•Conservation of momentum is one of the most fundamental and most useful
concepts of elementary physis
•Does it apply in special relativity?
•Consider the following collision process. All balls have mass m and
speed v before and after collision
m u1  0, v,0
•Surely this will conserve momentum Before

p  mu
p1   0, mv, 0 
p2   0, mv, 0 
p3   mv, 0, 0 
p4   mv, 0, 0 
After
m
u4   v,0,0 
p1  p2   0, 0, 0 
p3  p4   0, 0, 0 
u3   v,0,0
m u   0, v,0 
2
Momentum
is conserved
But what about other frames?
m

Momentum
in
another
frame
y’
m
u4   v,0,0 
m
v
u1   0, v,0
uy 
m
u3   v,0,0
m u   0, v,0 
2
u1   v,  v  , 0 
x’
u2   v,  v  , 0 
u3   0, 0, 0 

u4   2v 1  v 2 c 2  , 0, 0
p1  p2   2mv, 0, 0 

ux  v
ux 
,
2
1  ux v c
In this frame,
momentum is
not conserved!

2
2


p3  p4   2mv 1  v c  , 0, 0

uy
 1  u x v c
2

,
v
u1x 
  v,
1 0
v
v
u1y 
 ,
 1  0 

v  v
2v
u4 x 

,
2
2
2
2
1 v c 1 v c
0
u4 y 
 0,
2
2
 1  v c 
The problem with momentum
How can we explain this failure of conservation of momentum? Three choices:
1. This process is physically impossible; it cannot occur
2. Conservation of momentum is not a conservation law in S.R.
3. Our formula for momentum is wrong in S.R.
p  mu
x
px  mu x  m
t • Why is t given special privilege over other coordinates?
• We want something that is like time, but not coordinate
y
dependant in the denominator, like proper time 
p y  mu y  m
t  
t
z

x

x
pz  mu z  m
px  m
m
  mu x
t

t
mu
p   mu 
1  u 2 c2
Does this fix the problem? (1)
u1   v,  v  , 0 
p
u2   v,  v  , 0 
mv
mu
1  u 2 c2
 mv

1
1  v2 c2
mv
2




m

v  p2x
2
2
2
2
2
2
v
v
v
v v  v 
1 2  2 2
1  2  2 1  2  1  2
c  c
c
c c  c 

2
mv
1

2
mv
u3   0, 0, 0 

p4 x 

2
2
v
2 2
2
4
v


1

v
4
v
1


2
2v
1 2   2
c
2
2
2 2

u4   
,
0,
0

c 1  v c 
2
2
c 
c

1

v
c


2mv
2mv
2




2

mv
2
4
2
2v v
v
1 2  4
1 2
c
c
c
p1x 

Does this fix the problem? (2)
2

p1    mv,  mv, 0 
p2    mv,  mv, 0 
2
p3   0, 0, 0 
p4   2 2 mv, 0, 0 
p1  p2   2 2 mv, 0, 0 
p3  p4   2 2 mv, 0, 0 
Momentum is
conserved
What’s missing here . . .
 x y z 
•Hint – we’ve got all three dimensions . . .
p  m , , 
•But we live in four-dimensional spacetime
    
•Is there a conserved fourth component of momentum?
•These quantities are related by Lorentz boost, just as px and py are related by rotations
•If px, py and pz are conserved, then so must be pt
t
•Is it pt that is conserved, or cpt?
pt  m

•Actually it doesn’t matter; if pt is conserved, so is any multiple of it
•We will work with a different multiple
t
2
2
•The ERIC quantity
E  c pt  c m
  mc 2
•Eric is conserved

t  
E   mc 2
What is Eric, really?
E   mc 2
E1  E2  E1  E2
Consider low velocity limit:
m1 u1
m1 u1
m2 u
m2 u2
 1
m1c  m2 c  m1c  m2 c
2
2
2
2
2
2
Not careful enough . . .
u
  1 2
2c
2
2
2
2










u
u
u
u
2
2
2
2
1
2
1
2
m1c 1  2   m2c 1  2   m1c 1  2   m2c 1  2 
 2c 
 2c 
 2c 
 2c 
1
2
m1u12  12 m2u22  12 m1u12  12 m2u22
E is really
energy
Potential and Kinetic Energy
E   mc 2
•E is the total energy – it includes both potential and kinetic energy
•Total energy is always conserved
•Potential energy is the energy something has when it’s not moving
•Also called rest energy
E0  mc 2
•More about this later
•Kinetic energy is how much extra energy there is due to motion
Ekin  E  E0    1 mc2  12 mv 2  83 m v 4 c 2 
A sample problem . . .
In 2004, the total world consumption of energy was 15 TW. How much fuel
would have to be consumed each hour to power the world if we could extract
100% of its potential energy?
E0  mc 2
P  1.5 10 W  1.5 10 J/s
13
13
E  Pt  1.5 1013 J/s   3600 s   5.40 1016 J
E 5.40 1016 kg  m 2 /s 2
m 2 
 0.600 kg
2
8
c
3.00

