12.1 Chapter 12 Background The Lorentz force equation, F ≥ = q (E≥ + v≥c × B≥), (12.1) clearly can not be exact. Why? There must be a reaction to radiation emitted. This is assured by conservation of energy. Wheeler and Feynman wanted to get rid of self-energies in electrodynamics because they are infinite. For example, I showed in the class notes that the electrostatic self-energy of a point charge is linearly divergent (in quantum theory it's only logarithmically divergent). However, if one does this, one also throws out the force of radiative reaction! Dilemma was "solved" by them by assuming the force of radiative reaction was due to all the other particles in the universe! Let's see how this comes about. Imagine an isolated charge in an empty universe: e No radiation! 2 charges in the universe now:12.2 "emitter" "absorber" Radiation is allowed but only along this line! The basic idea is due to Tetrode (1922). Does to E&M what Mach's Principle does to gravity. Feynmann Propagator The EM wave equation has 2 types of solutions. From a HW problem (Ch.7, prob. 10): Gret.(x≥',t≥';x≥,t ) = 2c H(t-t') δ(R2 c2 -(t-t')2), (12.2) Gadv.(x≥',t';x≥,t) = 2c H(t'-t) δ(R2 c2 -(t-t')2), (12.3) δ(R2 c2 -(t-t')2) = "retarded" part ↓ 1 2 Rc { δ(Rc -(t-t')) + δ(Rc +(t-t'))}. (12.4) ↑ "advanced" part We also considered: = -kμxμ G+ ≠ 4π ∫ d3k (2π)3 ∫ dω (2π) eik≥.(x≥-x≥') -iω (t-t') k2 -ω2 c2 (1+iε) . (12.5)12.3 (12.5) defines the "Feynman propagator". It is evaluated in the complex ω-plane as shown (for the retarded part, one would go above both poles; see also Ch.7, prob. 11): x t-t' 0 x t-t' 0 => Re G+ = 12 (Gret. + Gadv.) = δ(R2 c2 -(t-t')2). (12.6) This turns out to be the crucial combination to use. To see this we have to look at the traditional calculation of radiation reaction. Radiation reaction calculation We know for a point charge (see Ch.11, prob.9; will adopt notation of DiBartolo, "Classical Theory of Electromagnetism" at this point): relativistic velocity ↓ fk(x≥,t) = qc Fkβ uβ. (β = 1,2,3,4) (12.7) ↑ ↑ force density field-strength12.4 Notice all indices are down. DiBartolo uses the "metric" xμ = (x≥, ict), => xμxμ = x≥2-c2t2. (12.8) Total self-force is (q uβ = jβ; ≈≥ ≡ x≥ -x≥') F k self(t) = 1c ∫ d3x Fkβ (x≥,t)|ret.jβ(x≥,t), (12.9) = 1 c2 ∫∫ d3x d3≈ 1 |≈≥| [∂jβ (x≥-≈≥,t') ∂xk -∂jk(x≥-≈≥,t') ∂xβ ]|t'= t-|≈ ≥ |/c jβ(x≥,t). (12.10) Adopt a model: jμ(x≥,t) = (0,icρ) , ρ(-x≥,t) = ρ(x≥,t), (12.11) which is true instantaneously at t. Picture: ≥x' ξ≥ \ x ≥ξ =x x' ≥ ≥ ≥ . -F k self(t) = Ik + IIk, Ik = -∫ d3x ρ(x≥,t) ∫ d3≈ 1 |≈ ≥ | (∂ρ(x≥ -≈ ≥ , t') ∂xk )|t'= t-|≈ ≥ |/c, (12.12)12.5 IIk = -1 c2 ∫ d3x ρ(x≥,t)∫ d3≈ 1 |≈ ≥ | (∂jk(x≥ -≈ ≥ , t') ∂t |t'= t-|≈ ≥ |/c. (12.13) A very long calculation follows. Just try to give you a flavor of the manipulations necessary. Start with a multipole expansion (here is where the NR assumption comes in) ρ(x≥ -≈ ≥ ,t') = ρ(x≥ -≈ ≥ ,t) -|≈| ≥c ∂ρ(x≥ -≈ ≥ ,t) ∂t + 12 |≈ ≥ |2 c2 ∂2ρ(x≥ -≈ ≥ ,t) ∂t2 -16 |≈ ≥ |3 c3 ∂3ρ(x≥ -≈ ≥ ,t) ∂t3 +... (12.14) Gives rise to (same notation as DiBartolo): Ik = k + k + k + k, k = -∫ ∫ d3x d3≈ ρ(x≥,t) 1 |≈| ≥ ∂ρ(x≥ -≈ ≥ , t) ∂xk , (12.15) = ∫ ∫ d3x d3≈ ρ(x≥,t) 1 |≈| ≥ ∂ ∂≈k ρ(x≥ -≈ ≥ , t), (12.16) = -∫ ∫ d3x d3≈ ρ(x≥, t) ρ(x≥ -≈ ≥ ,t) ∂ ∂≈k 1 |≈| ≥ , (12.17) \ even in ≥ξ (take x as origin ξ ≥ for integral) ≥ odd in ≥ξ-} ξξ ≥k 3 ⇒ k = 0. k = ∫ ∫ d3x d3≈ ρ(x≥,t) 1 |≈ ≥ | ∂ ∂≈k [|≈ ≥ | c ∂ρ(x≥ -≈ ≥ ,t) ∂t ], (12.18)12.6 ∂ρ ∂t = -∇≥.j ≥ = -∇≥(ρv≥) = -v≥.