• the Lorentz transformation equations• Once again is our frame Smoving at a speed v relative to S.

8.The Lorentz Velocity TransformationEquations Suppose two observers in relative motion with

respect to each other are both observing the motion of an object. Previously, we defined an event as occurring at an instant of time. Now, we wish to interpret the “event” as the motion of the object. We know that the Galilean velocity transformation is valid for low speeds. Which is

A particle moves through a displacement of magnitude dx along the x axis in a time interval dt as measured by an observer in S. the corresponding displacement dx measured by an observer in is S

• where frame is moving with speed S v in the x direction relative to frame S. Because dt= dt, we find that

• where ux and ux are the x components of the velocity of the particle measured by observers in S and , respectively. S

• We use the symbol u for particle velocity rather than v, which is used for the relative velocity of two reference frames.)

• This is the Galilean velocity transformation equation.

• How do the observers’ measurements of the velocity of the object relate to each other if the speed of the object is close to that of light? Once again is our frame moving at a speed S v relative to S.

• Suppose that an object has a velocity component ux measured in the frame, whereS

• Using Lorentz transformation equations

• We have

• Substituting these values into Equation

• We get

• But dx/dt is just the velocity component ux of the object measured by an observer in S, and so this expression becomes

• If the object has velocity components along the y and z axes, the components as measured by an observer in areS

• Note that u y and u z do not contain the parameter v in the numerator because the relative velocity is along the x axis.

• When v is much smaller than c (the nonrelativistic case), the denominator of Equation 39.16 approaches unity, and so

u x ≈ ux - v, which is the Galilean velocity transformation equation.

• In another extreme, when ux = c, Equation 39.16 becomes

• From this result, we see that a speed measured as c by an observer in S is also measured as c by an observer in S

independent of the relative motion of S and . S Note that this conclusion is consistent with Einstein’s second postulate—that the speed of light must be c relative to all inertial reference frames.

Furthermore, we find that the speed of an object can never be measured as larger than c. That is, the speed of light is the ultimate speed. We return to this point later.

Example:Two spacecraft A and B are moving in opposite directions, as

shown in Figure 39.15. An observer on the Earth measures the speed of craft A to be 0.750c and the speed of craft B to be 0.850c. Find the velocity of craft B as observed by the crew on craft A.

To finalize this problem, note that the negative sign indicates that craft B is moving in the negative x direction as observed by the crew on craft A. Note that the speed is less than c. That is, an object whose speed is less than c in one frame of reference must have a speed less than c in any other frame.

• Example:• A Klingon spacecraft moves away from the Earth at a

speed of 0.800c (Fig. P39.26). The starship Enterprise pursues at a speed of 0.900c relative to the Earth. Observers on the Earth see the Enterprise overtaking the Klingon craft at a relative speed of 0.100c. With what speed is the Enterprise overtaking the Klingon craft as seen by the crew of the Enterprise?

• Solution:

• Example:Two jets of material from the centre of a radio

galaxy are ejected in opposite directions. Both jets move at 0.750c relative to the galaxy. Determine the speed of one jet relative to the other.

• Solution:

• Example:• A spacecraft is launched from the surface of the

Earth with a velocity of 0.600c at an angle of 50.0° above the horizontal positive x axis. Another spacecraft is moving past, with a velocity of 0.700c in the negative x direction. Determine the magnitude and direction of the velocity of the first spacecraft as measured by the pilot of the second spacecraft.

• Solution:

9.RELATIVISTIC MOMENTUMProperly describing the motion of particles within

the framework of special relativity requires generalizing Newton’s laws of motion and the definitions of momentum and energy. These generalized definitions reduce to the classical (nonrelativistic) definitions when v is much less than c.

First, recall that conservation of momentum states that when two objects collide, the total momentum of the system remains constant, assuming that the objects are isolated, reacting only with each other

• In order to have momentum conservation in all inertial frames—even those moving at an appreciable fraction of c—the definition of momentum must be modified to read

• where v is the speed of the particle and m is its mass as measured by an observer at rest with respect to the particle. Note that when v is much less than c, the denominator of Equation 26.10 approaches one, so that p approaches mv. Therefore, the relativistic equation for momentum reduces to the classical expression when v is small compared with c.

• The relativistic force F acting on a particle whose linear momentum is p is defined as

• But and

• Where mo is the mass of the particle at rest and m is the particle mass after motion

• then

• This expression is reasonable because it preserves classical mechanics in the limit of low velocities and is consistent with conservation of linear momentum for an isolated system (F= 0) both relativistically and classically.

• under relativistic conditions, the acceleration a of a particle decreases under the action of a constant force, in which case a α (1 - u2/c2)3/2. From this proportionality, we see that as the particle’s speed approaches c, the acceleration caused by any finite force approaches zero.

• The mass of the particle will approach infinity For v=c, the dominator will reach zero which mean that the

mass will be infinity. Which is impossible physical condition.

• Hence, it is impossible to accelerate a particle from rest to a speed u≥c.

• This argument shows that the speed of light is the ultimate speed, as noted at the end of the preceding section.

EXAMPLEAn electron, which has a mass of 9.11 x 10-31 kg,

moves with a speed of 0.750c. Find the classical (nonrelativistic) momentum and compare it to its relativistic counterpart prel.

SolutionFirst, compute the classical (nonrelativistic)

momentum with v = 0.750c

Multiply this result by γ to obtain the relativistic momentum:

EXAMPLEAn unstable particle at rest breaks up into two

fragments of unequal mass. The mass of the lighter fragment is 2.50 x 10-28 kg, and that of the heavier fragment is 1.67x 10-27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?

SolutionRelativistic momentum of the system of fragments

must be conserved. For total momentum to be zero after as it was before, we must have, with subscript 2 referring to the heavier fragment, and subscript 1 to the lighter, p2 = p1

10.RELATIVISTIC ADDITION OF VELOCITIES

Imagine of a motorcycle rider moving with a speed of 0.80c past a stationary observer, as shown in Figure 26.12. If the rider tosses a ball in the forward direction with a speed of 0.70c relative to himself, what is the speed of the ball as seen by the stationary observer at the side of the road?

• Common sense and the ideas of Newtonian relativity say that the speed should be the sum of the two speeds, or 1.50c. This answer must be incorrect because it contradicts the assertion that no material object can travel faster than the speed of light.

• Let two frames or reference be labelled b and d, and suppose that frame d is moving at velocity vdb in the position x-direction relative frame b(the velocity with respect to stationary observer). If the velocity of an object a as measured in frame d is called vad (The velocity of the trajectory with respect to the moving reference)then the velocity of a as measured in frame b, vab, is given by

• We apply Equation 26.11 to Figure 26.13, which shows a motorcyclist, his ball, and a stationary observer. We are given

• and we want to find• v bo = the velocity of the

ball with respect to the stationary observer

• Thus,