E = E E = γ (E + v B), B. = B B = γ ( B v E/c 2), 2. Correct understanding of J = (ρc, j) and charge conservation.

Transcription

1 Lecture 8. Relativity and electromagnetism 1. Force per unit charge f = q(e + v B) Change frame: f f, hence: Transformation of electromagnetic field E = E E = γ (E + v B), B where S has velocity v in S. e.g. capacitor, particle beam = B B = γ ( B v E/c 2), 2. Correct understanding of J = (ρc, j) and charge conservation. 3. Fields due to a moving point charge: E = γqr 4πϵ 0 (γ 2 (x ) 2 + (y ) 2 + (z ) 2 ) 3/2, B = v E c 2.

2 Lecture 9. 4-vector potential; Maxwell s equations; introducing tensors 1. Lorentz covariance of Maxwell s equations 2. Scalar and vector potential B = A, E = ϕ A t automatically satisfy M2, M3. { A A + χ, 3. Gauge transformation: ϕ ϕ χ t 4. 4-vector potential ( ) ϕ/c A, Gauge transformation: A A + χ A 5. Lorenz gauge, Maxwell s equations in a manifestly covariant form: Maxwell s equations (!) 2 A = 1 c 2 ϵ 0 J, with A = General idea of 3-dimensional tensors such as conductivity, susceptibility, Outer product F = AB T F = ΛFΛ T. Also F B FgB

3 Maxwell equations and Lorentz force equation: E = ρ ϵ 0 B = 0 E = db dt c 2 B = j + de ϵ 0 dt, Change frame: ρc j x j y = Λ 1 j z f = q(e + v B) ρ c j x j y j z, (M1) (M2) (M3) (M4) E = E E = γ (E v B ), B = B B = γ ( B + v E /c 2). ( ) x ( ) y = ( ) t t x + ( ) x x x + ( ) y y x + ( ) z z x = etc.

4 Lecture 10. Tensor analysis and index notation 1. Basic idea: ϕ, A a, F ab, g ab, Λ a b 2. Summation convention: A ab X b means 3 A ab X b b=0... dummy index 3. Contravariant/covariant. A T gb = A T g B g = (Λ 1 ) T gλ 1 Contravariant: X ΛX Covariant: (gx) (Λ 1 ) T (gx) 4. Index lowering: F a g aµ F µ so U F = U λ F λ 5. Legal tensor operations: sum, outer product, contract. 6. Caution when comparing with matrix notation A aλ B λ A B but A λa B λ = B λ A λa B A 7. Invariants: contract down to a scalar. e.g. A λ B λ, T λ λ, T µν T µν 8. Differentiation. a x a = ( 1 c a. e.g. continuity equation λ J λ = 0 Product rule: a (U b V c ) = ( a U b )V c + U b ( a V c ) t, x, y, z ). Thus a = a = g aλ λ and a = a.

5 Lecture 11. Electromagnetic field theory via field tensor F 1. } { Lorentz covariance, 4-force = charge field 4-velocity simplicity F = qf U Field tensor 2. Pure force (F U = 0) F = F T 0 E x /c E y /c E z /c F = E x /c 0 B z B y E y /c B z 0 B x. E z /c B y B x 0 3. propose field equation F = µ 0 ρ 0 U, i.e. λ F λb = µ 0 ρ 0 U b 4. need a further equation; try F = A, i.e. F ab = a A b b A a The physical world? c F ab + a F bc + b F ca = 0. Implications 5. Antisymmetric F charge conservation: µ ν F µν = 0 λ J λ = 0 6. Invariants D 1 2 Fµν F µν = B 2 E 2 /c, α 1 4 F µν F µν = B E/c. e.g. orthogonal in one frame orthogonal in all (α = 0) purely magnetic in one frame not purely electric in another (D > 0). 7. Finding the frame (if there is one) in which B or E vanishes.

