28.7

The Relativistic Addition of Velocities

The velocity of an object relative to an observer plays a central role in special relativity, and to determine this velocity, it is sometimes necessary to add two or more velocities together. We first encountered relative velocity in Section 3.4, so we will begin by reviewing some of the ideas presented there. Figure 28-12 illustrates a truck moving at a constant velocity of vTG = + 15 m/s relative to an observer standing on the ground, where the plus sign denotes a direction to the right. Suppose someone on the truck throws a baseball toward the observer at a velocity of vBT = + 8.0 m/s relative to the truck. We might conclude that the observer on the ground sees the ball approaching at a velocity of vBG = vBT + vTG = 8.0 m/s + 15 m/s = + 23 m/s. These symbols are similar to those used in Section 3.4 and have the following meaning:

Although the result that vBG = + 23m/s seems reasonable, careful measurements would show that it is not quite right. According to special relativity, the equation vBG = vBT + vTG is not valid for the following reason. If the velocity of the truck had a magnitude sufficiently close to the speed of light, the equation would predict that the observer on the earth could see the baseball moving faster than the speed of light. This is not possible, since no object with a finite mass can move faster than the speed of light.

Figure 28-12 The truck is approaching the ground-based observer at a relative velocity of vTG = + 15 ;m/s. The velocity of the baseball relative to the truck is vBT = + 8.0 m/s.

For the case where the truck and ball are moving along the same straight line, the theory of special relativity reveals that the velocities are related according to

The subscripts in this equation have been chosen for the specific situation shown in Figure 28-12. For the general situation, the relative velocities are related by the velocity-addition formula:

| (28.8)

where all the velocities are assumed to be constant and the symbols have the following meanings:

The ordering of the subscripts in Equation 28.8 follows the discussion in Section 3.4. For motion along a straight line, the velocities can have either positive or negative values, depending on whether they are directed along the positive or negative direction. Furthermore, switching the order of the subscripts changes the sign of the velocity, so, for example, vBA = −vAB (see Example 12 in Chapter 3).
Equation 28.8 differs from the nonrelativistic formula (vAB = vAC + vCB) by the presence of the vACvCB/c2 term in the denominator. This term arises because of the effects of time dilation and length contraction that occur in special relativity. When vAC and vCB are small compared to c, the vACvCB/c2 term is small compared to 1, so the velocity-addition formula reduces to . However, when either vAC or vCB is comparable to c, the results can be quite different, as Example 10 illustrates.

| Example 10 | | | The Relativistic Addition of Velocities

Imagine a hypothetical situation in which the truck in Figure 28-12 is moving relative to the ground with a velocity of vTG = + 0.8c. A person riding on the truck throws a baseball at a velocity relative to the truck of vBT = + 0.5c. What is the velocity vBG of the baseball relative to a person standing on the ground?
Reasoning The observer standing on the ground does not see the baseball approaching at vBG = 0.5c + 0.8c = 1.3c. This cannot be because the speed of the ball would then exceed the speed of light. The velocity-addition formula gives the correct velocity, which has a magnitude less than the speed of light.
Solution The ground-based observer sees the ball approaching with a velocity of

| (28.8)

Example 10 discusses how the speed of a baseball is viewed by observers in different inertial reference frames. The next example deals with a similar situation, except that the baseball is replaced by the light of a laser beam.

| Conceptual Example 11 | | | The Speed of a Laser Beam

Figure 28-13 shows an intergalactic cruiser approaching a hostile spacecraft. The velocity of the cruiser relative to the spacecraft is vCS = + 0.7c. Both vehicles are moving at a constant velocity. The cruiser fires a beam of laser light at the enemy. The velocity of the laser beam relative to the cruiser is vLC = + c. (a) What is the velocity of the laser beam vLS relative to the renegades aboard the spacecraft? (b) At what velocity do the renegades aboard the spacecraft see the laser beam move away from the cruiser?

Figure 28-13 An intergalactic cruiser, closing in on a hostile spacecraft, fires a beam of laser light.

Reasoning and Solution

a. | Since both vehicles move at a constant velocity, each constitutes an inertial reference frame. According to the speed of light postulate, all observers in inertial reference frames measure the speed of light in a vacuum to be c. Thus, the renegades aboard the hostile spacecraft see the laser beam travel toward them at the speed of light, even though the beam is emitted from the cruiser, which itself is moving at seven-tenths the speed of light.
b. | The renegades aboard the spacecraft see the cruiser approach them at a relative velocity of vCS = + 0.7c, and they also see the laser beam approach them at a relative velocity of vLS = + c. Both these velocities are measured relative to the same inertial reference frame—namely, that of the spacecraft. Therefore, the renegades aboard the spacecraft see the laser beam move away from the cruiser at a velocity that is the difference between these two velocities, or +c − (+0.7c) = + 0.3c. The velocity-addition formula, Equation 28.8, is not applicable here because both velocities are measured relative to the same inertial reference frame (the spacecraft’s reference frame). The velocity-addition formula can be used only when the velocities are measured relative to different inertial reference frames.

Related Homework: Conceptual Question 12, Problem 40

It is a straightforward matter to show that the velocity-addition formula is consistent with the speed of light postulate. Consider Figure 28-14, which shows a person riding on a truck and holding a flashlight. The velocity of the light, relative to the person on the truck, is vLT = + c. The velocity vLG of the light relative to the observer standing on the ground is given by the velocity-addition formula as

Thus, the velocity-addition formula indicates that the observer on the ground and the person on the truck both measure the speed of light to be c, independent of the relative velocity vTG between them. This is exactly what the speed of light postulate states.

Figure 28-14 The speed of the light emitted by the flashlight is c relative to both the truck and the observer on the ground.

Check Your Understanding 4

Car A and car B are both traveling due east on a straight section of an interstate highway. The speed of car A relative to the ground is 10 m/s faster than the speed of car B relative to the ground. According to special relativity, is the speed of car A relative to car B (a) 10 m/s, (b) less than 10 m/s, or (c) greater than 10 m/s? (The answer is given at the end of the book.)
Background: The relativistic addition of velocities obeys the velocity-addition formula. However, the method of using subscripts to identify the moving objects, as introduced in Section 3.4, still applies.
For similar questions (including calculational counterparts), consult Self-Assessment Test 28.2, which is described next.

Test your understanding of the material in Sections 28.5, 28.6 and 28.7 :

· | Relativistic Momentum
· | The Equivalence of Mass and Energy
· | The Relativistic Addition of Velocities

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