Physics 2D Lecture Slides
Jan 15
Vivek Sharma
UCSD Physics
Watching an Inelastic Collision between two putty balls
Relativistic Momentum and Revised Newton’s Laws
Need to generalize the laws of Mechanics & Newton to confirm to Lorentz
Transform
G
G
and the Special theory of relativity: Example : p = mu
S
P = mv –mv = 0
P=0
Before
v
1
v
V=0
2
1
2
After
v1 − v
v2 − v
−2v
V −v
'
= −v
= 0, v2 =
=
, V '=
v =
2
v1v
v1v
V
v
1− 2
1− 2
1 − 2 1 + v2
c
c
c
c
−2mv '
'
, pafter = 2mV ' = −2mv
pbefore
= mv1' + mv2' =
2
v
1+ 2
S’
c
'
p'before ≠ pafter
'
1
1
v1’=0
Before
v2’
2
V’
1
2
After
Definition (without proof) of Relativistic Momentum
G
mu
G
G With the new definition relativistic
= γ mu momentum is conserved in all frames
p=
1 − (u / c) 2
of references : Do the exercise
New Concepts
Rest mass = mass of object measured
In a frame of ref. where object is at rest
γ=
1
1 − (u / c)
2
u is velocity of the object
NOT of a reference frame !
Nature of Relativistic Momentum
m
u
G
p=
G
mu
1 − (u / c) 2
G
= γ mu
With the new definition of
Relativistic momentum
Momentum is conserved in
all frames of references
Good old Newton
Relativistic Force & Acceleration
G
p=
G
mu
1 − (u / c)
2
G
= γ mu
Relativistic
Force
And
Acceleration
Reason why you cant
quite get up to the speed
of light no matter how
hard you try!
G
G dpG d 
mu
F =
= 
dt dt  1 − (u / c ) 2

 use


d
du d
=
dt dt du


m
mu
u
1
2
−
−
 du
F =
(
)(
)
+
×
 1 − (u / c ) 2 (1 − (u / c ) 2 )3 / 2
c 2  dt
2


 2
2
2 
mc
mu
mu
−
+
 du
F =
 c 2 (1 − (u / c ) 2 )3 / 2  dt




m
 du : Relativistic Force
F =
 (1 − (u / c ) 2 )3 / 2  dt


G
G du
Since A ccele ra tion a =
,
dt
G
G
F
2 3/ 2
1 − (u / c ) 
⇒ a=
m
G
Note: As u / c → 1, a → 0 !!!!
Its harder to accelerate when you get
closer to speed of light
A Linear Particle Accelerator
-
F
q
E
d
+
E= V/d
F=eE
V
Charged particle q moves in straight line
G
G
Under force, work is done
in a uniform electric field E with speed u
on the particle, it gains
G G
Kinetic energy
accelarates under force F=qE
G
G
G
2 3/ 2
2 3/ 2
New Unit of Energy
G du
F  u 
qE  u 
a =
=
1 − 2  =
1 − 2 
dt
m c 
m  c  1 eV = 1.6x 10-19 Joules
larger the potential difference V across
plates, larger the force on particle
A Linear Particle Accelerator
G
G eE
2 3/ 2
1 − (u / c) 
a=
m
PEPPEP-II accelerator schematic and tunnel view
Magnetic Confinement & Circular Particle Accelerator
G
V G
Classically
B
v2
F =m
r
v2
qvB = m
r
dp d (γ mu )
du
F=
=
=γm
= quB
dt
dt
dt
du u 2
(Centripetal accelaration)
=
dt
r
u2
γ m = quB ⇒ γ mu = qBr ⇒ p = qBr
r
G
FB
r
Charged Form of Matter & Anti-Matter in a B Field
Circular Particle Accelerator: LEP @ CERN, Geneve
Magnets Keep Circular Orbit of Particles
Inside A Circular Particle Accelerator @ CERN
Accelerating Electrons Thru RF Cavities
Test of Relativistic Momentum In Circular Accelerator
G
mu
G
G
p=
= γ mu
1 − (u / c ) 2
γ mu = qBr
qBr
γ =
mu
Relativistic Work Done & Change in Energy
W =
X2 , u=u
x2
∫
G G
F .dx =
x1
x1 , u=0
u
∫
0
u
W =
∫
G
dp G
.dx
dt
du
G
m
dp
dt
∴
=
, substitute i n W ,
2 3/2
dt

u 
1 − c 2 


u2
1− 2
c
∴W =
∫
x1
mu
p=
x2
du
udt
dt
2 3/2

u 
−
1
2 

c


m
m udu
3/2
=
(change in var x → u )
mc2
1/ 2
− mc2 = γ mc2 − mc2

u 
u2 
0
1 − c 2 
1 − c 2 




W o rk d one is change in Kinetic energy K
2
K = γ m c 2 − m c 2 or
Total Ener gy E= γ m c 2 = K + mc 2
Why Can’s Anything go faster than light ?
Lets accelerate a particle from rest, particle gains velocity & kinetic energy
K=
mc 2
1/ 2
 u 
1 − c 2 


