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Derivation

Relativistic Velocity Addition

D-095 Home PU-105 Threads light · symmetry Depends on Lorentz Transformation from the Two Postulates
Statement

Given two inertial frames \(S\) and \(S'\) in standard configuration, with \(S'\) moving at constant velocity \(v\) along the shared \(x\)-axis relative to \(S\), a particle whose velocity components in \(S'\) are \((u_x', u_y', u_z')\) has, in \(S\), the velocity components \[ u_x=\frac{u_x'+v}{1+\dfrac{u_x' v}{c^{2}}},\qquad u_y=\frac{u_y'}{\gamma\left(1+\dfrac{u_x' v}{c^{2}}\right)},\qquad u_z=\frac{u_z'}{\gamma\left(1+\dfrac{u_x' v}{c^{2}}\right)}, \] with \(\gamma=\left(1-v^{2}/c^{2}\right)^{-1/2}\). This composition law keeps \(c\) invariant and maps any pair of subluminal speeds to a subluminal speed.

Why it matters

The Galilean rule \(u_x=u_x'+v\) is the single most intuitive statement in kinematics, and it is wrong. It predicts that light chased at \(0.9c\) recedes at \(0.1c\), contradicting the second postulate and every measurement since Michelson–Morley. The relativistic law is the unique correction that reconciles velocity composition with a universal speed limit.

Beyond fixing light, it underpins the aberration of starlight, the relativistic Doppler effect, the Fresnel drag coefficient measured by Fizeau in flowing water decades before relativity, and the kinematics of particle collisions where lab and centre-of-mass frames must be reconciled.

Assumptions
The Lorentz transformation between \(S\) and \(S'\) holds.If we used the Galilean transformation instead, differentiation would return \(u_x=u_x'+v\) exactly, and \(c\) would not be invariant.
Relative motion is along a single axis (standard configuration).For a boost in an arbitrary direction the transverse/longitudinal split changes and one must decompose \(\vec{u}'\) parallel and perpendicular to \(\vec{v}\); the scalar formula for \(u_x\) alone no longer captures the full law.
The frames are inertial and \(v\) is constant.For accelerating frames the transformation is not a global Lorentz boost, and instantaneous velocity addition must be applied differentially along the worldline.
Spacetime is flat (special relativity).In curved spacetime there is no global notion of the relative velocity of distant objects, and comparing velocities requires parallel transport; the formula holds only locally.
Derivation
1
\[ x=\gamma\left(x'+v t'\right),\quad t=\gamma\left(t'+\frac{v x'}{c^{2}}\right),\quad y=y',\quad z=z' \]
The inverse Lorentz transformation from \(S'\) to \(S\); obtained from the assumed forward transformation by \(v\to -v\). A
2
\[ dx=\gamma\left(dx'+v\,dt'\right),\quad dt=\gamma\left(dt'+\frac{v\,dx'}{c^{2}}\right),\quad dy=dy',\quad dz=dz' \]
Differentiate each linear relation; the coefficients \(\gamma\), \(v\), \(c\) are constants so differentials transform identically to coordinates. A
3
\[ u_x=\frac{dx}{dt}=\frac{\gamma\left(dx'+v\,dt'\right)}{\gamma\left(dt'+\dfrac{v\,dx'}{c^{2}}\right)} \]
Velocity is the ratio of coordinate differentials, \(u_x=dx/dt\); substitute the transformed differentials from Step 2. A
4
\[ u_x=\frac{dx'+v\,dt'}{dt'+\dfrac{v\,dx'}{c^{2}}} \]
The common factor \(\gamma\) cancels between numerator and denominator, since \(\gamma\neq 0\) for \(|v|<c\). A
5
\[ u_x=\frac{\dfrac{dx'}{dt'}+v}{1+\dfrac{v}{c^{2}}\dfrac{dx'}{dt'}} \]
Divide numerator and denominator by \(dt'\) (nonzero for a timelike worldline). This isolates the frame-\(S'\) velocity \(u_x'=dx'/dt'\). A
6
\[ \boxed{\,u_x=\frac{u_x'+v}{1+\dfrac{u_x' v}{c^{2}}}\,} \]
Identify \(u_x'=dx'/dt'\); the symbolic rearrangement for the longitudinal component is complete, exact for all speeds. A
7
\[ u_y=\frac{dy}{dt}=\frac{dy'}{\gamma\left(dt'+\dfrac{v\,dx'}{c^{2}}\right)} =\frac{\dfrac{dy'}{dt'}}{\gamma\left(1+\dfrac{v}{c^{2}}\dfrac{dx'}{dt'}\right)} \]
Repeat with \(dy=dy'\) from Step 2; the numerator carries no \(\gamma\) because transverse lengths are unchanged, but the denominator retains it from \(dt\). B
8
\[ \boxed{\,u_y=\frac{u_y'}{\gamma\left(1+\dfrac{u_x' v}{c^{2}}\right)}\,},\qquad \boxed{\,u_z=\frac{u_z'}{\gamma\left(1+\dfrac{u_x' v}{c^{2}}\right)}\,} \]
Identify \(u_y'=dy'/dt'\); by the \(y\leftrightarrow z\) symmetry of the transformation the same result holds for \(u_z\). B
9
\[ c-u_x=\frac{(c-u_x')(c-v)}{1+\dfrac{u_x' v}{c^{2}}}\cdot\frac{1}{c}\,\cdot c \;\Longrightarrow\; u_x^2=c^2 \iff u_x'^2=c^2 \]
Closure/invariance check: rearranging shows \(c-u_x\) is a product of the non-negative factors \((c-u_x')\) and \((c-v)\) over a positive denominator, so \(u_x'<c\) and \(v<c\) force \(u_x<c\); and \(u_x'=c\) gives \(u_x=c\) for any \(v\). The null cone is preserved, as the second postulate demands. C
Result
\[ u_x=\frac{u_x'+v}{1+\dfrac{u_x' v}{c^{2}}},\qquad u_{y,z}=\frac{u_{y,z}'}{\gamma\left(1+\dfrac{u_x' v}{c^{2}}\right)} \]

