Electroweak Interactions a Lecture Course by W.B. von Schlippe

Transcription

1 Electroweak Interactions a Lecture Course by W.B. von Schlippe Lecture 3: Revision of Classical Electrodynamics This revision of classical electrodynamics serves a dual purpose. First it is to introduce notation and terminology, and second it is to emphasise those aspects of the theory which will be of particular importance in the following, namely the invariance under gauge transformations. This invariance, encountered first in electromagnetism and extended to quantum mechanics, is today understood to be a fundamental principle of physics. It is for this reason that the concepts of gauge transformation and gauge condition will be introduced very carefully. Maxwell s equations in SI units 1 are of the following form: curl H = j + D Ampére-Maxwell law (1) curl E = B Faraday-Lenz law (2) D = ρ Gauss law (3) B = 0 no magnetic charges (4) D = εe (5) B = µ H (6) Here H is the magnetic field strength, B is the magnetic flux density, E is the electric field strength and D is the electric displacement; these four vectors are functions of space and time. ρ and j are the charge density and current density, respectively. ε = ε r ε 0 is the permittivity, with ε r the relative permittivity (also called dielectric constant) and ε 0 the permittivity of free space or electric constant; ε r is dimensionless, ε 0 = pf m 1. µ = µ r µ 0 is the permeability with µ r the relative permeability (which is dimensionless) and µ 0 = 4π 10 7 Henry per meter the permeability of free space or magnetic constant. Note that 1/ ε 0 µ 0 = c (7) is the speed of light in free space. From the Ampére-Maxwell law together with Gauss law we get immediately the continuity equation: take the divergence of Eq. (1), use the vector identity ( H) = 0, and substitute the divergence of D from Eq. (3), hence 1 French: Système International d Unités ρ + j = 0 (8) 1

2 In free space ε r = µ r = 1, ρ = 0 and j = 0. We can therefor rewrite the Maxwell equations in terms of, for instance, H and D, eliminating also ε 0 and µ 0, thus curl H = D which can be shown to have plane wave solutions (9) curl D = 1 H (10) c 2 D = 0 (11) H = 0 (12) D = D 0 e i( k r ωt) and H = H 0 e i( k r ωt) (13) (14) with D H, D k and H k. The direction of D at different times t defines a plane, transverse to the direction of propagation of the wave. This plane is the plane of polarization, and the wave is said to be a transverse wave. The superposition of two waves travelling in the same direction with vectors D 1 and D 2, which are mutually orthogonal and whose phases differ by π/2, produces an elliptically polarised wave; in the particular case of equal amplitudes, D 01 = D 02, the wave is said to be circularly polarised. Exercise: Assuming that D is a plane wave, show that H is also a plane wave with the same phase as D, and that the electric and magnetic fields are orthogonal to each other and to the direction of propagation. We can deduce wave equations for D and H from the Maxwell equations, for instance by taking the curl of Eq. (9) and substituting the curl of D from (10), hence, applying the identity ( a( r)) = ( a( r)) 2 a( r) (15) and using Eq. (12), we get the wave equation ( 1 2 ) c H = 0 (16) substituting into which the plane wave function (14) we get the dispersion law of electromagnetic waves in free space, ω = ck (17) The wave equation for D is obtained similarly. 2

3 Electric and magnetic potentials. The electric potential φ and magnetic potential A are defined by E = φ A (18) B = A (19) Now, a fundamental theorem of vector calculus states that a vector field is uniquely defined if both its sources and vortices are known. In mathematical notation this means that a vector field V is uniquely defined if V and V are given. It follows then that the vector potential A is not completely defined by Eq. (19): we can add to it an arbitrary irrotational field. But such a field can always be represented as the gradient of a scalar field, i.e. we can carry out the transformation A A = A + χ (20) (where χ is an arbitrary function of r and t) without changing the magnetic flux density B. But then, in order to leave the electric field E unchanged, we must also transform the scalar potential φ according to φ φ = φ χ (21) Equations (20) and (21) together leave the fields, and hence the Maxwell equations, unchanged. This transformation is called gauge transformation, and the property of the Maxwell equations to remain unchanged under this transformation is called gauge invariance. We can derive a wave equation in terms of the vector potential. To do this we substitute the definition (19) into Eq. (9), hence, using also B = µ H and the identity (15), we get in free space 2 A 1 2 A c 2 2 = ( A + 1 c 2 φ It is straight forward to show that this equation is gauge invariant. However, it is not yet of the form of a wave equation, which would require the right-hand side to vanish. Here we can make use of the freedom of choosing the divergence of A which we have discussed above. We do this by carrying out a gauge transformation such that in terms of the new potentials 2 the expression in brackets on the right-hand side is equal to nought, i.e. ) (22) A + 1 c 2 φ = 0 (23) This is called the Lorentz condition, which gives us immediately the wave equation: 2 A 1 c 2 2 A 2 = 0 (24) The Lorentz condition is itself a gauge invariant equation if the transformation is done with a function χ that satisfies the constraint ( ) χ = 0 c which, for simplicity, we denote by the same symbols A and φ as the old ones 3

