Good Friday - no classes!
Due Monday, March 24 at 11 pm
Experiment 5: Spectroscopy
On its way next week...
Wednesday, March 19, 2008 23
1) Relativity Postulate : The laws of physics are the same in all
inertial reference frames.
→ all inertial frames are equally good.
2) Speed of Light Postulate : The speed of light in vacuum is
independent of the motion of the source of light or of the
observer.
v
Event 2
Event 1 Event 3
Wednesday, March 19, 2008 24

2
2
D
Clock moving at speed v s
D
Light arrives at time Δ t/2 s = c !
t / 2
Light arrives at time Δ t
1 3
Clock at rest
!
t
0
=
2 D c
“Proper” time
1 v !
t / 2 v !
t / 2 3
Because the speed of the light is not changed by the motion of the clock, the clock must appear to tick more slowly when it’s in motion as the light has to travel further.
proper time – events at same place
Time dilation:
!
t = dilated time – events at different places
!
!
t
0
1 − v 2 c 2
Wednesday, March 19, 2008 25
Time dilation: !
t = !
!
t
0
1 − v 2 c 2
Proper time, events at same place in observer’s frame
Length contraction: L = L
0
!
1 − v 2 c 2
Proper length, observer at rest relative to object
There is no “width contraction”
Wednesday, March 19, 2008 26

A B
L
0
L
0
A thought experiment : two identical ring-shaped spacecraft are travelling toward each other at relative speed v.
From ship A, the length of ship B seems to be contracted.
As no inertial frame is better than any other (relativity postulate), it must also be true that from ship B, the length of ship A seems to be contracted too.
Wednesday, March 19, 2008 27
1) Ship B is moving toward A.
From A, the length of B is reduced.
A
Assume for the moment that the radius of B is reduced too – that there is a “width contraction”.
L
0
B passes through A.
2) Can equally well say that ship A is moving toward B. So, from B, the length and radius of A are reduced.
A v
A passes through B.
L = L
Contradictory statements!
There can be no “width contraction”.
0
!
1 − v 2 c 2
Wednesday, March 19, 2008 v
L
0
B
L = L
0
!
1 − v 2 c 2
B
28

Prob. 28.15/43 : Twins who are 19 years old leave earth and travel to a planet 12 light years (LY) away. Assume that the planet and earth are at rest relative to one another.
The twins depart at the same time on different spaceships. One travels at speed v
1
= 0.9c, while the other travels at v
2
= 0.5c. a) What is the difference between their ages when they meet again at the planet?
b) Which twin is older?
• How long do their journeys take according to clocks on earth?
• Who measures the proper time for the journeys?
Wednesday, March 19, 2008 29
Prob. 28.14
: A rocket moves past a ramp with a relative speed of
0.730c in a direction parallel to the side x. What does a person aboard the rocket measure for the angle of the ramp?
• In what direction does the length contraction occur?
• By how much?
v = 0.73c
Wednesday, March 19, 2008 30

Observers in different inertial frames disagree on measurements of distance and time and velocity.
Consider conservation of momentum for colliding objects.
For momentum to be conserved in all inertial frames , the definition of momentum needs to be changed: p = !
mv
1 − v 2 c 2
“relativistic momentum”
Then, force F = Δ p/ Δ t, Newton’s 2nd law.
For v much less than c, p !
mv, so the new definition is consistent with the old at low speeds.
Wednesday, March 19, 2008 31
relativistic momentum nonrelativistic momentum p = !
mv
1 − v 2 c 2
The momentum becomes infinite when v = c.
!
It is impossible to accelerate an object to the speed of light.
Wednesday, March 19, 2008 32

Prob. 28.39/17 : At what speed is the magnitude of the relativistic momentum of a particle three times the magnitude of the nonrelativistic momentum?
Wednesday, March 19, 2008 33
In exothermal chemical and nuclear reactions energy is released when some of the mass is converted into energy.
The energy contained in a mass m at rest is:
E
0
= mc 2 ”Rest energy” of mass m.
When the object is moving, its energy becomes:
E = mc 2
!
1 − v 2 c 2
“total energy”
It takes an infinite amount of energy to accelerate a mass to the speed of light, so v < c always.
The kinetic energy is the difference between the total energy and the rest energy:
KE = E − E
0
= !
mc
1
2
− v 2 c 2
− mc
2
This is the work done by a force
F = !
mass.
p/ !
t that accelerates the
Wednesday, March 19, 2008 34

