1

Particle physics experiment

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.

Jiří Dolejší, Olga Kotrbová, Charles University in Prague

2

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 -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? Let‘s 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…

3

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.

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).

A

BC

Disclaimer: The collisions of a

euro and a dollar have neither

political nor economic meaning.

4

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

2121

2121

pppp

EEEE

m

pmvE

22

1 22 mEmvp 2

Both energy and momentum in an isolated system are conserved:

v1

The red coin is at rest at the beginning

v1'

m1

m2

v2= 0 v2'

.p

5

Let‘s work only with momenta:energy conservation... … and conservation of both momentum components

Let‘s play with momenta: We would like to get rid of p2 and and it is appealing to use for that purpose the relation

we can insert into energy equation:

2

22

1

21

1

21

222 m

p

m

p

m

p

sinsin0

coscos

21

211

pp

ppp

sinsin

coscos

21

211

pp

ppp

1sincos 22

We just rewrite the equations, square each one and sum them:

22

22

1

222

211

sinsin

coscos

pp

ppp

222

222

221

22111

21 sincossincoscos2 ppppppp

22p

22

2111

21 cos2 ppppp

2111

21

21

21

1

2 cos2 ppppppm

m

6

The discriminant is

is a quadratic equation for

and its solutions are

21p

.01cos211

22111

1

221

m

mppp

m

mp

2

1

2221

1

2

1

221

221 1cos4114cos4

m

mp

m

m

m

mppD

.1

1coscos

12

1cos2cos2

1

2

2

1

22

1

1

2

2

1

2211

1

mm

mm

p

mm

mm

pp

p

2111

21

21

21

1

2 cos2 ppppppm

m The last equation

7

Let‘s remember that we are looking for real non-negative solutions. The simplest case is the case of equal masses m1 = m2:

0or 90 for cos 2

coscos11

2

11

pppp

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:

0

0,2

0,4

0,6

0,8

1

1,2

0 30 60 90 120 150 180

1

1

p

p

21 mm 21 2mm

221

1 mm

1…projectile2…target

Only lighter particle can be scattered backwards

The deflection of the heavier particle is limited

8

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 m

pmvE

22

1 22

we should deal with total energy

.2242 cpcmE

We will insert into the energy conservation equation (all masses here are rest masses):

22p

222

222

221

421

22

221

421 cpcmcpcmcmcpcm

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:

0or

)(2sin

1

cos1

1

221

11

p

mEmp

pp

9

The difference between the non-relativistic and the relativistic calculation is visible on the graph:

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 ...

0

0,2

0,4

0,6

0,8

1

1,2

0 30 60 90

102

1 mc

E

10002

1 mc

E

Nonrelativistic case

1

1

p

p

Protons are accelerated to energiesmore than 1000 times larger than their rest energiesat Fermilab (USA)

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Unstable particle with unknown mass

The formula for relativistic energy

can be rewritten in the form

2242 cpcmE

.42222 cmcpE

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:

pE

,

11 , pE

22 , pE

21 EEE

21 ppp

Measure energies and momenta of decay products and ,then calculate

11 , pE

22 , pE

422221

221 )()( cMcppEE

M

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 ...

p

11

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.

12

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 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 is .jN

The standard unit for the cross section is 1 barn = 1 b = 10-28 m2.

13

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.

14

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 …)

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

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 = total - elastic

inelastic

0.5

1.0

1.5

2.0

0

2.5

Energy of the ball in apropriate units

ball-me

total

catch

hit injury deathAt this energy I catch best

At this energy I try to avoid getting hit by the ball

At this energy I am frozen from the fear

[m2]

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Every target catcherhas a cross section .There are Ntarget of them subjected to beam ofprojectiles. We assume that the individual catchersdo not collaborate anddo not shield others.So the number of catchesis Nevents = j .. Ntarget

Usually bunches of projectiles are scattered of many target particles. The situation may look like in our picture ...

beam target

Here are Nbeam incoming particles within a beam with the area Sbeam. The flux of incoming particles is then jbeam = Nbeam/ Sbeam

Nbeam

Sbeam

tnNt

t

S

NNN

S

NNjN targetbeam

beam

targetbeamtarget

beam

beamtargetbeamevents .........

Reality can be always looked at from different viewpoints ...

Density of target particles

t

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We arrived at an expression which says that the probability of an incoming particle interacting is

..ninteractio ofy probabilit targetbeam

events tnN

N

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.

beam target

t)0(beamN

)(beam tN

t measures the depth in the target

)(beam tN

t

)exp().0()( targetbeambeam tnNtN

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Let‘s try to find the dependence Nbeam(t). We can start from the equation

Expert pages! It‘s a challenge (but you don´t need to understand them)!

tnNN targetbeamevents ..and reinterpret it.