10
m/s 

p   mu
Good vs. Bad Notation
E   mc 2
Eric’s notation:
•m, called the mass, is a constant, and independent of velocity
•It does not increase with velocity
2
E

mc
•It does not approach infinity as u  c
•The famous formula E = mc2 is not true except when u = 0.
2
E

mc
0
•Einstein never wrote this formula
•The quantity m is then sometimes called the relativistic mass - but I
won’t use this term.
Satan’s notation:
•m  m0 is renamed the rest mass
•We then define m = m0 as the mass
•In this notation, mass m goes to infinity as u  c
•In this notation, E = mc2 is always true
•Do not use this notation/nomenclature in this class
Momentum and Energy
•For small velocities,
  1, and the momentum is
just the usual formula
•As u  c, the momentum and
energy both go to infinity
•This tells us we can’t
reach u = c.
E   mc 2 
p   mu 
mc 2
1 u c
2
mu
1  u 2 c2
E/mc2
p/mc
u/c
2
Particle physics units
F  qE
•Forces are usually induced by using electric fields
•Charges are usually the same as the electron charge (-e) or simple multiples of this
(+e, -2e, etc.)
•Electric potentials are measured in volts (V)
•Easy to work with units of electron charge times volts, or eV’s (called electron volts)
1 eV  1.602 1019 C  1 V   1.602  1019 J
•Metric multiples are also common
•Momentum has units of energy/velocity, so it is often
expressed in units of eV/c
p  0.980 TeV / c
•Mass has units of energy/velocity2, so it is often
expressed in units of eV/c2
m  0.938 GeV / c
2
1 keV  103 eV
1 MeV  10 eV
6
1 GeV  109 eV
1 TeV  1012 eV
A sample problem . . .
A Z-particle at rest (mass mZ = 91.19 GeV/c2) decays to a Bmeson and an anti-B meson (both mass mB = 5.279 GeV/c2).
What is the energy and velocity of the resulting B-mesons?
B
Z
B
•Since the initial Z is at rest, pZ = 0 and EZ = mZc2 = 91.19 GeV
•The B-meson and anti-B-meson must have equal & opposite momenta
•Since they have the same mass, this means they have equal speeds
•They therefore have equal energies
Eb  12 EZ  45.59 GeV
Eb   mb c 2    5.279 GeV 
45.59 GeV

 8.637
5.279 GeV
Note: Mass is not conserved!

1
1  v2 c2
v 2 c 2  1  1  2  0.9866
v  0.9933c  2.978 108 m/s
Two Useful Formulas
E   mc 2
p   mu
Divide them:
Written this way, both sides
are dimensionless
u
p  mu
 2

2
c
E  mc
Square and subtract them:
E  c p    mc

2
2
2
2

2 2
 c m u  

2
2
2
m2c 2  c 2  u 2 
1 u c
E  c p   mc
2
2
2
u pc

c
E
2

2 2
2
m c c  u
2 4

2
c2  u 2
2

Different types of particles
u pc

c E
E 2  c 2 p 2  m2 c 4
Must the combination E2 – c2p2 always be positive?
Maybe m2 needn’t always be positive
1. If m2 > 0 we call the particle a massive particle
• Electrons, protons, rockets, people
• E > cp, so u < c: Always slower than light
2. If m2 = 0 we call the particle a massless particle
• Photons (particles of light)
• E = cp, so u = c: Always at the speed of light
3. If m2 < 0 we call the particle a tachyon
• No known particles
• E < cp, so u > c: Always faster than light
Massless and nearly massless particles
u pc
E   mc
p   mu
E  c p   mc 

c E
2
2
2
2
2 2
•First two equations make no sense for massless particles – don’t use them
•If mass is very small (mc2 much smaller than E or cp), then often easier to treat as
massless
•Massless versions of these equations:
E c p
The Large Electron-Positron collider (LEP)
accelerates electrons (me = 0.511MeV/c2) to an
energy of E = 100 GeV. How fast are they moving?
u c
u  0.999999999987c
u c
Sample problem
A Z-particle at rest (mZ = 91.2 GeV/c2) decays to an electron (me =
0.511 MeV/c2) with energy 25.0 GeV, a positron (me) with energy 20.0
GeV moving at 71° angle compared to the electron, and an unknown
X particle. What is the momentum, energy, mass, and velocity of X?
e+ E2  20.0 GeV
•Electrons are nearly
massless
EZ  91.2 GeV
pZ  0
X
Z
71


p2c  20.0 iˆ cos 71  ˆj sin 71 GeV
 6.51iˆ  18.9 ˆj GeV

e-

E1  25.0 GeV
p1c  25.0iˆ GeV
Conservation of energy:
EX  EZ  E1  E2   91.2  25.0  20.0 GeV
Continued . . .
EX  46.2 GeV
Sample problem continued . . .
. . . What is the momentum, energy, mass, and velocity of X?
pZ  0
pX  pZ  p1  p2
Conservation of
momentum:
p1c  25.0iˆ GeV



  31.5iˆ  18.9 ˆj  GeV / c
 25.0iˆ  6.5iˆ  18.9 ˆj GeV / c

p2 c  6.51iˆ  18.9 ˆj GeV
u pc 31.5iˆ  18.9 ˆj
 0.682iˆ  0.409 ˆj

Velocity formula:

46.2
c E
u  0.795c
Mass formula:
 mc 
2 2
 E  c p   46.2 GeV    31.5 GeV   18.9 GeV 
2
2
2
 785 GeV
2
2
mX  28.0 GeV / c 2
2
2
Lorentz boost of four-momentum
Suppose I measure the momentum and energy of a particle or collection of
particles. What does a moving observer measure?
•Far easier than the book makes it out
•Difference in coordinates (x, y, z, t) satisfy same equations as (x, y, z, t)
x    x  vt 
px  mx 
y  y
z   z
p y  my 
t     t  vx c 2 
pz  mz 
E  mc 2 t 
px    px  vE c 2 
py  p y
pz  pz
E     E  vpx 
These formulas work for either a single particle or for the sum of all particles