Physics 2D Lecture Slides
Lecture 7
Jan. 22, 2007
Relativistic Force & Acceleration

p=

mu
1 − (u / c)2

= γ mu
Relativistic
Force
And
Acceleration
Reason why you cant
quite get up to the speed
of light no matter how
hard you try!

⎞
 dp d ⎛
mu
F=
= ⎜
⎟
dt dt ⎜⎝ 1 − (u / c)2 ⎟⎠
⎡
m
F=⎢
⎢
2
⎢⎣ 1 − (u / c)
⎤
⎥ du : Relativistic Force
3/ 2 ⎥
dt
⎥⎦

 du
Since Acceleration a =
,[rate of change of velocity]
dt
(
)

3/ 2
F

2
⎡1 − (u / c) ⎤
⇒ a=
⎦
m⎣

Note: As u / c → 1, a → 0 !!!!
Its harder to accelerate when you get
closer to speed of light
Linear Particle Accelerator : Parallel Plates With Potential Difference
-
F=-eE
e-
E
d
+
Parallel Plates
E= V/d
F= -eE
V
Charged particle q moves in straight line


in a uniform electric field E with speed u
 
accelarates under force F=qE


3/ 2
3/ 2

2
2
du
F ⎛
u ⎞
qE ⎛
u ⎞

a =
=
1− 2 ⎟ =
1− 2 ⎟
⎜
⎜
dt
m⎝
m ⎝
c ⎠
c ⎠
larger the potential difference V across
plates, larger the force on particle
Under force, work is done
on the particle, it gains
Kinetic energy
New Unit of Energy
1 eV = 1.6x10-19 Joules
1 MeV = 1.6x10-13 Joules
1 GeV = 1.6x10-10 Joules
Linear Particle Accelerator : 50 GigaVolts Accelating Potential

3/ 2
 eE
2
a = ⎡⎣1 − (u / c) ⎤⎦
m
PEP-II accelerator schematic and tunnel view
Magnetic Confinement & Circular Particle Accelerator
 B
V
Classically
v2
F=m
r
v2
qvB = m
r
B
F
r
dp d(γ mu)
du
=
=γm
= quB
dt
dt
dt
du u 2
=
(Centripetal accelaration)
dt
r
u2
γ m = quB ⇒ γ mu = qBr ⇒ p = qBr
r
F=
Charged Form of Matter & Anti-Matter in a B Field
Antimatter form of electron = Positron (e+)
Same Mass but opposite Charge
Positron curls the other way from electron in a B Field
Accelerating Electrons Thru RF Cavities
A Circular Accelerator : Using B Field to Confine the electron and RF cavity to power it
Circular Particle Accelerator: LEP @ CERN, Geneve
Accelerated electron through an effective voltage of 100 Billion Volts !
To be upgraded to 7 trillion Volts by 2007
circular track for accelerating electron
French
Border
Swiss
Border
Geneva Airport
In Tunnel 150m underground, 27km ring of Magnets Keep electron in Circular Orbit
Inside A Circular Particle Accelerator Tunnel : Monorail !
Test of Relativistic Momentum In Circular Accelerator

mu


p=
= γ mu
1 − (u / c)2
γ mu = qBr = p
qBr
γ =
mu
Proof That
p ≠ mu
Relativistic Work Done & Change in Energy
x2
W =
∫
 
