We would like to give a Lagrangian formulation of electrodynamics.

Transcription

1 Chapter 7 Lagrangian Formulation of Electrodynamics We would like to give a Lagrangian formulation of electrodynamics. Using Lagrangians to describe dynamics has a number of advantages It is a exceedingly compact notation of describing dynamics. Recall for example, that a symmetry of the Lagrangian generally leads to conservation law and this is one method of learning about what quantities are conserved. Also from the Lagrangian one can constrtuct the Hamiltonian and H is essential in doing quantum mechanics. In fact in Feynman path integral formulation of quantum mechanics, one can express q.m. using only the Lagrangian. In conventional Lagrangian analysis one sets up the action integral A = t 2 t 1 Ldt and the dynamics law arise from varyoing it:δa = 0 Now if we wish the laws of physics to be the same in any inertial frame, we need to have δa to be Poincare invariant and this will be accomplished if A is invariant under Poincare transformation: A = A; x µ = Λ µ αx α + a µ up to a time derivative in L (which does not contribute to the equation of motion), Let us first look at equations of motion for the point 51

2 charges. The form of A = t 2 t 1 Ldt does not look Lorentz invariant since dt = dx 0 /c is really the fourth component of a vector. As in the equations of motion, this suggest for Poincare inavriance we should use τ instead of t to write. Assume dτ = dτ A = τ2 τ 1 Ldτ; L = scalar (7.1) wher L L (for v/c << 1). Now we had that the Lorentz force depends on the 4-velocity u µ = dx µ (τ)/dτ and so we expect L to be linear in this. But we must form a scalar from this and the E.M. field. The only vector to dot this out is A µ, i.e., u µ A µ ; A µ = η µν A µ (7.2) Since the alternative involving F µν itself is: u µ u ν F µν = 0 (7.3) This we suggest L = a + bu µ A µ ; a, b = const. (7.4) and A = dτ(a + bu µ A µ ) (7.5). The a term involves and expanding for small v/c adτ = dt dτ dt a = dt c a (7.6) 2 adτ = adt(1 1 v 2 2 c ) (7.7) 2 Thus if we choose a = m (7.8) 52

3 This corresponds to a non-relativistic Lagrangian of L = m mv (7.9) which gives the cortrect kinetic energy. The constant m does not contribute to the equation of motion. Hence we have A = dtl (7.10) where and writing out L = m L = m + b(dxµ dt A µ) (7.11) + b(ca 0 + v i A i ) (7.12) where A 0 = φ. The Lagrange equation reads Recall d dt ( ) L v j x i ( ) L x j v i = 0 (7.13) A i = A i (x i (t), t), φ = φ(x i (t), t) (7.14) Hence we get or Hence d ( vj dt ( mc2 ) d dt ( mv j ) + b(v i Aj φ + ba j ) b( c x + A j vi i x ) = 0 (7.15) j + A i x i t d dt ( mv j ) + b(v i Aj v i A i φ ) + bc( 1 x i xj x + 1 j c v2 ) + bc φ x j bvi A i x j = 0 (7.16) A i t ) = 0 (7.17) 53

4 Now recall Hence we can write u µ = dxµ dτ = 1 (c, v i ) (7.18) d dt (muj ) = b c u µf jµ 2 (7.19) and finally since we get d dt = dτ dt d dτ = d dτ (7.20) m duj dτ = bu µf jµ (7.21) which is correct with the choice b = e c (7.22) Our Lagrangian is this L = m + e c dx µ dt A µ (7.23) which generates what about the last equation to get u 0 : m duj dτ = e c u µf jµ (7.24) m du0 dτ = e c u µf 0µ (7.25) we know that this is just a equation to get the relation between t and τ which can alternately gotten from integarting dτ = dt (7.26) Hence Eq.7.25 and Eq.7.26 gives us the full dynamics or alternately with Eq.7.17 with b = e/c does. 54

