Fundamental Concepts in high energy/Pa!icle Physics Lecture I : Introduction Géraldine SERVANT CERN-Th 1 Plan Lecture 1: intro to QFT: Relativity, kinematics, symmetries Lecture 2: Towards Gauge theories Lecture 3: Towards the Standard Model Lecture 4: Towards beyond the Standard Model 2 why high energies The Large Hadron Collider ➜ The LHC: the most gigantic microscope ever built Going to higher energies ➾ allows to study finer details L = 400 km L = 800 km Particle Physics: High energy physics study of short distances resolution limited by the de Broglie high resolution p>>m: wavelength λ =h/p necessitates large p relativistic regime 3 What is a particle? A smal, quantum and fa"-moving object Quantum Mechanics Special Relativity duality space-time wave-particle Heisenberg inequalities energy = mass energy non-conservation on time intervals ∆ t energy fluctuations ∆ E ∆t ∆E × ⇥ particle number non constant 4 Creation of matter from energy Chemistry : rearrangement of matter the different constituents of matter reorganize themselves CH4 +2O2 −→ CO2 +2H2O Particle physics : transformation energy ↔ matter W + e− µ+ u W − γ µ− d e+ E = mc2 d Equivalence between mass and energy (Einstein’s idea) plays a very fundamental role in particle physics 5 Natural units in high energy physics The fundamental units have dimension of length (L), mass (M) and time (T). All other units are derived from these. The two universal constants in SI units -h= 1.055×10-34 J s = 1.055×10-34 kg m2/s and c=3 108 m/s In particle physics we work with units ~ = c =1 Thus, velocity of particle is measured in units of the speed of light, very natural in particle physics where 0 ≤ v <1 for massive particles and v =1 for massless particles In the c=1 unit: [velocity]=pure number [energy]=[momentum]=[mass] notation: dimension of quantity P is [P] 6 Natural units in high energy physics ~ has dimension of [Energy]×[time]. ~ /mc~length (from uncertainty principle ∆ p ∆ x ~ / 2 ) or de Broglie’s formula λ =h/p ≥ ~ =1 ---> [length]=[mass]-1 Thus all physical quantities can be expressed as powers of mass or of length. e.g. energy density, E/L3~M4 e2 ↵ = pure number 4⇡~c We specify one more unit taken as that of the energy, the GeV. mass unit: M c2/c2 =1 GeV length unit: ~ c/M c2 =1 GeV-1=0.1975 fm time unit: ~ c/M c3 =1 GeV-1=6.59 10-25s. 7 Natural units in high energy physics 1 eV = 1.6 10-19 J --> -h c = 1.055×10-34 J s × 3 108 m/s= 1.978×10-7 eV m Using 1 fm = 10-15 m and 1 MeV = 106 eV: -h c = 197.8 MeV fm So in natural units: 1fm 1/(200 MeV) ⇡ -1 -25 -1 also, c=1 --> 1 fm ~ 3×10-24 s-1 --> GeV ~ 6 × 10 s -27 -h= 1.055×10-34 kg m2/s ----> GeV ~ 1.8 × 10 kg 8 The elementary blocks of matter Matter is made of molecules Molecules are built out of atoms Atoms are made of nuclei and electrons Nuclei are assemblies of protons and neutrons Protons and neutrons are quarks bound together... The volume of an atom corresponds to 1024 times the volume of an electron Classically, matter contains a lot of void Quantum mechanically, this void is populated by pairs of virtual particles 1 TeV = 1012 eV 1 electron volt The energy of an electron accelerated by an electric potential 19 ( eV) = difference of 1 volt. One eV is thus equal to ... 