Particle Notes. Ryan D. Reece

Transcription

1 Particle Notes Ryan D. Reece July 9, 2007

2 Chapter 1 Preliminaries 1.1 Overview of Special Relativity Lorentz Boosts Searches in the later part 19th century for the coordinate transformation that left the form of Maxwell s equations and the wave equation invariant lead to the discovery of the Lorentz Transformations. The boost transformation from one (unprimed) inertial frame to another (primed) inertial frame moving with dimensionless velocity β = v/c respect to the former frame, is given by ( ) ( ) ( ) c t γ γβ c t x = (1.1) γβ γ x Because a boost along one of the spacial dimensions leaves the other two unchanged, we can suppress the those two spacial dimensions. γ is the Lorentz Factor, defined by 1 γ = (1.2) 1 β 2 γ ranges from 1 to monotonicly in the nonrelativistic and relativistic limits, respectively. It is useful to remember that γ 1. The inverse transformation is given by ( ) ( ) ( ) c t γ γβ c t = x γβ γ x (1.3) 1

3 The differences betweeen two points in spacetime follow from the transformations: c t = γ c t γβ x (1.4) x = γβ c t + γ x (1.5) c t = γ c t + γβ x (1.6) x = γβ c t + γ x (1.7) Length Contraction and Time Dilation Consider a clock sitting at rest in the unprimed frame ( x = 0). Equation (1.4) and the fact that γ 1 implies that the time interval is dilated in the primed frame. t = γ t (1.8) Now consider a rod of length x in the unprimed frame. A measurement of the length in the primmed frame corresponds to determining the coordinates of the endpoints simultaneously in the unprimed frame ( t = 0). Then (1.7) implies that length is contracted in the primed frame. x = x (1.9) γ We call time intervals and lengths proper if they are measured in the frame where the subject is at rest (in this case, the unprimed frame). In summary, proper times and lengths are the shortest and longest possible, respectively Four-vectors Knowing that lengths and times transform from one reference frame to another, we wonder if there is anything is invariant. Consider the following, using equations (1.6) and (1.7). (c t) 2 ( x) 2 = (γ c t + γ β x ) 2 (γβ c t + γ x ) 2 (1.10) = γ [ 2 (c t ) 2 + 2β c t x + β 2 ( x ) 2 β 2 (c t ) 2 2β c t x ( x ) 2] (1.11) = γ 2 (1 β 2 ) }{{} γ 2 [ (c t ) 2 ( x ) 2] (1.12) = (c t ) 2 ( x ) 2 ( τ) 2 (1.13) 2

4 Which shows that τ has the same value in any frames related by Lorentz Transformations. τ is called the invariant length. Note that it is equal to the proper time interval. This motivates us to think of (t, x ) as a four-vector that transforms according to the Lorentz transformations, in a spacetime vector space, and there should be some kind of inner product, or contraction, of these vectors that leaves τ a scalar. This can be done by defining the Minkowski metric tensor as follows. g = (1.14) Four-vectors are indexed by a Greek index, x µ = (c t, x ) µ, ranging from 0 to 3 (x 0 = c t, x 1 = x, x 2 = y, x 3 = z). The contraction of a spacetime four-vector with itself, its square, is give by x µ g µν x ν x µ x µ = (c t) 2 x x = ( τ) 2 (1.15) giving the square of the invariant length between x µ and the origin. In equation (1.15), we have defined that the lowering of a four-vector index is done by multiplication by the metric tensor. Explicit matrix multiplication will show that the Minkowski metric has the same components regaurdless of the orientation of its indices. g µ ν g νλ g µλ = g µν (1.16) g µν g µσ g νλ g σλ = g µν (1.17) Anything that transforms according to the Lorentz Transformations like (ct, x ) is four-vector Another example of a four-vector is four-velocity, defined by η µ γ (c, v ) µ (1.18) One can show that the square of η µ is invariant as required. η µ η µ = γ 2 (c 2 v 2 ) (1.19) = 1 1 β (1 2 β2 ) c 2 (1.20) = c 2 (1.21) 3