5 We can construct the Hamiltonian from Eq.7.23 in the usual fashion. Writing Then the canonical momentum is and the Hamiltonian is L = m p i = L v i = + ea 0 + e c v. A (7.27) mvi e + 1 c A i (7.28) v2 H = p. v L (7.29) or H = p. v e c v. A + m + eφ (7.30) c2 Let us assume π i = p i e c Ai (7.31) Then Eq.7.28 reads and π i = H = π i v i + m mvi (7.32) + eφ (7.33) c2 In the Hamiltonian we eliminate v i in terms of p i. So we first solve Eq.7.32 backward. Thus squaring gives and hence π 2 = m2 v 2 (7.34) v 2 = π 2 π 2 + m 2 (7.35) 55

6 or and so = mc π2 + m 2 (7.36) v i c = π (7.37) π2 + m 2 Inserting Eq.7.36, Eq.7.37 into H of Eq.7.30 gives H = π 2 c π2 + m 2 c + m 2 c 3 + eφ (7.38) 2 π2 + m 2 c2 or H = π 2 + m 2 c 4 + eφ; π = p e c A (7.39) which is the Hamiltonian for a charge in the presence of an e.m. field characterized by A and φ. The second term is just the electrostatic energy while the first term coincides with the magnetic contribution. Expanding for π << mc gives H = m + π2 + eφ +.. (7.40) 2m which aside from the rest energy m is the usual non-relativistic Hamiltonian. In general, the Hamiltonian represents the energy of a particle when one is in an inertial frame. Let us look at the case of a free particle A µ = 0; H = E = p 2 + m 2 c 4 (7.41) The monemtim is then given by Eq.7.28 for the free particle p = m v ; v i = dx i dt (7.42) 56

7 Eliminating p in terms of v in Eq.7.41 gives E = m 2 v 2 + m 2 c 4 = mc2 We can write Eq.7.42 and Eq.7.43 in terms of 4-vector momentum. define (7.43) p µ = ( E c, pi ) (7.44) Then we have and p i = m dxi dt = m dx i ; dτ = dt (7.45) dτ and so p 0 = mc dx 0 = m dτ (7.46) p µ = m dxµ dτ = muµ (7.47) Thus p µ is a Lorentz 4-vector, i.e., energy and momentum in special relativity form a 4-vector. Note that the parameter m is constant associated with the particle, i.e., it is an intrinsic property of the particle (just as charge is). One recall m = rest mass of particle (7.48) and from Eq energy in frame at rest with respect to particle is just E rest = m (7.49) If one goes to a moving frame Then p µ transforms as a 4-vector p µ = Λ µ αp α (7.50) 57

8 and so the energy increases,i.e., if p µ : (mc, 0) then E = p 0 c = Λ 0 0p 0 c + Λ 0 ip i c = γm = mc2 (7.51) which is just Eq It is not the mass that changes, but rather the energy that changes as one goes to the moving frame. Similarly for the frame with relative velocity v: p i = Λ i 0p 0 + Λ i jp j = Λ i 0p 0 = γ( vi c )mc = γmvi = just as in Eq Note p µ p µ = m2 v 2 1 v2 m2 c4 1 v2 a free particle is represented by a plane wave mvi (7.52) = m 2 and m is invariant. Note that in q.m. ψ = e i (p.r Et) (7.53) we can write this as ψ = e i (p ix i E c ct) = e i (p ix i +p 0 x 0) ; p µ = η µν p ν = ( p 0, p i ) (7.54) Hence ψ = e i pµxµ (7.55) If we go to a new inertial frame Then x µ = Λ µ αx α + a µ = x µ x α xα + a µ ; p µ = (Λ 1 ) α µ p α; p µ = xβ x µ p β (7.56) ψ e i p µ x µ = e i p β xβ x µ x µ x α x α x +p β β x µ a µ now use xβ x α = δ β α Hence ψ e i pαxα e iφ, φ = p β x β x µ aµ (7.57) 58