6 Lecture 12. Introducing angular momentum, and some general tensor manipulations 1. Vector product, e.g. L = XP T PX T or L ab = X a P b X b P a. 2. Conservation of angular momentum and the motion of the centroid. i x c x ie i E tot 3. Transformation of an antisymmetric tensor: 0 a x a y a z F = a x 0 b z b y a y b z 0 b x, F = ΛFΛ T a z b y b x 0 a = a, a = γ(a + v b/c), b = b, = γ(b v a/c). b 4. Dual: F ab = 1 2 ϵ abµνf µν ; hence a b; b a. 5. Differentiation examples

7 Tips 1. Name your indices sensibly; make repeated indices easy to spot. 2. Look for scalars. e.g. F λµ A a b Fλµ is sa a b where s = F λµf λµ. 3. You can always change the names of dummy (summed over) indices; if there are two or more, you can swap names. 4. The see-saw rule A λ B λ = A λ B λ (works for any rank) 5. In the absence of differential operators, everything commutes.

8 Lecture 13. Wave equation and general solution of Maxwell s equations 1. Poisson equation and its solution (reminder) ( Green s method ): 2 ϕ = ρ ρ(r s ), ϕ(r) = ϵ 0 4πϵ 0 r r s dv s. 2. How to calculate 2 (1/r) 3. Wave equation and its ( retarded ) solution 1 c 2 2 ϕ t ϕ = N.B. source event, field event. ρ(r, t) ϵ 0, ϕ(r, t) = ρ(rs, t r r s /c) dv s. 4πϵ 0 r r s Hence A = 1 J(rs, t r sf /c) dv s. 4πϵ 0 c r sf 4. Potentials of an arbitrarily moving charged particle A = q 4πϵ 0 U/c ( R U). ( Liénard-Wiechart potentials)

9 Lecture 14. Electromagnetic radiation Fields of an accelerated charge: ( q n v/c E = + 4πϵ 0 κ 3 γ 2 r 2 B = n E/c ) n [(n v/c) a] c 2 r where n = r/r, κ = 1 v r /c = 1 n v/c In terms of the displacement r 0 = r vr/c from the projected position, ) q E = (γr 4πϵ 0 r0 3(γ2 cos 2 θ + sin 2 θ) 3/2 0 + γ3 c r [r 2 0 a] 1. General features: bound field, radiative field 2. Case of linear motion coming to rest 3. Radiation from slowly moving dipole oscillator A = q v 4πϵ 0 c 2 (r sf r sf v/c), far field: B = ω2 qx 0 sin θ sin(kr ωt) 4πϵ 0 c 3 r ˆϕ, E = cb 4. The half-wave dipole antenna. Emitted power = I 2 rms (73 ohm).

10 Lecture 15. Radiated power 1. Radiated power (Larmor) dp = Nr 2 dω = q2 a 2 sin 2 θ dω P 4πϵ 0 4πc 3 L = Linear particle accelerator: a 0 = f 0 /m = f/m, loss 0 3. Dipole oscillator q 2 q 2 a 2 0 4πϵ 0 c. 3 P L = 2 ω 2 x 3 4πϵ 0 c 3(γ3 0 cos ωt) 2, P L 1 3 4πϵ 0 c 3 ω4 x Headlight effect: received energy per unit time at the detector, per unit solid angle dp dω = q2 a 2 sin 2 θ for linear motion 4πϵ 0 c 3 (1 (v/c) cos θ) 6 q 2 5. Circular motion: a 0 = f 0 /m = γf/m = γ 2 a, E = q2 3ϵ 0 r γ4 (v/c) 3. (6. Synchrotron radiation: lighthouse pulses with frequency spread ω γ 3 ω 0.) (7. Self-force and radiation reaction)

11 Lecture 16. Spin; parity inversion symmetry 1. L ab (0) = L ab (R) + ( ) R a Ptot b R b Ptot a Total angular momentum J ab = S ab + L ab 2. Pauli-Lubanski vector W a J aλ P λ = 1 2 ϵ aλµνj µν P λ W a = (s p, (E/c)s) In the rest frame: (0, mcs 0 ) so W U = 0 and s = s 0, s = s 0 γ Hence for a photon, W is null and points along P. 3. Mirror reflection; polar and axial vectors 4. Parity inversion: x x, y y, z z x x, p p x p x p 5. Classical physics covariant under parity inversion 6. Parity non-conserving process

