Option J: Particle physics J2 Particle accelerators and detectors

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Presentation transcript:

Option J: Particle physics J2 Particle accelerators and detectors J.2.1 Explain the need for high energies in order to produce particles of large mass. J.2.2 Explain the need for high energies in order to resolve particles of small size. J.2.3 Outline the structure and operation of a linear accelerator and of a cyclotron. J.2.4 Outline the structure and explain the operation of the synchrotron. J.2.5 State what is meant by bremsstrahlung (braking) radiation. J.2.6 Compare the advantages and disadvantages of linear accelerators, cyclotrons and synchrotrons. J.2.7 Solve problems related to the production of particles in accelerators. © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Explain the need for high energies in order to produce particles of large mass. ●Recalling the mass-energy relationship we see that from E = mc2 we obtain ●Note that the bigger the mass of the particle we desire to produce, the larger E must be. mass-energy equivalence m = E/c2 EXAMPLE: Suppose you want to create a proton from energy. How much energy must you provide? SOLUTION: Particle creation must adhere to conservation laws. ●Since the baryon number of a proton is +1, and the baryon number of a photon is zero, an anti-proton must also be created. ●Thus we need at least E = 2(938 MeV) = 1876 MeV. © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Explain the need for high energies in order to produce particles of large mass. PRACTICE: Find the energy needed to create an electron. Then show that conservation of charge and conservation of lepton number are both satisfied. SOLUTION: ●The rest energy of an electron is 0.511 MeV. ●Thus we need E = 2(0.511 MeV) = 1.022 MeV to create an electron, anti-electron pair. © 2006 By Timothy K. Lund 2  e- + e+ CHARGE LEPTON NUMBER 0 -1 +1 0 +1 -1 FYI The minimum energy of any accelerator must be twice the rest energy of the heaviest particle it is designed to detect/create.

Option J: Particle physics J2 Particle accelerators and detectors Explain the need for high energies in order to produce particles of large mass. ●You may be asking yourself: “How do you get high energy photons?” ●You can “harvest” high energy photons from two sources: 1) The energy that is released by the annihilation of matter/anti-matter particles (that were created elsewhere). 2) The KINETIC ENERGY of the particles just before they collide. © 2006 By Timothy K. Lund FYI To maximize the amount of EK converted to the photons’ energy the particle and anti-particle are made to travel in opposite directions.

Option J: Particle physics J2 Particle accelerators and detectors Explain the need for high energies in order to produce particles of large mass. PRACTICE: Accelerators use E-fields and B-fields to control the speeds and paths of the charged particle beams they accelerate. Explain how this is advantageous when using matter, anti-matter beams traveling in opposite directions. SOLUTION: ●Matter and antimatter have opposite charges. ●Recall: FB = qvB sin , and FE = qE. ●Thus the directions of the magnetic force (given by the right hand rule) and the electric force depend on the signs of the charges the forces are acting on. ●Since matter and antimatter have opposite signs, the same fields will control both beams at once! © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Explain the need for high energies in order to resolve particles of small size. ●Recall that the wavelength  of a photon of energy E is given by E = hf = hc/. ●Clearly, the bigger the energy E the smaller the wavelength . ●Recall also that the smaller the wavelength, the better the resolution. ●Thus, the bigger the energy, the smaller the particle that can be resolved. wavelength – energy relationship  = hc/E © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Outline the structure and operation of a linear accelerator and of a cyclotron. ●As the name implies, a linear accelerator or linac is a straight tube. ●The longer it is, the more energy the accelerated particle will have when it reaches the end of the linac, at which point it smashes into a target at the end of the tube. ~ © 2006 By Timothy K. Lund FYI The accelerated particle is charged. The alternating p.d. must be timed so that the charge is repelled from the “behind” tube and attracted to the “in-front” tube.

Option J: Particle physics J2 Particle accelerators and detectors Outline the structure and operation of a linear accelerator and of a cyclotron. PRACTICE: Explain why the successive tubes increase in length in the linac. SOLUTION: ●Since the particle is accelerating, the distance covered in a fixed time interval increases. ●The tubes are designed so that the frequency of the alternating p.d. can remain fixed. ~ © 2006 By Timothy K. Lund FYI An alternate design might vary the frequency and keep all of the tube lengths equal.

