Particle physics experiment

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Particle physics experiment
Jirí Dolejší, Olga Kotrbová, Charles University
in Prague
We have standard ways of discovering properties
of things around us and to look inside of
objects (like the small alarm clock in the
photograph). But these methods may not be
applicable to atom and its components we do not
have a sufficiently small screwdriver and we even
cannot look at it with sufficient resolution.
With the best current microscopes we can only see
individual atoms, like in this picture from a
tunneling microscope, but not the inside of them.
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Already in the beginning of the 20th century E.
Rutherford and his collaborators developed a
novel method to study the inside of atoms. They
shot a-particles towards a thin gold foil and
discovered that what best corresponds to the
experimental results is the idea of atoms being
almost empty with small heavy nuclei and
electrons flying around. Today gold nuclei are
collided at RHIC (Relativistic Heavy Ion
Collider) and we expect to learn more about the
matter through collisions.
How can the methods based on collisons work?
Lets try!
A first attempt I will try to use a collision
instead of a screwdriver
I succeeded in getting inside! But not
sufficiently deep into the structure. Maybe I
need higher energy It would probably be better
to start with a simpler object
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Atoms, nuclei, particles Implicitly we look at
them as small balls. Instead of playing with
balls I suggest flicking coins and studying their
collisions. Put one coin on a smooth surface,
flick another coin against it and observe the
result.
Disclaimer The collisions of a euro and a dollar
have neither political nor economic meaning.
After playing a while with different coins you
will get some experience. Look at the following
situations and decide which of the coins is
heavier (arrows are proportional to velocities).
B
C
A
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I hope your answers are correct both coins have
the same mass in B, the blue one is lighter in A
and heavier in C. You can repeat the experiment
with coins and doublecoins glued together
with doublesided tape
What about describing the scattering of coins in
the usual manner in mechanics? Relevant variables
are the masses m1 and m2, the velocities v1 and
v2 before the collision and the velocities v1
and v2 after the collision. Important variables
are energy E and momentum
v1'
m1
v1
a
Both energy and momentum in an isolated system
are conserved
b
The red coin is at rest at the beginning
m2
v2'
v2 0
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Lets work only with momenta energy
conservation... and conservation of both
momentum components
Lets play with momenta We would like to get rid
of p2 and b and it is appealing to use for that
purpose the relation
We just rewrite the equations, square each one
and sum them
we can insert into energy equation
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The last equation
is a quadratic equation for
The discriminant is
and its solutions are
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Lets remember that we are looking for real
non-negative solutions. The simplest case is the
case of equal masses m1 m2
So in a collision of two coins with equal masses
coin 1 cannot be scattered backwards. It either
continues forward (straight or deflected) or
stops. We can see from the conservation
equations that if the coin stops, the target coin
takes over the whole energy and momentum. Try it
with billiard balls!!!
Instead of continuing with the detailed
discussion, we will plot the dependence of
momentum after the collision on the scattering
angle
1projectile 2target
Only lighter particle can be scattered backwards
The deflection of the heavier particle is limited
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Although we have played with coins, we used only
the most general mechanical laws. Our results
holds for collisions of any objects, including
subatomic particles. The only complication is
related to the pleasant fact that particles can
be accelerated to speeds close to the speed of
light. The effects predicted by the special
theory of relativity appear - particles have
bigger mass and unstable particles live longer.
We can quite easily modify our calculation
Instead of counting kinetic energy
we should deal with total energy
We will insert into the energy conservation
equation (all masses here are rest masses)
The calculation needs more time, more paper and
more patience than the last. We will only show
you the result of the simplest case m1 m2 m
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The difference between the non-relativistic and
the relativistic calculation is visible on the
graph
Nonrelativistic case
Protons are accelerated to energies more than
1000 times larger than their rest energies at
Fermilab (USA)
A quantitative understanding of the kinematics
of collisions enables us to compare masses of
coins or particles by just colliding them.
First success! Collisions are at least good for
something, not only for destroying everything ...
