7#,! tqtqL&&&6˰˰BxZ j Ί*8q"΢>/ Chapter 6 Plausibilty argument leading to Maxwell's equations Static equations (vacuum): electrostatics: blc{(s(l(o(,).o(E,) = 4pr, (6.1)),l(o(,) x o(E,) = 0. (6.2)))) magnetostatics: blc{(s(l(o(,).o(B,) = 0, (6.3)),l(o(,) x o(B,) = f(4p,c) o(J,). (6.4)))) Current conservation says: f(r,t) + o(,).o(J,) = 0. (6.5) By defn f(r,t) = 0 in electrostatics; I also argued that the same result held in magnetostatics where o(J,) = o(J,)(o(x,)) only. We now need to find the form of the field eqns when time variation of the sources and fields is considered. The following will be a plausibility argument only. (Argument largely due to Schwinger.) Let's start off by thinking about a static situation and then generalize to a moving frame. Let us say only electric fields are present in the "static" frame. We have simply f(do(E,),dt) = 0. (static form) (6.6) We will now examine the situation in a "moving" frame, applying the type of physics that we know is manifest at low velocities (i.e., Galilean transformations). In the moving frame, we will see a changing electric field, for example:  The above is drawn from the point of view of an observer at rest wrt the electric fields. The moving observer will see a changing electric field simply because of the spatial variation of o(E,). He or she describes the situation by, for example f(E2,t) = v1 f(E2,x1), (6.7) f(E1,t) = v1 f(E1,x1). (6.8) The generalization is f(o(E,),t) = b(o(v,).o(,))o(E,), (6.9) => b(f(,t) - o(v,).o(,)) o(E,) = 0. (6.10) b(f(,t) - o(v,).o(,)) is the "connective derivative". Can think of the - o(v,) as being the velocity of the charge distribution wrt the moving axes. We see that in going from one frame to another, we simply make the replacement: f(d,dt) . f(,t) - o(v,).o(,). (6.11) Notice that (o(v,)=constant) (assume o(,).o(E,) = 4pr in any frame) o(,) x b(o(v,) x o(E,)) = o(v,)b(o(,).o(E,)) - b(o(v,).o(,))o(E,), (6.12) = 4pr o(v,) - b(o(v,).o(,))o(E,). (6.13) Therefore, in the moving frame, we have f(1,c) f(o(E,),t) - 4pr f(o(v,),c) + o(,) x b( f(o(v,),c) x o(E,) ) = 0. (6.14) We recognize that the 4pro(v,) term is a current since it represents the displacement of charge. We confirm this as follows. In the rest frame we have f(dr,dt) = 0, (6.15) which, in the moving frame becomes f(r,t) - b(o(v,).o(,))r = 0, (6.16) f(r,t) + o(,).(-ro(v,)) = 0. (6.17) From which we identify o(J,) = - o(v,)r . (6.18) (This is correct from the point of view of the moving observer since the current flows in the -o(v,) direction.) Using this identification above, we have -o(,) x b(f(o(v,),c) x o(E,)) - f(1,c) f(o(E,),t) = f(4p,c) o(J,). (6.19) This is an equation now governing the time behavior of o(E,) in any frame of reference, forgetting about the origin of o(J,). If we insist that this be a generalization of (6.1) - (6.4), the only possibility, because of the presence of o(J,), is (6.4). The generalization is only possible if we set o(B,) = - f(o(v,),c) x o(E,), (6.20) (notice o(,).o(B,)=0) in which case we get o(,) x o(B,) - f(1,c) f(o(E,),t) = f(4p,c) o(J,). (6.21) What we are apparently seeing is the fact that a pure electric field, as seen by a moving observer, is no longer purely electric. This seems surprising, perhaps, if our attention is on the fields, but seems less so if we think of the source of o(E,), charge density, which is traveling in the - o(v,) direction for the moving observer. Although the above has been inferred in a special circumstance, we hypothesize that the above form is general. The added term does not spoil the linearity of the equations. Moreover, it is consistent with (6.1) above (which we have assumed is true in general) since o(,).(o(,) x o(B,) - f(1,c) f(o(E,),t)) = f(4p,c) o(,).o(J,), (6.22) => - f(1,c) f(,t) o(,) .o(E,) = f(4p,c) o(,).o(J,), (6.23) => f(r,t) + o(,).o(J,) = 0. (6.24) (This is the original derivation of continuity I did in the Introduction.) Next, imagine a situation with static magnetic fields only. Actually, the situation can not be as simple as this since we assume in classical electrodynamics that moving charges are the source of all magnetic fields. However, let us consider magnetic fields only in the region where o(E,) = 0. Example: the entire region outside a current loop. Then, as before with electric fields, f(do(B,),dt) = 0 "stationary" frame, (6.25) => f(o(B,),t) - b(o(v,).o(,))o(B,) = 0 "moving" frame. (6.26) As before (assume o(,).o(B,) = 0 continues to hold) o(,) x b(o(v,) x o(B,)) = o(v,)b(o(,).o(B,)) - b(o(v,).o(,))o(B,), (6.27) => f(1,c) f(o(B,),t) + o(,) x b(f(o(v,),c) x o(B,)) = 0. (6.28) If we insist again that this