Option J: Particle physics J2 Particle accelerators and detectors Outline the structure and operation of a linear accelerator and of a cyclotron. ●The cyclotron accelerates particles within two D-shaped hollow containers, like the ones shown here. ●Each D is connected to opposite terminals of an alternating p.d., which accelerates the charged particle from the center. ●A perpendicular magnetic field causes the particles to followed a curved spiral until reaching the outside circumference of the D’s. ●As the following slide will show, the period T of the alternating p.d. is independent of the radius of the particle’s trajectory… ~ + B © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Outline the structure and operation of a linear accelerator and of a cyclotron. ●In order for the particle of mass m and charge q to experience a centripetal force, it must satisfy F = mac. ●The force F is caused by the magnetic force F = qvB. ●But the centripetal acceleration ac = v2/r. ●Therefore F = qvB = mac so that qvB = mv2/r, or qB = mv/r = m(2r/T)/r = 2m/T. + ~ B © 2006 By Timothy K. Lund period of a cyclotron T = 2m/(qB) FYI Note that f (= 1/T) does not depend on the radius of the path. It depends only on m,q and B.

Option J: Particle physics J2 Particle accelerators and detectors Outline the structure and operation of a linear accelerator and of a cyclotron. ●To get an idea of the energy capabilities of a cyclotron consider the intermediate step from the previous slide: ●From qB = mv/r we get v = qBr/m so that EK = (1/2)mv2 = (1/2)m(qBr/m)2. + ~ B © 2006 By Timothy K. Lund energy of a cyclotron EK = q2B2r2/(2m) FYI The energy capabilities of the cyclotron are proportional to the square of its radius. Since creating large disks of vacuum within the D’s is very difficult, the largest cyclotrons are only able to produce energies in the MeV range.

Option J: Particle physics J2 Particle accelerators and detectors Outline the structure and explain the operation of the synchrotron. ●The synchrotron is similar to a linac in that acceleration occurs between tubes. ●The tubes in a synchrotron, however, are all the same length, and they are arranged in a circle. ●The charges are accelerated between the tubes by an electric field applied across two plates. ●The tubes are surrounded by a magnetic field that causes the charged particles to follow a curve. ~ electric field accelerates beam © 2006 By Timothy K. Lund magnetic field bends beam

Option J: Particle physics J2 Particle accelerators and detectors Outline the structure and explain the operation of the synchrotron. ●The alternating p.d. units are complex. They must be synchronized precisely to the accelerating particles (hence the name synchrotron). ●The magnetic field strength must also vary as the particles pick up speed. ●The advantage of a synchrotron over a linear accelerator is that the particle can make as many circuits as needed to accelerate it to any energy. ~ magnetic field bends beam electric field accelerates beam © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Outline the structure and explain the operation of the synchrotron. ●Another advantage of the synchrotron over either of the previous accelerators is that you can have a matter beam, and an antimatter beam, circulating at the same time in opposite directions. ●As we have already discussed, head-on collisions maximize the energy output of the collisions. proton antiproton © 2006 By Timothy K. Lund Fermilab Batavia, Illinois.

Option J: Particle physics J2 Particle accelerators and detectors Outline the structure and explain the operation of the synchrotron. ●The synchrotron’s maximum energy capability is limited by its curvature. The faster a particle travels, the larger the centripetal forces the magnetic field needs to provide. PLUS due to relativistic effects, the faster a particle goes, the more massive it becomes-thus adding to the difficulty of making it turn in a circle. ●The largest synchrotron in the world is at CERN, in Geneva, and has a circumference of 27 km. © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors State what is meant by bremsstrahlung (braking) radiation. ●When a charge is accelerated, it produces electromagnetic radiation called bremsstrahlung radiation. ●Given that an acceleration is a change in velocity and that velocity can change in its magnitude (across the plates) or its direction (the magnetic field tubes) we get radiation at both locations in the synchrotron. ●Since the radius of curvature is so large, the radiation from the tubes is relatively small compared to that of the plates. © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors State what is meant by bremsstrahlung (braking) radiation. ●The radiation at the plates is in the X-ray region of the spectrum, and is highly polarized. ●Because the X-ray radiation is very intense, very parallel, and very polarized, it is well-suited for X-ray diffraction and other experiments investigating the properties of materials. © 2006 By Timothy K. Lund FYI In fact, there are some synchrotrons that have been built solely for X-ray diffraction and materials physics.

