Chapter 2 Relativity II
Transcript of Chapter 2 Relativity II
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