Ch 26 1

Chapter 26

Relativity

© 2006, B.J. LiebSome figures electronically reproduced by permission of Pearson

Education, Inc., Upper Saddle River, New Jersey Giancoli, PHYSICS,6/E © 2004.

Ch 26 2

Galilian-Newtonian Relativity

•Relativity deals with experiments observed from different reference frames.

•Example: Person drops coin from moving car

a) In reference frame of car: coin is at rest and falls straight down

b) In “earth” reference frame, coin is moving with initial velocity and follows projectile path.

Ch 26 3

Inertial Reference Frames•Inertial Reference Frame: one in which Newton’s First Law (Law of Inertia is valid.)

•A reference frame moving with constant velocity with respect to an inertial reference frame is an inertial frame.

•If the car is moving with constant velocity relative to the earth, it is an inertial reference frame.

•Accelerated or rotating reference frames are noninertial reference frames.

•The earth is approximately inertial.

Ch 26 4

Relativity Principle

“the basic laws of physics are the same in all inertial reference frames”

•This principle was understood by Galileo and Newton.

•Space and time are absolute– different inertial reference frames measure the same length, time etc.

•All inertial reference frames are equivalent– there is no preferred frame.

Ch 26 5

Special Theory of Relativity

Einstein asked the question “What would happen if I rode a light beam?”

•Would see static electric and magnetic fields with no understandable source.

•Understanding electromagnetic radiation requires changing E and B fields.

•Concluded that:

• no one could travel at speed of light.

•No one could be in frame where speed of light was anything other than c.

•No absolute reference frame

Ch 26 6

Postulates of Special Theory of Relativity

•First: The laws of physics have the same form in all inertial reference frames

•Second: Light propagates through empty space with a definite speed c independent of the source and observer

•This means that an observer trying to catch a light beam and moving at 0.9c will measure the speed of that light as c and an observer on earth will also measure c.

•In order for this to be true, observers must differ on measurements of distance and time

•The special theory of relativity deals with reference frames that move at constant velocity with respect to an inertial reference frame.

•The General Theory deals with accelerated reference frames and is primarily a theory of gravity.

Ch 26 7

Relativistic Clock

•In the above clock, light is reflected back and forth between two mirrors and timer counts “ticks”

•This is an ideal clock because of special properties of light

•An observer at rest with respect to the clock concludes that the time for a tick is

c

Dt

20

•In order to study time dilation, we will places this clock in a spaceship moving past earth.

t0 is called proper or rest time because clock is at rest in spaceship (note, we don’t call it “correct” time)

Ch 26 8

Time dilation

•Spaceship moves by earth at speed v. (both observers agree that speed is v.)

222 LD Observer on earth sees light move distance

per tick.

tvL 2

•Observer on earth writes this equation for c

t

tvD

t

LDc

422 22222 /)(

•Observer on earth sees spaceship moving

Ch 26 9

Time dilation

The formulas on the previous slides can be combined to give

2

2

0

1cv

tt

Clocks moving relative to an observer are measured by that observer to run more slowly (as compared to clocks at rest).

•Clock is in spaceship so this measures the proper or rest time Δt0.•It is often convenient to write v as fraction of c, thus v = 3.0x107 m/s is written v = 0.10 c.•We call this effect time dilation because the time in the moving reference frame is always longer than the time in the proper reference frame

Ch 26 10

Time dilation factor

2

2

0

1cv

tt

Consider how this depends on v

v (m/s) v (c) t0 (sec) t (sec)

2.00x105 0.00067 c 1.00000 1.00000

3.00x106 0.01000c 1.00000 1.00005

3.00x107 0.10000c 1.00000 1.00500

2.00x108 0.66667c 1.00000 1.34200

2.97x108 0.99000c 1.00000 7.08900

Ch 26 11

Length ContractionA spaceship passes Earth and continues on to Neptune

•Earth and spaceship observers disagree on time, and they also disagree as to length L (distance to Neptune)

•Since earth is at rest, it measures the “proper” length L0.

•Both observers agree on the relative velocity v, so we can use the time dilation to derive length contraction equation

2

2

0 1c

vLL

Ch 26 12

Length Contraction

2

2

0 1c

vLL

The length of an object is measured to be shorter when it is moving relative to the observer than when it is at rest.

This contraction is only in the direction of the velocity. The drawing shows the changed shape of a picture when a person moves by horizontally.

Ch 26 13

Example 26-1 (7) Suppose you decide to travel to a star that is 85 light-years away at a speed that tells you the distance is only 25 light-years. How many years would it take you to make the trip?

We determine the speed from the length contraction. The light-year is a unit of length. The rest or proper length is L0 = 85 ly and the contracted length is L = 25 ly. 1

2 2

0 1 ;v

L Lc

1

2 2

25ly 85ly 1 ,v

c

0.956 .v c

25y26 y.

0.956

cdt

v c

which gives

We then find the time from the speed and distance:

22

185

25

c

v

Ch 26 14

Example 26-2: A muon is an elementary particle with an average lifetime of 2.2 μs. A muon is produced in the earth’s atmosphere with a velocity of 2.98 x108 m/s.

0tvL )1020.2)(/1098.2( 68 ssm m656

What is the distance traveled by the muon before it decays as measured by observers in the muon’s rest frame? We note that the muon’s reference frame is the proper frame for time and the earth is the proper frame for distance.