Option J: Particle physics J2 Particle accelerators and detectors Compare the advantages and disadvantages of linear accelerators, cyclotrons and synchrotrons. ●Since there are no curves in the linear accelerator, there is no radiation loss due to direction change as there is in the synchrotron. ●In the linac and the cyclotron, if you miss the collision at the end of the run, the projectiles are lost, whereas in the synchrotron the particles can continue to go around as many times as is needed to effect a collision. ●The difficulty in constructing large enough evacuated D’s and large-area magnetic fields prevents the cyclotron from being a serious contender in very high energy physics research. ●The cyclotron is the simplest to construct for low-energy applications. © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Solve problems related to the production of particles in accelerators. ●If a particle and an antiparticle collide head-on then we not only get the energy of annihilation (the rest mass energy of the two particles) but we harvest all of the original kinetic energy of the particles before their collision. ●If a particle collides with a stationary target not all of the energy can be converted to new particles. This is because to conserve momentum the new particles must move in the same direction as the original beam. ●Thus the energy needed to produce particles striking a stationary target must exceed the actual rest mass energy of the particles created. © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Solve problems related to the production of particles in accelerators. ●The minimum available energy Ea for particle creation obtained by the collision of a particle-projectile of rest mass m and total energy E with a stationary target of mass M is shown here: energy available in collision of a moving mass m with stationary mass M Ea2 = 2Mc2E + (Mc2)2 +(mc2)2 M = target rest mass. m = projectile rest mass. E = projectile total energy. © 2006 By Timothy K. Lund FYI Notice that each factor has the units of [energy]2. Don’t miss a square! Anywhere. In general, the numbers are more manageable if we keep the energies in MeV (or even GeV)…

Option J: Particle physics J2 Particle accelerators and detectors Solve problems related to the production of particles in accelerators. energy available in collision of a moving mass m with stationary mass M Ea2 = 2Mc2E + (Mc2)2 +(mc2)2 M = target rest mass. m = projectile rest mass. E = projectile total energy. EXAMPLE: How much energy would be available if a 15 GeV proton collided with a stationary proton? SOLUTION: ●m = M = 938 MeV c-2 = 0.938 GeVc-2. E = 15 GeV. Ea2 = 2Mc2E + (Mc2)2 +(mc2)2 Ea2 = 2(0.938)(15) + [0.938]2 +(0.938)2 [GeV]2 Ea2 = 29.899688 [GeV]2 Ea = 5.5 GeV. © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Solve problems related to the production of particles in accelerators. © 2006 By Timothy K. Lund ●The E-field accelerates the charged particles. ●The B-field makes the charged particles turn and follow a circular trajectory.

Option J: Particle physics J2 Particle accelerators and detectors Solve problems related to the production of particles in accelerators. ●As v increases so does m, and the centripetal force Fc = mac = mv2/r must also increase to keep the particles moving in the synchrotron’s circular path. © 2006 By Timothy K. Lund ●Since FB = qvB = Fc, mv2/r = qvB  qB = mv/r. ●Thus r = mv/qB. ●Since r = CONST for a synchrotron, B must increase with mv as the particle energy increases.

Option J: Particle physics J2 Particle accelerators and detectors Solve problems related to the production of particles in accelerators. © 2006 By Timothy K. Lund ●Each beam should supply E = 1120 MeV (half of 2240 MeV) in kinetic energy. ●We can use E = m0c2 + EK because the annihilation provides some energy. ●EK = E - m0c2 = 1120 – 938 = 182 MeV.

Option J: Particle physics J2 Particle accelerators and detectors Solve problems related to the production of particles in accelerators. © 2006 By Timothy K. Lund ●For a stationary target use Ea2 = 2Mc2E + (Mc2)2 + (mc2)2, where Ea = 2240, M = 938, and m = 938. Then 22402 = 2(938)E + (938)2 + (938)2 E = 1740 MeV (E is total energy). ●EK = E - m0c2 = 1740 – 938 = 802 MeV.

Option J: Particle physics J2 Particle accelerators and detectors Solve problems related to the production of particles in accelerators. ●EK = 182 MeV for proton, anti-proton head-on collision. ●EK = 802 MeV for anti-proton hitting stationary proton target. © 2006 By Timothy K. Lund ●This is 802/182 = 4.4 times more energy. ●Thus you only need about a quarter of the energy to produce the ,  pair using the head-on synchrotron collision.

