SERIES ARTICLE. Snippets of Physics. 2. Angular Momentum of Electromagnetic Field * T Padmanabhan 1 8¼ (E 2 + B 2 ); P = U =

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1 Snippets of Physics 2. Angular Momentum of Electromagnetic Field * T Padmanabhan works at IUCAA, Pune and is interested in all areas of theoretical physics, especially those which have something to do with gravity. * This is based on an article originally published by the author in Physics Education, Vol. 23, No.4, p.285, Keywords Angular momentum, electromagnetic field, magnetic flux, solenoid, momentum density. T Padmanabhan Electrom agnetic eldscarry notonly energy and m om entum but also angular m om entum. T he angular m om entum ofthe eld can lead to som e curious results like the one w hich is described here. Y o u w o u ld h a v e certain ly lea rn t th a t th e electrom a g - n etic eld p ossesses en ergy an d m om en tu m. T h e u su al expressions for energy per unit volum e (U )andmomentum per unit volum e (P )are U = 8¼ (E 2 + B 2 ); P = (E B ) () 4¼c For exam ple, the expression for energy density is used in elem en tary cou rses to stu d y th e en ergy stored in a cap acitor or in a solen oid, w h ile th e ex p ression for electrom agn etic m om en tu m is requ ired to stu d y th e rad iation p ressu re of th e electrom agn etic w aves an d related phenom ena. W h at is n ot stressed ad eq u ately in tex tb ooks is that the electrom agnetic elds { and pretty sim ple ones at that { also p ossess an gu lar m om en tu m. Ju st as th e electrom agnetic eld can exchange its energy and m om entum w ith charged particles, it can also exchange its angular m om en tu m w ith a sy stem of ch arged p articles, often lea d in g to rath er su rp risin g resu lts. In th is in stallm en t, w e sh all exp lore on e su ch ex am p le. A sim ple con guration in w hich exchange ofangularm o- m entum occurs is show n in Figure, d iscu ssed in V olum eiioffeynmanlecturesinphysics[ ]. A p la stic d isk, locatedinthex {y p lan e, is free to rotate ab ou t th e vertical z -ax is. O n th e d isk is em b ed d ed a th in m etallic 08 RESONANCE February 2008

2 Solenoid with magnetic flux Φ Figure. The initial configuration of solenoid and a plastic disk. Ring of total charge Q rin g of rad iu s a carryin g a u n iform ly d istrib u ted ch arge Q.Alongthez-ax is, th ere is a th in lon g cu rren t-carry in g solenoid producing a m agnetic eld B con trib u tin g a total ux. T hisinitialcon guration iscom pletely static w ith a m a g n etic eld B con ned w ithin the solenoid and an electric eld E produced by the charge located on the ring. Let us suppose that the current source is disconnected leading the m agnetic eld to die dow n. T he ch an ge in th e m agn etic u x w ill lead to an electric eld w h ich w ill act tan gen tial to th e rin g of ch arge th ereby giving it a torque. O nce the m agnetic eld dies dow n, th is torqu e w ill resu lt in th e d isk sp in n in g ab ou t th e z - ax is w ith a n ite an gu lar m om entu m. T h e qu estion is w here does this angular m om entum com e from? F ey n m an p resen ts a d etailed d iscu ssion ab ou t th is p rob - lem butitis obviousthatthe angular m om entum in the in itia l eld is w h a t a p p ea rs a s th e m ech a n ica l a n g u la r m om entum ofthe rotating disk in the nalstage. W hat is really im p ortan t an d in terestin g is to w ork th is ou t and explicitly verify thatthe angularm om entum isconserved (w hich Feynm an unfortunately doesn't do!). I w ill d escrib e th is calcu lation as w ell as som e in terestin g issu es w h ich arise from it in th is in stallm ent[2]. The question is where does this angular momentum come from? RESONANCE February

