PARTICLE CREATION
RELATIVISTIC COLLISIONS
Ken Cheney
March 26.2, 2006
ABSTRACT
The creation of particles will be investigated using pictures taken of pi
mesons in a liquid hydrogen bubble chamber. Relativistic conservation of
energy and momentum will be used to deduce what particles were created in
the collisions.
IMPORTANCE
We can actually “see” particles being created, something quite inconceivable
until fairly recently.
We can use relativity to calculate something real.
SYMBOLS
m
q
B
T
E
Eelec
p
vc
v
Mass, rest or proper mass
Charge
Magnetic field strength, Kgauss
Tesla, SI units for magnetic field strength = Webbers/ m2
Energy, Mev
Electric field
Momentum, Mev/c
Velocity, m/s ; conventional
Velocity in vc/c
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PARTICLES
0
pi meson, zero charge

p+
pi meson plus or minus
proton
N0
neutron

mass= 135.0 Mev
NUMBERS
For this experiment
p of incoming pi mesons 921 Mev/c
KE of incoming pi mesons 790 Mev
B of magnetic field in the bubble chamber 12.2 Kgauss
Universal values
c
q
2.9979x108 m/s or 1in units of v/c
electron charge
1.6x10-19C
m 
139.6 Mev pion, pi meson
m 0
135.0 Mev pion, pi meson
m p+
938.3 Mev proton
m N0
939.6 Mev neutron
m e-
0.511 Mev electron
m
106 Mev
muon
EXPERIMENT AT BROOKHAVEN
Protons were accelerated in Brookhaven’s Alternating Gradient
Synchrotron, AGS.
The protons were collided with metal atoms to produce a shower of
particles.
Magnetic and electrical filters were used to select only pi mesons with
energy of 790 Mev, and momentum of 921 Mev/c.
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The pi mesons then collide with protons in a bubble chamber.
The charged particles entering and leaving the collision deposit energy in the
“super heated” liquid hydrogen which boils along the path of the particles
leaving a track of bubbles.
The possible particles leaving the collision are limited by the energy
available and various conservation laws, e.g. conservation of charge. The
more common possibilities are:
(0.1)
   p     p
Elastic




0
(0.2)
  p   p 
Inelastic



0

(0.3)
  p   N 
Inelastic
Notice that the inelastic cases both have neutral particles produced. These
neutral particles will not leave a trace in the bubble chamber.
Elastic collisions do not change the number of particles, hence do not change
the mass (rest or proper mass) of the particles involved.
Inelastic collisions have incoming KE go into mass of new particles, or
possible vice versa!
There is enough energy to produce electrons (only a half Mev), muons (106
Mev), neutrinos (hardly any ev!), or gammas (any Mev) but evidently these
are less common. You will see small spirals in the pictures, these are
electrons or positrons.
There is a magnetic field through the bubble chamber, 12.2 Kgauss; this
field will bend the paths of the charged particles. From the radius of
curvature the momentum and the sign of the charge of the particle can be
calculated.
The photographs were selected to show paths perpendicular to the camera
and the magnetic field.
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PARTICLE SELECTION
Electric Fields
The particles are sent through a curved channel, radius R, with a
perpendicular electric field Eelec . The only particles that pass through the
channel are those for which the force due to the electric field produces a
centrifugal force, which gives a path with a radius R:
Centrifical Force = Electrostatic Force
mvc 2 / R  qEelec
So
1
KE= mvc 2  RqEelec
2
Magnetic Fields
Like a mass spectrometer:
Centripital force = qv  B force
mvc 2 / R  q v c  B
momentum = p  RqB
(0.4)
assuming vc  B
RELATIVISTIC COLLISION THEORY
DIMENSIONS / UNITS
We want mass and energy in units of Mev.
Mass is the rest or proper mass.
An electron volt is defined as the energy an electron charge gets going
through a one volt potential difference:
1ev  Energy = qV  1.6  1019 C  1v  1.6  1019 J
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(0.5)
A Mev is a million ev.
Momentum then will be in units of Mev/c.
B will be in units of Kgauss. = 103 gauss
1 Tesla = 104 gauss, this is the SI unit of magnetic field strength
R will be in meters
To convert p=qBR so we can put in an electron charge, Kgauss, and meters
and get out Mev/c we multiply by several conversion factors:
1T
103 gauss
1ev
1Mev
p  1.6  1019 C  B( Kgauss )  ( 4
)(
)(
)( 6 ) (
19
10 gauss
1Kg
1.6  10 J
10 ev
 30 BR
(0.6)
The handout is incorrect in giving p=3x103 BR.
SUMMERY OF EQUATIONS
E 2  m2  p2
p  30 BR
Energy is conserved
Momentum is conserved
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3  108
c
m
s 2 ) R( m)
ROUGH PROCEDURE
The manual you can check out from the stockroom (29 pages!) gives
important and much more detailed instructions.
For each of three photographs:
 Select a clean collision!
 Measure the angles between incoming and outgoing trails; see good
advice on page 23 of the handout.
 Measure the curvature of the paths, see vital hints in the handout, page
21.
 Calculate the correction factor (scaling) using the fiducial marks on
the pictures, see page 19 of the handout.
 Calculate how well energy and momentum are conserved for each
hypothetical type of collision. We know the initial energy and
momentum, py=0. Calculate the energy and momentum of visible
final particles. Does the missing mass, if any, match a probable
collision?
 Reach a decision as to what type of collision occurred for each
photograph.
KC
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