Chapter 3 Dynamis of the Eletromagneti Fields 3.1 Maxwell Displaement Current In the early 1860s (during the Amerian ivil war!) eletriity inluding indution was well established experimentally. A big row was going on about theory. The warring amps were divided into the Ation-at-a-distane advoates and the Field-theory advoates. James Clerk Maxwell was firmly in the field-theory amp. He invented mehanial analogies for the behavior of the fields loally in spae and how the eletri and magneti influenes were arried through spae by invisible irulating ogs. Being a onsumate mathematiian he also formulated differential equations to desribe the fields. In modern notation, they would (in 1860) have read: ρ.e = Coulomb s Law ɛ 0 B E = Faraday s Law (3.1).B = 0 B = µ 0 j Ampere s Law. (Quasi-stati) Maxwell s stroke of genius was to realize that this set of equations is inonsistent with harge onservation. In partiular it is the quasi-stati form of Ampere s law that has a problem. Taking its divergene µ 0.j =. ( B) = 0 (3.) (beause divergene of a url is zero). This is fine for a stati situation, but an t work for a time-varying one. Conservation of harge in time-dependent ase is ρ.j = not zero. (3.3) 55
The problem an be fixed by adding an extra term to Ampere s law beause ρ E.j + =.j + ɛ 0.E =. j + ɛ 0 (3.4) Therefore Ampere s law is onsistent with harge onservation only if it is really to be written with the quantity (j + ɛ 0 E/) replaing j. i.e. E E B = µ 0 j + ɛ 0 = µ 0 j + ɛ 0 µ 0 (3.5) The term ɛ 0 E/ is alled Maxwell s Displaement Current. There was, at the time, no experimental evidene for this term; still, the equations demanded it, as Maxwell saw. The addition makes the soure-free forms (where ρ and j are zero) beautifully symmetri B E E = ; B = µ 0 ɛ 0. (3.6) Meaning a hanging B-field indues an E-field (indution) but also a hanging E-field indues a B-field. The extra term is also responsible, as we shall shortly see, for eletromagneti waves. Despite their elegane, the Maxwell equations did not settle the argument. The equations an be rewritten in integral form, similar to what the ation-at-a-distane advoates wanted. However the ation, it now beame lear, ould not be regarded as instantaneous. Moreover EM waves were not deteted unequivoally for another 3 years! 3. Field Dynamis, Energy and Momentum 3..1 Introdution Suppose we take a apaitor and harge it up using a power supply. During harging, a B I V I E V Stored Energy. Figure 3.1: Energy obtained from the power supply in harging up a apaitor or indutor is stored in the eletromagneti field. Q urrent I is flowing and at any instant potential is V = C dq Q = Idt, I =. (3.7) dt 56
Thus the power supply does work at rate IV (per se) and total work done is dq IV dt = V dt = V dq (3.8) dt Q Q = dq = Starting at Q = 0). (3.9) C C Where has this energy gone? Answer: into the eletri field. The E-field within the apaitor stores energy with a volumetri energy density we shall alulate Now onsider harging an indutor with self-indutane L. di LI V = L. Work done V Idt = LIdI = (3.10) dt Where has this energy gone? The magneti field. 3.. Poynting s Theorem: Energy Conservation How do we know, or show, that EM fields store and transport energy? Formally from a theorem derived from Maxwell s Equations. Energy dissipation is the rate of doing work by fields on partiles. Energy is transferred from fields to partiles (and then often turned by some randomizing proess into heat ). Magneti field does no work on partiles beause F v. Eletri field rate of doing work on a single partile is Average rate of doing work on all partiles in a volume V is F.v = qe.v (3.11) j in V q j E (x j ).v j (3.1) so for an elemental volume dv suh that E an be taken uniform aross the volume, rate of work is E. q j v j = E. qnv dv (3.13) j where n is the number/unit volume = density. The average qnv is just the urrent-density, j. Hene the energy dissipation (= work done on partiles) rate (i.e. power) per unit volume is E.j (3.14) [Of ourse this is a loalized form of the iruit equation P = V I]. We have the energy dissipation rate density E.j, but now we an express this in terms of the fields by using Ampere s law: 1 E E.j = E. B µ 0 ɛ 0 (3.15) µ0 Now we onvert the form of the first term, E.( B) using a vetor identity. (E B) = B. ( E) E. ( B) (3.16) 57