The sun radiates electromagnetic energy at the rate 3.92
" 10 26 W. This is possible because hydrogen nuclei fuse together to form helium, with the conversion of some of the mass (that is, rest energy, E
0
= mc 2 ) into electromagnetic radiation.
In 1 s, the sun produces 3.92
" 10 26 J of electromagnetic radiation.
The mass that is converted each second by fusion is given by:
Δ E = Δ mc 2 , so Δ m = Δ E/c 2 = (3.92
" 10 26 J)/(3 " 10 8 m/s) 2 = 4.36
" 10 9 kg.
4.36 billion kilograms per second converted by nuclear fusion!
Wednesday, March 19, 2008 35
TNT explosive : releases 4.2
" 10 6 J of energy per kg of TNT.
Mass converted, Δ m = Δ E/c 2 = (4.2
" 10 6 )/(3x10 8 ) 2
= 4.7
" 10 -11 kg per kg
Fraction of mass converted: Δ m/m = 4.7
" 10 -11 or 4.7
" 10 -9 %
Fission of 235 U : Δ E !
200 MeV per fission and E = mc 2 !
235,000 MeV
So Δ m/m !
200/235,000 = 8.5
" 10 -4 or 0.085%
Nuclear conversion of mass is much more efficient than chemical conversion – 1 kg of nuclear fuel produces same energy as about
800,000 tonnes of coal.
Wednesday, March 19, 2008 36

Prob. 28.26/25 : A nuclear power reactor generates 3 " 10 9 W (3 GW) of power. In one year, what is the change in mass of the nuclear fuel due to the energy being taken from the reactor?
Wednesday, March 19, 2008 37
KE = mc
2

 #
1
1 − v
2 c
2

− 1  →
1
2 mv
2 when v << c.
Useful approximation formula: ( 1 + x ) n !
1 + nx (when |
!
1 − v 2
"
− 1 / 2 c 2
" 1 +
So, KE !
mc 2
!
1 +
1
2 v 2 c 2
1
2 v 2 c 2
− 1
"
= mv 2
2
(n = -1/2, x = -v 2 /c 2 ) x | << 1)
So, KE = mv 2
2 when v is much less than c
Wednesday, March 19, 2008 38

Prob. 28.27
: How much work must be done on an electron to accelerate it from rest to a speed of 0.990c?
The mass of an electron is 9.11
" 10 -31 kg.
Wednesday, March 19, 2008 39
The relationship between relativistic momentum and total energy is:
E 2 = p 2 c 2 + m 2 c 4
Verify by substituting for E and p :
( mc 2 ) 2
1 −
So, ( mc 2 ) v 2 c 2
2
=
= (
( mv ) 2 c 2
+ m
2 c
4
1 − v 2 c 2 mv ) 2 c 2 + ( 1 − v 2 c 2
) m 2 c 4
E = mc 2
!
1 − v 2 / c 2
, p = !
mv
1 − v 2 / c 2
Therefore, m 2 c 4 = m 2 v 2 c 2 + m 2 c 4 − m 2 v 2 c 2
Correct!
So E 2 = p 2 c 2 + m 2 c 4 is correct
If m = 0 (example, photon of light), E = pc !
light carries momentum!!
Wednesday, March 19, 2008 40

Prob. 28.42/28 : An object has a total energy of 5 " 10 15 J and a kinetic energy of 2 " 10 15 relativistic momentum?
J. What is the magnitude of the object’s
Wednesday, March 19, 2008 41
Time dilation: !
t = !
!
t
0
1 − v 2 c 2
Proper time
Length contraction: L = L
0
!
1 − v 2 c 2
Proper length p = !
mv
1 − v 2 c 2 relativistic momentum
E = mc 2
!
1 − v 2 c 2 total energy
KE = mc 2
 
$
1
1 − v 2 c 2
− 1 
E = mc
2
= “ rest energy ”
E
2 = p
2 c
2 + m
2 c
4
Wednesday, March 19, 2008 42