∇≥ρ = 0, (12.19) ↑ v≥ = 0 at t. ⇒ k = 0. k = -12 ∫ ∫ d3x d3≈ ρ(x≥,t) 1 |≈| ≥ ∂ ∂xμ [ |≈| ≥ 2 c2 ∂2ρ(x≥ -≈ ≥ ,t) ∂t2 ].(12.20) Remaining mathematical steps: 1) Change ∂ ∂xk into -∂ ∂≈k . 2) Use integrate by parts to get the ∂ ∂≈k on 1 |≈| ≥ . 3) Use equation of continuity to change ∂2ρ ∂t2 into (∂v≥ ∂t.∇≥≈) ρ(x≥ -≈ ≥ ,t). (12.21) 4) Use integrate by parts to put ∂v ≥ ∂t.∇ ≥ ≈ on ≈k |≈|3 ≥ from step 2. 5) Use ∫ d3≈ ≈i ≈k |≈|≥ 3 x (even function in ≈ ≥ ) = 13 δik ∫ d3≈ |≈| ≥ x (even function in ≈ ≥ ) (12.22) ⇒ k = 1 3c2 ∂vk ∂t ∫ ∫ d3x d3≈ ρ(x≥ ,t) ρ(x≥ -≈ ≥ ,t) |≈| ≥ (12.23)12.7 = v •k ≠ 2 Uel (electrostatic self-energy) Similar steps for k give: k = -1 3c3 v••k ∫ ∫ d3x d3≈ ρ(x≥,t)ρ(x≥ -≈ ≥ ,t) (12.24) = -q2 3c3 v••k, ⇒ Ik = 2 3c2 v•k Uel -q2 3c3 v••k. (12.25) Similarly, IIk = -2 c2 v•k Uel + q2 c3 v••k. (12.26) Wrong factor of 43 (caused by not accounting for all selfforces) ↓ ⇒ F k self(t) = -4 3c2 v•k Uel + 23 q2 c3 v••k. (12.27) ↑ ↑ contribution to mass (ultimately a quantum effect) gives rise to Larmor radiation form (for periodic motion) The above used the retarded G.f. from the very start. Advanced would use t' = t + |≈| ≥c . (12.28) We then find (problem):12.8 Ret: F k self = -43 v•k Uel c2 + 23 q2 c3 v••k, (12.29) Adv: F k self = -43 v•k Uel c2 -23 q2 c3 v••k, (12.30) 12 (Ret. + Adv.): f k self = -43 v•k Uel c2, (12.31) 12 (Ret. -Adv.): f k self = 23 q2 c3 v••k. (12.32) That the combination 12 (Ret. -Adv.) gives only the radiation reaction term was apparently first noticed by Dirac. Breakdown according to Feynman and Wheeler: Ret. = 12 (Ret. + Adv.) + 12 (Ret. -Adv.) (12.33) it's own field "emitter" contributed by other "absorber" particles Some qualitative pictures may help in understanding this:12.9 absorber (relative to emitter) emitter 12 (Ret.-Adv) 12 (Ret. + Adv) outgoing radiation want to have this situation The advanced field of the absorber just before the source emits: Equivalent to emitter advanced wave and just after the source emits: Equivalent to emitter retarded wave So, the absorber advanced wave is a linear combination of Adv., Ret. relative to emitter. Which one? External fields must be continuous at emitter source:12.10⇒ 12 Adv.|absorber = 12 (Ret. -Adv.)|emitter, (12.34) which is just Dirac's combination which gives the correct radiation reaction force, which is now assigned to the other particles in the universe. The radiation fields have the forms: A≥ret.(x≥,t) = 1c ∫d3x'dt' Gret.(x',t';x,t) J≥(x≥',t') = 1c ∫ d3x' J≥(x≥',t + Rc) |x≥ -x≥'| , (12.35) harmonic radiation eikr r ∫ d3x' 1c J ≥ (x≥')e-ikn^.x ≥ '. ↑ outgoing A≥adv.(x≥, t) = 1c ∫ d3x'dt' Gadv.(x',t';x,t) J≥(x≥',t') = 1c ∫ d3x' J≥(x≥',t-Rc) |x≥ -x≥'| , (12.36) same e-ikr r ∫ d3x' 1c J≥(x≥') eikn^.x ≥ '. ↑ incoming Atnthe source: B≥ret. = ikr^ × A≥ret., B ≥adv.= -ikr^ × A≥adv., (12.37) E≥ret. = B ≥ret.× r^ = -E≥adv. (12.38) Gives rise to the quasi-ID picture (after Cramer):12.11 atom emitter excited state absorber atom time space Thick line and thin lines add! Thick and thin lines cancel! Thick and thin lines cancel! Thick wavy line: emitter 12 (Ret. + Adv.) (E≥ or B≥ fields) Thin wavy line: absorber 12 (Ret. + Adv.) (E≥ or B≥ fields) Notice the advanced fields of the absorber, admitter cancel before the source "acts", and that the retarded fields of the two also cancel after the sink "absorbs". Between emission and absorption the fields add to the ususal retarted field. Comments: 1. Self-fields (rad. part) are zero at source location, consistent with zero self-force. Notice sign differences in (12.37), (12.38). 