represents a generalization of the static Maxwell eqns, the only possibility is (6.2) above. This identification is complete of we set o(E,) = k f(o(v,),c) x o(B,). (6.29) Compare with Lorentz force ("q" a test charge): o(F,)' = qo(E,) = kq f(o(v,),c) x o(B,) <=> o(F,) = q f(o(v,),c) x o(B,) => k=1, => o(,) x o(E,) + f(1,c) f(o(B,),t) = 0. (6.30) Physical interpretation: moving observer sees an electric field also. Full set: o(,).o(E,) = 4pr, (6.31) o(,) x o(E,) + f(1,c) f(o(B,),t) = 0, (6.32) o(,) x o(B,) - f(1,c) f(o(E,),t) = f(4p,c) o(J,), (6.33) o(,).o(B,) = 0. (6.34) We are momentarily pleased with ourselves for this deduction. However, even if we acknowledge that o(E,) and o(B,) are not absolute fields, but depend upon the motion of observers, we still have a problem with 19th century physics. These equations do not have the same form in a moving coordinate system, i.e. they are not form invariant ("covariant") under a Galilean transformation. That is, if we start with the above and make the changes (ignore the quantities in parentheses in (6.35) and (6.38) for now) o(,) . o(,) b( + f(o(v,),c2) f(,t) ), (6.35) f(,t) . f(,t) + o(v,).o(,) , b(s(assuming f(d,dt') = f(d,dt),l(=> f(,t') - o(v,).o(,) = f(,t)),,l(=> f(,t') = f(,t) + o(v,).o(,) ))), (6.36) blc{(a(l(o(E,) . o(E,) + f(o(v,),c) x o(B,)),l(o(B,) . o(B,) - f(o(v,),c) x o(E,)))) (6.37) blc{(a(l(r . r (- f(o(v,),c2).o(J,))),l(o(J,) . o(J,) - o(v,)r ))) (6.38) We do not get back the same form in the moving frame. However, if we add or subract the two quantities in brackets as indicated, we do get form invariance, at least if we consider only terms first order in o(v,)/c. What is the physical significance of this? Also, what is the general transformation that Maxwell's eqns are covariant under? You probably already know the answers to these questions; well be talking more about such transformations next semester. Faraday's law There are a wealth of new phenomena accounted for when these 2 new terms are introduced. ( Historically, the f(1,c) f(o(B,),t) term, the "displacement current", was introduced by Maxwell.) Consider any closed loop in space:  Integrate o(,) x o(E,) equation over the surface: i(s,, )da o(n,^).(o(,) x o(E,) + f(1,c) f(o(B,),t)) = 0. (6.39) Use Stokes thm. on first term: o(,o) o(E,).do(l,) = - f(1,c) f(d,dt) i(,, )da o(n,^).o(B,). (6.40) Now let the loop be a thin, conducting wire. Define "electromotive force" (not a force at all!) eo(=,-) o(,o) o(E,).do(l,). (6.41) The o(E,) field is forming a closed loop; o(E,) o(=,/) -o(,)F. Define magnetic flux: F o(=,-) i(,, )da o(n,^).o(B,), (6.42) => e = - f(1,c) f(dF,dt). "Faraday's law" (6.43) True for any surface, S. The minus sign is all-important. The induced current (if the wire is conducting) is in such a direction as to oppose the change in flux in the circuit (Lenz' Law). This is the basic law necessary to understand diamagnetism; electron orbits play the role of resistanceless current loops. Model introducing macroscopic Maxwell equations Next: macroscopic Maxwell eqns. (This model is from Schwinger). We've already seen the appropriate generalizations in the static cases. To do this, we replaced summations over atoms by integrals over density functions. In other words, we are trying to build an effective classical model of the microscopic world. This is of course impossible; does not mean it will not be useful, however! As before, the following should be regarded as a plausibility argument only. Consider an assembly of charges (electrons + heavy nucleus) which is electrically neutral: ISU(k,, )ek = 0. (6.44) The net force on it is o(F,)atom = ISU(k,, )[ek o(E,)(o(x,)k) + ek f(o(v,)k,c) x o(B,)(o(x,)k)]. (6.45) [o(F,)k = f(1,c) i(,, )d3x'o(J,)k(o(x,)',t) x o(B,)(o(x,)'); o(J,)k(o(x,)',t) = qk f(do(x,)k,dt) d (o(x,)'-o(x,)k(t)); => o(F,)k = f(qk,c)o(v,)k x o(B,)(o(x,)k); Lorentz force law.] If the system is small enough (implying small changes in the effective fields o(E,)(o(x,)k), o(B,)(o(x,)k)), we can expand o(E,)(o(x,)k) and o(B,)(o(x,)k) about the C of M of the charge distribution, which let's say, lies at o(R,). Then o(E,)(o(x,)k) = o(E,)(o(R,)) + [(o(x,)k - o(R,)).o(,)]o(E,)(o(R,)) + ... , (6.46) o(B,)(o(x,)k) = o(B,)(o(R,)) + [(o(x,)k - o(R,)).o(,)]o(B,)(o(R,)) + ... , (6.47) (1) => o(F,)atom = isu(k,, )[ek o(E,)(o(R,)) + ek f(o(v,)k,c) x o(B,)(o(R,))] (2) small for atomic systems (3) + isu(k,, )ek [(o(x,)k - o(R,)).o(,)]o(E,)(o(R,)) + isu(k,, )ek f(o(v,)k,c) x [[(o(x,)k - o(R,)).o(,)]o(B,)(o(R,))]+... . (6.48) In (2) identify o(p,) = isu(k,, )ek(o(x,)k - o(R,)) = isu(k,, )ek o(x,)k, (6.49) (consistent with