12 Lecture 17. Lagrangian mechanics (symmetry again!) 1. Reminder of Least Action and Euler-Lagrange equations Lagrangian Action S[q(t)] = L = L({q i }, { q i }, t) T V q2,t 2 q 1,t 1 L(q, q, t)dt stationary for path satisfying Euler-Lagrange equations: Hamiltonian: L q i = generalized force, H(q, p, t) n p i q i L(q, q, t) i d dt ( ) L = L. q i q i L q i = canonical momentum 2. Special Relativity (version 1). Freely moving particle: L = mc 2 /γ, p = γmv Particle in an e-m field: L = mc 2 /γ + q( ϕ + v A), p = γmv + qa Taking a derivative along the worldline: da dt = A + (v )A t ( ) 1/2 Hamiltonian H = γmc 2 + qϕ = ( p qa) 2 c 2 + m 2 c 4 + qϕ. 3. Special Relativity (version 2), using τ instead of t in the action: L(X, U) = mc( U U) 1/2 + qu A, S[X(τ)] = d dτ L U = L a X a, Pa = mu a + qa a, m du dτ (X2 ) (X 1 ) L(X, U, τ)dτ, = q( A) U

13 Use of a parameter to minimize the action. Integrating with respect to proper time means the value of τ at the end event is different for each path. Problem!: calculus of variations needs fixed start and end values. Introduce a parameter λ: L(X, Ẋ, τ)dτ = λ2 Now the Lagrangian is L = L dτ dλ = L1 c λ 1 ( g µν dx µ (using dτ 2 = dt 2 (dx 2 + dy 2 + dz 2 )/c 2 ) giving Euler-Lagrange equations in which Ẋ = dx/dλ. d L L = dλ Ẋa X a dλ Now pick λ = τ along the solution worldline. L dτ dλ dλ. dx ν ) 1/2. dλ For that worldline, and for that worldline only (but it is the only one we are interested in from now on), we must then find dτ/dλ = 1 and L = L and Ẋ a = U a.

14 Now so da a dτ = U λ λ A a or m du a dτ dp dτ = q (( a A λ ) ( λ A a )) U λ = q( A) U

15 Lecture 18. Energy-momentum flow; stress-energy tensor 1. Conservation of energy: u t = N + E j. energy density u = 1ϵ 2 0E 2 + 1ϵ } 2 0c 2 B 2 Poynting vector N = ϵ 0 c 2 follow from Maxwell s equations E B. 2. Transfer of 4-momentum per unit volume from fields to matter: Let W = (E j/c, ρe + j B) then W b = λ T λb [ W = T, and one finds (by using Maxwell s equations): ( ) u N/c T ab = where σ ij = uδ ij ϵ 0 (E i E j + c 2 B i B j ) N/c σ ij This can also be written ( T ab = ϵ 0 c 2 F aµ Fµ b 1 ) 2 gab D, where D = 1F 2 µνf µν. ( [ i.e. T = ϵ 0 c 2 F F 1 ) 2 gd. ] Conservation of 4-momentum of matter and field together ( ) ( ) ( ) 1 u N/c E j/c, ρe + j B = c t, N/c σ

16 Poynting s argument (John Henry Poynting ( )): We want to find expressions for u and N, such that Using M4 to express j in terms of the fields: u t = N + E j. but for any pair of vectors E, B, E j = ϵ 0 c 2 E ( B) ϵ 0 E E t. (E B) = B ( E) E ( B). so E j = ϵ 0 c 2 (E B) + ϵ 0 c 2 B ( E) t ( 1 2 ϵ 0E E). Now use M3: E j = ϵ 0 c 2 (E B) t ( 1 2 ϵ 0c 2 B B ϵ 0E E ) Which shows that a possible assignment is u = 1 2 ϵ 0c 2 B ϵ 0E 2, N = ϵ 0 c 2 E B.