Particle physics experiment

1. Jiří 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 asufficiently small screwdriver and we even cannot look at it with sufficient resolution. Particle physics experiment 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.

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 a-particles towards a thin gold foil and discovered that what best corresponds to the experimental results is the idea of atomsbeing 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. 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

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 andm2, the velocities v1 andv2 before the collision and the velocities v1’ andv2’after the collision. Important variables are energyE 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

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 ofp2 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:

6. The last equation is a quadratic equation for The discriminant is and its solutions are

7. Let‘s remember that we are looking for real non-negative solutions. The simplest case is the case of equal massesm1 = 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: 1…projectile 2…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 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:

9. 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 ...

10. 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 ofE and above alwaysequals 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 themass 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 ...

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 sfor 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.

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. 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 catchbest Energy of the ball in apropriate units The cross-sections for proton-proton interactions are displayed on this plot: Elastic crosssection - 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

15. Usuallybunches 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 areNtarget 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 areNbeam 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 ...

16. 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 - Nbeamdepends on the depth in the target. target beam t t measures the depth in the target

17. Expert pages! It‘s a challenge (but you don´t need to understand them)! Let‘s 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 inNbeam, which we will call -DNbeam. The minus sign reminds usthat 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 Nbeamont.

18. Expert pages! It‘s a challenge (but you don´t 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:

19. Expert pages! It‘s a challenge (but you don´t need to understand them)! The probability of survival to deptht 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 reallythe 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 tafter penetrating into this deptht: With this result we can rewrite the previous formulas in the nice form

20. Expert pages! It‘s a challenge (but you don´t need to understand them)! 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 ... 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 it‘s possible to look out window? 2. How much is a beamwhich passed 10 absorption lengths reduced? 3. How many absorption lengths are needed to kill the beam completely?

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 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?

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 a? … We willlook 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 areaDS=2pbDb will be scattered into the angular interval arotated around the axis, i.e. to solid angle DW=2p 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:

23. 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

24. 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! It‘s 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, isn´t it? What about more complicated models?

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 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.

26. Expert pages! It‘s a challenge (but you don´t need to understand them)! 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: 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 – 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: ze a a r b The momentum change is caused by the Coulomb force: Coulomb force Ze

27. Expert pages! It‘s a challenge (but you don´t 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 L=mr2w 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). Now we are ready to put everything together...

28. Expert pages! It‘s a challenge (but you don´t need to understand them)! To calculate the differential cross section we need b(a): Then as before

29. Expert pages! It‘s a challenge (but you don´t 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 smallerb) will suffer stronger Coulombic repulsion and will be deflected more than a particle flying farther awayfrom 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?

30. Expert pages! It‘s a challenge (but you don´t need to understand them)! To work with this formula quantitatively, let‘s 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 = 1,60×10-13 J. Universal constants e = 1,60×10-19 C e0 = 8,85×10-12 Fm-1 Charges of scattered objects ZAu = 79 za = 2 What is the probability that a particle will be deflected by more than 1 degree? Substitution

31. Expert pages! It‘s a challenge (but you don´t 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) ?

32. Rutherford´s 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 chargedmatter with grains of negatively charged light electrons. +2e - 6e a (He++) Thomson´s view of carbon: 6 negative light electrons and amassive 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 Thomson´s 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 needa 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?

33. 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 graduatedcircular 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 foiland 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°.

34. 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

35. 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 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.

36. To be continued