3 The angular momentum in the initial field is related to the mechanical angular momentum in the final stage. T he angular m om entum ofthe nalrotating disk is easy to com pute. T he rate ofchange ofangular m om entum d L /dt due to the torque acting on the ring of charge is along the z -ax is an d so w e on ly n eed to com p u te its m a g n itu d e. T h is is g iv en b y d L d t = aq E = Q 2¼ I E d l = : (2) HereE is the tangential electric eld generated due to th e ch a n g in g m a g n etic eld a n d th e la st eq u a lity fo llo w s from F arad ay 's law. In tegratin g th is equ ation an d n oting that the initialangular m om entum ofthe disk and the nalm agnetic ux are zero,w e get L = Q 2¼c in itia l: (3) It is interesting that the nal angular m om entum depends only on the total ux and not on other con guration al d etails. W e n ow n eed to sh ow th at th e in itial static electrom agnetic con guration had this m uch ofstored angular m o- m entum. I w ill rst do this in a rather unconventional m annerand then indicate the connection w ith the m ore fam iliar approach. To do this, let us recall that the canonical m om entum of a charge q located in a m agnetic eld is given by p (q=c)a,wherea is the vector p oten tial related to th e m agn etic eld b y B = r A and p is th e u su al k in em atic m om en tu m. T h is su ggests th at on e can asso ciate w ith ch arges lo cated in a m agnetic eld, a m om entum (q=c)a. F or a d istrib u tion of charge,w ith a charge density ½,the eld m om entum per unit volum e w illbe (=c)½a. H en ce, to a ch arge d istribution located in a region of vector p otential A,we can attrib u te an an gu lar m om en tu m L A = c Z d 3 x ½ (x )[x A (x )]: (4) 0 RESONANCE February 2008

4 In ou r p rob lem, th e ch arge d istrib u tion is con n ed to a rin g of rad iu s a a n d th ere is n eg lig ib le m a g n etic eld in th e lo cation of th e ch arge. B u t th e vector p oten tial w ill exist ou tsid e th e solen oid an d th e ab ove exp ression can be non-zero. To com pute this, let us use a cylindrical coord in ate sy stem w ith (r;μ;z) as the coordinates. W e w ill ch o o se a g a u g e in w h ich th e v ector p o ten tia l h a s only the tangentialcom ponent; that is, only A μ is nonzero. U sing I A d l = ; (5) The derivation gives a physical meaning to the field momentum (q/c)a which is somewhat mysterious in conventional approaches. where isthetotalmagnetic ux,weget2¼ra μ = for a line integral of A around any circle. H ence A μ = = (2¼r). T h is can b e w ritten in a n ice vectorial form as A = (^z r); (6) 2¼r2 where^z is the unit vector in the z-direction. W hen we su b stitu te th is ex p ression in eq u ation (4) an d calcu late the angular m om entum, the integral gets contribution only from a circle ofradius a. U sin g fu rth er th e id en tity, r (^z r) = ^z r 2,wegettheresultthat L A = Q 2¼c in itia l^z ; (7) w hich is exactly the nal angular m om entum com puted in equation (3). R ather nice! that we T h is elem en tary d erivation, as w ell as th e ex p ression for electrom agnetic angular m om entum in equation (4) raises several intrigu in g issu es. O n th e p ositive sid e, it m akes vector potential a very tangible quantity, som e- thing w hich w e learnt from relativity and quantum m e- chanics but could never be clearly dem onstrated w ithin th e contex t of classicalelectrom agn etism. In th e p ro cess, italsogivesaphysicalmeaningtothe eldmomentum (q=c)a w hich is som ew hat m ysterious in conventional approaches. O n the ip side,one should note thata,by RESONANCE February 2008

5 We would like to have a definition of electromagnetic angular momentum which is gauge invariant. the very de nition, is gauge dependent and one w ould have preferred a de nition of electrom agnetic angular m om en tu m w h ich is p rop erly gau ge in variant. It is, of cou rse, p ossib le to w rite d ow n an oth er ex p ression for th e electrom agn etic an gu lar m om en tu m w h ich is m ore con ven tion al. G iven th e d en sity of electrom agn etic momentum,p,w e can de ne the corresponding angular m om entum density asx P. Integratin g it over allsp ace sh ou ld give th e an gu lar m om en tu m asso ciated w ith th e electrom agnetic eld. Since the m om entum density P involves on ly th e electric an d m agn etic eld s, th e resu lting expressions are autom atically gauge invariant. T his leads to a de nition ofangular m om entum given by L EM = 4¼c Z d 3 x [x (E B )]; (8) w hich justreplacesthem om entum density ½ A =c in eq u a - tion (4) b y (E B =4¼c). It is trivial to verify th at, as m om entum densities,these two expressions are unequal in general.butwhatisrelevant,asfarasourcomputation go es, is th e in tegral over th e w h ole sp ace of th ese tw o ex p ression s. If th ese tw o ex p ression s d i er b y term s w h ich van ish w h en in tegrated over w h ole sp ace, th en w e have an equivalent gauge invariant de nition of eld angular m om entum. It tu rn s ou t th at th is is in d eed th e case in an y static con guration ifw e choose to describe the m agnetic eld in a gauge w hich satis es r A = 0. O ne can then show th at 4¼ (E B ) = 4¼ (E (r A )) = ; (9) wherev isa com plicated second rank tensorbuiltout of eld variables. W e are using the convention that repeated G reek indices (like in th e a b o v e la st term o f the above expression) are sum m ed over,2,3. W hile 2 RESONANCE February 2008