4 and we note that the wave equation is also gauge invariant if this constraint is imposed. We find also a wave equation for the scalar potential if we substitute Eq. (18) into (11), using also D = ε E, and again applying the Lorentz condition (23), hence ( ) φ = 0 (25) c 2 2 Covariant form of the Maxwell equations. We have seen that the magnetic flux density can be represented as the curl of a vector potential, B = A. This definition, written in components, reads B x = A z y A y z and similarly for B y and B z. In other words, the components of B have the form of an antisymmetric second-rank tensor. Now, since Maxwell s equations connect the E field with the B field in a symmetric way, one can ask the question of whether the components of E can also be represented as the components of a second-rank tensor. This is indeed the case as can be seen if we introduce four-dimensional notation. Thus, let us define the four-vector potential A µ as A µ = ( A 0, A ) (26) with A 0 = φ/c. We also note that we can define the four-dimensional operator µ = ( 0, 1, 2, 3) = ( ) c, hence E x = A x φ x = c ( 0 A 1 1 A 0) and similarly for E y and E z. Therefore we can indeed define an antisymmetric four-dimensional second-rank tensor F µν : F µν = µ A ν ν A µ (27) such that F µν = 0 E x /c E y /c E z /c E x /c 0 B z B y E y /c B z 0 B x E z /c B y B x 0 The tensor F µν is called the field strength tensor. We note that the definition of F µν automatically satisfies the homogeneous Maxwell equations (1) and (3). The inhomogeneous Maxwell equations (2) and (4) are now combined in the form of (28) µ F µν = j ν (29) 4

5 where we have also defined the four-vector current density j µ = (cρ, j). Equation (29) immediately yields the continuity equation. Indeed, taking the four-dimensional divergence of (29), we get ν µ F µν = ν j ν = 0 (30) since on the left-hand side we have the contraction of the the symmetric tensor ν µ with the antisymmetric tensor F µν, which is identically equal to nought. We can also write the covariant Maxwell equation (29) in terms of the vector potential A µ : µ F µν = µ ( µ A ν ν A µ ) = µ µ A ν ν ( µ A µ ) = j ν Here µ µ is the d Alambertian operator, and we note that µ A µ = 0 by virtue of the Lorentz condition (23). Thus finally we get µ µ A ν = j ν (31) or in free space, i.e. in the absence of charges and currents, µ µ A ν = 0 (32) which is the covariant form of the wave equation. Comparing this equation with the Klein- Gordon equation (see Lecture 1), we can say that each component of the four-vector potential A µ satisfies a Klein-Gordon equation of a particle of zero mass. The need for four components of the electromagnetic potential is connected with the two states of polarization, discussed previously, or equivalently with the spin of the photon, the quantum of the electromagnetic field. To see this we substitute into the wave equation a plane wave solution, A µ = Nε µ e ip x (33) where N is a normalization factor and ε µ is the polarization vector, whichn can be chosen as a unit vector. Differentiating (33) we get ν A µ = p ν A µ and differentiating again we get ν ν A µ = p ν p ν A µ = 0 (34) hence p ν p ν = 0 (35) which is the dispersion law found previously, and which we can also interpret as the relativistic energy-momentum relation of a zero-mass particle. If we substitute the plane wave expression into the Lorentz condition, we get µ A µ = p µ A µ = 0 (36) or p µ ε µ = 0 (37) which means that the four components of the polarization vector are not independent. But we have seen before that the Lorentz condition is itself gauge invariant, if the gauge function χ 5

6 satisfies an equation of the form of µ µ χ = 0. We can make use of this remaining freedom and choose, in the absence of charges and currents, for instance A 0 = 0. This modifies the Lorentz condition to A = 0 (38) which is called the Coulomb gauge. The time-like component of the polarization vector vanishes in the Coulomb gauge, ε 0 = 0, and we get p ε = 0 (39) which implies that the polarization vector is perpendicular to the direction of propagation of the wave, given by p. Moreover, the latter relation also implies that the polarization vector has only two independent components. Thus, if the plane wave propagates in the z direction, we can choose one of the unit vectors in the x or y direction or any linear combination of them as the polarization vector. Such a wave is said to be plane polarised. A polarization vector of the form of ε (1) = (1/ 2)(1, i, 0) or ε (2) = (1/ 2)(1, i, 0) describes a circularly polarized wave. The two states of circular polarization correspond to two orientations of the photon spin, one in the direction of motion and the other in the opposite direction. A third projection of spin, which one expects from a spin-1 particle, is absent in the case of the photon because of its zero mass. 6