ct$ and $x^{\prime}=-ct^{\prime}$, 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

\[x^{\prime}=A(x-vt), \qquad t^{\prime}=A\left(t-\frac{vx}{c^2}\right).\]

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

\[x=A(x^{\prime}+vt^{\prime}), \qquad t=A\left(t^{\prime}+\frac{vx^{\prime}}{c^2}\right).\]

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

\[\boxed{x^{\prime}=\gamma(x-vt)},\qquad \boxed{t^{\prime}=\gamma\left(t-\frac{vx}{c^2}\right)},\] \[\boxed{y^{\prime}=y},\qquad \boxed{z^{\prime}=z}.\]

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

\[\boxed{x=\gamma(x^{\prime}+vt^{\prime})},\qquad \boxed{t=\gamma\left(t^{\prime}+\frac{vx^{\prime}}{c^2}\right)}, \qquad y=y^{\prime},\quad z=z^{\prime}.\]

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

Interval check

Direct substitution gives

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

Including the unchanged transverse coordinates,

\[\boxed{c^2(t^{\prime})^2-(x^{\prime})^2-(y^{\prime})^2-(z^{\prime})^2 =c^2t^2-x^2-y^2-z^2}.\]

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 $x^{\prime}\simeq x-vt$ and $t^{\prime}\simeq t$.

Solved Problems

1. Transforming an event and checking its interval

Frame $S^{\prime}$ moves along $+x$ at $v=0.600c$ relative to $S$. An event has $x=600\ \mathrm m$ and $t=3.00\ \mu\mathrm s$ in $S$. Find $(x^{\prime},t^{\prime})$ and verify the spacetime interval. Use $c=3.00\times10^8\ \mathrm{m\,s^{-1}}$.

For $\beta=0.600$,

\[\gamma=\frac{1}{\sqrt{1-0.600^2}}=1.25.\]

The direct transformation gives

\[\begin{aligned} x^{\prime} &=\gamma(x-vt)\\ &=1.25\left[600-(0.600)(3.00\times10^8)(3.00\times10^{-6})\right]\mathrm m\\ &=\boxed{75.0\ \mathrm m}, \end{aligned}\]

and

\[\begin{aligned} t^{\prime} &=\gamma\left(t-\frac{vx}{c^2}\right)\\ &=1.25\left[3.00-\frac{(0.600)(600)}{300}\right]\mu\mathrm s\\ &=\boxed{2.25\ \mu\mathrm s}. \end{aligned}\]

In $S$, $ct=900\ \mathrm m$, so

\[c^2t^2-x^2=(900)^2-(600)^2=4.50\times10^5\ \mathrm{m^2}.\]

In $S^{\prime}$, $ct^{\prime}=675\ \mathrm m$, and

\[c^2(t^{\prime})^2-(x^{\prime})^2=(675)^2-(75)^2 =4.50\times10^5\ \mathrm{m^2}.\]

The positive interval is unchanged, so the event lies inside the future light cone of the common origin. As $v\to0$, $x^{\prime}\to x$ and $t^{\prime}\to t$.

2. Two collinear Lorentz boosts

Frame $S^{\prime}$ moves at $v_1=0.600c$ relative to $S$, and $S^{\prime\prime}$ moves at $v_2=0.800c$ relative to $S^{\prime}$, both along $+x$. Find the velocity and Lorentz factor of $S^{\prime\prime}$ relative to $S$ by composing the transformations.

Write $\beta_i=v_i/c$. Applying the two position transformations gives

\[\begin{aligned} x^{\prime\prime} &=\gamma_2(x^{\prime}-v_2t^{\prime})\\ &=\gamma_1\gamma_2 \left[(1+\beta_1\beta_2)x-(v_1+v_2)t\right]. \end{aligned}\]

Factor the coefficient of $x$:

\[x^{\prime\prime}=\gamma_1\gamma_2(1+\beta_1\beta_2) \left[x-\frac{v_1+v_2}{1+v_1v_2/c^2}t\right].\]

It has the standard Lorentz form, hence

\[\beta=\frac{\beta_1+\beta_2}{1+\beta_1\beta_2} =\frac{0.600+0.800}{1+(0.600)(0.800)} =\boxed{\frac{35}{37}\simeq0.94595}.\]

Also,

\[\gamma =\gamma_1\gamma_2(1+\beta_1\beta_2) =\left(\frac54\right)\left(\frac53\right)\left(\frac{37}{25}\right) =\boxed{\frac{37}{12}\simeq3.0833}.\]

Both velocities were defined positive, so the resultant is along $+x$ and remains below $c$. If $v_2\to0$, the result becomes $v_1$; when both speeds are small compared with $c$, the denominator tends to one and the Galilean sum is recovered.

Descriptive Questions

1. Derive the Lorentz transformation from linearity, the two light rays $x=\pm ct$, and reciprocity between inertial frames.
2. Explain why the positive root of the normalization factor is selected when deriving $\gamma$.
3. Obtain the inverse Lorentz transformation and show explicitly that it recovers the original event coordinates.
4. Prove that the spacetime interval is invariant under a standard Lorentz boost and state its physical significance.

Numerical Problems

1. An event has $ct=500\ \mathrm m$ and $x=500/3\ \mathrm m$ in $S$. Apart from $v=0$, find the boost along $+x$ for which its transformed coordinate time is unchanged, $t^{\prime}=t$, and then find $x^{\prime}$. Answer: $v=+0.600c$ and $x^{\prime}=-500/3\ \mathrm m\simeq-166.7\ \mathrm m$.
2. For $v=0.600c$, an event has $x^{\prime}=120\ \mathrm m$, $t^{\prime}=1.00\ \mu\mathrm s$. Use the inverse transformation. Answer: $x=375\ \mathrm m$, $t=1.55\ \mu\mathrm s$.
3. Calculate the Lorentz factor at $v=0.960c$. Answer: $\gamma=25/7\simeq3.571$.
4. A light event has $x=900\ \mathrm m$, $t=3.00\ \mu\mathrm s$. Transform it to a frame moving at $0.800c$ along $+x$. Answer: $x^{\prime}=300\ \mathrm m$, $t^{\prime}=1.00\ \mu\mathrm s$, so $x^{\prime}=ct^{\prime}$.
5. Two events satisfy $c\Delta t=500\ \mathrm m$ and $\Delta x=300\ \mathrm m$. Find $\Delta s^2$ and the proper time. Answer: $\Delta s^2=1.60\times10^5\ \mathrm{m^2}$, $\Delta\tau=1.333\ \mu\mathrm s$.
6. An event on the moving origin has $x=480\ \mathrm m$ at $t=2.00\ \mu\mathrm s$ in $S$. Find the velocity of $S^{\prime}$ relative to $S$. Answer: $v=x/t=+0.800c$.

The derivations and all problem values are checked in the Maxima worksheet; every printed residual is zero.

References

1. “Lorentz transformation,” Wikipedia.
2. A. P. French, Special Relativity, 1st ed., MIT Introductory Physics Series, W. W. Norton, 1968, Chapter 3, “Einstein and the Lorentz–Einstein Transformations.”
3. David J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press, 2017, Chapter 12, §12.1, “The Special Theory of Relativity.”