5 which is obviously invariant. Any equation where all of the factors are scalars (with no indices or contracting indices), or are four-vectors/tensors, with matching indices on the other side of the equal sign, is called manifestly invariant Momentum and Energy The Classically conserved definitions of momentum and energy, being dependent on the coordinate frame, will not be conserved in other frames. We are motivated to consider the effect of defining momentum with the four-velocity instead of the classical velocity. The rest (invariant) mass of a particle, m, being an intrinsic property of the particle, must be a Lorentz scalar. Therefore, the following definition of the four-momentum is manifestly a four-vector. The square of which is p µ m η µ = γ m (c, v ) µ (1.22) p µ p µ = m 2 η µ η µ = m 2 c 2 (1.23) Now let s give some interpretation to the components of the four-momentum. To consider the nonrelativistic limit, let us expand γ in the β 0 limit. γ β β4 + (1.24) Then the leading order term of the space-like components of the four-momentum is just the Classical momentum. p = m v + (1.25) We therefore, interpret the space-like components of the four-momentum as the relativistic momentum. p = γ m v (1.26) The expansion of the time-like term gives m c m v2 + (1.27) 4

6 We can now recognize the second term as the Classical kinetic energy. The first term is evidently the rest mass energy, energy present even when v = 0. Higher order terms give relativistic corrections. E = γ m c 2 (1.28) We can therefore write the four-momentum in terms of the relativistic energy, E, and relativistic momentum, p. p µ = (E, p ) µ (1.29) The four-momentum is the combination of momentum and energy necessary to transform according to Lorentz Transformations. Both E and p are conserved quantities in any given frame, but they are not invariant; they transform when going to another frame. Scalar quantities, like mass, are invariant but are not necessarily conserved. Mass can be exchanged for kinetic energy and vice versa. Charge is an example of a scalar quantity that is also conserved. Taking the ratio of equations (1.26) and (1.28) gives the following interesting relation. p E = v c 2 (1.30) which leads to p c E = β (1.31) Note that in order for energy and momentum to remain finite, velocity has an upper bound at c, because γ diverges as v c in equations (1.26) and (1.28). The only way around the prohibition against light speed is for a particle to have zero mass. Conversely, any massless particle must move at speed c, since otherwise it would have no energy, no momentum no existence at all. For massless particles, (1.31) reduces to E = p c (1.32) Also note the relationship to the equation from quantum mechanics for the energy of a photon, E = ω p = ω = k, consistent with c de Broglie s relation. 5

7 1.1.5 Relationship Between Energy, Momentum, and Mass We now can use equations (1.28) and (1.31) to eliminate the velocity dependence to get a frame independent relationship. E = γ m c 2 (1.33) = (1 β 2 ) 1 2 m c 2 (1.34) ( ( ) ) p c = 1 m c 2 (1.35) E E 2 = p 2 c 2 + m 2 c 4 (1.36) which is consistent with the zero mass case (1.32). Another way to arrive at equation (1.36) is to apply our interpretation of the components of p µ to equation (1.23). 1.2 Units Natural Units p µ p µ = E 2 p 2 c 2 = m 2 c 4 (1.37) Factors of c were explict in the above review of special relativity. From now on, we will use a form of natural units, where certain natural constants are set to one by using units derived from the God-given scales in Nature. = c = ε 0 = 1 (1.38) From = GeV s = 1, it follows that if we choose to measure energy in units GeV, then time can be measured in units GeV 1. From c = m/s = 1, it follows that 1 GeV 1 = s (1.39) 1 = ( m ) ( GeV ) = m GeV (1.40) Summarizing the dimensionality: 1 GeV 1 = m (1.41) time = length = 6 1 energy (1.42)

8 1.2.2 Barns We will later see later that when calculating cross sections, the conventional unit of area in particle physics is a barn. From (1.41), it can be shown that Electromagnetism Finally, from ε 0 = 1 and c = 1 1 barn (10 fm) 2 = cm 2 (1.43) 1 mb 10 3 barns = cm 2 (1.44) c = 1 GeV 2 = mb (1.45) 1 µ0 ε 0 µ 0 = 1 (1.46) giving Maxwell s equations the following form. Field Tensor: F µν µ A ν ν A µ (1.47) Homogeneous: Inhomogeneous: 1.3 Relativistic Kinematics µ F νλ + ν F λµ + λ F µν = 0 (1.48) Lorentz Invariant Phase Space Mandelstam Variables 1.4 Perturbation Theory ν F µν = J µ (1.49) 7