Reading. The Galilean numerator \(u_x'+v\) survives, but is suppressed by the denominator \(1+u_x' v/c^{2}\). When both speeds are small compared with \(c\) the denominator is \(\approx 1\) and the ordinary rule returns; when either approaches \(c\) the denominator grows just enough to keep the composite below \(c\). Transverse velocities are not frame-invariant: they inherit the same denominator plus a factor \(1/\gamma\) from time dilation, so a purely transverse motion in \(S'\) is both slowed and tilted in \(S\).

Units check. Numerator \(u_x'+v\) has units of \(\mathrm{m\,s^{-1}}\). The product \(u_x' v/c^{2}\) has units \((\mathrm{m\,s^{-1}})^2/(\mathrm{m\,s^{-1}})^2\), i.e. dimensionless, so the denominator is dimensionless and \(u_x\) comes out in \(\mathrm{m\,s^{-1}}\). Likewise \(\gamma\) is dimensionless, so \(u_y\) is a velocity. Consistent.

Limiting cases
  • Non-relativistic limit \(u_x',v\ll c\): the denominator \(\to 1\), \(\gamma\to 1\), recovering \(u_x=u_x'+v\), \(u_y=u_y'\) (Galilean addition).
  • Light ray \(u_x'=c\): \(u_x=(c+v)/(1+v/c)=c\) for any \(v\) — the speed of light is invariant, as required.
  • Two speeds near \(c\): \(u_x'=v=c-\epsilon\) gives \(u_x\to c^{-}\); subluminal composed with subluminal stays strictly subluminal.
  • Equal and opposite \(u_x'=-v\): \(u_x=0\); the particle is at rest in \(S\), as symmetry requires.
  • Rapidity form: writing \(u=c\tanh\phi\), \(v=c\tanh\eta\) gives \(\phi_{\text{total}}=\phi+\eta\) — collinear boosts add linearly in rapidity.
Breaks when
  • Non-collinear velocities. If \(\vec{u}'\) is not parallel to \(\vec{v}\), the scalar formula for \(u_x\) is insufficient; one must apply the full vector law, and successive non-collinear boosts fail to commute (a Thomas–Wigner rotation appears). The simple "add and divide" picture breaks.
  • Non-inertial frames or gravity. For accelerating \(S'\) or in curved spacetime there is no single \(v\) and \(\gamma\); velocities can only be composed differentially along a worldline, and distant relative velocity is not even well defined.
  • Superluminal inputs. Feeding \(u_x'>c\) can drive the denominator through zero at \(u_x' v=-c^2\), producing infinite or negative \(u_x\); tachyonic inputs violate the derivation's premises.
  • Massless-frame limit \(v\to c\). \(\gamma\to\infty\) and there is no rest frame of light; the transformation is singular, so one cannot use a photon's frame as \(S'\).
Failure modes
  • Sign of \(v\) confusion. Using \(u_x=(u_x'-v)/(1-u_x'v/c^2)\) (the forward transform) when the problem asks for the object's velocity in \(S\) given its velocity in \(S'\). The denominator sign always matches the numerator's added-velocity sign.
  • Forgetting the \(1/\gamma\) on transverse components. Writing \(u_y=u_y'/(1+u_x'v/c^2)\) without the \(\gamma\). The transverse velocity carries both the denominator and time dilation.
  • Applying it to speeds, not components. Plugging total speeds \(|\vec u'|\) and \(v\) into the collinear formula when the motion is at an angle. The formula is per-component in standard configuration.
  • Using the wrong velocity in \(\gamma\). \(\gamma\) is built from the boost speed \(v\) between frames, never from the particle speed \(u'\).
  • Assuming velocities clamp at \(c\). Believing \(0.9c\oplus 0.9c=c\) exactly; the correct result is \(0.994c\), strictly below \(c\) but not pinned to it.
  • Closing-speed confusion. Calling a \(1.8c\) closing rate between two objects (measured in one frame) a violation of relativity; it is legal because it is not any single body's speed in another's rest frame.
Discussion

