Lorentz Transformation from the Two Postulates

Derivation

Lorentz Transformation from the Two Postulates

D-090 Home PU-105 Threads light · symmetry Depends on galilean-invariance-of-newtonian-mechanics, maxwell-equations-in-vacuum

Statement

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 \[ x' = \gamma\,(x - v t), \qquad t' = \gamma\left(t - \frac{v x}{c^2}\right), \qquad y'=y,\quad z'=z, \qquad \gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}. \]

Why it matters

These four equations are the operational content of special relativity: every kinematic consequence — length contraction, time dilation, the relativity of simultaneity, the velocity-addition law, and the invariance of the interval \(s^2 = c^2t^2 - x^2\) — is read off directly from them. Deriving the transformation from postulates, rather than positing it, shows that the entire structure follows from two experimentally grounded statements and nothing else.

The derivation also exposes exactly where Galilean kinematics is replaced. The Galilean rule \(t'=t\) is not a low-order approximation that happens to be convenient; it is the one assumption the light postulate directly contradicts. Watching \(t'=\gamma(t-vx/c^2)\) emerge makes the loss of absolute simultaneity a theorem rather than a philosophical claim.

Assumptions

Spacetime is homogeneous and space is isotropic.If dropped, transformation coefficients could depend on position or time and the map need not be linear; straight inertial worldlines could bend, and no single \(\gamma\) would exist.

Principle of relativity: the laws of physics take the same form in every inertial frame.If dropped, the inverse transformation need not have the same functional form with \(v\to -v\), and one frame becomes physically preferred — reintroducing an ether rest frame.

Invariance of \(c\): a light pulse in vacuum has speed \(c\) in every inertial frame, independent of the source.If dropped, the closing condition \(c^2=\gamma^2(c^2-v^2)\) never arises and the coefficients are left undetermined — one recovers Galilean or any ad hoc linear map.

Inertial frames map inertial (force-free) worldlines to inertial worldlines.If dropped, linearity fails at its root: uniform straight-line motion in \(S\) need not be uniform straight-line motion in \(S'\), so no boost of the above form can be defined.

Spatial parity/reflection symmetry across the boost axis and reciprocity of relative velocity.If dropped, the transverse directions could contract asymmetrically and the relative speed of \(S\) seen from \(S'\) need not equal \(v\); the sign and equality of coefficients used below would not hold.

Derivation

\[ x' = A\,x + B\,t, \qquad t' = D\,x + E\,t \]

Homogeneity forces constant coefficients, and mapping straight worldlines to straight worldlines forces linearity, so the most general map is affine; coincident origins at \(t=t'=0\) kill the constant terms. B

\[ y' = y, \qquad z' = z \]

By isotropy about the \(x\)-axis a transverse length can only scale, \(y'=\kappa(v)\,y\); applying relativity to the reciprocal boost gives \(\kappa(v)\kappa(-v)=1\), and reflection symmetry gives \(\kappa(v)=\kappa(-v)\), so \(\kappa=1\). C

\[ x'=0 \;\Rightarrow\; x = v t \;\Rightarrow\; B = -Av \;\Rightarrow\; x' = A\,(x - v t) \]

The spatial origin of \(S'\) is the worldline \(x'=0\); in \(S\) it moves as \(x=vt\) by definition of the boost speed, fixing \(B\) in terms of \(A\). Write \(A\equiv\gamma\). A

\[ x = \gamma\,(x' + v t') \]

By the principle of relativity the inverse boost has the identical form with \(v\to -v\); reciprocity makes the multiplicative factor the same \(\gamma\). B

\[ x = ct,\quad x'=ct' \quad\Longrightarrow\quad ct' = \gamma\,t\,(c-v),\qquad ct = \gamma\,t'\,(c+v) \]

Emit a light pulse from the common origin at \(t=t'=0\); the light postulate makes its speed \(c\) in both frames. Substitute each into the forward and inverse relations of Steps 3–4. A

\[ (ct')(ct) = \gamma^2\,(t t')\,(c-v)(c+v) \;\Rightarrow\; c^2 = \gamma^2\,(c^2 - v^2) \]