o(p,) = i(,, )d3x'o(x,)'r(o(x,)'); r (o(x,)') = isu(k,,) ek d(o(x,)'-o(x,)k + o(R,)).) In (1) above, we recognize isu(k,, )ek o(v,)k = f(do(p,),dt). (6.50) Setting aside the third term temporarily, we have o(F,)atom = (o(p,).o(,)) o(E,)(o(R,)) + f(1,c) f(do(p,),dt) x o(B,)(o(R,)) + (3) +... . (6.51) However (either f(,t) or f(d,dt) will do in (6.52) and (6.53)) f(1,c) f(do(p,),dt) x o(B,) = f(1,c) f(,t) (o(p,) x o(B,)) - o(p,) x f(1,c) f(o(B,),t), = f(1,c) f(,t) (o(p,) x o(B,)) - o(p,) x (-o(,) x o(E,)) (o(p,) = const. in space) = f(1,c) f(,t) (o(p,) x o(B,)) + o(,)(o(E,).o(p,)) - (o(p,).o(,))o(E,), (6.52) => o(F,)atom = o(,) (o(p,).o(E,)) + f(1,c) f(,t) (o(p,) x o(B,)) + (3) +... . (6.53) An approximate rearrangement of (3) is isu(k,, )ek f(o(v,)k,c) x [[(o(x,)k - o(R,)).o(,)] o(B,)(o(R,))]  o(-,~) f(d,dt) {isu(k,, )ek f(1,c) (o(x,)k - o(R,))x[[(o(x,)k - o(R,)).o(,)] o(B,)(o(R,))]} -isu(k,, )ek f(1,c) (o(x,)k - o(R,)) x [(o(v,)k.o(,))] o(B,)(o(R,)), (6.54) where we are neglecting terms involving o(v,) f(do(R,),dt) because we assume that |o(v,)| << |o(v,)k| << c. (6.55) We are also neglecting terms involving f(do(B,),dt). (Says the effective magnetic field varies only slightly in a characteristic atomic time.) Neglecting the first term (it represents a contribution to the non-traceless quadrupole tensor) on the right of the above, we have that isu(k,, )ek f(o(v,)k,c) x[[(o(x,)k - o(R,)).o(,)] o(B,)(o(R,))]o(-,~) -isu(k,, )ek f(1,c) (o(x,)k - o(R,))x[(o(v,)k.o(,))]o(B,)(o(R,)). (6.56) An average of the two forms, since [(o(x,)k - o(R,)) x o(v,)k]x o(,) = -(o(x,)k - o(R,))(o(v,)k.o(,)) + o(v,)k[(o(x,)k - o(R,)).o(,)], (6.57) gives isu(k,, )ek o(v,)k x[[(o(x,)k - o(R,)).o(,)] o(B,)(o(R,))] -> f(1,2) isu(k,, )f(1,c) ek {[(o(x,)k - o(R,)) x o(v,)k] x o(,)} x o(B,)(o(R,)). (6.58) We then identify o(m,) = f(1,2c) isu(k,, )ek (o(x,)k - o(R,)) x o(v,)k. (6.59) [It follows from o(m,) = f(1,2c) i(,, )d3x'o(x,)'x o(J,)(o(x,)'), where o(J,)(o(x,)') = isu(k,, )ek o(v,)k d(o(x,)'-o(x,)k+o(R,)). (neglecting f(do(R,),dt) term) Notice, however, i(,, )d3x o(J,)(o(x,)') = isu(k,, )ek o(v,)k = f(do(p,),dt) 0, so the lowest moment does not vanish. Therefore, o(m,), as defined above, is dependent upon our choice of the origin; this is a failure of the classical model. However, we assume from the dynamics of heavy nuclei and light electrons, that the center of mass in our classical model is most appropriate place to center our coordinates.] So, we have for the third term: (3) o(-,~) (o(m,) x o(,)) x o(B,)(o(R,)) = o(,) (o(m,).o(B,)), (6.60) uses o(,).o(B,)=0 constant in space also => o(F,)atom = o(,)b(o(p,).o(E,) + o(m,).o(B,)) + f(1,c) f(d,dt) (o(p,) x o(B,)) +... . (6.61) [A similar argument gives the microscopic torque as o(N,)atom o(-,~) o(p,) x o(E,)(o(R,)) + o(m,) x o(B,)(o(R,)), (6.62) as we expect.] Now do the same thing as before, introduce a density of atoms, n(o(x,)), and calculate the bulk force: o(F,)bulk = i(,, )d3x n(o(x,)){isu(i,, )[pi o(,) Ei + mi o(,) Bi] + f(1,c) f(,t) (o(p,) x o(B,))}, (6.63) s(l(by parts), =) i(,, )d3x isu(i,, )[-(o(,)Pi)Ei - (o(,)Mi)Bi] + f(1,c) i(,, )d3x f(,t)(o(P,) x o(B,)). (6.64) where as usual o(P,)(o(x,)) = n(o(x,))o(p,), M(o(x,)) = n(o(x,))o(m,) and I am neglecting surface integrals. Now consider (take integral over entire bulk volume, including the surface) - f(1,c) o(P,) x f(o(B,),t) = o(P,) x (o(,) x o(E,)), (6.65) = isu(i,, )Pio(,) Ei - (o(P,).o(,))o(E,), (6.66) = o(E,)(o(,).o(P,)) - isu(i,, )Eio(,)Pi + o(,)(o(P,).o(E,)) - isu(i,, )i (Pio(E,)). (6.67)  . surface integrals This gives - i(,, )d3x Isu(i,, )Eio(,)Pi = i(,, )d3x[-(o(,).o(P,)) o(E,) - f(1,c) o(P,) x f(o(B,),t)], (6.68) = i(,, )d3x[-(o(,).o(P,)) o(E,) + f(1,c) f(o(P,),t) x o(B,)] - f(1,c) i(,, )d3x f(,t)(o(P,) x o(B,)), (6.69)  cancels with other term => o(F,)bulk = i(,, )d3x[-(o(,).o(P,))o(E,) + f(1,c) f(o(P,),t) x o(B,)]+ i(,, )d3x Isu(i,, )[-(o(,) Mi)Bi]. (6.70) Have shown earlier that (see 5.247; I continue to ignore surface contributions of the sample) i(,, )d3x Isu(i,, )Mio(,) Bi = i(,, )d3x(o(,) x o(M,)) x o(B,). (6.71) Therefore o(F,)bulk = i(,, )d3x[-(o(,).o(P,))o(E,) + f(1,c) f(o(P,),t) x o(B,) + (o(,) x o(M,)) x o(B,)]. (6.72) Identify reff = -o(,).o(P,), (6.73) o(J,)eff = co(,) x o(M,) + f(o(P,),t). (6.74) o(|,^)new term They are consistent since we have identically that f(o(r,)eff,t) + o(,).o(J,)eff = 0. (6.75) Our effective field equations, which are taking the average electronic properties of the material into account, are o(,) x o(B,) - f(1,c) f(o(E,),t) = f(4,c) (o(J,)free + f(o(P,),t) + co(,) x o(M,)), (6.76) o(,).o(E,) = 4(rfree - o(,).o(P,)), (6.77) o(,) x o(E,) + f(1,c) f(o(B,),t) = 0, (6.78) o(,).o(B,) = 0. (6.79) Introduce, as before o(D,) = o(E,) + 4o(P,), (6.80) o(H,) = o(B,) - 4o(M,). (6.81) Then, our macroscopic eqns are o() x o(H,) - f(1,c) f(o(D,),t) = f(4,c) o(J,)free, (6.82) o(,).o(D,) = 4rfree, (6.83) o(,) x o(E,) + f(1,c) f(o(B,),t) = 0, (6.84) o(,).o(B,) = 0. (6.85) Points: 1. o(E,),o(B,) more fundamental than o(D,), o(H,). 