Adding Velocities the Relativistic Way
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You should be comfortable with the Lorentz transformation equations and the definition of the Lorentz factor \(\gamma\). Familiarity with the relativity of simultaneity is also assumed.
Why this matters
Galilean velocity addition fails when speeds approach the speed of light, necessitating a new composition law.
In classical mechanics, velocities simply add. If a train moves at \(v\) and a ball is thrown forward at \(u'\), the ground observer sees \(u = v + u'\). However, Maxwell's equations predict a constant speed of light \(c\) for all inertial observers. If we applied Galilean addition to light, an observer moving towards a light source would measure a speed greater than \(c\), violating the second postulate of special relativity. We must therefore derive a velocity addition formula that ensures no massive object exceeds \(c\) and that light always travels at \(c\).
The Intuition Behind Relativistic Addition
Imagine two observers, Alice and Bob. Alice is stationary on Earth, and Bob is in a spaceship moving at a significant fraction of the speed of light relative to Alice. Bob throws a ball forward. In Bob's frame, the ball moves at speed \(u'\). In Alice's frame, the ball moves at speed \(u\). If we used simple addition, \(u = v + u'\), we might predict the ball exceeds \(c\). Relativity prevents this by "dilating" the time and length measurements between the frames. The denominator in the relativistic formula acts as a correction factor that becomes significant only when \(v\) or \(u'\) are close to \(c\). At low speeds, this correction is negligible, and we recover the familiar Galilean result. This ensures that the speed of light remains the universal speed limit, acting as an asymptote that massive particles can approach but never cross.
Deriving the Longitudinal Component
To find the velocity in frame \(S\), we start with the definition of velocity \(u_x = \frac{dx}{dt}\). Using the Lorentz transformations for space and time, we express \(dx\) and \(dt\) in terms of the primed coordinates. The spatial transformation is \(dx = \gamma(dx' + v dt')\) and the temporal transformation is \(dt = \gamma(dt' + \frac{v}{c^2} dx')\). Dividing these expressions gives the ratio \(\frac{dx}{dt}\).
Dividing the numerator and denominator by \(dt'\) allows us to substitute \(u_x' = \frac{dx'}{dt'}\). The \(\gamma\) factors cancel out, leaving the standard relativistic velocity addition formula for the parallel component.
This equation shows that if \(u_x' = c\), then \(u_x = \frac{c+v}{1+v/c} = c\). The speed of light is invariant. If \(u_x' < c\) and \(v < c\), the result is always less than \(c\).
Transverse Velocity Components
Velocities perpendicular to the direction of relative motion also change, but in a different way. For the \(y\)-component, we use \(u_y = \frac{dy}{dt}\). Since the Lorentz transformation does not mix \(y\) and \(t\) directly in the same way, we have \(dy = dy'\). However, \(dt\) still contains the \(dx'\) term. Substituting \(dt = \gamma(dt' + \frac{v}{c^2} dx')\) into the velocity definition yields a more complex expression.
Recognizing that \(\frac{dy'}{dt'} = u_y'\) and \(\frac{dx'}{dt'} = u_x'\), we arrive at the transverse velocity transformation. Note the presence of \(\gamma\) in the denominator and the coupling with the longitudinal velocity \(u_x'\).
The same logic applies to the \(z\)-component. This result implies that even if an object moves purely in the \(y\)-direction in its own frame (\(u_x' = 0\)), an observer in \(S\) will see a reduced \(y\)-velocity due to time dilation (\(u_y = u_y'/\gamma\)).
Consistency with the Speed of Light
A crucial test of any relativistic formula is its behavior for light. Consider a photon moving in the \(x\)-direction in frame \(S'\), so \(u_x' = c\) and \(u_y' = u_z' = 0\). Applying the longitudinal formula:
The result is exactly \(c\), independent of \(v\). This confirms that the speed of light is invariant under Lorentz transformations. For massive particles, where \(u' < c\), the denominator \(1 + \frac{u' v}{c^2}\) is always greater than 1 (assuming \(v\) and \(u'\) are in the same direction), ensuring that \(u\) remains less than \(c\). This mathematical structure enforces the causal structure of spacetime, preventing information or matter from traveling faster than light.
The derivation
Relativistic Velocity Addition
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.
A common misconception is that velocity addition is symmetric in the sense that \(u = v + u'\) and \(u' = u - v\). In relativity, the inverse transformation is not simply subtracting \(v\). To find \(u'\) from \(u\), you must use the formula with \(v\) replaced by \(-v\), which changes the sign in the denominator. This asymmetry arises because the time coordinate mixes with space differently in forward and inverse transformations. C
Worked example
Problem. A spaceship moves at \(v = 0.8c\) relative to Earth. It fires a probe forward at a speed of \(u' = 0.6c\) relative to the spaceship. What is the speed of the probe as measured by an observer on Earth?
- Identify the given values: \(v = 0.8c\), \(u' = 0.6c\).
- Use the relativistic velocity addition formula for the longitudinal component: \(u = \frac{u' + v}{1 + \frac{u' v}{c^2}}\).
- Substitute the values into the numerator: \(u' + v = 0.6c + 0.8c = 1.4c\).
- Substitute the values into the denominator term: \(\frac{u' v}{c^2} = \frac{(0.6c)(0.8c)}{c^2} = 0.48\).
- Calculate the full denominator: \(1 + 0.48 = 1.48\).
- Divide the numerator by the denominator: \(u = \frac{1.4c}{1.48}\).
- Perform the division: \(u \approx 0.946c\).
Units check. The result is a fraction of \(c\), which is dimensionless in natural units but represents a velocity. If expressed in \(\mathrm{m/s}\), it would be \(0.946 \times 3.00 \times 10^8 \approx 2.84 \times 10^8 \ \mathrm{m/s}\). This is less than \(c\), as required.
Three quick tests
- If \(v = 0.5c\) and \(u' = 0.5c\), what is \(u\)?
Answer
\(u = \frac{0.5c + 0.5c}{1 + 0.25} = \frac{1.0c}{1.25} = 0.8c\). - Why does the transverse velocity \(u_y\) decrease when observed from \(S\)?
Answer
Due to time dilation, the time interval \(dt\) in \(S\) is longer than \(dt'\) in \(S'\), reducing the ratio \(dy/dt\). - Can two subluminal velocities add to produce a superluminal velocity?
Answer
No, the denominator ensures the result is always less than \(c\).
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.