so { } 1 E E.j =. (E B) + B. ( E) µ 0 ɛ 0 E. (3.17) µ 0 Then, using Faraday s law E = B/ we have { } 1 B E E.j =. (E B) + B. + µ 0 ɛ 0 E. (3.18) µ 0 Note that if we had used the auxiliary fields D and H (whih in vauo are D = ɛ 0 E and H = B/µ 0 ) we would have obtained { } B D E.j =. (E H) + H. + E. (3.19) whih is entirely equivalent in the vauum but subtly different for dieletri and magneti media in the way energy aounting is done. Notie that this an be written (for vauum) { } E.j =. (E 1 B/µ0 ) + (B.B/µ 0 + ɛ 0 E.E) (3.0) Although it may not be obvious, this is now in the form of an energy onservation law. The physial meaning an be made obvious by onsidering an arbitrary volume V, with surfae A. 1 E.jd 3 x =. (E B/µ 0 ) + [B.B/µ 0 + ɛ 0 E.E] d 3 x (3.1) V V 1 = (E B/µ 0 ).da B /µ 0 + ɛ 0 E d 3 x (3.) A V This says that the total rate of work on partiles in V is equal to minus the rate of hange V s = E ^B/µ 0 da Figure 3.: Integral of Poynting flux over surfae of V. of the integral over V of 1 [ B ] /µ 0 + ɛ 0 E (3.3) minus the flux of the vetor E B/µ 0 aross the surfae A. Physially this says that 1 [B /µ 0 + ɛ 0 E ] must be onsidered to be the eletromagneti energy density in the volume 58
V and the quantity E B/µ 0 must be onsidered to be the energy flux density (aross any surfae). s = E B/µ 0 = E H is alled the Poynting Vetor. If we write w 1 [B /µ 0 + ɛ 0 E ] the energy density, then Poynting s theorem is: w E.j =.s. (3.4) The signifiane of the indentifiation of field energy density and energy flux density is immense. Even today we tend to think of eletri power as being arried some how in the onduting ables. But if we understand Poynting s theorem and EM theory we realize the power is arried by the fields as represented by E B/µ 0. Not by the eletrons in the ondutor, although they do arry the urrent. 3..3 Momentum Conservation The rate at whih fields transfer momentum to partiles is equal to the EM fore q(e+v B) and the fore density is f = ρe + j B. (3.5) Use Maxwell s equations to eliminate ρ and j in favor of fields: 1 E ρ = ɛ 0.E ; j = B µ 0 ɛ 0 µ0. (3.6) Then E 1 f = ɛ 0 E (.E) + B B ( B) ɛ 0 µ 0 [ { } B 1 1 ] = (ɛ 0 E B) + ɛ 0 E + E (.E). (B.B) (B. ) B ɛ0 µ 0 1 1 = (ɛ 0 E B) + ɛ 0 [( E) E + E (.E)] µ0 B (B. ) B 1 = (ɛ 0 E B) + ɛ 0 (E. ) E (E.E) + E (.E) (3.7) 1 1 + (B. ) B (B.B) + B (.B) µ 0 Now the last two terms an be written as the divergene.t of a tensor quantity 1 1 1 T ɛ 0 EE E I + BB B I (3.8) µ0 where I denotes the unit tensor. In suffix notation 1 1 1 T ij = ɛ 0 E i E j E δ ij + B i B j Bδ ij µ0. (3.9) 59
T is alled the Maxwell Stress Tensor. And the fore equation (Momentum onservation) beomes f = (ɛ 0 E B) +.T (3.30) whih is, like Poynting s theorem, in onservation form. Just as before, the physial meaning is seen by integration over a volume, finding that 1 1 ɛ 0 E B = E B/µ 0 = s (3.31) is the volumetri field momentum density, and T is the fore per unit area at a surfae, i.e. 1 the stress. Eletromagneti fields thus arry momentum density that is times their energy 1 flux density, where we have used the fat that ɛ 0 µ 0 =. By onentrating on E and B, leaving all harges and urrents expliit in ρ and j we exlude all mehanial energy and momentum assoiated e.g., with motion of or polarization of atoms or moleules or their onstituent parts. (Even though that energy or momentum might be eletromagneti if we were dealing with E and B averaged aross all atoms). Most onfusion with energy and momentum in EM problems arises from not being lear about what ounts in the EM energy/momentum versus partile energy/momentum. 3.3 Indutane, Energy, and Magnet Stresses Poynting s theorem formalizes the observation we already made that the energy required to harge up an indutane (i.e. raise the urrent in it) is stored in the magneti field. We now know that the energy density is B /µ 0 (in vauuo) or 1 B.H in a linear magneti medium (but most magneti materials are not linear). Similarly the energy density stored in the eletri field of a apaitor is ɛ 0 E /, or 1 ED. We also found that the fore density assoiated with B was governed by a tensor 1 BB 1 B I the seond term of whih is µ 0 of the same form as the energy density B /µ 0. There is a fundamental reason for this relationship that we an show by thinking about fores on magnets. 