2. However, in the lab, o(H,) is more useful than o(B,). 3. All fields above are average, effective fields. This averaging process has not been specified (i.e., how to get the actual form of the n(x) function for a given material.) Second-order formulation of Maxwell's equations This is a first order formulation. We can eliminate the last two equations and obtain a second order formulation as follows. Define o(B,) = o(,) x o(A,). (6.86) (Always possible since o(,).o(B,)=0.) Then we have o(,) x(o(E,) + f(1,c) f(o(A,),t)) = 0, (6.87) => o(E,) + f(1,c) f(o(A,),t) = -o(,)F, (6.88) or o(E,) = -o(,)F - f(1,c) f(o(A,),t). (6.89) Restrict attention to vacuum form of Maxwell eqns. Substitute these potentials: o(,).o(E,) = 4r becomes 2F + f(1,c) f(,t) (o(,).o(A,)) = 4r. (6.90) o(,) x o(B,) - f(1,c) f(o(E,),t) = f(4,c) o(J,) becomes 2o(A,) - f(1,c2) f(2o(A,),t2) - o(,)(o(,).o(A,) + f(1,c) f(F,t)) = - f(4,c) o(J,). (6.91) Now since o(A,) -> o(A,) + o(,)L, (6.92) F -> F - f(1,c) f(L,t), (6.93) leaves o(B,) and o(E,) unchanged, choose o(,).o(A,) + f(1,c) f(F,t) = 0. ("Lorentz condition") (6.94) Then 2F - f(1,c2) f(2F,t2) = 4r, (6.95) 2o(A,) - f(1,c2) f(2o(A,),t) = - f(4,c) o(J,). (6.96) These are called wave equations. Always possible since if o(,).o(A,)0 + f(1,c) f(F0,t) = F o(=,/) 0, (6.97) we can then choose o(A,) = o(A,)0 + o(,)L, (6.98) F = F0 - f(1,c) f(L,t), (6.99) such that 2L - f(1,c2) f(2L,t2) = - F. (6.100) Notice, called it Lorentz condition, not gauge. Gauge is not completely fixed. Can still have o(A,) -> o(A,) + o(,)L, (6.101) F -> F - f(1,c) f(L,t), (6.102) if 2L - f(1,c2) f(2L,t2) = 0. (6.103) Another way of seeing the consistency of the Lorentz condition is by considering - (2 - f(1,c2) f(2,t2)) (o(,).o(A,) + f(1,c) f(F,t)) = 0, (6.104) => f(4,c) (o(,).o(J,) + f(r,t)) = 0. (6.105) Another choice: (radiation or Coulomb gauge) o(,).o(A,) = 0. (6.106) (Earlier showed it was always possible.) Then 2F = - 4r, (6.107) 2o(A,)- f(1,c2) f(2o(A,),t2) = - f(4,c) o(J,) + f(1,c) f(o(,)F,t). (6.108) Also consistent in the same sense: - b( 2 - f(1,c2) f(2,t2) ) o(,).o(A,) = 0, (6.109) => f(4p,c) b( o(,).o(J,) - f(1,4p) f(,t ) 2F ) = 0. (6.110)  -4p f(r,t ) Will discuss solutions of wave equations next semester. Energy and forces in magnetostatics Let us discuss some aspects of energy in magnetostatics. Natural question: why wait until the chapter on the full Maxwell eqns to discuss a supposedly static quantity? Answer: for a full understanding, it is necessary to consider time-variable initial situations to pass to the final, static situation. First, will give the general argument, then will give another argument that will not be as general, but will bring out an important aspect of magnetostatic energy. First, the general picture. Just like assembling charge distributions out of some small, imaginary elementary charge, imagine assembly macroscopic current distributions (circuits or volume distributions of currents) out of some small, elementary current loops. Also imagine that each little circuit comes equipped with it's own tiny battery whose mission in life is to keep the current flowing at a constant rate. Those elementary circuits are resistanceless. We can then compute the energy necessary to establish a given magnetic field configuration, W, by calculating the total energy expended by the tiny batteries. (Each tiny battery may have infinite energy reservoirs, as we'll see.) See Fig. 5.20 for an illustration of the final configuration, assembled from circuits at . Now consider a single one of these current loops imagining for the moment it has a resistance. Change the exterior field slowly. The current is given by Ohm's law: e0 + e = IR. (6.111) e0: source of emf (battery) e : induced emf (due to changing magnetic flux) Work done by the battery in moving a charge dq through the circuit is dW e0 dq = e0 I dt = -eI dt + I2R dt. (6.112) Using Faraday's law and setting R = 0 now gives dW = f(1,c) I dF. (6.113) We are imagining that the dF above corresponds to some external sources. If we imagine it applies universally, including the flux change induced by the circuit itself, we will see that the batteries have to have infinte energy. (This is the problem of magnetic