Deriving the Lorentz Transformation
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You should be comfortable with the two postulates of special relativity and the concept of inertial reference frames. Basic algebraic manipulation of linear equations is required.
Why this matters
The Lorentz transformation is the mathematical bridge between the two postulates and the observable phenomena of special relativity.
Galilean transformations fail to preserve the speed of light, leading to contradictions with Maxwell's equations. By deriving the Lorentz transformation, we establish the correct symmetry group of flat spacetime. This derivation is not merely a calculation; it is the logical consequence of demanding that the laws of physics, specifically the constancy of \(c\), hold true for all inertial observers. Understanding this derivation provides the foundation for interpreting time dilation, length contraction, and the relativity of simultaneity as geometric properties of spacetime rather than mechanical effects.
The Geometric Necessity
Imagine two observers, Alice and Bob, moving relative to each other at a constant speed. In classical mechanics, we assume that time is absolute and space is separate. If Bob throws a ball, Alice calculates its speed by simply adding Bob's speed to the ball's speed relative to Bob. This is the Galilean transformation. However, light does not behave this way. If Alice shines a laser, she measures the speed as \(c\). If Bob, moving towards the light source, measures the same beam, he must also measure \(c\), not \(c + v\). This requirement forces us to abandon the idea that time and space are independent. Instead, they must mix together in such a way that the quantity \(x^2 + y^2 + z^2 - c^2t^2\) remains invariant for both observers. The Lorentz transformation is the specific mathematical rule that achieves this mixing, ensuring that the "speed of light" is a universal constant rather than a relative velocity.
Linearity and Standard Configuration
To derive the transformation, we assume the "standard configuration": frame \(S'\) moves at speed \(v\) along the \(x\)-axis of frame \(S\), and their origins coincide at \(t=t'=0\). Because space and time are homogeneous (no preferred point in space or time), the transformation must be linear. A linear transformation preserves the straight-line motion of inertial particles. If the transformation were non-linear, a particle moving at constant velocity in \(S\) would appear to accelerate in \(S'\), violating the principle of relativity. Therefore, we can write the general form as \(x' = Ax + Bt\) and \(t' = Cx + Dt\). By symmetry, the transverse coordinates \(y\) and \(z\) must remain unchanged (\(y'=y, z'=z\)) because the motion is only along the \(x\)-axis and there is no preferred direction perpendicular to the motion.
Applying the Postulates
We now apply the two postulates to determine the coefficients \(A, B, C,\) and \(D\). First, the origin of \(S'\) is at \(x'=0\). In frame \(S\), this origin moves at \(x=vt\). Substituting into the linear form \(0 = A(vt) + Bt\), we find \(B = -Av\). Thus, \(x' = A(x - vt)\). Second, we use the invariance of the speed of light. A light pulse emitted from the origin at \(t=0\) satisfies \(x=ct\) in \(S\) and \(x'=ct'\) in \(S'\). Substituting these into the transformation equations allows us to solve for \(A\) in terms of \(v\) and \(c\). The result is that \(A\) must be the Lorentz factor \(\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}\). This factor is always greater than or equal to 1, and it diverges as \(v\) approaches \(c\), reflecting the impossibility of reaching the speed of light for massive objects.
The Time Component
Determining the time transformation \(t'\) is the subtle part. We cannot simply assume \(t'=t\). Using the inverse transformation (swapping \(v\) for \(-v\)) and the requirement that the transformation be invertible, we find that time must also mix with space. The coefficient for the \(x\) term in the time equation is determined by the condition that the inverse transformation must recover the original coordinates. This leads to the term \(-\frac{v x}{c^2}\) in the expression for \(t'\). The final form is \(t' = \gamma\left(t - \frac{v x}{c^2}\right)\). This mixing of space and time is the direct mathematical origin of the relativity of simultaneity. Two events that are simultaneous in \(S\) (\(\Delta t = 0\)) will generally have \(\Delta t' \neq 0\) in \(S'\) if they are separated in the \(x\)-direction. This is not a paradox but a fundamental feature of the geometry of spacetime.
The derivation
Lorentz Transformation from the Two Postulates
For two inertial frames \(S\) and \(S'\) in standard configuration — \(S'\) moving at constant speed \(v\) along the shared \(x\)-axis, origins coinciding at \(t=t'=0\) — the two postulates (the principle of relativity and the invariance of the vacuum speed of light \(c\)) force the coordinate transformation to be linear, transverse coordinates to be unchanged, and the boost to take the form \[ x' = \gamma\,(x - v t), \qquad t' = \gamma\left(t - \frac{v x}{c^2}\right), \qquad y'=y,\quad z'=z, \qquad \gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}. \]
A common misconception is that the Lorentz factor \(\gamma\) represents a physical "stretching" of time or "shrinking" of space due to mechanical forces. In reality, \(\gamma\) is a geometric projection factor. Just as the shadow of a rod depends on its angle relative to the light source, the measured length and time interval depend on the observer's velocity relative to the event. There is no absolute "true" length or time; there are only projections of the four-dimensional spacetime interval onto the observer's specific coordinate axes. C
Worked example
Problem. A spaceship moves at \(v = 0.6c\) relative to Earth. In the spaceship's frame, a laser pulse is emitted from the rear (\(x'=0\)) and received at the front (\(x'=L'\)) where \(L' = 30\ \mathrm{m}\). Calculate the time interval \(\Delta t\) measured by an observer on Earth between the emission and reception events.
- Identify the events: Event 1 (emission) at \(x'=0, t'=0\). Event 2 (reception) at \(x'=30\ \mathrm{m}\). Since light travels at \(c\), \(t' = x'/c = 30/c\ \mathrm{s}\).
- Calculate the Lorentz factor: \(\gamma = \frac{1}{\sqrt{1 - 0.6^2}} = \frac{1}{\sqrt{0.64}} = \frac{1}{0.8} = 1.25\).
- Use the inverse Lorentz transformation for time: \(t = \gamma\left(t' + \frac{v x'}{c^2}\right)\). Note: We are transforming from \(S'\) to \(S\), so the sign of the \(vx/c^2\) term is positive.
- Substitute values: \(t = 1.25 \left( \frac{30}{c} + \frac{0.6c \cdot 30}{c^2} \right) = 1.25 \left( \frac{30}{c} + \frac{18}{c} \right) = 1.25 \left( \frac{48}{c} \right)\).
- Compute final value: \(t = \frac{60}{c}\ \mathrm{s}\). Using \(c \approx 3 \times 10^8\ \mathrm{m/s}\), \(t = 2 \times 10^{-7}\ \mathrm{s}\).
Units check. The term \(vx'/c^2\) has units \((\mathrm{m/s} \cdot \mathrm{m}) / (\mathrm{m^2/s^2}) = \mathrm{s}\). The term \(t'\) is in seconds. The sum is in seconds, multiplied by dimensionless \(\gamma\), yielding seconds.
Three quick tests
- Why must the Lorentz transformation be linear?
Answer
To preserve the straight-line motion of inertial particles and ensure the principle of relativity holds for all observers. - If \(v \to 0\), what does the Lorentz transformation reduce to?
Answer
It reduces to the Galilean transformation: \(x' = x - vt\) and \(t' = t\). - What is the physical significance of the term \(-\frac{v x}{c^2}\) in the time transformation?
Answer
It represents the relativity of simultaneity, showing that time measurements depend on position in the direction of motion.
Problem set for PU-105 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.