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The formula for relativistic energy can be
rewritten in the form
Energy and momentum have different values in
different reference frames a bottle in my hand
in the train has no kinetic energy with reference
to the train but it may have quite a significant
energy with reference to the ground. But the
special combination of E and above always
equals the square of the particle rest mass times
c4 (independent of the reference frame), i.e. is
constant. This feature offers a surprisingly
simple way to measure the mass of an unstable
particle
Measure energies and momenta of decay products
and , then calculate
Unstable particle with unknown mass
You have got the mass M !
You will find words like energy-momentum
four-vector, invariant mass, etc. in advanced
textbooks. They refer to the same things as
above. You can learn more ...
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To measure the masses of particles is clearly not
enough. How to get some deeper insight? How to
understand structure, interactions, etc.? Maybe
the way is to look at everything that can happen
and how probable it is. We can start again with
our macroworld.
What is the probability that I will catch a ball
shot at me? The best solution to answer this
question is to do an experiment. Being exposed
to randomly placed moderately fast shots, I
caught all the yellow shots. After several
repetitions of this experiment I succeeded
catching everything inside the area with the
yellow boundary.
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My ability to catch balls is characterized by
the yellow area - it is about 2 m2 and it means
that of the flow of shots with a density 10
shots per square meter I expect to have 20
catches.
My catching ability is characterized by an
effective area which I can cover. Physicists
use a special name for this quantity - the cross
section and typically use the letter s for
it. If a flux of incoming particles with a
density j hits the target, then the number of
interesting events N with the cross section s is
The standard unit for the cross section is 1
barn 1 b 10-28 m2.
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We can express a lot of information in terms of
cross-section. For example the shots I am
catching can vary in velocity (and energy) of the
ball. I can easily catch the slow balls but I
will probably try to hide myself from hard shots.
So for hard shots my cross section of catching
will be zero and the cross section of a ball
hitting me will be close to the area of my
silhouette ( the band around accounting for the
diameter of the ball). One may consider the
special (or partial) cross section for a ball
breaking my glasses etc. All the possible
processes can be summarized in the total cross
section.
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sball-me
The energy dependence of different cross-sections
relating to the interaction of me and the ball
could look like this graph (and betray a lot
about me )
At this energy I try to avoid getting hit by the
ball
m2
At this energy I am frozen from the fear
total
catch
hit
injury
death
At this energy I catch best
Energy of the ball in apropriate units
The cross-sections for proton-proton interactions
are displayed on this plot
Elastic cross section - colliding particles stay
intact and they only change the direction of
their flight
inelastic
At this energy the colliding protons have enough
energy to create a new particle a pion. This
is one example of a process contributing to the
inelastic cross section stotal - selastic
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Usually bunches of projectiles are scattered of
many target particles. The situation may look
like in our picture ...
target
beam
Every target catcher has a cross section s. There
are Ntarget of them subjected to beam
of projectiles. We assume that the individual
catchers do not collaborate and do not shield
others. So the number of catches is Nevents j
.s. Ntarget
Nbeam
Sbeam
Here are Nbeam incoming particles within a beam
with the area Sbeam. The flux of incoming
particles is then jbeam Nbeam/ Sbeam
t
Density of target particles
Reality can be always looked at from different
viewpoints ...
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We arrived at an expression which says that the
probability of an incoming particle interacting is
In the case when the beam particles are caught
(experts speak about absorp- tion) the beam
intensity is gradually reduced - Nbeam depends on
the depth in the target.
target
beam
t
t measures the depth in the target
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Expert pages! Its a challenge (but you dont
need to understand them)!
Lets try to find the dependence Nbeam(t). We can
start from the equation

• and reinterpret it.
• Nevents means the number of absorbed
  particles, i.e. the change in Nbeam, which we
  will call -DNbeam. The minus sign reminds us
  that Nbeam is decreasing.
• t in this equation means the thickness of the
  target. Now we should consider the number of
  absorbed particles in some layer of the target
  with a thickness Dt, where t describes the depth
  in the target.
• After all these changes we have

t
or with infinitesimal Dt
where we stressed the dependence of Nbeam on t.
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Expert pages! Its a challenge (but you dont
need to understand them)!