Multiply the two relations from Step 5 and cancel the common nonzero factor \(t t'\); this is the closing condition the light postulate imposes on \(\gamma\). B

\[ \gamma^2 = \frac{c^2}{c^2 - v^2} = \frac{1}{1 - v^2/c^2} \;\Rightarrow\; \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \]

Solve algebraically for \(\gamma\); choose the positive root so the transformation reduces to the identity as \(v\to 0\). A

\[ x = \gamma\big(\gamma(x-vt) + v t'\big) \;\Rightarrow\; \gamma v\,t' = x\,(1-\gamma^2) + \gamma^2 v\,t \]

Insert the forward relation \(x'=\gamma(x-vt)\) into the inverse relation \(x=\gamma(x'+vt')\) and solve for \(t'\); this eliminates \(x'\) and leaves \(t'\) in terms of \(x,t\). B

\[ 1-\gamma^2 = -\,\frac{v^2/c^2}{1-v^2/c^2} = -\,\frac{v^2}{c^2}\,\gamma^2 \;\Rightarrow\; \frac{1-\gamma^2}{\gamma v} = -\,\frac{v}{c^2}\,\gamma \]

Substitute \(\gamma^2=1/(1-v^2/c^2)\) into the coefficient of \(x\) and simplify; pure algebra using the definition of \(\gamma\). B

\[ t' = \gamma\left(t - \frac{v x}{c^2}\right) \]

Divide the Step 8 identity by \(\gamma v\) and insert the simplified coefficient from Step 9; this is the time equation of the boost. A

Result

\[ x' = \gamma\,(x - v t),\qquad t' = \gamma\left(t - \frac{v x}{c^2}\right),\qquad y'=y,\quad z'=z,\qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \]

Reading. A boost mixes space and time linearly. The spatial equation says positions measured in \(S'\) are the Galilean \(x-vt\) rescaled by \(\gamma\ge 1\); the time equation adds a position-dependent term \(-vx/c^2\), so two events simultaneous in \(S\) (equal \(t\), different \(x\)) are not simultaneous in \(S'\). Simultaneity is frame-dependent, and \(\gamma\) governs how strongly clocks and rods disagree. As \(v/c\to 0\), \(\gamma\to 1\) and the map collapses to the Galilean \(x'=x-vt,\ t'=t\).

Units check. \(vt\) has units of length \((\text{m/s})(\text{s})=\text{m}\), matching \(x\); \(\gamma\) is the dimensionless function of \(v^2/c^2\). In the time equation, \(vx/c^2\) has units \(\tfrac{(\text{m/s})(\text{m})}{(\text{m/s})^2} = \text{s}\), matching \(t\). Both sides of each equation are dimensionally homogeneous, and \(1-v^2/c^2\) is dimensionless as required inside the square root.

Limiting cases

Low speed \(v\ll c\): \(\gamma\to 1\) and \(vx/c^2\to 0\), giving \(x'=x-vt,\ t'=t\) — the Galilean boost is recovered as the leading-order limit.
Zero boost \(v=0\): \(\gamma=1\) and the transformation is the identity, as any self-consistent kinematics requires.
Ultrarelativistic \(v\to c^-\): \(\gamma\to\infty\); rods contract toward zero length and moving clocks freeze, signalling that massive frames cannot reach \(v=c\).
Small first-order effects: expanding \(\gamma\approx 1+\tfrac12 v^2/c^2\) shows time dilation and length contraction are second order in \(v/c\), while the simultaneity shift \(-vx/c^2\) is first order and dominates at moderate speeds.
Interval invariance: substituting the boost gives \(c^2t'^2-x'^2 = c^2t^2-x^2\) for all \(v\), the geometric statement equivalent to the two postulates.

