# 5.6 Relativistic velocity transformation  (Page 2/4)

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## Velocity transformation equations for light

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

## Strategy

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

## Solution

1. Identify the knowns: $v=0.500c;u\prime =c.$
2. Identify the unknown: u .
3. Express the answer as an equation: $u=\frac{v+u\prime }{1+\frac{vu\prime }{{c}^{2}}}.$
4. Do the calculation:
$\begin{array}{cc}\hfill u& =\frac{v+u\prime }{1+\frac{vu\prime }{{c}^{2}}}\hfill \\ & =\frac{0.500c+c}{1+\frac{\left(0.500c\right)\left(c\right)}{{c}^{2}}}\hfill \\ & =\frac{\left(0.500+1\right)c}{\left(\frac{{c}^{2}+0.500{c}^{2}}{{c}^{2}}\right)}\hfill \\ & =c.\hfill \end{array}$

## Significance

Relativistic velocity addition gives the correct result. Light leaves the ship at speed c and approaches 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 earthbound.

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.

## Relativistic package delivery

Suppose the spaceship in the previous example approaches Earth at half the speed of light and shoots a canister at a speed of 0.750 c ( [link] ). (a) At what velocity does an earthbound observer see the canister if it is shot directly toward Earth? (b) If it is shot directly away from Earth?

## Strategy

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

## Solution for (a)

1. Identify the knowns: $v=0.500c;u\prime =0.750c.$
2. Identify the unknown: u .
3. Express the answer as an equation: $u=\frac{v+u\prime }{1+\frac{vu\prime }{{c}^{2}}}.$
4. Do the calculation:
$\begin{array}{cc}\hfill u& =\frac{v+u\prime }{1+\frac{vu\prime }{{c}^{2}}}\hfill \\ & =\frac{0.500c+0.750c}{1+\frac{\left(0.500c\right)\left(0.750c\right)}{{c}^{2}}}\hfill \\ & =0.909c.\hfill \end{array}$

## Solution for (b)

1. Identify the knowns: $v=0.500c;u\prime =-0.750c.$
2. Identify the unknown: u .
3. Express the answer as an equation: $u=\frac{v+u\prime }{1+\frac{vu\prime }{{c}^{2}}}.$
4. Do the calculation:
$\begin{array}{cc}\hfill u& =\frac{v+u\prime }{1+\frac{vu\prime }{{c}^{2}}}\hfill \\ & =\frac{0.500c+\left(-0.750c\right)}{1+\frac{\left(0.500c\right)\left(-0.750c\right)}{{c}^{2}}}\hfill \\ & =-0.400c.\hfill \end{array}$

## Significance

The minus sign indicates a velocity away from Earth (in the opposite direction from v ), which means the canister is heading toward 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 Earth faster, but at less than the vector sum of the velocities, which would give 1.250 c . In part (b), the canister moves away from Earth at a velocity of $-0.400c,$ which is faster than the −0.250 c expected classically. The differences in velocities are not even symmetric: In part (a), an observer on Earth sees the canister and the ship moving apart at a speed of 0.409 c , and at a speed of 0.900 c in part (b).

Check Your Understanding Distances along a direction perpendicular to the relative motion of the two frames are the same in both frames. Why then are velocities perpendicular to the x -direction different in the two frames?

Although displacements perpendicular to the relative motion are the same in both frames of reference, the time interval between events differ, and differences in dt and $dt\prime$ lead to different velocities seen from the two frames.

## Summary

• With classical velocity addition, velocities add like regular numbers in one-dimensional motion: $u=v+u\prime ,$ where v is the 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.
• Velocities cannot add to be greater than the speed of light.
• Relativistic velocity addition describes the velocities of an object moving at a relativistic velocity.

## Problems

If two spaceships are heading directly toward each other at 0.800 c , at what speed must a canister be shot from the first ship to approach the other at 0.999 c as seen by the second ship?

Two planets are on a collision course, heading directly toward each other at 0.250 c . A spaceship sent from one planet approaches the second at 0.750 c as seen by the second planet. What is the velocity of the ship relative to the first planet?

0.615 c

When a missile is shot from one spaceship toward another, it leaves the first at 0.950 c and approaches the other at 0.750 c . What is the relative velocity of the two ships?

What is the relative velocity of two spaceships if one fires a missile at the other at 0.750 c and the other observes it to approach at 0.950 c ?

0.696 c

Prove that for any relative velocity v between two observers, a beam of light sent from one to the other will approach at speed c (provided that v is less than c , of course).

Show that for any relative velocity v between two observers, a beam of light projected by one directly away from the other will move away at the speed of light (provided that v is less than c , of course).

(Proof)

in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
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there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
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40cm=40.0×10^-2m =400.0×10^-3m =400mm.
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