# 28.4 Relativistic addition of velocities  (Page 2/7)

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The second postulate of relativity (verified by extensive experimental observation) says that classical velocity addition does not apply to light. Imagine a car traveling at night along a straight road, as in [link] . If classical velocity addition applied to light, then the light from the car’s headlights would approach the observer on the sidewalk at a speed $\text{u=v+c}$ . But we know that light will move away from the car at speed $c$ relative to the driver of the car, and light will move towards the observer on the sidewalk at speed $c$ , too.

Either light is an exception, or the classical velocity addition formula only works at low velocities. The latter is the case. The correct formula for one-dimensional relativistic velocity addition    is

$u=\frac{\mathrm{v+u}\prime }{1+\frac{v\text{u}\prime }{{c}^{2}}},$

where $v$ is the relative velocity between two observers, $u$ is the velocity of an object relative to one observer, and $u\prime$ is the velocity relative to the other observer. (For ease of visualization, we often choose to measure $u$ in our reference frame, while someone moving at $v$ relative to us measures $u\prime$ .) Note that the term $\frac{vu\prime }{{c}^{2}}$ becomes very small at low velocities, and $u=\frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}$ gives a result very close to classical velocity addition. As before, we see that classical velocity addition is an excellent approximation to the correct relativistic formula for small velocities. No wonder that it seems correct in our experience.

## Showing that the speed of light towards an observer is constant (in a vacuum): the speed of light is the speed of light

Suppose a spaceship heading directly towards the Earth at half the speed of light sends a signal to us on a laser-produced beam of light. Given that the light leaves the ship at speed $c$ as observed from the ship, calculate the speed at which it approaches the Earth.

Strategy

Because the light and the spaceship are moving at relativistic speeds, we cannot use simple velocity addition. Instead, we can determine the speed at which the light approaches the Earth using relativistic velocity addition.

Solution

1. Identify the knowns. $\text{v=}0\text{.}\text{500}c$ ; $u\prime =c$
2. Identify the unknown. $u$
3. Choose the appropriate equation. $u=\frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}$
4. Plug the knowns into the equation.
$\begin{array}{lll}u& =& \frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}\\ & =& \frac{\text{0.500}c+c}{1+\frac{\left(\text{0.500}c\right)\left(c\right)}{{c}^{2}}}\\ & =& \frac{\left(\text{0.500}+1\right)c}{1+\frac{\text{0.500}{c}^{2}}{{c}^{2}}}\\ & =& \frac{\text{1.500}c}{1+\text{0.500}}\\ & =& \frac{\text{1.500}c}{\text{1.500}}\\ & =& c\end{array}$

Discussion

Relativistic velocity addition gives the correct result. Light leaves the ship at speed $c$ and approaches the Earth at speed $c$ . The speed of light is independent of the relative motion of source and observer, whether the observer is on the ship or Earth-bound.

Velocities cannot add to greater than the speed of light, provided that $v$ is less than $c$ and $u\prime$ does not exceed $c$ . The following example illustrates that relativistic velocity addition is not as symmetric as classical velocity addition.

## Comparing the speed of light towards and away from an observer: relativistic package delivery

Suppose the spaceship in the previous example is approaching the Earth at half the speed of light and shoots a canister at a speed of $0.750c$ . (a) At what velocity will an Earth-bound observer see the canister if it is shot directly towards the Earth? (b) If it is shot directly away from the Earth? (See [link] .)

Strategy

Because the canister and the spaceship are moving at relativistic speeds, we must determine the speed of the canister by an Earth-bound observer using relativistic velocity addition instead of simple velocity addition.

Solution for (a)

1. Identify the knowns. $\text{v=}0.500c$ ; $u\prime =0\text{.}\text{750}c$
2. Identify the unknown. $u$
3. Choose the appropriate equation. $\text{u=}\frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}$
4. Plug the knowns into the equation.
$\begin{array}{ll}u& =& \frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}\\ & =& \frac{0.500\text{c +}0.750c}{1+\frac{\left(0.500c\right)\left(0.750c\right)}{{c}^{2}}}\\ & =& \frac{1.250c}{1+0.375}\\ & =& 0.909c\end{array}$

Solution for (b)

1. Identify the knowns. $\text{v}=0.500c$ ; $u\prime =-0.750c$
2. Identify the unknown. $u$
3. Choose the appropriate equation. $\text{u}=\frac{\mathrm{v+u}\prime }{1+\frac{v\text{u}\prime }{{c}^{2}}}$
4. Plug the knowns into the equation.
$\begin{array}{ll}u& =& \frac{\mathrm{v+u}\prime }{1+\frac{vu\prime }{{c}^{2}}}\\ & =& \frac{0.500\mathrm{c +}\left(-0.750c\right)}{1+\frac{\left(0.500c\right)\left(-0.750c\right)}{{c}^{2}}}\\ & =& \frac{-0.250c}{1-0.375}\\ & =& -0.400c\end{array}$

Discussion

The minus sign indicates velocity away from the Earth (in the opposite direction from $v$ ), which means the canister is heading towards the Earth in part (a) and away in part (b), as expected. But relativistic velocities do not add as simply as they do classically. In part (a), the canister does approach the Earth faster, but not at the simple sum of $1.250c$ . The total velocity is less than you would get classically. And in part (b), the canister moves away from the Earth at a velocity of $-0.400c$ , which is faster than the $-0.250c$ you would expect classically. The velocities are not even symmetric. In part (a) the canister moves $0.409c$ faster than the ship relative to the Earth, whereas in part (b) it moves $0.900c$ slower than the ship.

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