Breaks when

Non-inertial frames. If \(S'\) accelerates, \(v=v(t)\) and the coefficients are no longer constant; linearity in Step 1 fails and the global boost is replaced by a local (instantaneous) transformation valid only along a worldline.
Curved spacetime / strong gravity. With a nonzero gravitational field the metric is not globally Minkowskian; the interval \(c^2t^2-x^2\) is not preserved and the Lorentz transformation holds only in a local inertial frame (the tangent space), not across finite regions.
Media with \(n\ne 1\). Light in a medium travels at \(c/n<c\), and that phase speed is not frame-invariant; the light postulate as used in Step 5 applies to the vacuum speed \(c\), so the closing condition must not be applied to the in-medium speed.
Non-collinear or rotating boosts. The single-axis derivation assumes the boost is along \(x\); composing boosts in different directions introduces a rotation (Wigner rotation) that the standard-configuration form does not display.

Failure modes

Keeping \(t'=t\). Carrying the Galilean time equation and only rescaling \(x\); this violates the light postulate and makes the two relations in Step 5 inconsistent.
Assuming the form of \(\gamma\) at the start. Writing \(\gamma=1/\sqrt{1-v^2/c^2}\) before Step 6 hides the physics; \(\gamma\) must emerge from the light postulate, not be imported.
Sign error in the time term. Writing \(t'=\gamma(t+vx/c^2)\) confuses the forward boost with its inverse; the forward boost (from \(S\) to \(S'\) moving with \(+v\)) carries a minus sign.
Contracting transverse lengths. Setting \(y'=\gamma y\) by analogy with \(x\); the reciprocity/reflection argument of Step 2 forces \(y'=y\).
Mismatched frame arguments. Substituting \(x=ct\) but \(x'=vt'\) (or vice versa) in Step 5; the light postulate requires \(x'=ct'\) in the primed frame too.
Dropping the sign of \(v\). Ignoring the sign of \(v\) when writing the inverse boost breaks the \(v\to -v\) symmetry that fixes the inverse coefficient in Step 4.

Discussion

The derivation is really a statement about symmetry. Homogeneity and isotropy reduce the allowed maps to a two-parameter family of linear transformations; the requirement that the composition of two boosts be another boost of the same family (a group) forces the existence of a single invariant speed. The two postulates fix that invariant to be \(c\). Had experiment given an infinite invariant speed, the same machinery would have produced the Galilean group instead — so relativity and Newtonian kinematics are two faces of one group-theoretic structure, distinguished only by the value of one constant.

Geometrically, the boost is a hyperbolic rotation in the \((ct,x)\) plane. Introducing the rapidity \(\phi\) with \(\tanh\phi = v/c\), the transformation reads \(ct'=ct\cosh\phi - x\sinh\phi\) and \(x'=x\cosh\phi - ct\sinh\phi\), so \(\gamma=\cosh\phi\) and \(\gamma v/c=\sinh\phi\). Collinear boosts then add by adding rapidities, which is why velocities add nonlinearly: \(\tanh\) of a sum is not the sum of \(\tanh\)s. This is the clean statement behind the velocity-addition law.

The invariant that survives every boost is the interval \(s^2=c^2t^2-x^2-y^2-z^2\). The Lorentz transformations are exactly the linear maps preserving this Minkowski quadratic form, just as ordinary rotations are the linear maps preserving \(x^2+y^2+z^2\). Special relativity is thus the geometry of a spacetime with signature \((+,-,-,-)\); the two postulates are the physical inputs that select this metric.

A deeper route bypasses light altogether. Assuming only homogeneity, isotropy, the relativity principle, and that boosts form a one-parameter group, one shows the transformation must contain a universal invariant speed \(K\) with \(\gamma=1/\sqrt{1-v^2/K^2}\); the value of \(K\) is then left to experiment, and the constancy of light merely measures \(K=c\). In this view the second postulate is not logically indispensable — causal structure and group closure already demand a finite invariant speed, and light is simply the massless field that happens to travel at it. This is why photons, gravitational waves, and gluon-mediated causal limits all share the same \(c\): it is a property of spacetime, not of light.