F .dx =
x1
X2 , u=u
x1 , u=0
x2
∫
x1

dp 
.dx
dt
du

m
mu
dp
dt
p=
∴
=
, substitute in W,
3/ 2
2
2
dt
u
⎡
u ⎤
1− 2
⎢1 − 2 ⎥
c
c ⎦
⎣
du
u
m
udt
dt
∴W = ∫
(change in var x → u)
3/ 2
0 ⎡
u2 ⎤
⎢1 − 2 ⎥
c ⎦
⎣
u
W =
∫⎡
mudu
3/ 2
=
mc 2
1/ 2
− mc 2 = γ mc 2 − mc 2
⎡
u ⎤
u ⎤
1
−
1
−
⎢
⎢
2 ⎥
2 ⎥
c
c
⎣
⎦
⎣
⎦
Work done is change in Kinetic energy K
0
2
2
K = γ mc 2 − mc 2 or
Total Energy E= γ mc 2 = K + mc 2
Relativistic Kinetic Energy & Newtonian Physics
Relativistic KE K = γ mc 2 − mc 2
⎡ Remember Binomial Theorem
⎤
⎢
⎥
nx
n(n-1)
⎢ for x << 1; (1+x) n = (1+
+
x 2 +smaller terms) ⎥
⎢⎣
⎥⎦
1!
2!
−
1
2
⎡ u2 ⎤
1 u2
∴ When u << c, ⎢1- 2 ⎥
≅ 1+
+ ...smaller terms
2
2c
⎣ c ⎦
1 u2
1 2
2
2
so K ≅ mc [1 +
]
−
mc
=
mu (classical form recovered)
2
2c
2
Total Energy of a Particle E = γ mc 2 = KE + mc 2
For a particle at rest, u = 0 ⇒ Total Energy E= mc 2
Mass is Energy, Energy is Mass : Mass-Energy Conservation
Examine Kinetic energy Before and After Inelastic Collision: Conserved?
K=0
S
K = mu2
Before
v
1
v
V=0
2
1
2
After
Mass-Energy Conservation: sum of mass-energy of a system of particles
before interaction must equal sum of mass-energy after interaction
=
E before
mc 2
2
+
mc 2
2
=
Eafter
Mc 2 ⇒ M =
2m
2
> 2m
u
u
u
1
−
1
−
c2
c2
c2
Kinetic energy has been transformed into mass increase
1−
⎛
⎞
⎟
2K 2 ⎜⎜ mc 2
2⎟
ΔM = M - 2m = 2 = 2
− mc
2
⎜
⎟
c
c
u
⎜ 1− 2
⎟
⎝
⎠
c
Kinetic energy is not lost,
its transformed into
more mass in final state
Creation and Annihilation of Particles
γ
µ-
µ+
Sequence of events in a matter-antimatter collision:
e + e →γ → µ + µ
+
−
+
−
Relativistic Kinematics of Subatomic Particles
Reconstructing Decay of a π Meson
The decay of a stationary π + → µ +υ happens quickly,
υ is invisible, has m ≅ 0; µ + leaves a trace in a B field
Pν , Εν
µ + mass=106 MeV/c 2 , KE = 4.6 MeV
What was mass of the fleeting π + ?
ν
π+
Energy Conservation:
At rest
µ+
Eπ = Eµ + Eν ⇒ m π c 2 = (mµ c 2 )2 + pµ2 c 2 + pυ c
Momentum Conservation : pµ = pν
⇒ mπ c 2 = (mµ c 2 )2 + pµ2 c 2 + pµ c
Pµ , Εµ
Relativistic Kinematics of Subatomic Particles
Energy Conservation:
Eπ = Eµ + Eν ⇒ m π c 2 = (mµ c 2 )2 + pµ2 c 2 + pυ c
Pν , Εν
Momentum Conservation : pµ = pν
⇒ mπ c 2 = (mµ c 2 )2 + pµ2 c 2 + pµ c
ν
also pµ2 c 2 = Eµ2 − (mµ c 2 )2 = (K µ + mµ c 2 )2 − (mµ c 2 )2
π+ At rest
µ+
=K µ2 + 2K µ mµ c 2
Substituting for pµ2 c 2 ⇒
m π c 2 = mµ2 c 4 + K µ2 + 2K µ mµ c 2 + K µ2 + 2K µ mµ c 2
Pµ , Εµ Put in all the known #s
⇒ mπ c 2 = 111MeV + 31MeV = 141MeV
⇒ mπ = 141MeV / c 2
Conservation of Mass-Energy: Nuclear Fission
M1
M
Mc 2 =
M1c 2
1−
2
1
2
u
c
<1
+
+
M2
M 2c2
1−
<1
2
2
2
u
c
+
+
M 3c 2
1−
2
3
2
u
M3
Nuclear Fission
⇒ M > M1 + M 2 + M 3
c
<1
Loss of mass shows up as kinetic energy of final state particles
Disintegration energy per fission Q=(M – (M1+M2+M3))c2 =ΔMc2
236
92
90
1
U → 143
Cs+
Rb+3
n
55
92
0
(1 AMU= 1.6605402 × 10-27 kg = 931.49 MeV)
Δm=0.177537u=2.9471 × 10-28 kg = 165.4 MeV= energy release/fission =peanuts
What makes it explosive is 1 mole of Uranium = 6.023 x 1023 Nuclei !!
Energy Released by 1 Kg of Fissionable Uranium
1 Mole of Uranium = 236 gm, Avagadro''s # = 6.023 × 1023 Nuclei
6.023 × 1023
So in 1 kg N =
× 1000g = 2.55 × 1024 nuclei
236g / mole
1 Nuclear fission = 165.4 MeV ∴ 103 g = 2.55 × 1024 × 165.4 MeV