Instead of being frightened by solving this
differential equation remember which function is
almost identical to its derivative - surely you
remember the exponential function. It is only
necessary to fix the constant in it Assume and
put it into the differential equation
Clearly we need
We still have a free constant C in our solution.
But this solution should obey the initial
condition - at the surface of the
target at . So we at last arrive at the
result
We will play with this result for a while The
ratio tells us the
probability of the incoming particle to survive
path t in the target
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Expert pages! Its a challenge (but you dont
need to understand them)!
The probability of survival to depth t and of
interaction at this depth is
If the target is thick enough (infinite), the
beam should die out - the probability to interact
anywhere ( ?integral from 0 to infinity) should
be 1.
This was just a consistency check, that our
formulas correspond to common sense. What is
the mean interaction (absorption) length (let us
use a special symbol for it t)? Is it really
the average distance at which the beam particle
is absorbed? We can calculate an average depth of
interaction by weighting t with the probability
that a particle interacts at t after penetrating
into this depth t
With this result we can rewrite the previous
formulas in the nice form
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Expert pages! Its a challenge (but you dont
need to understand them)!
Lets continue our games. The exponential
function may be unfamiliar to some people - why
work with such a funny number like e? We can
replace e in the formula by any other number ...
We may introduce a half depth or a half
thickness t1/2 t ln2 instead of the mean
interaction length t - after passing t1/2 the
intensity of the beam is reduced to one half.
Surely you noticed that the formulas describing
the absorption of a particle beam by the target
matter are close to the formulas describing the
decay of unstable particles. We can switch
between these two meanings and learn from both t
is either the depth or time, t is either the mean
interaction length or the mean lifetime, t1/2 is
either the half-thickness or the half-life
There are more interesting questions, can you
answer them? 1. Assume that atoms (considered as
black disks) have a cross section for absorbing
visible light of about 10-19 m. How large is the
mean absorption length? Does it make sense? How
its possible to look out window? 2. How much is
a beam which passed 10 absorption lengths
reduced? 3. How many absorption lengths are
needed to kill the beam completely?
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If the ball is not caught it may hit me and
continue in flight in some direction. Even this
is important for a goalkeeper (a ball ends in the
net or outside). Let me imagine the (very
unpleasant) situation that the ball hits my head
and is deflected by an angle a from its original
direction
a
A less painful and more simple model for the same
situation is the elastic scattering of small
light balls (like peas) from a hard heavy sphere.
We may expect that the number of balls scattered
to any precise angle will be negligible. We
should look at some interval of angles Da
?a
I took off my glasses ...
a
a?a
Db
Unlike to my head, the hard sphere is completely
symmetric so the scattering from it will be
symmetric around this axis
b
b-Db
Balls flying further from the axis are deflected
less
Where is the cross section?
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The cross section was introduced as an effective
area of the flow of balls (or particles) causing
some effect. What is the effective area
corresponding to balls being scattered to the
interval of angles ?a? We will look at it from
the point of view of incoming balls ...
?a
Solid angle 2p sina Da
Db
All balls flying through this ring with the area
DS2pbDb will be scattered into the
angular interval ?a rotated around the axis, i.e.
to solid angle DW2p sina Da
b
Area 2pbDb
In this situation we speak about the differential
cross section
The cross section corresponding to particles
being scattered in all directions we will get by
integration
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To calculate the differential cross section we
need to know the relation between r and a. We
will try to derive this relation for the simple
case of a small ball elastically scattering off
a heavy hard sphere
Elastic scattering means that balls are scattered
from the surface at the same angle they hit it.
For the small green triangle and for angles a and
b we can write Eliminating b and keeping a
We have what we need!
b
a
b
R
b
b
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The minus sign reflects the fact that the angle a
decreases with growing r. We need only the
absolute values for calculating the cross
section. So we arrive at the result for the
differential cross section for elastic scattering
of a small ball on a big hard heavy sphere
Oh, we are so good!!!
Hurrah!!!