3.3.1 Relation between energy density and magneti pressure in a solenoid Consider a solenoid formed by a urrent flowing azimuthally. It is easy to show that the EM fore j B is always outward. Atually we an t just take the total urrent and multiply by internal field to get the fore, beause j and B vary through the magnet. B = 0 outside. Instead of doing the integral of jb, let s alulate the fore by the method of virtual work. This involves imagining a small inremental motion, alulating energy hanges and putting them equal to the work done F.dx. So ignore the thikness of ondutor, and onsider an expansion of the initial radius a by a small inrement da. The stored magneti energy (per unit axial length) a B o πrdr hanges beause a hanges, and beause B (possibly) hanges. µ 0 Atually, whether or not B hanges depends on the external iruit attahed to the solenoid through whih the urrent flows. Let us hoose to suppose that that iruit ats to keep the 60
j a B j B ^ Figure 3.3: Fore on a solenoid magnet. urrent onstant so that B is onstant. We need to alulate how muh energy the iruit provides to the indutane. This requires the voltage during the expansion. Remembering Faraday s law, Φ V = E.dl = (3.3) If B is onstant, Φ = πa B so dφ da = Bπa (3.33) dt dt Hene the voltage indued in a single turn is da V = Bπa. (3.34) dt The urrent per unit length needed to give B is µ 0 J = B. So the work done by the external iruit is (per unit length) B da V Jdt = Bπa dt (3.35) µ 0 dt B dw iruit = πada (3.36) µ 0 (for a small inremental da). The hange in stored magneti energy is Then energy onservation is that B B d 3 x = πada = dw magneti. (3.37) µ 0 µ 0 dw iruit = dw magneti + dw mehanial (3.38) where dw mehanial = πada.p and P is the fore per unit length axially, per unit length azimuthally, i.e. the fore per unit area or pressure. Substituting B B πada = πada + P πada (3.39) µ 0 µ 0 61
Hene B P = (3.40) µ 0 is the outward pressure exerted by the B-field on the magnet. Notie that we never invoked any fore law suh as the Maxwell stress tensor, we only used our knowledge of magneti energy density. Also, the final result does not depend on our assumption about the external iruit. We ould have assumed anything we liked. If we did the energy ounting orretly, we would get the same fore result. Also, if we assume the B-field distribution in the ondutor does not hange, then we don t need to know what it is to obtain this result. So the total fore/area on the magnet doesn t depend on the urrent/field distribution in the magnet, provided the magnet is thin so that any energy stored in the magneti field in the thikness of the magnet is small. Magneti pressure is large for high fields. B P = µ0 = 4.0 10 5 B Pa = 4B bar. (3.41) 1T magneti field pressure is 4 bar ( atmospheres). 10T B-field pressure is 40MPa (.f. yield strength of hard opper 300MP a). For a thin ylinder, the stress (hoop stress) indued by a pressure P is a P (3.4) t a = radius, t = thikness. High field magnets have to be hunky and even then soon run t a θ σt Pθ σt Figure 3.4: The hoop stress, σ in a thin ylinder balanes the outward pressure, P. into stress limits. 3.4 Potentials for Time Varying Fields Eletrostati and Magnetostati problems are most easily solved using the potentials φ and A. These potentials are also ritial in time varying situations and general equations an be found as follows for the omplete Maxwell equations..b = 0 means that B = A (3.43) 6
is still a valid representation. Then B A E = = A = (3.44) So A E + = 0. (3.45) Therefore an be written as the gradient E + Ȧ (3.46) E + Ȧ = φ or E = φ A. (3.47) Then Coulomb s law beomes ρ A =.E =. φ + = φ (.A) (3.48) ɛ 0 and Ampere s Law E 1 1 φ A µ 0 j = B = ( A) + + 1 φ 1 A =.A + A + (3.49) Now remember that there is an arbitrariness to our hoie of A sine only its url is equal to B. In point of fat we an hoose.a to be whatever we want. One hoie, Coulomb gauge, was.a = 0. Lorentz Gauge : 1 φ.a =. (3.50) Then φ A 1 φ ρ = ɛ0 (3.51) 1 A = µ 0 j (3.5) Maxwell s equations are ompletely equivalent to these wave equations with soures. (Plus Lorentz gauge ondition.) Considered in this gauge, we see that the EM influene of ρ or j does not at instantaneously at-a-distane. Instead the influene has to propagate from the soures at the speed of light,. 63