• Nevents means the number of absorbed particles, i.e. the change in Nbeam, which we

will call -Nbeam. 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 t, where t describes the depth in the target.

After all these changes we have tnNN targetbeambeam ..

or with infinitesimal t

targetbeambeam

targetbeambeam

.).(d

)(d

d..d

ntNt

tN

tnNN

where we stressed the dependence of Nbeam on t.

t

)0(beamN )(beam tN beamN

t

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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 ).exp(.)(beam tDCtN

targetbeambeam .).(d

)(dntN

t

tN

..).exp()exp(. targetnDtCDtCD Clearly we need .. targetnD

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

)..exp()0()( targetbeambeam tnNtN

)0(beamN 0t

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:

)0(/)( beambeam NtN

)..exp()0(

)()(survive y toprobabilit target

beam

beam tnN

tNt

Expert pages! It‘s a challenge (but you don´t need to understand them)!

19

The probability of survival to depth t and of interaction at this depth is

.d.)..exp(d).at n interactio ofty (probabili targettarget tntntt

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.

.1d)exp(d.)..exp(d).at n interactio ofty (probabili00

targettarget

0

xxtntntt

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 – )? 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:

0target0

targettarget

0

d)exp(.

1d.)..exp(.d).at n interactio ofty (probabili. xxx

ntntntttt

target.

1

n With this result we can rewrite the

previous formulas in the nice form )exp()0()( beambeam t

NtN

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20

Let‘s 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 ...

.2

1)0()

2ln

2lnexp()0()exp()0()(

2ln

beambeambeambeam

t

Nt

Nt

NtN

We may introduce a “half depth” or a “half thickness” t1/2 = ln2 instead of the mean interaction length - 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, 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 it‘s 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?

Expert pages! It‘s a challenge (but you don´t need to understand them)!

21

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 from its original direction:

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

b b-b

b

Balls flying further from the axis are deflected less

Unlike to my head, the hard sphere is completely symmetric so the scattering from it will be symmetric around this axis

I took off my glasses ...

Where is the cross section?

22

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 ? … We willlook at it from the point of view of incoming balls ...

In this situation we speak about the differential cross section:

b

Area =bb

All balls flying through this ring

with the area S=bb will be scattered into the angular

interval rotated around the axis,

i.e. to solid angle =sin

b

Solid angle =sin

.dsin2

d2

d

d

bb

The cross section corresponding to particles being scattered in all directions we will get by integration:

.dsin2d

d

0

23

b

To calculate the differential cross section we need to know the relation between r and We will try to derive this relation for the simple case of a small ball elastically scattering off a heavy hard sphere:

Rb

For the small green triangle and for angles and we can write

Eliminating and keeping

We have what we need!

.2,sin R

b

.2

cos2

sin

RRb

.4sin4

2cos

2sin2

2

1

2sin

sin2

cos

d

d

sindsin2

d2 22 R

RRRbbbb

Elastic scattering means that ballsare scattered from the surfaceat the same angle they hit it.

d

d

24

4d

d 2R

The minus sign reflects the fact that the angle 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

The cross section corresponding to particles being scattered to all directions we will get by the integration:

.cos2

dsin2

dsin2d

d 20

2

0

2

0

RRR

This result corresponds to common sense - the effective area of a hard sphere is R2… Unbelievable! It‘s not completely wrong!The differential cross section does not depend on angle 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, isn´t it? What about more complicated models?

Oh, we are so good!!!

Hurrah!!!

25

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

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.

.

2sin

1

8 4

2

20

2

mv

Zze

d

d For the derivation see the following expert pages if you have enough courage ...

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Let‘s 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:

b

Ze

ze

Coulomb force

2

2

04

1)(

r

ZzerF

The path of the incoming particleis 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 .

r

The first step in the calculation could be to look at energy and momentum conservation – let‘s 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:

vmp

vmp

p

2sin2

mvp

The momentum change is caused by the Coulomb force:

.)(

trFp d

Expert pages! It‘s a challenge (but you don´t need to understand them)!

27

The anglechanges from -(-)/2 at t=- to (-)/2 at t=+.

r

There are vectors in this equation, but we can avoid them. The easiest way is to project the equation onto the direction of p

trFp d)(

.dcos)(

trFp

What about using the angleto parameterize the path instead of time t ? To do this we only need to change the integration variable from t to

2

2 dd

dcos))((d)(cos))((

t

rFtttrFp

What is d /dt ? … angular velocity ! And moreover, the angular momentum L=mr2 is conserved for a particle in the central field! We can also calculate the angular momentum from the initial condition L=mvb (angular momentum = momentum×radius).

mvbmrL 2 2r

vb

Now we are ready to put everything together...