The cross section corresponding to particles
being scattered to all directions we will get by
the integration
This result corresponds to common sense - the
effective area of a hard sphere is pR2
Unbelievable! Its not completely wrong! The
differential cross section does not depend on
angle a. This means that scattered balls fly with
equal probability in any direction like light
radiates from an ideal (perfectly isotropic)
source. Calculating the differential cross
section in this model is an easy game, isnt it?
What about more complicated models?
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An interesting model could be a scattering of
charged particles on a heavy charged object. A
calculation a little bit more complicated than
what we have done up to now gives the result
For the derivation see the following expert pages
if you have enough courage ...
The differential cross section for this model is
different from that for the case of elastic
balls! It is not isotropic it is strongly
peaked in the forward direction!
Model predictions can be compared to experiment
and in such way we can learn which model is a
more appropriate description of the nature.
In this way we can even discover what is inside
the atoms, as we wanted from the begining of our
story. Ernest Rutherford and his collaborators
Geiger and Marsden used this method for the first
time almost a century ago.
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Expert pages! Its a challenge (but you dont
need to understand them)!
Lets try to calculate the differential cross
section for the scattering of a particle with
mass m, velocity v and charge ze on an infinitely
heavy target with charge Ze (e is the elementary
charge). The scattering is almost the same as the
case of hard spheres, only the particles
interact, not in contact, but at a distance
according to the Coulomb law
The path of the incoming particle is curved - it
is a hyperbola. In the far past it is close to
the red line at the distance b from the center
of the target. In the far future the path is
close to the line at the angle a.
The first step in the calculation could be to
look at energy and momentum conservation lets
try that and see where it goes. The mass of the
target is infinite, so it will absorb any
momentum without moving and will not carry away
any energy (a good exercise prove it!). The
energy of the incoming particle is therefore
conserved but the momentum is changed
ze
a
a
r
b
The momentum change is caused by the Coulomb
force
Coulomb force
Ze
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Expert pages! Its a challenge (but you dont
need to understand them)!
There are vectors in this equation, but we can
avoid them. The easiest way is to project the
equation onto the direction of
a
The angle ? changes from -(p-a)/2 at t-? to
(p-a)/2 at t?.
r
What about using the angle ? to parameterize the
path instead of time t ? To do this we only need
to change the integration variable from t to ?.
What is d? /dt ? angular velocity w! And
moreover, the angular momentum Lmr2w is
conserved for a particle in the central field! We
can also calculate the angular momentum from the
initial condition Lmvb (angular momentum
momentumradius).
Now we are ready to put everything together...
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Expert pages! Its a challenge (but you dont
need to understand them)!
To calculate the differential cross section we
need b(a) Then as before
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Expert pages! Its a challenge (but you dont
need to understand them)!
Can one guess the result? Does it correspond to a
common sense? Do we understand the formulae
qualitatively?
Thanks to Rupert Leitner for stimulating this page
We start with the relation between the distance
b (called the impact parameter) and the
scattering angle a.
Since the Coulomb interaction has the infinite
range, b can be infinite and still some
(infinitely small) deflection happens.
A particle flying closer to the target (with
smaller b) will suffer stronger Coulombic
repulsion and will be deflected more than a
particle flying farther away from the target.
The limiting case is b 0 when the particle
clearly bounces back (a p).
For simplicity and beauty we may expect a smooth
function connecting limiting regions ... Which
function it could be?
Dealing with angles makes us think about
trigonometric functions for b (a). Which of them
touches infinity? Which of them grows to
infinity at an angle of zero? At which angle
does it go through zero? How can we modify the
argument to accommodate this?
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Expert pages! Its a challenge (but you dont
need to understand them)!
To work with this formula quantitatively, lets
consider a-gold scattering
Instead of substituting here the mass and
velocity of the incoming a, we may use its
kinetic energy mv2 2 Ek A typical kinetic
energy of a from an radioactive source is about
1 MeV 1,6010-13 J.
Charges of scattered objects ZAu 79 za 2
Universal constants e 1,6010-19 C e0
8,8510-12 Fm-1
What is the probability that a particle will be
deflected by more than 1 degree?
Substitution
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Expert pages! Its a challenge (but you dont
need to understand them)!