9 Thus ψ changes by just a phase at most which does not change the physical content of the wavefunction. Thus the conventional plane wave of q.m. generalizes to Lorentz covariant form. Let us now return to our charges interacting with the e.m. field. We have obtained a Lagrangian for the particle and its interaction with the e.m. field. To complete prescription we had Lagrangian for the e.m. field itself. To see how that might go we can rewrite our interaction between the charge and the field in a more covariant looking form. We had from Eq.7.23: and so L = m + e c dx µ dτ c A 2 µ (7.58) A = dtl = m dt + e c dt dxµ dτ c A 2 µ = A part + A int (7.59) where A int = e c dτu µ A µ (x i (t), t(τ)) (7.60) we can rewrite this as A int = e c d 4 xdτu µ (τ)a µ (x α )δ 4 (x α x α (τ)) (7.61) Recall we had j µ (x) = e dτu µ (τ)δ 4 (x x(τ)) (7.62) So we have A int = 1 c d 4 xl int = dt d 3 xl int (7.63) 59

10 where L int = 1 c jµ (x)a µ (x) = interaction Lagrangian density (7.64) Eqs.7.63,7.64 suggest what the form of the action should be for a field system, i.e., it should be the volume integral of the Lagrangian density: A = 1 c d 4 xl (7.65) we saw that d 4 x is a scalar and so if A is to be invariant, i.e., A = A (7.66) so that the equations of motion look the same in all frame, we require L int = scalar (7.67) This precisely what L int is (which gives the coupling of the e.m. field to the charges). Of course when we vary A we must get from the requirement that δa vanishes, the field eqns. for the fields. Let us first therefore investigate the general form Lagrange s eqns. will take for a field system. We assume in general there are a set of fields Φ A (x) = fields (7.68) and an action where A = t2 d 3 x dtl(x) (7.69) t 1 L(x) = L(φ A (x), µ (φ A (x)) (7.70) 60

11 f df t 1 t 2 X Figure 7.1: We assume that under variations of φ A δφ A (x) = arbitrary with boundary condition; δφ A 0(r ), δφ A (r, t i ) 0(t i t 1, t 2 ) (7.71) That the action is an extrema, i.e., δa = 0 (7.72) To vary A explicitly gives δa = t2 d 3 x dt[ L(x) δφ A + L(x) t 1 φ A ( µ φ A ) µδφ A ] (7.73) Now L(x) ( µ φ A ) µδφ A = µ ( L(x) ( µ φ A ) δφ A) x ( L(x) µ ( µ φ A ) )δφ A (7.74) The first term vanishes since d 3 x dt µ F µ = dt d 3 x.f + t2 d 3 1 x t 1 c F 0 dt (7.75) t and both terms vanish by our b.c. in Eq Thus δa = 1 c [ L(x) φ A µ L(x) ( µ φ A ) ] φ A (7.76) 61

12 and since δa = 0 for arbitrary δφ A = 0 we get L(x) φ A µ L(x) ( µ φ A ) = 0 (7.77) These are the Lagranages equations for the system. We wish now to choose L such that Eq.7.76 yields Maxwell equation ν F µν = 1 c jµ (7.78) F µν = µ A ν ν A µ (7.79) Eqs.7.78,7.79 are first order differential equations. Since Eq.7.76 contains single derivative, L can be at most first order in field derivative and linear in these derivatives and at most quadratic in the fields since Eqs.7.78,7.79 are linear. Thus L can be constructed from F µν, A µ, µ A ν, λ F µν (7.80) Finally L must be Poincare scalar, so all the indices must be dotted out. The possibilities are F µν F µν ; F µν A µ A ν ; F µν µ A ν ; A µ A µ (7.81) i.e., other possibities are either quadratic in derivative (e.g.,( λ F µν )( λ F µν ) )or cubic or higher (e.g., A λ µ A ν λ F µν ) Now the second possibility vanishes identically F µν A µ A ν 0 (7.82) and the last possibility can be written as 1 2 F µν ( µ A ν ν A µ ) F µν ( µ A ν + ν A µ ) (7.83) (The last term is 0) Hence there are only to possibilities are (we neglect A µ A µ as it is not gauge inavriant); L em = 1 4 F µν F µν 1 2 F µν ( µ A ν ν A µ ) (7.84) 62