9 Chapter 2 Quantum Electrodynamics 2.1 The Dirac Equation The Dirac Equation is the equation of motion for fermion (spin 1/2) fields. It is written (iγ µ µ m)ψ = 0 (2.1) It is actually four coupled equations for the four components of the fermion field, ψ. The four components are that of a spinor space, describing the spin, like the usual two component spinor from non-relativistic quantum mechanics, and also describing the particle/antiparticle nature of particles. Each γ, indexed by a four-vector index, µ, is a 4 4 matrix, operating in the spinor space. The mass of the fermion, m, is multiplied by an assumed identity matrix in the spinor space The γ-matrices The γ-matrices satisfy the following defining anticommutation relation {γ µ, γ ν } = 2g µν (2.2) where again, multiplied by g µν, there is an assumed identity matrix in the spinor space. Be careful to distinguish between spacetime and the spinor space. Unfortunately, they both have four components. Note that g µν, a component of the Minkowski metric, is just a number (+1, 1, or 0). From the anticommutation relation, one can derive the following properties: γ 0 = γ 0 (2.3) 8

10 and which generalize as γ k = γ k (2.4) γ µ = γ 0 γ µ γ 0 (2.5) Like four-vectors, the space-index parts of γ µ change sign when lowering the index. γ 0 = γ 0, γ k = γ k (2.6) There are several representations (Pauli-Dirac, Weyl, Majorana... ) of the γ-matrices, where each γ µ is represented by a specific 4 4 matrix, but it is often not necessary to assume a specific representation to do a calculation. Often, all that is needed is the anticommutation relation and its resulting properties Bar and Slash Notation We will see that it is convenient to define some compact notation. adjoint (row) spinor, said psi-bar, is defined The ψ ψ γ 0 (2.7) The Feynman Slash abreviates the contraction of γ-matrices with a fourvector. It is defined by /a γ µ a µ γ µ a µ (2.8) where a is any four-vector The Adjoint Equation and Conserved Current Written in a more explicit notation notation, the Dirac Equation is Taking the adjoint gives iγ 0 ψ t ψ + iγk mψ = 0 (2.9) xk i ψ t γ0 i ψ x k ( γk ) mψ = 0 (2.10) 9

11 To restore the covariant form, we need to remove the minus sign of γ k while leaving the first term unchanged. Since γ 0 γ k = γ k γ 0, this can be done by multiplying on the right by γ 0. Using the bar notation, we have i ψ t γ0 γ 0 + i ψ x k γ0 γ k + mψ γ 0 = 0 (2.11) i µ ψγ µ + m ψ = 0 (2.12) We can now show that the fermion field conserves charge current, as it should in order to describe particles like electrons. Multiplying the Dirac Equation (2.1) from the left by ψ and adding it to the adjoint equation (2.12) multiplied by ψ from the right gives i ψγ µ µ ψ m ψψ + i µ ψγ µ ψ + m ψψ = 0 (2.13) ψγ µ µ ψ + µ ψγ µ ψ = µ ( ψγ µ ψ) = 0 (2.14) Multiplying by the charge of an electron ( e), we interpret j µ = e ψγ µ ψ (2.15) as the charge current density four-vector of electrodynamics, satisfying the continuity equation: µ j µ = 0 (2.16) Antiparticles and Free Particle Spinors The Dirac Equation has plane-wave solutions of the form ψ = u(p) e ip x (2.17) where u(p) is a four-spinor. Plugging equation (2.17) into the Dirac Equation (2.1) gives (γ µ p µ m) u(p) = 0 or (/p m) u = 0 (2.18) where we have used the Feynman slash notation. 10

12 We will now assume the Pauli-Dirac Representation of the γ-matrices in order to study the components of solutions to the Dirac Equation in some detail. It is given by γ 0 = ( I 0 0 I ), γ i = ( 0 σi σ i 0 ) (2.19) where σ i are the Pauli Matricies and I is the 2 2 identity matrix. Let u (1) ( ) ( ) ( ) u = u (2) u (3) = ua u (1) u (3), u u A = B u (2), u B = u (4) u (4) (2.20) Plugging in the γ-matrices into (2.18), we have ( ) ( ) E m σ p ua (p) = 0 (2.21) σ p E m u B (p) We can see right away that in the limit p 0, solutions have E = ±m. Does this mean that the Dirac Equation has negative energy solutions?! No. We interpret the negative energy solutions as antiparticles with positive energy. Antiparticles are like their corresponding particles with all signs of charge-like quantities (electric charge, baryon number, lepton number... ) flipped. Therefore, the conserved currents from antiparticles can be described as that of particles going backward in time (t t). The phase factor of the field, then is unchanged because the negative sign from the time cancels the negatives sign of the energy. e i( E)( t) = e iet (2.22) Solutions u (1) and u (2) correspond to particles while solutions u (3) and u (4) correspond to antiparticles. 2.2 Scattering Interaction with the Electromagnetic Field e µ e µ 11