Note 1 MeV = 4.45 × 10-20 kWh
If the power plant has conversion efficiency = 40%
Energy Transformed = 748 × 106 kWh
⇒ 1 100W lamp can be lit for 8500 years !
Nuclear Fission Schematic : Tickling a Nucleus
Excited U
Absorption of Neutron
Unstable
Nucleus
Oscillation
Deforms Nucleus
Sustaining Chain Reaction: 1st three Fissions
Average # of Neutrons/Fission = 2.5
Neutron emitted in fission of one U
Needs to be captured by another
To control reaction => define factor K
Supercritical K >> 1 in a Nuclear Bomb
Critical
K = 1 in a Nuclear Reactor
Schematic of a Pressurized-Water Reactor
Water in contact with reactor core serves as a moderator and heat transfer
Medium. Heat produced in fission drives turbine
Lowering Fuel Core in a Nuclear Reactor
First Nuke Reactor :Pennsylvania
1957
Pressure Vessel contains :
14 Tons of Natural Uranium
+ 165 lb of enriched Uranium
Power plant rated at 90MW,
Retired (82)
Pressure vessel packed with
Concrete now sits in Nuclear Waste
Facility in Hanford, Washington
Nuclear Fusion : What Powers the Sun
Opposite of Fission
Mass of a Nucleus < mass of its component protons+Neutrons
Nuclei are stable, bound by an attractive "Strong Force"
Think of Nuclei as molecules and proton/neutron as atoms making it
Binding Energy: Work/Energy required to pull a bound system (M) apart leaving its
components (m) free of the attractive force and at rest:
n
Mc 2 +BE=∑ m ic 2
i=1
2
1
H
Deuterium
+
2
1
H
=
4
2
He + 23.9 MeV =
Deuterium = Helium + Released Energy
Think of energy released in Fusion as Dissociation energy of Chemistry
Sun's Power Output = 4 × 1026 Watts ⇒ 1038 Fusion/Second !!!!
Nuclear Fusion: Wishing For The Star
• Fusion is eminently desirable because
– More Energy/Nucleon
• (3.52 MeV in fusion Vs 1 MeV in fission)
• 2H + 3H  4He + n + 17.6 MeV
– Relatively abundant fuel supply, No danger like nuclear reactor going
supercritical
• Unfortunately technology not commercially available
– What’s inside nuclei => protons and Neutrons
– Need Large KE to overcome Coulomb repulsion between nuclei
• About 1 MeV needed to bring nuclei close enough together for Strong
Nuclear Attraction  fusion
• Need to
– heat particle to high temp such that thermal energy E= kT ≈ 10keV 
tunneling thru coulomb barrier
– Implies heating to T ≈ 108 K ( like in stars)
– Confine Plasma (± ions) long enough for fusion
» In stars, enormous gravitational field confines plasma
Inertial Fusion Reactor : Schematic
Pellet of frozen-solid Deuterium & tritium bombarded
from all sides with intense pulsed laser beam with
energy ≈106 Joules lasting 10-8 S
Momentum imparted by laser beam compresses
pellet by 1/10000 of normal density and heats it to
temp T ≈ 108 K for 10-10 S
Burst of fusion energy transported away by liquid Li
World’s Most Powerful Laser : NOVA @ LLNL
Size of football field, 3 stories tall
Generates 1.0 x 1014 watts (100 terawatts)
10 laser beams converge onto H pellet (0.5mm diam)
Fusion reaction is visible as a starlight lasting 10-10 S
Releasing 1013 neutrons
ITER: The Next Big Step in Nuclear Fusion
Visit www.iter.org for Details of this mega Science & Engineering Project
This may be future of cheap, clean Nuclear Energy for Earthlings