13 where we have chosen the constants as we will see to correctly give Maxwell equations. The total part of the Lagrangian is then L = L em + L int = 1 4 F µν F µν 1 2 F µν ( µ A ν ν A µ ) + 1 c jµ A µ (7.85) Our set of fields for Eq.7.76 are then {φ A } = {F µν, A µ } (7.86) and so we get the equations and L F L µν λ ( λ F µν ) = 0 (7.87) L L µ A ν ( µ A ν ) = 0 (7.88) we first consider Eq.7.87 and note that L does not depend on λ F µν so that we have To take the derivative of L w.r.t F µν, i.e., L = 0 (7.89) F µν F µν = η µα η νβ F αβ (7.90) we take the derivative by first varying w.r.t F µν δ F L = 1 4 δf µν F µν F µν η µα η νβ δf αβ 1 2 δf µν ( µ A ν ν A µ ) (7.91) In the second term we can write since all indices are summed F µν η µα η νβ δf αβ = F αβ η αµ η βν δf µν = F µν δf µν (7.92) So that the the first two terms combine to give δ F L = 1 2 δf µν F µν 1 2 δf µν ( µ A ν ν A µ ) (7.93) 63

14 The derivative is L F µν = δ F L δf µν = 1 2 (F µν ( µ A ν ν A µ )) (7.94) and hence Eq.7.87 reads F µν = µ A ν ν A µ (7.95) which is correct. To calculate Eq.7.88, we need first and to get the second term we first vary: L A ν = 1 c jν (7.96) δ A L = 1 2 F µν (δ( µ A ν ) F µν (δ( ν A µ ) (7.97) Hence and putting into Eq.7.88 gives L ( µ A ν ) = F µν (7.98) 1 c jν + µ F µν = 0 (7.99) or µ F νµ = 1 c jν (7.100) which is correctly Eq We can now generalize to a system with an arbitrary number of charged particles. The action that characterizes the dynmaics is then A = dt a ( m a a(t) )+ 1 c d 4 xj µ (x)a µ + 1 c d 4 x[ 1 4 F µν F µν 1 2 F µν ( µ A ν ν A µ )] (7.101) where j µ = c a e a dτu µ a(τ)δ 4 (x x a (τ)) (7.102) 64

15 In general we know that Maxwell s theory is gauge invariant and a gauge transformation is given in 4-vector form by A µ A µ + µ Λ(x); F µν F µν F µν (7.103) It is clear from Eq that the term µ A ν ν A µ is gauge invariant since µ A ν ν A µ ( µ A ν + µ ν Λ) ( ν A µ + µ ν Λ) = µ A ν ν A µ (7.104) But what about the interaction term 1 c d 4 xj µ A µ 1 c d 4 xj µ A µ + 1 c d 4 xj µ µ Λ (7.105) The last term can be written as (2) = 1 c d 4 x µ (j µ Λ) 1 c d 4 x( µ j µ )Λ (7.106) (2) = 1 c d 3 xj(r, t) 0 Λ(r, t)] t=t 1 1 t=t 2 c d 4 x( µ j µ )Λ (7.107) We are free to choose Λ to be arbitrary at any interior point and vanish at end point. Then (2 ) = 1 c d 4 x( µ j µ )Λ (7.108) But µ j µ = 0 is conservation of charge. Hence 1 c d 4 xj µ A µ = gauge invariant (7.109) for any arbitrary gauge transformation for t 2 < t < t 1. Gauge invariance is thus a consequence of conservation of charge. 65