The deepest way to read the law is through rapidity. Define \(\phi\) by \(u=c\tanh\phi\). Then the collinear composition \(u_x=(u_x'+v)/(1+u_x'v/c^2)\) is exactly the addition formula for \(\tanh\), so \(\tanh\phi_{\text{tot}}=\tanh(\phi'+\eta)\), i.e. rapidities add. Velocities look awkward precisely because \(\tanh\) is nonlinear; the underlying group of collinear boosts is simply the additive real line. The speed \(c\) is the image of \(\phi\to\infty\), which is why it can never be reached by finite composition — you can keep adding rapidity forever and \(\tanh\) only asymptotes to \(1\).

The invariance of \(c\) is not an accident of the algebra; it is built into the Lorentz transformation, which was constructed to preserve the interval \(c^2 t^2-x^2\). Velocity addition is the tangent-space expression of that same symmetry. This is the "symmetry" thread: the composition law is the group law of boosts, and the "light" thread: \(c\) is the fixed point of that group law. The two postulates enter exactly once — through the Lorentz transformation assumed in Step 1 — and everything else is calculus.

The cancellation of \(\gamma\) in Step 4 carries a physical message: the longitudinal composition law knows nothing about time dilation per se. The same \(\gamma\) governs energy and momentum, yet the longitudinal velocity law is a pure ratio and is scale-free in \(\gamma\). Historically, the Fizeau experiment (1851) measured the speed of light in water flowing at speed \(v\) and found \(u\approx c/n + v(1-1/n^2)\), the Fresnel drag coefficient; Fresnel fitted it with an ad hoc ether-drag hypothesis, whereas relativity derives it in two lines by composing \(u'=c/n\) with \(v\) and expanding to first order.

For non-collinear boosts the story is richer: the composition of two Lorentz boosts in different directions is not a pure boost but a boost combined with a spatial rotation, the Thomas–Wigner rotation. This is why the set of pure boosts does not form a subgroup of the Lorentz group, and it is the kinematic origin of Thomas precession in atomic spin–orbit coupling, contributing the famous factor of \(1/2\). Geometrically, subluminal velocities with this composition rule form the Beltrami–Klein model of hyperbolic space on the ball \(|\vec u|/c<1\); the non-commutativity is exactly the negative curvature of that velocity space. The collinear formula derived here is the flat, one-dimensional geodesic case.

Common misconceptions. The formula is often stated as "you can't go faster than light by adding speeds," but the sharper statement is that the subluminal region \(|u|<c\) is closed under this composition and \(|u|=c\) is a fixed boundary. It is not that velocities are "capped"; the map simply never pushes a subluminal input across the light cone. Also, superluminal phase velocities or projected spots (e.g. a laser dot sweeping across the Moon) are not composed by this law and carry no energy or information.

Worked examples
1
\[ u_x'=0.60c,\qquad v=0.80c \]
A ship moves at \(v=0.80c\) relative to Earth and fires a probe forward at \(u_x'=0.60c\) relative to the ship. Find the probe's speed in the Earth frame. Collinear, so use Step 6. A
2
\[ u_x=\frac{u_x'+v}{1+\dfrac{u_x'v}{c^2}}=\frac{0.60c+0.80c}{1+(0.60)(0.80)}=\frac{1.40c}{1.48} \]
Substitute numbers; the dimensionless product \(u_x'v/c^2=(0.60)(0.80)=0.48\). A
\[ u_x=0.9459\,c\approx 2.84\times10^{8}\ \mathrm{m\,s^{-1}} \]

Reading. Two large subluminal speeds compose to \(0.946c\), close to but strictly below \(c\). The Galilean answer \(1.40c\) is unphysical; the denominator supplies the essential correction.

Units check. The bracket is dimensionless; \(0.9459\times c=0.9459\times 2.998\times10^8\ \mathrm{m\,s^{-1}}\) is a velocity. Consistent.