2
− mc ⇒
2
( K + mc )
2 2
 u2 
⇒ 1 − 2  = m 2c 4  K + mc 2 
 c 
⇒ u = c 1− (




2
mc

=
  u 2 1/ 2 
 1 − 2  


c




−2
K
K
−2
1)
(Parabolic
in
u
Vs
)
+
2
2
mc
mc
Non-relativistic case: K =
1 2
2K
mu ⇒ u =
2
m
2
Relativistic Kinetic Energy
When Electron Goes Fast it Gets “Fat”
E = γ mc
2
v
As → 1, γ → ∞
c
Apparent Mass approaches ∞
Relativistic Kinetic Energy & Newtonian Physics
Relativistic KE = γ mc 2 − mc 2
−
1
2
2
 u2 
1u
When u << c, 1- 2  ≅ 1 −
+ ...smaller terms
2
2c
 c 
2
1
1
u
2
2
2
so K ≅ mc [1 −
]
−
mc
=
mu
(classical form recovered)
2
2c
2
Total Energy of a Particle
E = γ mc = KE + mc
2
2
For a particle at rest, u = 0
⇒ Total Energy E= mc
2
E = γ mc 2 ⇒ E 2 = γ 2 m 2 c 4
Relationship between P and E
p = γ mu ⇒ p 2 c 2 = γ 2 m 2 u 2 c 2
⇒ E 2 − p 2 c 2 = γ 2 m 2 c 4 − γ 2 m 2u 2 c 2 = γ 2 m 2 (c 2 − u 2 )
2 4
m 2c 2 2
m
c
2
2
2 4
2
=
(
)
(
)
−
=
−
=
c
u
c
u
m
c
2
2
2
u
c −u
1− 2
c
E 2 = p 2 c 2 + ( mc 2 ) 2 ........important relati on
For p articles with zero rest mass like photon (EM waves)
E
E= pc or p =
c
(light has momentum!)
Re lativistic Invariance : E 2 − p 2 c 2 = m 2 c 4 : In all Ref Frames
Rest M ass is a "finger print" of the particle
Mass Can “Morph” into Energy & Vice Verca
• Unlike in Newtonian mechanics
• In relativistic physics : Mass and Energy are the same
thing
• New word/concept : Mass-Energy
• It is the mass-energy that is always conserved in every
reaction : Before & After a reaction has happened
• Like squeezing a balloon :
– If you squeeze mass, it becomes (kinetic) energy &
vice verca !
• CONVERSION FACTOR = C2
Mass is Energy, Energy is Mass : Mass-Energy Conservation
Examine Kinetic energy Before and After Inelastic Collision: Conserved?
S
K=0
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
=
E after
=
2
Mc
⇒ M =
2m
2
> 2m
u
u
u
1− 2
1− 2
1− 2
Kinetic energy is not lost,
c
c
c
Kinetic energy has been transformed into mass increase its transformed into

 more mass in final state


2 K 2  mc 2
− mc 2 
∆M = M - 2m = 2 = 2

c
c 
u2
 1− 2

c


Conservation of Mass-Energy: Nuclear Fission
M1
M
Mc 2 =
M 1c 2
u12
1− 2
c
<1
+
+
M 2c 2
u22
1− 2
c
M2
+
<1
+
M 3c 2
u32
1− 2
c
M3
Nuclear Fission
⇒ M > M1 + M 2 + M 3
<1
Loss of mass shows up as kinetic energy of final state particles
Disintegration energy per fission Q=(M – (M1+M2+M3))c2 =∆Mc2
236
92
1
U → 14553 Cs+ 90
R
b
+3
92
0n
∆m=0.177537u=2.9471× 10-28 kg = 165.4 MeV
What makes it explosive is 1 mole U = 6.023 x 1023 Nuclei !!
Relativistic Kinematics of Subatomic Particles
Reconstructing Decay of a π Meson