KAHIBARO
Table of Contents
Why ordinary velocity addition fails
In everyday life, if a car moves at $20 \, \text{m/s}$ inside a train moving at $30 \, \text{m/s}$, we simply add the speeds and get $50 \, \text{m/s}$. This works well when all speeds are much smaller than the speed of light.
In special relativity, this simple rule cannot always be correct. If it were, then combining two large speeds could give a result greater than the speed of light $c$, which would contradict one of the central results of relativity. The speed of light must be the same for all inertial observers, and no material object can be accelerated to a speed greater than $c$.
So, at high speeds, velocities must combine in a different way.
The basic idea
Imagine two reference frames. Frame $S'$ moves with speed $v$ relative to frame $S$ along the $x$ axis. An object moves in $S'$ with speed $u'$. We want to know its speed $u$ measured in $S$.
In classical physics, we would write
$$
u = u' + v
$$
but in special relativity the correct formula is
$$
u = \frac{u' + v}{1 + \frac{u'v}{c^2}}
$$
This is the relativistic velocity addition formula for motion along the same line.
For velocities in the same direction along one axis,
$$
u = \frac{u' + v}{1 + \frac{u'v}{c^2}}
$$
This replaces the classical rule $u = u' + v$ at high speeds.
What the formula means
The denominator is what makes relativity different. When $u'$ and $v$ are both small compared with $c$, the term $\frac{u'v}{c^2}$ is tiny, so the denominator is almost $1$. Then the formula becomes approximately
$$
u \approx u' + v
$$
which agrees with ordinary experience.
But when the speeds are large, the denominator matters. It reduces the final result so that the combined speed never exceeds $c$.
A simple example
Suppose a spaceship moves at
$$
v = 0.80c
$$
relative to Earth. Inside the spaceship, a probe is launched forward at
$$
u' = 0.70c
$$
relative to the spaceship. The speed of the probe relative to Earth is not $1.50c$. Instead,
$$
u = \frac{0.70c + 0.80c}{1 + (0.70)(0.80)}
= \frac{1.50c}{1.56}
\approx 0.962c
$$
So the probe still moves at less than the speed of light.
No matter how you combine subluminal velocities, the result remains less than $c$.
If $u' < c$ and $v < c$, then $u < c$.
If one of the velocities is the speed of light
Now suppose light is emitted in the forward direction inside the spaceship. In the spaceship frame,
$$
u' = c
$$
Then the observed speed in Earth frame is
$$
u = \frac{c + v}{1 + \frac{cv}{c^2}}
= \frac{c + v}{1 + \frac{v}{c}}
$$
Multiplying top and bottom appropriately gives
$$
u = c
$$
So Earth observers also measure the light speed as $c$.
This is exactly what relativity requires.
If $u' = c$, then the relativistic addition formula gives $u = c$.
This preserves the invariance of the speed of light.
Motion in the opposite direction
If the object moves backward in $S'$, then $u'$ is negative. The same formula still works:
$$
u = \frac{u' + v}{1 + \frac{u'v}{c^2}}
$$
For example, if a spaceship moves at $v = 0.80c$ and a probe is launched backward at $u' = -0.50c$, then
$$
u = \frac{-0.50c + 0.80c}{1 + (-0.50)(0.80)}
= \frac{0.30c}{0.60}
= 0.50c
$$
So Earth sees the probe still moving forward, but more slowly than the spaceship.
Signs are important. Positive and negative velocities tell us direction.
Comparing classical and relativistic addition
The difference becomes important only at very high speeds.
| Case | Classical result | Relativistic result |
|---|---|---|
| $u' = 10 \, \text{m/s}$, $v = 20 \, \text{m/s}$ | $30 \, \text{m/s}$ | almost $30 \, \text{m/s}$ |
| $u' = 0.70c$, $v = 0.80c$ | $1.50c$ | $0.962c$ |
| $u' = c$, $v = 0.80c$ | $1.80c$ | $c$ |
Relative velocity in the reverse direction
Sometimes we want to solve the formula for $u'$ instead of $u$. Rearranging gives
$$
u' = \frac{u - v}{1 - \frac{uv}{c^2}}
$$
This is useful when one observer measures the object's velocity and we want the velocity in another moving frame.
Inverse form of relativistic velocity addition:
$$
u' = \frac{u - v}{1 - \frac{uv}{c^2}}
$$
Velocities in perpendicular directions
When motion is not purely along one line, the transformation is more complicated. If frame $S'$ moves at speed $v$ along the $x$ axis, and an object has components $u_x'$ and $u_y'$ in $S'$, then in frame $S$:
$$
u_x = \frac{u_x' + v}{1 + \frac{u_x'v}{c^2}}
$$
and
$$
u_y = \frac{u_y'}{\gamma \left(1 + \frac{u_x'v}{c^2}\right)}
$$
where
$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
The parallel component transforms differently from the perpendicular component. This is a distinctly relativistic effect.
We do not need to explore all consequences here, but it is important to see that velocity transformation in relativity is a component-by-component rule, not simple vector addition.
Physical interpretation
Relativistic velocity addition tells us that space and time are connected. Different observers do not just disagree about distance. They also disagree about time intervals, and this changes how velocity, which is distance divided by time, must transform.
That is why the formula is not a simple sum.
The rule also shows that $c$ acts like a limiting speed. The closer an object's speed is to $c$, the harder it is for another added motion to increase it by much.
A visual picture
Key points to remember
Relativistic velocity addition replaces ordinary addition when speeds are comparable to the speed of light. The one dimensional formula is
$$
u = \frac{u' + v}{1 + \frac{u'v}{c^2}}
$$
It reduces to the classical formula for small speeds. It guarantees that no observer measures a material object moving faster than light, and it ensures that light is measured to travel at speed $c$ in every inertial frame.
Important conclusions:
$$
u = \frac{u' + v}{1 + \frac{u'v}{c^2}}
$$
For $u', v \ll c$, this becomes approximately $u \approx u' + v$.
If $u' = c$, then $u = c$.
If $u' < c$ and $v < c$, then $u < c$.