Expert pages! It‘s a challenge (but you don´t need to understand them)!

28

2

2 2

20

2 dcos

1

42sin2

rvbr

Zzemv

To calculate the differential cross section we need b():Then as before

2

2

04

1)(

r

ZzerF

2d

d

r

vb

t2

sin2

mvp

2

2 dd

dcos))(cos((

t

rFp

2cos

4dcos

82sin

20

22

2

20

2

mv

Zze

mv

Zzeb

2

cos22

sin2sindcos2

2

2

2

.24 2

0

2

cotanmv

Zzeb

.

2sin2

2cotan

2cos

2sin2

1

42

sin2

2cotan

sin

1

4d

d

sind

d

2

2

20

2

2

2

20

2

mv

Zze

mv

Zzebb

2sin

1

8d

d

4

2

20

2

mv

Zze

Expert pages! It‘s a challenge (but you don´t need to understand them)!

29

Dealing with angles makes us think about trigonometric functions for b (). 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?

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 ( = ).

2cotan.const

b

Can one guess the result? Does it correspond to a “common sense“? Do we understand the formulae qualitatively?

Since the Coulomb interaction has the infinite range, b can be infinite and still some (infinitely small) deflection happens.

We start with the relation between the distance b (called the impact parameter) and the scattering angle .

angle scattering)(180

00

bparameter impact

For simplicity and beauty we may expect a smooth function connecting limiting regions ...Which function it could be?

Thanks to Rupert Leitner for stimulating this page

Expert pages! It‘s a challenge (but you don´t need to understand them)!

30

Charges of scattered objectsZAu = 79z= 2

To work with this formula quantitatively, let‘s consider -gold scattering:

Universal constantse = 1,60×10-19 C0 = 8,85×10-12 Fm-1

Instead of substituting here the mass and velocity of the incoming , we may use its kinetic energy … mv2 =2 Ek

A typical kinetic energy of from an radioactive source is about 1 MeV = 1,60×10-13 J.

What is the probability that particle will be deflected by more than 1 degree?

180

1 3

2180

1 4

2180

1 4

21

2sin

2d

2c

...8

2sin

d2

c2

sin2...2

2sin

dsin2...

ososa

22290sin

5.0sin2

290sin

5.0sin3

2 m1033.52

1...8

d...8

xx

x

2sin

m1023.3

2sin

1

8d

d

4

227

4

2

20

2

mv

Zze

Substitution x2

sin

Expert pages! It‘s a challenge (but you don´t need to understand them)!

31

What is the distance b (experts speak about the impact parameter) corresponding to the deflection angle 1°?

m 101.30m0.5cotan .1014.12

cotan4

11-132

0

2

mv

Zzeb

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) ?

Expert pages! It‘s a challenge (but you don´t need to understand them)!

32

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 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 particle passes through the distributed charge, it is deflected much less! How it could be?

Rutherford´s experiments scattering 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.

Thomson´s view of carbon: 6 negative light electrons and a massive matter with a positive charge of 6 units distributedmore 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 particles (helium atom without electrons) were slightly deflected when passed through materials. This was rather strange from the point of view of Thomson´s model and inspired further study … he and Geiger did their famous experiment ...

+2e- 6e

(He++)

33

An experiment carried out by Geiger (1911) and Geiger & Marsden (1913)

9. By rotating the platform A the particles scattered in different directions could be observed on the zinc sulphide screen. Observations were taken for scattering anglesbetween 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°.

1. The apparatus consisted in essence of a strong cylindrical metal box B

2. … which contained a source of particles R.

3. … the gold scattering foil F

4. … and a zinc sulphide screen Sattached rigidly to a microscope M

6. The box was fixed to a graduated circular platform A, which could be rotatedin the airtight joint C.

7. The microscope and zinc sulphide screen rotated with the box, while the goldfoil and source remained fixed.

8. The box was closed by a ground glass plate Pand could be evacuated through the tube T.

5. The source of particles was an ray tube filled with radon. A narrow beam of particles from the source R was directed through the diaphragm D to fall normally on the scattering foil F.

34

An excerpt from Radiations from Radioactive Substances by Sir Ernest Rutherford, James Chadwick and C. D. Ellis, published by Cambridge University Pres,s, 1930

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:

10

100

1000

10000

100000

1000000

0 30 60 90 120 150 180

1

10

100

1000

0 10 20 30

The gold data and prediction compared

Angle

Angle

Number of counts

Number of counts

35

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 Rutherford´s 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.

+

Light negative electrons

Positively charged nucleus with almostall mass of the atom.

A lot ofemptyspace

Ernest RutherfordHans Geiger and his atom

36

To be continued