Relativistic Velocity Addition
Relativistic Velocity Addition
In everyday life, velocities add in the obvious way. If you walk at 5 kilometers per hour on a train moving at 100 kilometers per hour, a person on the platform sees you moving at 105 kilometers per hour. This intuitive rule — called Galilean velocity addition — works perfectly at low speeds but fails dramatically near the speed of light.
The problem is clear: if Galilean addition were correct, two rockets each moving at 0.6c relative to each other would see each other at 1.2c — faster than light. But Einstein's second postulate says nothing can exceed c. Something must give, and what gives is the simple addition rule itself.
Einstein derived the relativistic velocity addition formula, which replaces simple addition with a formula that ensures the result never exceeds c. If an object moves at velocity u-prime in a reference frame that is itself moving at velocity v relative to a stationary observer, then the stationary observer measures the combined velocity as u equals u-prime plus v, all divided by one plus u-prime times v over c-squared.
This formula has remarkable properties. When both u-prime and v are much smaller than c, the denominator is essentially one and the formula reduces to the familiar u equals u-prime plus v. But when either velocity approaches c, the denominator grows and limits the result to at most c. If u-prime equals c — for instance, if someone on the moving frame shines a flashlight — then the result is exactly c regardless of v. Light always travels at c in every reference frame, just as Einstein's postulate demands.
The relativistic velocity addition formula is built from nothing more than multiplication, addition, and division. It is perhaps the most elegant example in all of physics of how a simple algebraic formula can encode a profound truth about the structure of spacetime.
If an object moves at velocity $u'$ in a frame that moves at velocity $v$ relative to a stationary observer, the observed velocity is:
$$u = \frac{u' + v}{1 + \dfrac{u' v}{c^2}}$$
This guarantees $u \le c$ whenever $u' \le c$ and $v \le c$.
Two spaceships approach each other, each moving at $0.7c$ relative to Earth. What is their speed relative to each other?
Step 1: Set $u' = 0.7c$ and $v = 0.7c$.
Step 2: Apply the formula:
$$u = \frac{0.7c + 0.7c}{1 + \frac{(0.7c)(0.7c)}{c^2}} = \frac{1.4c}{1 + 0.49} = \frac{1.4c}{1.49} \approx 0.9396c$$
Classical result: $0.7c + 0.7c = 1.4c$ (impossible!)
Relativistic result: $\approx 0.94c$ (safely below $c$) ✓
A spaceship moves at $0.9c$. It fires a probe forward at $0.5c$ relative to the ship. What speed does Earth observe for the probe?
$$u = \frac{0.5c + 0.9c}{1 + \frac{(0.5)(0.9)c^2}{c^2}} = \frac{1.4c}{1 + 0.45} = \frac{1.4c}{1.45} \approx 0.9655c$$
Not $1.4c$, but $0.966c$. The speed limit holds.
A ship at $0.8c$ shines a flashlight forward. What speed does Earth measure for the light?
$$u = \frac{c + 0.8c}{1 + \frac{(c)(0.8c)}{c^2}} = \frac{1.8c}{1 + 0.8} = \frac{1.8c}{1.8} = c$$
Light always travels at $c$, exactly as Einstein's second postulate demands. ✓
At low speeds ($v = 30$ m/s, $u' = 20$ m/s), show the formula reduces to simple addition.
$$u = \frac{20 + 30}{1 + \frac{20 \times 30}{(3 \times 10^8)^2}} = \frac{50}{1 + 6.67 \times 10^{-15}} \approx 50 \text{ m/s}$$
The correction is negligible — classical addition works fine at everyday speeds.
Practice Problems
Show Answer Key
1. $u = (0.5+0.5)/(1+0.25) = 1.0/1.25 = 0.8c$.
2. $v/c \approx 9.26 \times 10^{-8}$. Denominator $\approx 1 + 8.57 \times 10^{-15} \approx 1$. Result $\approx 200$ km/h. ✓
3. $u = 1.98/(1+0.9801) = 1.98/1.9801 \approx 0.99995c$. Still below $c$. ✓
4. $u = (c+v)/(1+v/c) = (c+v)/((c+v)/c) = c$. ✓
5. $u = (0.8+0.8)/(1+0.64) = 1.6/1.64 \approx 0.976c$.
6. $u = (0.6c - 0.3c)/(1 - 0.18) = 0.3c/0.82 \approx 0.366c$ (forward).
7. Let $u' = \alpha c$, $v = \beta c$ with $\alpha, \beta < 1$. Then $u/c = (\alpha+\beta)/(1+\alpha\beta)$. Since $(\alpha+\beta) < (1+\alpha\beta)$ when $\alpha,\beta < 1$ (verify: $1+\alpha\beta - \alpha - \beta = (1-\alpha)(1-\beta) > 0$), we have $u < c$. ✓
8. $0.9 = (0.5+v)/(1+0.5v)$. Solve: $0.9 + 0.45v = 0.5 + v$, so $0.55v = 0.4$, $v \approx 0.727c$.
9. $u = (0.6-0.4)/(1-0.24) = 0.2/0.76 \approx 0.263c$.
10. Galilean addition allows $u > c$, violating the principle that the speed of light is the same in all frames. The relativistic denominator prevents this.