Common misconceptions. \(\gamma\) is not a speed and does not "add to" velocities; it is a dimensionless clock/rod-comparison factor. Length contraction and time dilation are not optical illusions or signal-delay artifacts — they are what synchronized rulers and clocks actually record, and they are reciprocal (each frame sees the other's rods shortened). Finally, "the speed of light is constant" refers to the vacuum speed measured locally; light slowed in glass does not falsify the postulate.

Worked examples

Muon lab-frame lifetime. A muon moves at \(v=0.98\,c\). Its proper lifetime is \(\tau_0 = 2.2\,\mu\text{s}\). Creation and decay occur at the same place in the muon frame \(S'\) (so \(\Delta x'=0\), \(\Delta t'=\tau_0\)). Find the lab-frame lifetime \(\Delta t\).

\[ \Delta t = \gamma\left(\Delta t' + \frac{v\,\Delta x'}{c^2}\right) = \gamma\,\Delta t' \quad(\Delta x'=0) \]

Use the inverse time equation \(t=\gamma(t'+vx'/c^2)\); the two decay events share a location in \(S'\). A

\[ \gamma = \frac{1}{\sqrt{1-(0.98)^2}} = \frac{1}{\sqrt{1-0.9604}} = \frac{1}{\sqrt{0.0396}} = 5.025 \]

Evaluate \(\gamma\) numerically from \(\beta=0.98\). A

\[ \Delta t = 5.025 \times 2.2\,\mu\text{s} = 11.1\,\mu\text{s} \]

Multiply through; units of microseconds carry through since \(\gamma\) is dimensionless. A

\[ \Delta t = \gamma\,\tau_0 \approx 11.1\ \mu\text{s} \]

Reading. The moving muon lives about five times longer in the lab than in its own frame, which is why cosmic-ray muons reach the ground. Units check: \(\gamma\) dimensionless \(\times\) \(\mu\text{s}\) gives \(\mu\text{s}\).

Relativity of simultaneity on a train. In the ground frame \(S\), two lamps flash simultaneously (\(\Delta t = 0\)) at the ends of a platform separated by \(\Delta x = 300\ \text{m}\). A train moves at \(v = 0.6\,c\) along \(x\). Find the time gap \(\Delta t'\) the train frame assigns to the two flashes.

\[ \Delta t' = \gamma\left(\Delta t - \frac{v\,\Delta x}{c^2}\right) = -\,\gamma\,\frac{v\,\Delta x}{c^2}\quad(\Delta t = 0) \]

Apply the forward time equation to the coordinate differences; simultaneity in \(S\) sets \(\Delta t=0\). A

\[ \gamma = \frac{1}{\sqrt{1-(0.6)^2}} = \frac{1}{\sqrt{0.64}} = 1.25 \]

Evaluate \(\gamma\) from \(\beta=0.6\). A

\[ \Delta t' = -\,(1.25)\,\frac{(0.6\,c)(300\,\text{m})}{c^2} = -\,(1.25)\,\frac{(0.6)(300\,\text{m})}{c} = -\,(1.25)\,\frac{180\,\text{m}}{3\times10^8\,\text{m/s}} \]

Insert \(v=0.6c\) and cancel one factor of \(c\); note \(v\Delta x/c^2 = 0.6\,\Delta x/c\). B

\[ \Delta t' = -\,1.25 \times 6.0\times10^{-7}\ \text{s} = -\,7.5\times10^{-7}\ \text{s} = -0.75\ \mu\text{s} \]

Arithmetic; the negative sign means the leading-\(x\) flash occurs later in \(S'\). A

\[ \Delta t' = -\,\gamma\,\frac{v\,\Delta x}{c^2} \approx -0.75\ \mu\text{s} \]

Reading. Events simultaneous on the platform are \(0.75\ \mu\text{s}\) apart for the train; simultaneity is not absolute. Units check: \(\tfrac{(\text{m/s})(\text{m})}{(\text{m/s})^2}=\text{s}\), and \(\gamma\) is dimensionless.