1.6 10− J 1 kg of sugar = 4000 kCalories= 17 millions of Joules ≈ 1014 TeV but 1 kg sugar ≈ 1027 protons --> 0.1 eV / protons To accelerate each proton contained in 1 kg of matter at 14 TeV, we would need the energy of of 1014 kg of sugar* i.e. 1% of the world energy production *world annual production of sugar=150 millions of tons≈1011 kg How impressive is tis? 24 energies involved at CERN: 1 TeV = 1000 billions of eV= 10 − kg compared with the kinetic energy of a mosquito 10−3 J ∼ 1016 eV ∼ 104TeV ...however, in terms of energy density... this corresponds to the mass of the Earth concentrated in a 1 mm3 cube ! Classical versus quantum Collision Compton wave length λ =h/mc strawberry : m~30 g ~1025 GeV/c2 ➲ λ~10-40 m λ << R classical: e- : m~9.1x10-31 kg ~0.5 MeV/c2 ➲ λ~4x10-13 m λ >> R p : m~1.6x10-27 kg ~1 GeV/c2 ➲ λ~10-16 m quantum : 11 Why relativity Particle physics is all about creating and annihilating particles. This can only occur if we can convert mass to energy and vice-versa, which requires relativistic kinematics A bit of historyContemplating the unusual invariance of Maxwell’s equations under Lorentz transformation, Einstein stated that Lorentz invariance must be the invariance of our space and time. -> completely changed our view of space and time, so intertwined that it is now called spacetime, leading to exotic phenomena such as - time dilation -length contraction -prediction of antimatter when special relativity is married with quantum mechanics 12 Relativistic transformations The two postulates - Speed of light is the same in all reference frames of Special relativity - Laws of physics are unchanged under a galilean transformation, i.e. in all reference frames moving at constant velocity with respect to each other x x’ Look for coordinate transformations that S’ satisfy thesect requirements0 ct0 γ γ γβγβ ctct = = −− z 0 !z0 ! γβγβ γ γ !! zz !! Unique choice: − − S z’ Lorentz transformations v v β =β = z c c ct0 γ γβ ct 1 = − γ = 1 z0 ! γβ γ ! z ! γ = v2 y y’ − 1 2 2 1 −v c q− c2 q 13 1 1 1 ct0 γ γβ ct = − z0 γβ γ z ! − ! ! ct0 γ γβ ct = − v Implicationsz 0of Lorentzγβ transformation γ z β = ! − ! ! c v 1 ct0 γ γβ ct β = γ = = − v2 z γβ γ z c 1 2 0 ! ! ! − c − q 1 consider time interval ⌧ = γ = t 0 t 0 in S’, the rest frame Time dilation 2 1 v2 −1 2 of a particle located at z 0 = − z 0 c =0 . 1q 2 then in frame S where the particle is moving: t t = ⌧ 2 − 1 -->The observed lifetime of a particle is γ ⌧ so it can travel over a distance βcγ⌧⇥ 6 -->muons which have a lifetime ⌧ 2 10 − s produced by reaction of ⇠ ⇥ cosmic rays with atmosphere at 15-20 km altitude can reach the surface length an object at rest in S’ has length L = z0 z0 0 2 − 1 contraction It measures in S z z = L /γ 2 − 1 0 --> densities increase ⇢0 =∆n/(∆x0∆y0∆z0) ⇢ =∆n/(∆x∆y∆z)=⇢0 is invariant ∆x∆y∆z∆t 14 1 1 1 ct0 γ γβ ct = − z0 γβ γ z ! − ! ! v β = c 1 γ = 4-vectors 1 v2 − c2 q Time and space get mixed-up under Lorentz ct x0 1 0 transformations. They are