self-energy; the above eqn is analagous to dW = Vdq for conducting surfaces.) Now imagine we have many of these tiny, discrete circuits making up some macroscopic current distribution: dW isu(i,, )dWi = f(1,c) isu(i,, )IidFi, (6.114) dFi = i( si,, )daio(n,^)i.do(B,) = o(,o(o,ci))do(l,,)i.do(A,), (Stokes) (6.115)  o(,) x do(A,) => dW = f(1,c) isu(i,, )o(,o(o,ci))Ii do(l,,)i.do(A,). (6.116) Go from a discrete sum over elementary circuits to an integral over a current density: isu(i,, )o(,o(o,Ci)) . i( V,, ), (6.117) Ii do(l,,)i . o(J,)(o(x,)) d3x, (6.118) => dW = f(1,c) i(,, )d3x o(J,)(o(x,)).do(A,)(o(x,)). (6.119) Assemble the circuits slowly: o(,) x o(H,) = f(4p,c) o(J,) + f(1,c) f(o(D,),t), (6.120) => dW = f(1,4p) i(,, )d3xb(o(,) x o(H,)).do(A,) d3x. (6.121) Integrate by parts: b(o(,) x o(H,)).do(A,) = isu(i,j,k,, )eijk dAijHk, (6.122) = isu(i,j,k,, )eijk j(dAiHk) - isu(i,j,k,, )eijk (jdAi)Hk, (6.123)  = do(B,).o(H,), (6.124) => dW = f(1,4p) i(,, )d3x o(H,).do(B,). (6.125) Similar to electrostatic situation for dW = i(,, )d3x o(E,).do(D,); can't go any further without specific assumptions about the medium. Assume ( ~ I.9 of Jackson) Ba = isu(b,, )mab Hb (linear but not isotropic), (6.126) => dBa = isu(b,, )mab dHb, (6.127) => isu(a,, )HadBa = isu(l(a,b),, )mab HadHb, (6.128) isu(a,, )BadHa = isu(l(a,b),, )mab HbdHa, (6.129) => do(B,).o(H,) = o(B,).do(H,) if mab = mba, (6.130) = f(1,2) db(o(B,).o(H,)). (6.131) Then, for linear media, W = f(1,8p)i( ,, )d3x o(B,).o(H,) s( =,isotropic) f(1,8p) i( ,, )d3x m(o(x,)) o(H,)2(o(x,)). (6.132) Integrate by parts on the first form of (6.132): W = f(1,2c) i( ,, )d3x o(J,).o(A,)s( =, o(gauge,Coulomb)) f(1,2c2)i( ,, )d3xd3x' f(o(J,)(o(x,)).o(J,)(o(x,)'),|o(x,)-o(x,)'|). (6.133) Apply to circuits: W = isu(l(i,j),, )f(IiIj,2c2) o(,o(o,i))o(,o(o,j)) f(do(x,)i.do(x,)j,|o(x,)i-o(x,)j|). (6.134) Can show (see section 5.17 of Jackson) that this is equivalent to W = f(1,2c) isu(i,, )IiFi, (6.135) where Fi is magnetic flux Fi = i(si,, )dai o(n,^).o(B,) = isu(j,, )f(Ij,c) o(n,^)i.o(,o(o,Cj)) do(x,)j x f((o(x,)i-o(x,)j),|o(x,)i-o(x,)j|3). (6.136) Can also argue this form for W directly from dW = f(1,c) isu(i,, )IidFi, (6.137) if we assume a linear relationship between Ii and Fi: brc}(a(l(Ii(a) aIi), l(dFi Fida))) a: 0 . 1 => dWa f(1,c) isu(i,, )IiFi ada, (6.138) W = i( 0, 1, )da dWa = f(1,2c) isu(i,, )IiFi. (6.139) Interaction energy for two circuits (can be positive or negative): W12 = f(I1I2,c2) o(,o(o,1)) o(,o(o,2)) f(do(x,)1.do(x,)2,|o(x,)1-o(x,)2|). (6.140) The above process for assembling circuits, by asembling small circuits with their own batteries, is clearly not the only way of calculating the field energy, W. Can imagine assembling such resistanceless circuits at fixed flux by disconnecting the batteries. This is in direct analogy with electrostatics where one may imagine bringing in conductors at fixed voltage (analagous to fixed current case) or fixed charge (analagous to fixing the magnetic flux). Actually there are an infinite number of weays of imaging how the currents or charges are assembled and any one of them will do to get the field energy, as long as we are consistent. More on magnetostatic energies Let's now try to derive this result from an alternate point of view. This will be confirmatory rather than independent; however, we will learn something important that we haven't seen already. Two circuits:  From previous discussion, we have the force: (force of 2 on 1) o(F,)12 = - f(I1I2,c2) o(,o(o,1)) o(,o(o,2)) f((o(x,)1-o(x,)2)do(x,)1.do(x,)2,|o(x,)1-o(x,)2|3). (6.141) Move (1) by o(R,) and use new coordinate system:  o(F,)12(o(R,)) = - f(I1I2,c2) o(,o(o,1)) o(,o(o,2)) f((o(x,)1-o(x,)2+o(R,))do(x,)1.do(x,)2,|o(x,)1-o(x,)2+o(R,)|3). (6.142) Can be written (see Jackson, prob.5.33) o(F,)12(o(R,)) = I1I2o(,)R M12(o(R,)), (6.143) M12(o(R,)) f(1,c2) o(,o(o,1)) o(,o(o,2)) f(do(x,)1.do(x,)2,|o(x,)1-o(x,)2+o(R,)|). (6.144) (M12 is the "mutual inductance".) Notice s(2 ,R)M12(o(R,)) = 0. (6.145) as long as the circuits don't overlap. (I found this useful in prob.5.34(c) of Jackson.) So, the mechanical energy of assembly is Wmech = -i( , 0, )o(F,)12(o(R,)).do(R,), (6.146) => Wmech = - f(I1I2,c2) o(,o(o,1)) o(,o(o,2)) f(do(x,)1.do(x,)2,|o(x,)1-o(x,)2|). (6.147) ? We have pumped energy into the batteries as well as the field: battery energy dIWmech = dIW12 + dIWb. (6.148) keeping I constant Under a small change in position, (dIW12 = dIW; W includes self-energies, which, however, do not change since I is constant) W = f(1,2c) isu(i,, )IiFi, (6.149) => dIW = f(1,2c) isu(i,, )IidFi. (6.150) However, for the batteries, as before (change in battery energy is negative the work it's done.) dIWb = - isu(i,, )e0idqi = isu(i,, )eidqi. (6.151) The battery's emf (ei) opposes the induced emf (e0i) to keep the current constant. Now from Faraday: ei = - f(1,c) f(dFi,dt), (6.152) => dIWb = - isu(i,, )f(1,c) dFi f(dqi,dt), (6.153) Ii => dIWb = - 2 dIW. (6.154) Analogous to the electrostatic situation: dVWb = -2 dVW. For the total change in this situation then dIWmech = dIW12 + dIWb, (6.155)  -2dIW12 => dIW12 = -dIWmech. (6.156) Theres our minus sign! Other implication from the above argument: o(F,) = - f(,o(x,)) (W+Wb)I = + b(f(W,o(x,)))I. (6.157) ! This is analagous to the force at constant voltage seen in Ch.4. Similarly, when calculating the magnetostatic force at fixed flux there will be no battery energy and then one expects to have a minus sign in the equation analagous to (6.157), just as in the electrostatic case (fixed charge). Let's make sure (6.157) is correct in the simpliest possible situation: parallel wires (already calculated force directly in Ch.5). Interaction energy:  W = i(,, )dz1dz2 f(I1I2,c2) f(o(n,^)1.o(n,^)2,|o(x,)1-o(x,)2|), (6.158) (|o(x,)1-o(x,)2| = r(,r2+(z1-z2)2) ) => W = f(I1I2,c2) o(n,^)1.o(n,^)2 i( -L, L, )i( -L, L, ) f(dz1dz2,r(,r2+(z1-z2)2)). (6.159) As before z' f(z1+z2,2), (dz1dz2 = dz'dz") (6.160) z" z1-z2, (6.161) => i(,, )i(,, )f(dz1dz2,r(,r2+(z1-z2)2)) = i( -L, L, )dz'i( -2L, 2L, ) f(dz",r(r2+z"2)), (6.162) o(~,-,L>>r) i( -L, L, )dz'2ln(f(4L,r)). (6.163) => f(W,2L) o(~,-) f(2I1I2,c2) o(n,^)1.o(n,^)2[ln f(4L,r)], (6.164) s(l(force),l(of 1 on 2),l(per length)) => f(Fr,2L) = +f(,r) b(f(W,2L)) - f(2I1I2,c2) f(o(n,^)1.o(n,^)2,r). (6.165) Already saw that electrostatic self-energy of a point charge is infinite. What about the magnetostatic analog? For a single current loop, Wself = f(1,2c2) i(,, )d3x d3x' f(o(J,)(o(x,)).o(J,)(o(x,)'),|o(x,)-o(x,)'|), (6.166) circular loop: o(J,)(o(x,)) = Io(e,^)f d(z)d(r-a), (6.167) Wself = f(I2,2c2) i(,, ) f(dfdf'o(e,^)f.o(e,^)f',|o(x,)-o(x,)'|)|o(z=z'=0,r=r'=a). (6.168) Now |o(x,)-o(x,)'||o(z=z'=0,r=r'=a) = a[(cosf-cosf')2+(sinf-sinf')2 ]1/2, = r(,2)a [1-cos(f-f')]1/2, (6.169) o(e,^)f.o(e,^)f' = (-o(i,^)sinf + o(j,^)cosf).(-o(i,^)sinf + o(j,^)cosf'), = cos(f-f'). (6.170) In the integration, f1 f(1,2)(f'+f'), limits: 0,2p f2 f - f', limits: -2p,2p => Wself = f(I2,2r(,2)ac2) i( 0,2p, )df1 i(-2p,2p, )df2 f(cosf2,[1-cosf2]1/2). (6.171) Near f2 o(~,~) 0, (e > 0) i(-2p,-e, ) df2 f(cosf2,[1-cosf2]1/2) + i(e,2p, ) same o(~,~) 2r(,2) i(e,2p, ) f(df2,f2). (6.172) (logarithmically divergent) Of course, self-energy of a real current loop is not infinite. A more realistic model would use a space distributed o(J,)(o(x,)), in which case we would in general get a finite result. Similar to situation in electrostatics where Wself = f(1,2) i(,, )d3xd3x' f(r(o(x,)')r(o(x,)),|o(x,)- o(x,)'|), (6.173) is infinite (linearly) for a point charge but finite for continuous distributions. Bulk forces on magnetic materials: theory and examples Lets compute the bulk forces on magnetic materials analogously to the discussion in Ch.4 on dielectrics. Energy Method. Start with W = W - W0, (6.174) W: final magnetostatic field energy W0: Initial magnetostatic field energy (material at or some convenient location) W = f(1,2c) i(,, )d3x o(J,).o(A,). (6.175) Keep o(J,) fixed. There now results (see p.214 of Jackson) DW = f(1,8) i( DV,, )d3x (f(1,m0) - f(1,m))o(B,).o(B,)0, (6.176) If m0 = 1, DW = f(1,2) i( DV,, )d3x o(M,).o(B,)0. (6.177) DV is the introduced bulk volume. Can then use such an expression in Fx = +(f(dW,dx))I to find the force. (This is usually the easy, approximate way.) Example:  Neglect edge effects and assume a thick rod (solenoid length L). o(M,) = f(1,4) (o(B,) - o(H,)) = f(o(B,),4) (1 - f(1,m) ) (o(B,) = o(B,)(x)), (6.178) W o(-,~) f(1,2) i(,, )d3x f(o(B,).o(B,)0,4) (1 - f(1,m) ) = f(o(B,)s(2,0) LAx(1 - f(1,m)),8[L + x(f(1,m) - 1)]), (6.179) => Fx = + bbc(f(W,x))I = f(o(B,)s(2,0) AL2,8) f((1 - f(1,m)),[L + f(x,L) (f(1,m) - 1)]2) . (6.180) blc{(a(m > 1, pulled in (paramagnetic),m < 1, pushed out (diamagnetic))) Another approximate solution results if one assumes the rod is so thin that o(H,) is a constant field everywhere inside the solenoid (HW prob.). Example 2:  Expect, from W = f(1,2) i(,, )d3x o(M,).o(B,)0, that as it is pulled in from infinity, s(l(Paramagnetic: m>1 =>force toward loop),l(Diamagnetic: m<1 =>force away from loop)) Qualitatively, this is the same as we found in Ch.5. (Im whimping out on actually calculating the force.) Force Method The above gives no idea as to where the forces arise. Will derive some explicit