Problems

A spaceship passes Earth at \(v=0.8\,c\). How long does a \(10\ \text{s}\) shipboard interval last in Earth's frame?
Solution
\(\gamma = 1/\sqrt{1-0.64}=1/\sqrt{0.36}=1.667\). The two events (start and stop) share a location on the ship, so \(\Delta t_{\text{Earth}}=\gamma\,\Delta t_{\text{ship}} = 1.667\times 10\ \text{s} = 16.7\ \text{s}\).
A rod of proper length \(L_0 = 2.0\ \text{m}\) lies along \(x\) and moves at \(v=0.6\,c\). Find its length in the lab. (A length is measured at fixed lab time, \(\Delta t=0\), so \(\Delta x = L_0/\gamma\).)
Solution
\(\gamma = 1/\sqrt{1-0.36}=1/0.8=1.25\). Length contraction: \(L = L_0/\gamma = 2.0/1.25 = 1.6\ \text{m}\). The moving rod is shorter along the direction of motion.
In frame \(S\), event A is at \((t_A, x_A)=(0,\,0)\) and event B is at \((t_B, x_B)=(2\ \mu\text{s},\ 900\ \text{m})\). For a boost \(v=0.5\,c\), find \(\Delta t'\) and \(\Delta x'\).
Solution
\(\gamma=1/\sqrt{1-0.25}=1.1547\). \(\Delta x = 900\ \text{m}\), \(\Delta t = 2\times10^{-6}\ \text{s}\). \(\Delta x' = \gamma(\Delta x - v\Delta t) = 1.1547\,(900 - 0.5\cdot 3\times10^8\cdot 2\times10^{-6}) = 1.1547\,(900-300)=692.8\ \text{m}\). \(\Delta t' = \gamma(\Delta t - v\Delta x/c^2) = 1.1547\,(2\times10^{-6} - 0.5\cdot 900/(3\times10^8)) = 1.1547\,(2\times10^{-6}-1.5\times10^{-6}) = 5.77\times10^{-7}\ \text{s}=0.577\ \mu\text{s}\).
Show that the interval \(s^2 = c^2 t^2 - x^2\) is invariant under the boost, i.e. \(c^2 t'^2 - x'^2 = c^2 t^2 - x^2\).
Solution
\(c^2t'^2 = c^2\gamma^2(t-vx/c^2)^2 = \gamma^2(c^2t^2 - 2vxt + v^2x^2/c^2)\) and \(x'^2 = \gamma^2(x-vt)^2 = \gamma^2(x^2 - 2vxt + v^2t^2)\). Subtract: \(c^2t'^2 - x'^2 = \gamma^2\big[c^2t^2 + v^2x^2/c^2 - x^2 - v^2t^2\big] = \gamma^2\big[c^2t^2(1-v^2/c^2) - x^2(1-v^2/c^2)\big] = \gamma^2(1-v^2/c^2)(c^2t^2-x^2)\). Since \(\gamma^2(1-v^2/c^2)=1\), the cross terms cancel and \(c^2t'^2 - x'^2 = c^2t^2 - x^2\).
Two particles move along \(x\): one at \(u=0.9\,c\) in frame \(S\), observed from frame \(S'\) moving at \(v=0.9\,c\). Use the boost to derive the velocity-addition law \(u' = (u-v)/(1-uv/c^2)\) and evaluate \(u'\).
Solution
From \(x'=\gamma(x-vt)\) and \(t'=\gamma(t-vx/c^2)\), take differentials: \(dx'=\gamma(dx-v\,dt)\), \(dt'=\gamma(dt-v\,dx/c^2)\). Then \(u'=\dfrac{dx'}{dt'}=\dfrac{dx-v\,dt}{dt-v\,dx/c^2}=\dfrac{dx/dt-v}{1-v\,(dx/dt)/c^2}=\dfrac{u-v}{1-uv/c^2}\). Numerically \(u'=\dfrac{0.9c-0.9c}{1-(0.9)(0.9)}=\dfrac{0}{0.19}=0\): the particle is at rest in \(S'\), as expected since it comoves with \(S'\). As a check with \(u=0.5c,\ v=0.9c\): \(u'=(0.5-0.9)c/(1-0.45)=-0.4c/0.55=-0.727c\), still below \(c\) in magnitude.