3.4.1 General Solutions We want to find the general solution to these equations. We work just on the φ equation beause it is salar. Its solution will generalize immediately. First let s disuss the homogeneous equation 1 φ φ = 0 (3.53) whih is satisfied wherever ρ = 0 (in vauo). Also remember we an add any solution of this equation to a solution of the inhomogeneous eq. One solution type is plane waves, i.e. things that vary in only one diretion. If we hoose axes suh that = = 0 then equation is y z 1-d: φ 1 φ x = 0 (3.54) The general solutions of this equation are φ (x, t) = f (x ± t). (3.55) That is, arbitrary shaped funtions that move toward either inreasing or dereasing x, preserving their shape. For our problem the more interesting ase is spherially symmetri t=0 or t t x Figure 3.5: Arbitrary solution of the one-dimensional wave equation. waves. That is, in spherial oordinates (r, θ, χ) solutions suh that = = 0. Then θ χ Make the substitution then so 1 φ 1 φ φ = r r r r = (3.56) u φ = (3.57) r 1 φ 1 u 1 u 1 u r r r = r r r r r r = r u r r r r 1 u u u 1 u = + r = (3.58) r r r r r r u 1 u + = 0 (3.59) r 64
So u satisfies the 1-d (planar) wave equation, with general solution f(r ± t). Hene φ = f (r ± t) (3.60) r is the general solution of the homogeneous wave equation that is spherially symmetri. Expanding (- sign) or onverging (+ sign) radial waves. Atually this spherially symmetri solution doesn t satisfy the homogeneous equation at r = 0 beause of the singularity there. And in fat we already know from the stati problem that 1 r = 4πδ (r) = 4πδ (x x ) (3.61) (taking the enter of oordinate spae to be x ). Therefore, if f is everywhere non-singular, then 1 f(r ± t) = 4πf(±t)δ (x x ) (3.6) r So our solution φ = f(r±t) is really the solution of the (mathematial) problem of alulating r the potential of a time-varying point harge at position x, that is, of a harge density ρ = q(t)δ(x x ), where the harge is related to f by f (±t) = q (t) and hene f (r ± t) = q (t ± r/). (3.63) 4πɛ 0 4πɛ 0 In short, the potential at position x due to a time-varying harge of magnitude q(t) at x is φ(x, t) = q(t ± x x /) 4πɛ 0 x x (3.64) By onsidering the ase where the harge is a delta-funtion in time, as well as spae, q(t) = δ(t t ), we see that the Green Funtion G(x, t x, t ) (in time and spae) for the operator 1 L, (3.65) namely the funtion that solves LG(x, t x, t ) = δ(x x ) δ(t t ), is t ± x x t G (x, t x, t ) = δ (3.66) 4π x x and that the general solution of the eletrostati potential equation, 1 φ ρ φ = ɛ0 (3.67) is therefore 1 ρ x, t ± x x φ (x, t) = d 3 x. (3.68) 4πɛ 0 x x 65
3.5 Advaned and Retarded Solutions Notie we still have this ± sign in our potential solution. If we take the sign, then the integrand is ρ x, t x x. (3.69) x x This says that the ontribution to our potential at x, t from a harge density at x depends only on the value of that harge at the time t = t x x (3.70) This is earlier by the time it takes the EM influene to propagate from x to x, i.e. by x x. A potential based on this sign is alled the retarded potential beause the influene arrives later than the harge: retarded. The time t is alled retarded time very often, despite being earlier. [In English retarded delayed.] If we were to take the + sign we would have the very peuliar result that the influene (i.e. potential) would depend on the harge density at a later time. The advaned solution from ρ x, t + x x x x f x, t x x 1 ρ (x, t ) 4πɛ 0 x x d3 x µ 0 j (x, t ) d 3 x 4π x x (3.71) thus violates our ideas of ausality. We generally hold that an effet (potential) an only arise from a ause (harge) if the ause is earlier in time. For that reason, the advaned potential is disarded as unphysial but the justifiation for this hoie is mysterious, bound up in philosophial disussions of the arrow of time. Having obtained the general solution for the salar wave equation with soures, we an immediately apply it to eah vetor omponent of the equation for A; so in summary: φ (x, t) A (x, t) = = (3.7) (3.73) with t = t x x /. Often the notation [[f]] is used to denote evaluation of any funtion f at retarded time. You must be extremely areful with taking differentials of retarded quantities beause there is dependene on x not only in the spae but also in the time argument. Thus, for example = [[f]] = [[ f]] (3.74) the retarded value of a gradient is not equal to the gradient of the retarded value. 66
In general { } f x, t x x = [[ f]] + t x x f (3.75) x x f = [[ f]] + = [[f]] (3.76) x x and [[v]] = [[ x x v]] + x x v. (3.77) 67