What is the distance b (experts speak about the
impact parameter) corresponding to the deflection
angle 1?
These quantitative results have the meaning which
we will discuss further.
Experts need some more training to get a working
knowledge of the subject. Here are some
additional tasks or questions 1. Could you prove
the statements on energy and momentum balance in
the case of the scattering of a light particle
on an infinitely heavy target? 2. Could you
explain, why angular momentum is conserved in a
central field? 3. Could you translate the
calculated value of the differential cross
section for scattering into angles greater than
1 into an effective area, and compare this area
to the geometrical cross section of a sphere
with a diameter of 10-10 m? 4. Could you repeat
the calculation and find the cross section for
particles being deflected to angle less than
1? 5. We have got the result for a point-like
central charge (or one within a sphere less than
b). What would happen if b were smaller than the
radius of the central charge (If the projectile
were aimed at the target better) ?
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Rutherfords experiments scattering a particles
on a thin gold foil came at a time when the atom
was considered (due to J. J. Thomson) as
something like a plum pudding of positively
charged matter with grains of negatively charged
light electrons.
2e
- 6e
a (He)
Thomsons view of carbon 6 negative light
electrons and a massive matter with a positive
charge of 6 units distributed more or less
uniformly over the volume of the atom. The
whole atom is neutral and has a diameter of
about 10-10 m.
Rutherford realized that a particles (helium atom
without electrons) were slightly deflected when
passed through materials. This was rather
strange from the point of view of Thomsons model
and inspired further study he and Geiger did
their famous experiment ...
The calculation (if interested, see the expert
pages) shows that to get some non-negligible
deflection you need a big enough charge and a
small enough distance of flight to the
scattering center the a particle should be
aimed at a distance of 1/10 of the atomic radius
and feel the whole positive charge of a gold
atom to be deflected by 1. But in the Thomson
model the positive charge is spread over the
whole volume of the atom! If the a particle
passes through the distributed charge, it is
deflected much less! How it could be?
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An experiment carried out by Geiger (1911) and
Geiger Marsden (1913)
1. The apparatus consisted in essence of a
strong cylindrical metal box B
3. the gold scattering foil F
2. which contained a source of a particles R.
4. and a zinc sulphide screen S attached
rigidly to a microscope M
5. The source of a particles was an a ray tube
filled with radon. A narrow beam of a particles
from the source R was directed through the
diaphragm D to fall normally on the scattering
foil F.
6. The box was fixed to a graduated circular
platform A, which could be rotated in the
airtight joint C.
7. The microscope and zinc sulphide screen
rotated with the box, while the gold foil and
source remained fixed.
8. The box was closed by a ground glass plate
P and could be evacuated through the tube T.
9. By rotating the platform A the a particles
scattered in different directions could be
observed on the zinc sulphide screen.
Observations were taken for scattering
angles between 5 and 150 and both silver and
gold foils were used as scattering material. Two
sets of experiments were carried out, the first
comparing angles from 15 to 150 and the second
angles from 5 to 30.
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Most a-particles scatter at small angles but
there were also large deflections visible in the
experiment. Rutherford suggested a model with
small heavy nucleus and electrons around. Look at
the comparison of the data by Geiger Marsden
with the model
Number of counts
The gold data and prediction compared
Number of counts
Angle
Angle
An excerpt from Radiations from Radioactive
Substances by Sir Ernest Rutherford, James
Chadwick and C. D. Ellis, published by Cambridge
University Pres,s, 1930
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Ernest Rutherford
and his atom
Hans Geiger
A lot of empty space

Light negative electrons
Positively charged nucleus with almost all mass
of the atom.
The data agreed well with the Rutherford model.
During many subsequent years physicists
discovered how electrons behave in atoms and what
the nucleus is composed of and All
experiments performed until now agree well with
Rutherfords model. As usual in physics, we
speak frequently about reality (atom is composed
from the nucleus and ) and not about the best
available model. Please, remember what we have
done since we first discussed the idea of a
scattering experiment. Later we will meet more
variants of this type of experiment.
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To be continued