6 one can provide a proof of equation (9) using vector id en tities (you sh ou ld try it ou t!), it is a lot faster an d neater to use four-dim ensionalnotation and specialrelativ ity to get th is resu lt. S u ch a d erivation is ou tlin ed in th e ap p en d ix for th ose w h o are fam iliar w ith th e fou r- dim ensionalnotation. G iven the result in (9),it is easy toseethatinourexamplewewillgetthesameresult irresp ective of w h eth er w e u se L A or L EM.Thisisbecause, w hen w e integrate the expressions in (9) over all sp ace, th e term in volv in g V can be converted to a surface term at in nity w hich does not contribute. A ppendix L et m e b rie y o u tlin e th e d eriv a tio n o f (9 ) fo r th o se w h o a re fa m ilia r w ith th e fo u r-d im en sio n a l n o tatio n. W e b e- gin w ith the expression for the m om entum density of th e electrom agn etic eld in term s of th e stress ten sor T ab of the electrom agnetic eld (w ith the convention th at L atin letters ran ge over 0,,2,3). T o sim p lify th e expressions w e w ill also use the i (@=@x i ), etc. T he T 00 com p on en t of th is ten sor is p rop ortion al to th e en ergy d en sity of th e electrom agn etic eld, w h ile th e T 0 is p rop ortion al to th e m om en tu m d en sity P. M ore precisely, T 0 = 4¼ (E B ) = cp : (0) O n the otherhand,theelectrom agneticstresstensorcan b e w ritten in term s of th e fou r-d im en sion al eld ten sor F ab in th e fo rm T 0 = (=4¼ )F F 0. W e w illm anipulate this expression using the facts that (i) the con guration is static an d (ii) th e vector p oten tial satis es th e gauge condition r A A = 0, to p rove (9). U sin g th e d e n ition of th e eld ten sor in term s of th e fou r-vector p oten tial, F ij i A i A j,we can write RESONANCE February

7 SERIES ARTICLE Suggested Reading [] R P Feynman, R B Leighton and M Sands, Feynman Lectures in Physics, Narosa Publishing House, New Delhi, Vol.II, Section 7-4, 986. [2] There is large literature on this problem not all of which is illuminating; one good place to start the search, J M Agqirregabiria and A Hernandez, Eur.J.Phys., Vol.2, p.68, 98. Address for Correspondence T Padmanabhan IUCAA, Post Bag 4 Pune University Campus Ganeshkhind Pune 4 007, India. paddy@iucaa.ernet.in nabhan@iucaa.ernet.in T 0 = 4¼ F F 0 = 4¼ (@ A )F 0 ; = 4¼ (@ A )F 0 + (F 0 A ) F 0 ; 4¼ = 4¼ A 0 ) + (F 0 A ) F 0 : 4¼ () T o arrive at th e secon d lin e w e h ave d on e an integration by p arts an d to ob tain th e th ird lin e w e h ave u 0 A = 0 sin ce th e con gu ration is tim e in d ep en d en t. W e n ext u se th e resu F 0 = r E = 4¼ ½ in th e last term an d an oth er integration by p arts in th e rst term, u sin g th e gau ge con d ition r A A = 0. T h is gives T 0 = ½ A + [A A A 0 ]: (2) W e th u s n d th at cp = ½ A V ; V 4¼ [A 0@ A A 0 ] (3) w h ich p roves th e equ ivalen ce b etw een th e tw o ex p ression s for electrom agn etic m om en tu m d en sity (cp an d ½ A ) w h en u sed in integrals over all sp ace, p rovid ed th e secon d term van ish es su ± cien tly fast. F or th e case w e are d iscu ssin g, th is is in d eed tru e. Scientific reasoning does not differ from ordinary everyday thinking in kind, but merely in degree of refinement and accuracy, more or less as the performance of the microscope differs from that of the naked eye. To be sure, when the pioneer in science sends forth the groping feelers of his thoughts, he must have a vivid intuitive imagination, for new ideas are not generated by deduction, but by an artistically creative imagination. Max Planck 4 RESONANCE February 2008