Chapter 2

Relativity II

Chapter Outline

• 2.1 Relativistic Momentum and the

Relativistic Form of Newton’s Laws

• 2.2 Relativistic Energy

• 2.3 Mass as a Measure of Energy

• 2.4 Conservation of Relativistic

• Momentum and Energy

• 2.5 General Relativity

Relativistic Momentum

• In classical mechanics, momentum is defined as p = mu ,where u is the speed of the particle and m its mass.

• in collision, the momentum is conserved, that is momentum before collision = momentum after collision in all reference frames.

• If the collision is described in S , then the

momentum is conserved

• If the velocities in S’ are calculated by using Lorentz transformation and the classical definition of momentum p = mu is used, then the momentum is not conserved !!!.

• What is the correct relativistic definition of momentum?

• To show the failure of the classical definition

p = mu

• Let us consider the collision between two identical particel

p before = mv + m(–v) = 0 p after = 0

That is: Momentum is conserved in S

In elastic collision - S’

• Using

We have

Now

It is clear that

• If we use p = mu, the momentum is not conserved in S’ , But according to Einestine ( relativistic postulate) all the physical laws are the same in all IRFs’.

• momentum is conserved in both S and S’, if we redefine momentum as

where u is the velocity of the particle and m is the proper mass, that is, the mass measured by an observer at rest with respect to the mass

Equation 2.1 is often written as

where

The relativistic force is

Note that this has the same functional form as in the Lorentz transformation, but here contains u, the particle speed, while in the Lorentz transformation, contains v, the relative speed of the two frames

EXAMPLE 2.1 An electron, which has a mass of 9.11 x 1031 kg, moves with a speed of 0.750c. Find its relativistic momentum and compare this with the momentum calculated from the classical expression

Solution:

The incorrect expression gives

EXAMPLE 2.2 The Measurement of the Momentum of a High-Speed Charged Particle

• Suppose a particle of mass m and charge q is injected with a relativistic velocity u into a region containing a magnetic field B. The magnetic force F on the particle is given by

the magnitude of the force

on the particle is

Substituting

2.2 RELATIVISTIC ENERGY

• The work- energy theorem, W=K in relativistic mechanics =?

• We use the definition

where we have assumed that the force and the motion are along x- axis,

Now

The result is

Because the initial kinetic energy is zero, we conclude that the work W in Eq. 2.7 is equal to the relativistic kinetic energy K, that is

2

2

22

1

1,

c

umcmc

using

Low speed limit:

which agrees with the classical result. A graph comparing the relativistic and nonrelativistic expressions for u as a function of K is given in Figure 2.2. Note that in the relativistic case, the particle speed never exceeds c.

then

• The constant term mc2, which is independent of the speed, is called the rest energy of the particle. The term mc2, which depends on the particle speed, is therefore the sum of the kinetic and rest energies. We define mc2 to be the total energy E, that is,

2

2

22

1

1,

c

umcmcK

It can be written as

Energy momentum relation

mupandmcE 2

22222222 , umPcmE

squaring the above two equations,

We get

222

2

2

22

1

1, ucm

c

ucP

It is useful to relate the total energy E to the relativistic

momentum p. Using the expressions

Multiplying the second equation by c2

2

2

222

2

2

222

2

2

222 )(

1

1)(

1

1c

c

ucm

c

ucm

c

ucPE

When the particle is at rest, p = 0, and so we see that E = m c2. That is, the total energy equals the rest energy. For the case of particles that have zero

mass, such as photons (massless, chargeless particles of light), we set m= 0 in Equation 2.11, and find

This equation is an exact expression relating energy and momentum for photons, which always travel at the speed of light

Subtracting from E2

The energy is measured by eV (electron volt)

• For example, the mass of an electron is me = 9.11x 10-31 kg. Hence, the rest energy of the electron is

EXAMPLE 2.3

• An electron has a speed u = 0.850c. Find its total energy and kinetic energy in electron volts.

• Solution Using the fact that the rest energy of the electron is 0.511 MeV together with E= mc2 gives

The kinetic energy:

EXAMPLE 2.4

• The total energy of a proton is three times its rest energy.

• (a) Find the proton’s rest energy in electron volts

Solution:

(b) With what speed is the proton moving?

Solution Because the total energy E

is three times the rest energy

Solving for u gives

(c) Determine the kinetic energy of the proton in electron volts. Solution:

2.3 MASS AS A MEASURE OF ENERGY

. The sum of the mass–energy of a system of particles before interaction must equal the sum of the mass–energy of the system after interaction where the mass–energy of the ith particle is defined as the total relativistic energy

•In the simple inelastic collision illustration:

In classical theory, the momentum is conserved but the kinetic energy is

not. When does the energy loss?

using the relativistic mass–energy conservation law, we must have

M is the composite mass

,The composite mass M>2m

What is the increase of mass?

The mass increase is,

That is M equals 2/c2 of the incident kinetic energy. This means that

the kinetic energy is not lost in inelastic collision but shows as an

increase in the mass of the final composite object.

Conclusion :

•In relativistic theory, both relativistic momentum and mass- energy are

conserved .

The famous examples are : • The fission of heavy radioactive nucleus into several lighter

particles that are emitted with large kinetic energy

• The second example is the fusion, which is the reaction between

two deuterium nuclei to combine to form a helium nucleus

Show that momentum is conserved by using the relativistic definition

2

2

1

2

c

v

mM

Example:

2

2

1c

u

mup

and

Proof: From side 6 we have

2

22

1

2'

c

v

vv

2

2

2

2

2

2

2

22

2

2

2

2

2

2

2

22

2

2

4

4

2

2

2

2'

2

2

22

2

2

2

2

2

4

2

2

2

22

2

2

2

2

2

2'

2

2

22

2

2

2

2

2

2'

2

2

2

2

2

2

2

2

2

22'

2

)1(

)1(

)1()1(

)1)(1(

)1()1(

21

1

)1()1(

421

)1()1(

4

11

)1()1(

4

)1)(1(

4

)1(

4

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

c

v

v

c

v

vv

squaring

2

2

2

2

2

2

2

2

2

2'

22

2

2

2'

2

'

2

2

2'

1

'

1

1

2

)1(

)1(

.

1

2

1

1.

1

)2(0

11

'

c

v

mv

c

v

c

v

c

v

mv

c

v

c

v

vm

c

v

mv

c

v

mvP before

Now

2

2

2

2

2

2

2

2

2

2

2

1

2

)1(

2'

1

2,

1'

1

''

c

v

mv

c

v

mvP

c

v

mM

c

v

Mv

c

v

MvP

after

after

and

EXAMPLE 2.5 Calculate the mass increase for a completely inelastic head-on collision of two 5.0-kg balls each moving toward the other at 450m/s in opposite directions solution:

H.W (Ch2)

Assigments Problems:

2,13,15,21