What is the lifetime of the muon as measured by observers in the earth reference frame?

2

1

c

v 2

8

8

/1000.3

/1098.21

sm

sm 115.0

2

2

0

1c

v

tt

sss 0.19100.19

115.0

1020.2 66

Ch 26 15

Example 26-2 (continued): A muon is an elementary particle with an average lifetime of 2.20 μs. A muon is produced in the earth’s atmosphere with a velocity of 2.98 x108 m/s.

What is the distance traveled by the muon as measured in the earth reference frame?

L2

0 1

c

vL

20

1

cv

LL

mm 3107.5

115.0

656

Use the above results to calculate the velocity of the muon in each reference frame.

sms

mvmuon /1098.2

102.2

656 86

sms

mvearth /1098.2

100.19

107.5 86

3

Notice that observers in the two reference frames disagree as to distance and time but agree as to velocity.

Ch 26 16

Four-Dimensional Space-Time

Since observers in different reference frames often do not agree on time measurements as well as length measurements, time is treated as a coordinate in our coordinate system along with the three spatial coordinates X,Y and Z. Thus we speak of a four-dimensional space time.

Space

Time

The red line represents a light ray and the blue line represents the “world-line” of an object through space-time.

Ch 26 17

Mass IncreaseThe mass of an object increases as the speed of the object increases:

22

0

1 cv

mm

/

•Mass increase is observed in particle accelerators

•m0 is the “rest mass”, the mass as measured in a reference frame at rest with respect to the object.

•This mass increase can be seen as the reason that objects can’t travel at the speed of light.

Ch 26 18

Relativistic Kinetic Energy

Einstein used the work-energy theorem to derive a new, relativistic equation for kinetic energy

20

2 cmmcKE

When v << c, this equation can be approximated by

KE = (1/2) m v2,

so this equation can still be used for non-relativistic speeds.The equation for kinetic energy is the difference between two quantities which have units of energy

Ch 26 19

Total Energy

EKcmE 20

2mcE

This is the total energy of the object. It implies that there is energy in mass and mass can be converted into energy and energy can be converted into mass

20cm

This is the rest energy of the object. We can thus write the total energy as

Ch 26 20

Units of Mass-Energy

•Earlier in the course we defined the electronvolt (eV) as a unit of energy where 1 eV = 1.6 x 10-19J.

•We will also use keV (103) and MeV (106)

•From E=mc2, we can solve for m = E / c2 and thus MeV / c2 is a unit of mass.

•Example: electron mass is

me = 0.511 MeV/c2

•It is useful to use c2 = 931.5 MeV / u.

Ch 26 21

Additional Relativistic Equations•The Special Theory of Relativity is a revision of the laws of mechanics for objects with velocity close to the speed of light.

•The relativistic equations all can be approximated by the non-relativistic equations when v << c.

•For problems where v 0.10 c, the relativistic correction is less than 1 percent and the non-relativistic equations are used.

420

222 cmcpE

Momentum:

Energy-Momentum:

2

2

0

1cv

vmp

Ch 26 22

Example 26-3. An electron is accelerated through an electrical potential difference of 2.00x106 V. Calculate the mass energy, kinetic energy, total energy and veloctiy of the electron.

MeVeVKE 00.21000.2 6

( One eV is the energy gained by electron accelerated through 1.0 V.

J142831 1020.8)/103)(1011.9( smkg

eVJ

eVMeVeV

19

6

14

106.1

)100.1)(1()1020.8( J MeV511.0

EKcmEcm 20

2

20 cm

0414.012

2

c

v

MeVMeVMeV 51.200.2511.0

2

2

0

1c

vm

m

2

0

2

2

2

1

m

m

c

v 2

2

2

0

cm

cm 2

51.2

511.0

MeV

MeV

smcv 8109.2979.0

Ch 26 23

Example 28-4. Neutrons outside of the nucleus are unstable and decay with a half-life of 10.4 mins into a proton, electron and a neutrino. If a neutron decays at rest, calculate the energy released by this decay. This energy is shared by the proton, electron and neutrino.

Note: The neutrino mass is only a few eV/c2 so we will assume it is zero. Table 30-1 on page 838 contains the masses we need.

Mass of electron = me = 0.00054858 u = 0.511 MeV/c2

Mass of proton = mp = 1.007276 u = 938.27 MeV/c2 Mass of neutron = mn = 1.008665 u = 939.57 MeV/c2

)( epn mmmm

uuuum 00084.0)00054858.0007276.1(008665.1

MeVuMeVuE 78.0)/5.931)(00084.0(

2cmE

Ch 26 24

Alternative Solution

Example 28-4. Neutrons outside of the nucleus are unstable and decay with a half-life of 10.4 mins into a proton, electron and a neutrino. If a neutron decays at rest, calculate the energy released by this decay. This energy is shared by the proton, electron and neutrino.

Solution in MeV/c2:me = 0.511 MeV/c2

mp = 938.27 MeV/c2 mn = 939.57 MeV/c2

(slight difference due to round-off)

2222 /79.0)/511.0/27.938(/57.939 cMeVcMeVcMeVcMeVm

)( epn mmmm

2cmE MeVccMeV 79.0/79.0 22