considered as µ 0 x 1 0 x 1 x x = = 2 = y x ~x ! different components of a single object, a B C B C B z C B x3 C four-component spacetime vector: B C B C @ A @ A µ 2 02 2 By construction: x x = x = x ~x is invariant µ − d⌧ = dt2 d~x2 is invariant − Lorentz invariant p action built with the S = m d⌧ = dt proper time d ⌧ . − L Z Z { = m 1 x˙ 2 L −@ − p~ = L p= mγβ~ p~ = Eβ~ @x˙ E = p~ . x˙ = mγ −L 15 1 Energy-momentum four-vector 2 2 2 so we find: m = E p~ − µ E This suggests to define the four-vector p =( ,p ,p ,p ) c x y z m =0 β =1 γ = ⌧ = ! ! 1! 1 a massless particle cannot decay 16 Conservation of energy-momentum Consider collision between A and B Define center of mass (CM) frame as where p~ A + p~ B =0 Energy available in center of mass frame 2 2 ptot = E is an invariant:sqrt(s)=E* = EA + EB ⇤ 1) Collision on fixed target B is at rest in lab frame, EB =mB and EA is energy of incident particle 2 2 2 E = mA + mB +2mBEA ⇤ 2) Colliding beams A and B travel in opposite directions 2 2 2 E = mA + mB + 2(EAEB + pA pB ) 4EAEB ⇤ | || | ⇡ if mA, mB << EA, EB So for fixed target machine E 2mBEA ⇤ ⇠ while for colliding beam accelerators E 2E p⇤ ⇠ To obtain 2 TeV in the CM with a fixed proton target accelerator the energy of a proton beam would need to be 2000 TeV! 17 ct0 γ γβ ct = − z0 ! γβ γ ! z ! Conservation of energy-momentum− v β = ct0 γ γβ ct c Consider the interaction e + p -> =e + p due to− exchange of electromagnetic field z0 ! γβ γ ! z ! − 1 γ = v 1 v2 β = − c2 p1 p3 a photon is massless, however, for a shortc amountq of time, an “exchanged” photon γ (virtual photon) q 1 ct x0 γ =⇤ can have a mass (Heisenberg inequalities)2 1 0 µ 0v x 1 0 x 1 x x 1= c2 = 2 = p p − y x ~x p! p q B C B C 2 4 p1 + p2 = p3 + p4 B C B 3 C B z C B x C ct x0@ A @ A q = p p = p 1 p 0 1 µ 30 x 1 40 x 1 2 x x = = 2 = − y −x ~x !m q is the transfer energy-momentumB C Bfour-vectorC p = B C B 3 C 2 ~ B z C B x C 0 ! 2 2 @ A @ A q = mγ : invariant, can be computed in any frame ⇤ m E4 E = m + T in frame where proton is at rest: p2 = ~ p4 = 4 0 ! p~4 ! kinetic energy 2 2 2 2 2 T p /2m q =(p4 p2) = p + p 2p4.p2 ⇡ − 4 2 − 2 2 2 q = m + m 2E4 m = 2mT < 0 virtual photon − − 18 1 1 Range of an interaction ~c R = m p1 p3 | ⇤| q reminder: ~c 200 MeV fm ⇠ p p2 p4p To probe the proton, we need 1 R R 1fm− ⌧ proton ⇠ m 200 MeV | ⇤| 19 Next step : marry quantum mechanics and relativity (Schrodinger’s equation cannot account for creation/annihilation of particles) non relativistic ∂ 2 Schrödinger Equation (1926): i + ∆ V ⇥ =0 ∂t 2m − p2 ⇥ E = + V classical ↔ quantum ∂ ∂ E i & p i 2m correspondance → ∂t → ∂x relativistic 1 ∂2 m2c2 Klein-Gordon Equation (1927): ∆ + ⇥ =0 c2 ∂t2 − 2 E2 ⇥ = p2 + m2c2 c2 2 2 1/2 negative energies E = (p + m ) and ± does not admit a positive probability density and does not describe fermions 20 Antimatter and Dirac equation µ mc Dirac Equation (1928): iγ ⇥µ Ψ =0 − + p2c2 + m2c4 matter ⇥ E = γµ,γ⌫ =2⌘µ⌫ ⌅p2c2 + m2c4 antimatter { } ⇤ − ⇥ ⌅ i(p.x Et)/ plane wave solution (x, t)=u(p)e − ~ a particle of energy -E travelling backward in time -> antiparticle + positron (e ) discovered by C.