1
\[ u_x'=\frac{c}{n},\qquad n=1.33,\qquad v=5.0\ \mathrm{m\,s^{-1}} \]
Fizeau's experiment: light travels at \(c/n\) in still water (frame \(S'\)); the water flows at \(v=5.0\ \mathrm{m\,s^{-1}}\) relative to the lab (frame \(S\)). Find the light speed in the lab and identify the Fresnel drag coefficient. B
2
\[ u_x=\frac{\dfrac{c}{n}+v}{1+\dfrac{v}{nc}} \approx\left(\frac{c}{n}+v\right)\left(1-\frac{v}{nc}\right) \approx\frac{c}{n}+v-\frac{v}{n^2} \]
Since \(v/(nc)\sim 10^{-8}\ll 1\), expand the denominator to first order in \(v\); the \((c/n)\cdot v/(nc)\) and higher cross terms are second order and dropped. B
3
\[ u_x\approx\frac{c}{n}+v\left(1-\frac{1}{n^2}\right),\qquad 1-\frac{1}{n^2}=1-\frac{1}{1.33^2}=0.435 \]
Factor the drag coefficient \(1-1/n^2\) and evaluate with \(n=1.33\), so \(1/n^2=0.565\). B
\[ \Delta u = v\left(1-\frac{1}{n^2}\right)=5.0\times 0.435\approx 2.2\ \mathrm{m\,s^{-1}} \]

Reading. The light is dragged forward by only \(2.2\ \mathrm{m\,s^{-1}}\), not the full \(5.0\ \mathrm{m\,s^{-1}}\) of the water. The Fresnel drag coefficient \(1-1/n^2\) emerges directly from relativistic velocity addition, matching Fizeau's 1851 measurement decades before relativity existed.

Units check. \(1-1/n^2\) is dimensionless; \(v\) is in \(\mathrm{m\,s^{-1}}\), so \(\Delta u\) is a velocity. Consistent.

Problems
  1. Two particles head toward each other, each at \(0.90c\) in the lab frame. What is the speed of one particle in the rest frame of the other?
    Solution Boost to ride with one particle at \(v=0.90c\); the other approaches at \(u'=0.90c\): \[ u=\frac{0.90c+0.90c}{1+(0.90)(0.90)}=\frac{1.80c}{1.81}=0.9945c. \] The lab "closing speed" \(1.80c\) is legal (it is not any single object's speed); the physical relative speed is \(0.994c<c\).
  2. A frame \(S'\) moves at \(v=0.60c\). An object in \(S'\) moves backward at \(u'=-0.60c\). Find \(u\) and interpret.
    Solution \[ u=\frac{-0.60c+0.60c}{1+(-0.60)(0.60)}=\frac{0}{1-0.36}=0. \] The object is at rest in \(S\): its backward motion in \(S'\) exactly cancels the frame's forward motion, as symmetry demands.
  3. An object in \(S'\) has velocity \((u_x',u_y')=(0,\,0.50c)\), purely transverse, while \(S'\) moves at \(v=0.60c\). Find its velocity components in \(S\) and its speed.
    Solution Here \(\gamma=1/\sqrt{1-0.36}=1/0.8=1.25\) and the denominator \(1+u_x'v/c^2=1\): \[ u_x=\frac{0+0.60c}{1}=0.60c,\qquad u_y=\frac{0.50c}{1.25\times 1}=0.40c. \] Speed \(=\sqrt{0.60^2+0.40^2}\,c=\sqrt{0.52}\,c=0.721c\). The transverse component is reduced by \(1/\gamma\), and the motion is tilted toward the boost axis.
  4. Show algebraically that if \(|u'|<c\) and \(|v|<c\) then \(|u|<c\) (collinear case).
    Solution Compute \[ c-u=c-\frac{u'+v}{1+u'v/c^2}=\frac{c+u'v/c-u'-v}{1+u'v/c^2}=\frac{(c-u')(c-v)/c}{1+u'v/c^2}. \] Since \(c-u'>0\), \(c-v>0\), and the denominator \(1+u'v/c^2>0\), we have \(c-u>0\). The mirror computation gives \(c+u>0\). Hence \(|u|<c\). \(\square\)
  5. A ship moves at \(v=0.50c\); it launches a shuttle forward at \(0.50c\) relative to itself, and the shuttle launches a pod forward at \(0.50c\) relative to the shuttle. Find the pod's speed in the original frame, and check with rapidities.
    Solution First composition: \[ u_1=\frac{0.50c+0.50c}{1+0.25}=\frac{1.0c}{1.25}=0.80c. \] Second composition with \(0.50c\): \[ u_2=\frac{0.50c+0.80c}{1+(0.50)(0.80)}=\frac{1.30c}{1.40}=0.9286c. \] Rapidity check: \(\phi_0=\operatorname{artanh}(0.5)=0.5493\) each; total \(3\phi_0=1.6479\); \(c\tanh(1.6479)=0.9286c\). Agreement confirms rapidities add. Three \(0.5c\) boosts give \(0.93c\), not \(1.5c\).