Option J: Particle physics J2 Particle accelerators and detectors cyclotron linac detail © 2006 By Timothy K. Lund synchrotron

Option J: Particle physics J2 Particle accelerators and detectors Particle detectors J.2.8 Outline the structure and operation of a bubble chamber, the photomultiplier and the wire chamber. J.2.9 Outline international aspects of research into high-energy particle physics. J.2.10 Discuss the economic and ethical implications of high-energy particle physics research. © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Particle detectors Outline the structure and operation of a bubble chamber, a photomultiplier and a wire chamber. ●The particles created in collisions must somehow be detected. The bubble chamber was one of the first detectors for such particles. ●When you pop the top of a soda, pressure is suddenly released with a fizzing sound. ●If pressure is released the liquid reaches its boiling point, and bubbles form. In the case of soda, the bubbles are CO2. ●Each bubble forms at the site of an impurity - if there were no impurities, bubbles would not form automatically. ●Instead of water and CO2, a bubble chamber uses hydrogen that has been cooled to the liquid state just below its boiling point. © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Particle detectors Outline the structure and operation of a bubble chamber, a photomultiplier and a wire chamber. ●The bubble chamber consists of a liquid-hydrogen-filled cylinder having one of its faces made of glass. ●Incoming particles strike the target and produce new particles. ●The piston moves outward, lowering the liquid pressure. ●Bubbles form in the depres- surized liquid hydrogen along the path of the particles. ●A magnetic field passing through the bubble chamber ensures that the particles will have curved trajectories if they are charged. Light Camera © 2006 By Timothy K. Lund Piston Stationary Target

Option J: Particle physics J2 Particle accelerators and detectors Particle detectors Outline the structure and operation of a bubble chamber, a photomultiplier and a wire chamber. ●Some of the products of collision experiments are gamma particles, which are of course high-energy photons. ●Gamma particles are very adept at ionizing matter, but the act of ionization absorbs the photon, removing it from the picture. ●Detection of a single photon being absorbed is difficult, but it is made possible by a device called a photomultiplier. ●A single photon enters a photomultiplier through a small window and is absorbed by a photosensitive plate, releasing an electron according to the photoelectric effect. photo-sensitive material 50V 100V 150V 200V 250V dynode © 2006 By Timothy K. Lund photo-multiplier

Option J: Particle physics J2 Particle accelerators and detectors Particle detectors Outline the structure and operation of a bubble chamber, a photomultiplier and a wire chamber. ●In the photomultiplier are a cascade of dynodes, each at a higher potential than the previous. ●Because of the acceleration caused by the p.d., more electrons are released during each step in the cascade. ●The end of the cascade has enough electrons to create measurable current. photo-sensitive material 50V 100V 150V 200V 250V dynode © 2006 By Timothy K. Lund FYI The Geiger counter uses a photomultiplier. For very high-energy photons a photomul- tiplier doesn’t work. Instead a scintillator is used. A scintillator has a screen of phosphorescent material that glows when struck by a very high energy gamma photon. photo-multiplier

Option J: Particle physics J2 Particle accelerators and detectors Particle detectors Outline the structure and operation of a bubble chamber, a photomultiplier and a wire chamber. ●If high-energy particles pass through a gas, the gas particles become ionized. Think of the high-energy particle as “knocking” electrons off of the gas atoms as it passes by. ●If the gas is between two wires which have a p.d. applied to them, these freed electrons travel to the positive wire and away from the negative wire, setting up a current. ●The voltage across a resistor which the current is made to pass through can then be digitally recorded. ●This device is called a wire chamber. wire chamber A V R © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Particle detectors Outline the structure and operation of a bubble chamber, a photomultiplier and a wire chamber. ●So how does one get a 3D picture of particles? Simply set up an array of wire chambers, and record not only the place, but the time the chambers detect a particle. ●The blue array tells us the left- right coordinate of the particle. ●The red array tells us the up-down coordinate of the particle. ●The timing from one double-grid to the next tells us the forward-backward coordinate of the particle. (1,1,t1) (3,2,t2) 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 © 2006 By Timothy K. Lund double grid double grid FYI 3D images can then be computer generated!

Option J: Particle physics J2 Particle accelerators and detectors Particle detectors Geiger counter © 2006 By Timothy K. Lund bubble chamber Large Hadron Collider

Option J: Particle physics J2 Particle accelerators and detectors Particle detectors Outline international aspects of research into high-energy particle physics. ●Because of the extreme expense in building and operating high energy particle physics installations, the larger ones have to be an international collaboration to help share the costs. ●Because of the international aspect of particle research, there is a tendency of the academic edifices of the world to come together, even in times of war. ●The research is also transparent-no country can be the “clearinghouse” of the information generated by any international facility. © 2006 By Timothy K. Lund

Option J: Particle physics J2 Particle accelerators and detectors Particle detectors Discuss the economic and ethical implications of high-energy particle physics research. ●People ask whether the cost of HEP is worth it. ●Theoretical physics stagnates without experimental verification. ●Curiosity is a fundamental part of the human mind. ●Sharing large research costs among many countries encourages cooperation between different cultures. ●Synchrotron radiation has a large range of applications, including biology, medicine, and technology. © 2006 By Timothy K. Lund