Electromagnetism Notes, NYU Spring 2018

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1 Eletromgnetism Notes, NYU Spring 208 April 2, 208 Ation formultion of EM. Free field desription Let us first onsider the free EM field, i.e. in the bsene of ny hrges or urrents. To tret this s mehnil system one must first identify the generliztion oordintes. As usul there re different possibilities: one is to onsider s oordintes the vlues of the field t every point in spe thus hving infinite number of generlized oordintes, the other is to expnd the field in set of modes nd view the expnsion oeffiients s the generlized oordintes. The ltter is the one we will follow, in prtiulr sine we lredy know how to expnd the free-field in terms of the hrmonis osilltor expnsion of EM wves. Thus, we use the vetor potentil Fourier oeffiients A k s oordintes, whih obey A k + ω 2 A k = 0 ω k = k To write the Lgrngin for these oordintes we use the ft tht mode by mode i.e. eh k seprtely is hrmoni osilltor, thus the Lgrngin for mode k reds: L k = A k ω 2 A k A 2 Indeed, the Euler-Lgrnge equtions red d L dt L A k A = 0 = d A k dt k ω k 2 A k = A k + ω2 A k k 3 To write down the totl Lgrngin, we just dd mode by mode nd hoose n overll normliztion to obtin the orret Hmiltonin, s derived before from the energy, we will double hek this lter L = V A k 2 A k 2 k 2 A k A k 4 with V the volume of the box where the Fourier modes re defined. Sine A k = 0 A = 0, E k = A, k B k = i A k, nd then E k E k = 2 A k A k, B k B k = k A k k A = k A k A k k A k k A = k 2 A k A k. Thus sine E k = E nd the sme for B k, we n write L = V E k E k B k B k = V E k 2 + B 2 k 5

2 From Prsevl theorem we n just write this in rel spe L = ˆ [ E 2 x B x] 2 d 3 x 6 i.e. E 2 d 3 x =, V e i+ r d 3 r = V E k E Note tht the potentils dispper, so is being written in terms of physil guge invrint EM fields. Let us now see tht we n derive the soure-free Mxwell s equtions from the orresponding tion. ˆ S = L dt = ˆ E 2 B 2 d 3 x dt 7 We now onsider E & B in terms of Φ & A: E = A Φ, B = A 8 nd vry Φ & A we do not fix the guge, for resons tht will beome ler lter, doing so does not led to hnge in the tion tht hnges the EOM s. Vrying Φ by δφ first: δ E = δφ &δ B = 0 9 δs = ˆ 2E δe d 3 x dt = ˆ 4π where the lst eqution using integrte by prts. Then, setting the vrition to zero, δs = 0 E δφ d 3 x dt = ˆ 4π EδΦ d 3 x dt 0 δ A E = 0 Similrly A A + δa, δe =, δ B = δa, then: δs = ˆ [ ] E 4π B δa d 3 x d t 2 Thus, setting δ A = 0, we hve: B = The remining Mxwell s equtions, B = 0 & E = of E, B in terms of φ, A. We n now disuss the Hmiltonin. The generliztion momentum is E B re just onsequene of the definition p k = L A = 2 V A 2 k 3 k 2

3 where 2 omes from nd in sum ontribute to. H = k p k A k L = V 2 = V 2 2 A k A k V A k k 2 A k A 2 k 2 A k A 4 whih gin we n write in rel spe s H = ˆ V E 2 + B 2 d 3 x 5 s we expet from U with orret normliztion, inherited from normliztion of L s pointed out bove..2 Chrged prtile desription Let us now onsider the motion of hrge prtile in given EM field nd derive the Lgrngin nd Hmiltonin desription. The eqution of motion is F = m for the Lorentz fore: m v = q E + v B 6 whih we n write in terms of potentils, m v =q Φ = q Φ + q v A d dt A + q v A q + v A m v + q A = q Φ v A 7 8 where we used v A v A, v = 0 nd + v = d dt. This is the Lorentz fore eqution in terms of potentils. Now, we wnt this to be the Euler-Lgrnge eqution of motion for the Lgrngin of hrged prtile, i.e. d L dt v L r = 0 9 Whih, nothing tht v v A = A nd r v A = v A We n reprodue the A dependent terms in the EOM, then L = 2 mv2 qφ + q v A 20 3