2. Actually, nothing oscillates on the light-cone.12.12 3. A similar diagram for electrostatic fields would show all fields adding everywhere. Further random comments: 1. Notice a standing wave is produced, but in time, not space. Thus there is no time evolution. Things only appear to evolve as we move along the time axis. 2. Picture has been extended to NR QMs by Cramer: "Transactional QMs". Picture above was taken from Cramer's review paper, listed below. 3. The differential equation including radiation reaction predicts either "run-away" solutions or "preacceleration", which, however, is unobservable in the lab. (It is the "witness" to the noncasual propagation.) 4. Implications (??): • Predestination ? • Closed universe ? 5. Is there physical content to this picture which can be tested, or is it merely a reinterpretation ? Ultimately, Feynman and Wheeler's theory failed because the self-forces are real and should not be eliminated. The Lamb shift in hydrogen demonstrates this. As Kramer points out, however, the FW assumption of noninteration of a particle with it's own field is ad hoc and can be removed from the theory. What remains is apparently a consistent, alternate theory of time develoment in classical electrodynamics. We have gone as far as possible with classical tools; must leave Clasical Electrodynamics (CED) for quantum electrodynamics (QED) to penetrate deeper. The understanding12.13 is worth the effort. May God's great and profound wisdom guide you in your journey!12.14 "Trust in the Lord with all your heart and lean not on your own understanding; in all your ways acknowledge Him and He will make your paths straight." -Proverbs 3:5,6 (NIV) "It is the glory of God to conceal a matter; to search out a matter is the glory of kings." -Proverbs 25:2 (NIV) References for Ch.12: J. D. Jackson, "Classical Electrodynamics", 3rd Ed., Ch.16. Di Bartolo, "Classical Theory of Electromagnetism", Ch.9. Wheeler & Feynman, Rev. Mod. Phys. 17, 157 (1945). J. Cramer, Rev. Mod. Phys. 58, 647 (1986).12.15 Problems 1. Finish the reduction of the term 4 k in the multipole expansion calculation of the self force, Fself. Show that 4 k = ∫ ∫ d3x d3ξ ρ(x≥,t) 1 |ξ≥ | ∂ ∂xk[ 1 6c3|ξ≥ |3∂3ρ(x≥-ξ≥ ,t) ∂t3 ], can be written as, 4 k = -e2 3c3 v.. k. 2. Consider the next term 5 k in the multipole expansion calculation of the "I" self force term. Let 5 k ≡ -∫ ∫ d3x d3ξ ρ(x≥,t) 1 |ξ≥ | ∂ ∂xk[ 1 24c2|ξ≥ |4 ∂4ρ(x≥-ξ≥ ,t) ∂t4 ]. This term can be shown to vanish. Show that you understand this by reducing it appropriately. 3. Let's say that we broke down the Ret. propagator in the Feynman Wheeler picture as Ret. = 12 (Ret. + X Adv.) + 12 (Ret. -X Adv.), it's own field "emitter" contributed by other "absorber" particles where X is some number. Would this be an acceptable change?12.16 4. Show that the advanced propagator contribution gives (12.30). 5. In your own words, explain why or why not this picture changes the expressions of classical E&M.