4 is the Lgrngin. For the Hmiltonin we proeed s usul. The onjugte momentum is p = L v = m v + q A 2 H = p v L = m v + q A v 2 mv2 qφ + q v A 22 But in terms of p = 2 mv2 + qφ H = p q A 2m 2 + qφ 23 from whih you n hek Hmilton s eqution leds to Lorentz fore eqution bove.3 Interting hrge-fields desription We n now put things together, dding the Lgrngin for olletive of hrged prtiles in given field, to the Lgrngin of the free EM fields: L = [ 2 m v 2 q Φ r, t + q ] v A r, t + ˆ E 2 B 2 d 3 x 24 where sums over the hrge in the system. We n lso write this in ontinuous form by noting ρ r = q δ r r, j r = q v δ D r r whih leds to: L = ˆ 2 m v 2 + j A ρφ d 3 x + ˆ E 2 B 2 d 3 x 25 where the first term is free prtile, the seond term is intertion prtile-fields nd the third term free fields. While it is obvious the intertion term generte the Lorentz fore for the hrges, it is not obvious t this stge tht it lso ounts for the retion of EM fields by hrges nd urrents. To hek this, let s derive the equtions of motion from this L. First, wht re the generlized oordintes? As we did before, we tke them to be the positions of the hrges nd the potentils Φ & A s oordintes, we expnded A in Fourier modes nd worked with the oeffiients A k s generlized oordintes we worked in Coulomb guge where A = 0 nd φ = 0 sine ρ = 0 for the free fields. Now we proeed in different fshion using Φ r, t nd A r, t s oordintes whih lso helps us s wrm up for the ovrint formultion we shll disuss lter. First, note tht the Euler-Lgrnge equtions for the degrees of freedom orresponding to the hrges sme s before, sine using the first expnding of L bove it is ler tht r nd v only pper in re the the free & intertion terms, whih re identil to the se of hrges in given field. Then d dt leds s before to the Lorentz fore eqution for hrge q : d [ m v + q A r dt ] [ = q Φ r v ] A r 4 L r = L r 26

5 To work out the Euler-Lgrnge equtions for the fields Φ r, t nd A r, t we now note tht these generlized oordintes depend on two prmeters : r nd t not just t like r t; i.e. they re fields tht obey prtil differentil equtions in spe & time insted of ordinry differentil eqution for time evolution like r t. Then the Euler-Lgrnge equtions re in this se d L dt F + L i i F = L F summtion over i =, 2, 3 27 A where F is either Φ, or the omponent of A. To rry out the lultion, we rewrite L in terms of Φ & L = ˆ 2 m v 2 + j A ρφ d 3 x + ˆ Φ + 2 A A 2 d 3 x 28 Note tht s onsequene of this, the homogeneous Mxwell equtions re utomtilly stisfied, i.e. B = 0 & E = B. The E-L eqution for Φ is L L = 0, Φ i φ = 2 Φ + A = 4π E L i, φ = ρ 29 i s expeted it s Guss s lw. For A we hve L = A 2 Φ + 4π E = ρ, E = 4πρ 30 A = E 4π 3 nd L A = j. Then we obtin from E-L eqution L = A 2 ial i A l d El + dt 4π 4π ɛ lij i B j = jl = 4π A A i A l = B 4π ê jɛ jkm k A m i A k = B 4π ê j ɛ jkm δ ki δ lm = 4π B jɛ jil = 4π ɛ lijb j B = 4π j + E 4π + 4π B = j E

6 whih is Ampere-Mxwell. Then indeed the intertion term in the Lgrngin lso ounts for the retion of EM fields by ρ & j. Note: In ll the lultions bove, when tking derivtives of the Lgrngin with respet to fields we hve used tht you n think of the integrl over spe s sum over the degrees of freedom position r, nd only those orresponding to the field t the position we re lulting the derivtive ontribute s if it were sum. Formlly this n be written s the so-lled funtionl derivtive, but there is no need to introdue tht for our purpose..4 Guge invrine nd hrge onservtion Suppose we mke guge trnsformtion, Φ Φ ψ, A A + ψ 35 where ψ is rbitrry. As we sw, this mintins the physil fields E & B, so they re guge-invrint. Let s look t the effet of this guge trnsformtion on L in two wys: first in terms of the disretized version where the intertion Lgrngin where Φ & A reside is sum over prtiles, then in the ontinuous version where ρ nd j show up. So, first, we hve δl = q ψ + v ψ r, t = d q dt ψ r, t 36 where we used d dt ψ r, t = ψ + d r dt ψ i.e. totl time derivtive, whih hnges the tion δs = δl dt by boundry term nd does not hnge the E-L eqution of motion. Tht is the reson why we did not fix the guge before in doing our lultions, beuse the E-L eqution re not ffeted by suh hoie. Inidentlly, this onfirms wht we lredy know, i.e. the vrition we see here only oming from the intertion term in L desribes the Lorentz fore s we sw when we onsidered hrge in given EM field nd this is lerly guge invrint sine F = q E + v B depends on guge-invrint E, B. Now, to see the onnetion to hrge onservtion ρ + j = 0 we must use the ontinuous version, beuse the ontinuity eqution is trivilly stisfied otherwise i.e. for ρ = q δ D r r, j = q ρ v δ D r r, = δ d r D r r dt = q v δ D r r = j. So, in this se δs = ˆ ˆ dt d 3 x ρ ψ + j ψ = ˆ [ ρψ dt d 3 x + jψ ˆ 37 ρ dt d 3 x ψ + j where we used integrl by prts. The first term gin is boundry term, does not hnge EOM s. Sine ψ is rbitrry, we thus see tht guge invrine requires hrge onservtion ρ + j = 0. 6