formulas for o(F,), analogous to electrostatic case. Remember that o(F,)bulk = f(1,c) i(,, )d3x o(J,)eff x o(B,), (6.181) external field o(J,)eff = co(,) x o(M,). (6.182) If we want to break this up into explicit volume and surface pieces, o(F,)bulk = i( v,, )d3x(o(,) x o(M,)) x o(B,) + f(1,c) i( s,, )da o(K,)eff x o(B,), (6.183) we can proceed as follows (more arguments for the form for o(K,)eff):  i(,, )da o(t,^).b(c o(,) x o(M,)) = i(,, )da o(t,^).o(J,)eff, (6.184) => -co(M,).o(l,)21 Dl = o(K,)eff.o(t,^)Dl, (6.185) b(c o(M,) x o(n,^)21).o(t,^) = o(K,)eff.o(t,^), (6.186) => o(K,)eff = co(M,) x o(n,^). (6.187) Another way of seeing this: o(n,^) x (o(B,)2-o(B,)1) = f(4p,c) b(o(K,)free + o(K,)eff), (6.188) (o(B,)2 = o(H,)2) o(B,)1 = o(H,)1 + 4po(M,) b (o(n,^) x (o(H,)2 - o(H,)1) = f(4p,c) o(K,)free ), (6.189) => o(K,)eff = f(c,4p) o(n,^) x -4p o(M,) = co(M,) x o(n,^), (6.190) => o(F,)bulk = i(v,, )d3xb(o(,) x o(M,))x o(B,) + i( s,, )da(o(M,) x o(n,^)) x o(B,). (6.191) When we have a linear isotropic medium o(M,) = f(o(B,),4p) b(1 - f(1,m(o(x,))) ), (6.192) m = const.=> o(J,)eff = co(,) x o(M,) = f(c,4p) b(1 - f(1,m) ) o(,) x o(B,) = b(1 - f(1,m) ) o(J,)tot,  (6.193) f(4p,c) o(J,)tot => o(J,)eff = (m-1)o(J,)free (volume currents). (6.194) (o(J,)tot = o(J,)free + o(J,)eff) If m=const. and o(J,)free = 0 in V, then we are simply left with the surface term. Then in general: do(F,)|surface = f(1,4p) da(o(K,)free + o(K,)eff) x o(B,), (6.195) = f(1,4p) da[o(n,^) x (o(B,)2 - o(B,)1)] x o(B,). (6.196) o(B,): external field or f((o(B,)1 + o(B,)2),2) again. o(B,)1self, o(B,)2self tang. to surface In the latter case: do(F,)bulk = f(1,8p) da[o(n,^) x (o(B,)2 - o(B,)1)]xb(o(B,)1 + o(B,)2), (6.197) = f(1,8p) [- o(n,^)(o(B,)s(2,2) - o(B,)s(2,1)) + (o(B,)2 - o(B,)1)(B1n + B2n)], (6.198) => fn f(do(F,)bulk . o(n,^),da) = f(1,8p) [-o(B,)s(2,2) + o(B,)s(2,1) + (Bs(2,2n)-Bs(2,1n))], (6.199) fn = f(1,8p) [Bs(2,1||) - Bs(2,2||)]. (6.200) In the case where (2) is vacuum and o(B,)1 = mo(H,)1, (use | |o(H,) B.C.) fn . f(Bs(2,2||),8p) (m2-1). (6.201) Again, we see for our situation,  that m > 1 means attraction and m < 1 means repulsion. Our general expression can also be written o(F,)bulk = i( v,, )d3xb(o(,) x o(M,))x o(B,)e + i( s,, )da b(o(M,) x o(n,^))x o(B,)e, (6.202) when free currents = 0. Apply this when the external current distribution does not overlap with the bulk material being studied. We will see this is equivalent to an electrostatic problem. What are the effective volume and surface "charges"? Let's begin by looking first at the volume term above. The following is an identity: o(,)b(o(M,).o(B,)e)= b(o(M,).o(,))o(B,)e + b(o(B,)e.o(,))o(M,) + o(B,)e x b(o(,) x o(M,)) + o(M,) x b(o(,) x o(B,)e). (6.203) In the region of interest, and because we are considering a static situation, o(,) x o(B,)e = 0, (6.204) => b(o(,) x o(M,))x o(B,)e = b(o(M,).o(,))o(B,)e + b(o(B,)e.o(,))o(M,) - o(,)b(o(M,).o(B,)e). (6.205) Since we are now dealing with volume and surface terms separately, must be careful about not dropping such terms here. With this in mind, consider (1) (2) i(,, )d3xb(o(,) x o(M,)) x o(B,)e = i(,, )d3xb(o(M,).o(,))o(B,)e + i(,, )d3x b(o(B,)e.o(,))o(M,) (3) -i(,, )d3x o(,)b(o(M,).o(B,)). (6.206) (1) = i(,, )d3x isu(i,, )i(Mio( B,)e) - i(,, )d3x b(o(,).o(M,))o(B,), (6.207) = i(,, )dab(o(n,^).o(M,))o(B,)e - i(,, )d3x b(o(,).o(M,))o(B,)e, (6.208) (2) = i(,, )d3x isu(i,, )ib(o(B,)ei o(M,)) - i(,, )d3xb(o(,).o(B,)e)o(M,), (6.209) = i(,, )dab(o(n,^).o(B,)e)o(M,), (6.210) (3) = - i(,, )d3x o(,)b(o(M,).o(B,)e) = -i(,, )dab(o(M,).o(B,)e)o(n,^). (6.211) From the original surface term in o(F,)bulk we get i(,, )da (o(M,) x o(n,^)) x o(B,)e = - i(,, )da (o(n,^).o(B,)e) o(M,) + i(,, )da (o(M,).o(B,)e) o(n,^). (6.212) Putting all the pieces together yields o(F,)bulk = i(,, )d3x(-o(,).o(M,)) o(B,)e + i(,, )da(o(n,^).o(M,)) o(B,)e, (6.213)   rm sm rm = -o(,).o(M,), sm = o(M,).o(n,^). The effective interaction is as if the magnetic material possessed elementary magnetic charge in it's interior and surface of -o( ,).o(M,) and o(M,).o(n,^), respectively. (This has been prob. 5.20(a) of Jackson.) Magnetic charge and the macroscopic Maxwell equations Switch gears again. Usual assumption: moving charge is the source of all magnetic fields. Let's go back to the fundamental, microscopic form of Maxwell's eqns. Let us add magnetic charge and currents in the obvious, symmetrical manner. o(,).o(E,) = 4re, (6.214) o(,) x o(B,) - f(1,c) f(o(E,),t) = f(4,c) o(J,)e, (6.215) o(,).o(B,) = 4rm, (6.216) o(,) x o(E,) + f(1,c) f(o(B,),t) = - f(4,c) o(J,)m. (6.217) Are these really different? Consider: xe = xe'cos x + xm'sin x, (6.218) xm = -xe'sin x + xm'cos x, (6.219) where xe,m o(=,-) re,m or o(J,)e,m. Now redefine fields in the same way, o(E,) = o(E,)'cos x + o(B,)'sin x, (6.220) o(B,) = -o(E,)'sin x + o(B,)'cos x. (6.221) We then get a new, primed set of equations which have the same form as before. This implies that electric or magnetic charges or currents are not absolutes, but can be rotated to different values. (Their value depends on some convention.) However, also rm2 + re2 = rm'2 + re'2, (6.222) o(J,)m2 + o(J,)e2 = o(J,)m'2 + o(J,)e'2, (6.223) so that the sum of squares can not change. The appropriate picture:  Moral: if all the particles in the universe have the same ratio rm/re, we can always rotate to rm = 0. What if they have different ratios?   By applying a universal rotation, can no longer get rm = 0 for all particles. Let us consider the case of a purely magnetic particle interacting with a pure electric one. Then we can not use the Maxwell eqns with rm, o(J,)m = 0. This is inconvenient. However, there is a formulation of magnetic charge tht uses the standard macroscopic Maxwell eqns with rm, o(J,)m = 0. To begin to understand it, let's go back to the discussion involving space-distributed magnetic dipoles:  Then we had o(,) x do(H,)m = 0, (6.224) do(H,)m = -o(,)dFm = -o(,) b(f(do(m,).(o(x,)-o(x,)'),|o(x,)-o(x,)'|3)), (6.225) and o(,) x do(B,)m = f(4p,c) do(J,)m, (6.226) do(J,)m = -cdo(m,) x o(,) d(o(x,)-o(x,)'), (saw this in a HW problem) (6.227) => do(B,)m = o(,) x do(A,)m = o(,) x b(f(do(m,) x (o(x,)-o(x,)'),|o(x,)-o(x,)'|3) ). (6.228) For a volume distribution, we let do(m,) . o(M,)(o(x,)) d3x'. (6.229) Now, however, let us consider a line of magnetization made of magnetic dipoles. (End it at the origin.)  Choose (g > 0 to be consistent with the picture) do(m,) . -g do(x,)'. (6.230) We then have o(A,)(o(x,)) = - gi( L,, )do(x,)' x f((o(x,)-o(x,)'),|o(x,)-o(x,)'|3) = gi( L,, )do(x,)'x o(,) f(1,|o(x,)-o(x,)'|), (6.231) [This is consistent with the Coulomb gauge: o(A,)(o(x,)) = -go( ,) x i(,, )do(x,)'f(1,|o(x,)-o(x,)'|), => o(,).o(A,) = 0. ] => o(B,)(o(x,)) = gi( L,, )o(,)x(do(x,)'x(o(,)f(1,|o(x,)-o(x,)'|))). (6.232) Some rearrangement is needed: (BAC-CAB) o(,)x(do(x,)'x(o(,)f(1,|o(x,)-o(x,)'|)))= isu(i,, )bbc[(i(do(x,)'if(1,|o(x,)-o(x,)'|))- i(dxi'o(,) f(1,|o(x,)-o(x,)'|))), (6.233) = do(x,)'2 f(1,|o(x,)-o(x,)'|) - (do(x,)'.o(,))(o(,)f(1,|o(x,)-o(x,)'|)), (6.234) o(,)-> -o(,)' = -4 do(x,)'d(o(x,)-o(x,)') + do(x,)'.o(,)'(o(,)f(1,|o(x,)-o(x,)'|)). (6.235) Can now do the integrals: o(B,)(o(x,)) = gi( L,, )do(x,)'.o(,)'(o(,)f(1,|o(x,)-o(x,)'|)) - 4gi( L,, )do(x,)'d(o(x,)-o(x,)'), (6.236) i( L,, )do(x,)'.o(,)'(o(,)f(1,|o(x,)-o(x,)'|)) = o(,)f(1,|o(x,)-o(x,)'|)|s(|o(x,)'|=,|o(x,)'|=0) = f(o(x,),|o(x,)|3), (6.237) => o(B,)(o(x,)) f(go(x,),|o(x,)|3) - 4gi(L,, )do(x,)'d(o(x,)-o(x,)'). (6.238) a directed line delta function The new term is necessary to maintain o(,).o(B,) = 0: o(,).o(B,) = g2f(1,|o(x,)|) - 4gi( L,, )(do(x,)'.o(,)) d(o(x,)-o(x,)'), (6.239) -o(,)' o(,).o(B,) = 4gd(o(x,)) + 4gi( L,, )(do(x,)'.o(,)') d(o(x,)-o(x,)'). (6.240) Formally (see later comments) i( L,, )(do(x,)'.o(,)')d(o(x,)-o(x,)') = d(o(x,)-o(x,)')|s(|o(x,)'|=,|o(x,)'|=0) =- d(o(x,)), (o(x,) finite) (6.241) => o(,).o(B,) = 0. (6.242) The extra term represents an intense inward line distribution of o(B,) that cancels off the outgoing field (g > 0) so that the total flux on any closed surface enclosing the origin is zero: F = i( s,, )da o(n,^).o(B,) = gi( s,, )daf(o(x,).o(n,^),|o(x,)|3) - 4gi( L,, )i( s,, )dao(n,^).do(x,)'d(o(x,)-o(x,)'). (6.243)   dW (subtended ~ dV (includes o(x,)') at origin) Picture:  => F = 4g - 4g = 0. (6.244) Given o(B,), can always get o(H,) from o(H,) = o(B,) - 4o(M,). We get o(M,)(o(x,)), which is a line distribution, by adding the contributions from each do(m,)i: (Im being vague about the limit) o(M,)(o(x,)) = lim (isu(i,, )do(m,)i d(o(x,)-o(x,)i)), (6.245) o(M,)(o(x,)) = lim (-gisu(i,, )do(x,)i d(o(x,)-o(x,)i)) => -gi(,, )do(x,)'d(o(x,)-o(x,)'). (6.246) Therefore o(B,)(o(x,)) = f(go(x,),|o(x,)|3) + 4o(M,)(o(x,)), (6.247) => o(H,) = f(go(x,),|o(x,)|3). (monopole field) (6.248) No singularity along L! If we say that it is the o(H,) field which is observable, then there are no effects of the string. To confirm the consistency of this picture, go back and work out o(H,) directly: Fm(o(x,)) = -gi(,, )f(do(x,)'.(o(x,)-o(x,)'),|o(x,)-o(x,)'|3), (6.249) f(do(x,)'.(o(x,)-o(x,)'),|o(x,)-o(x,)'|3) = do(x,)'.o(,)f(1,|o(x,)-o(x,)'|), (6.250) => Fm(o(x,)) = f(-g,|o(x,)-o(x,)'|)|s(|o(x,)'|=,|o(x,)'|=0) = f(-g,|o(x,)|), (6.251) => o(H,)(o(x,)) = f(g(o(x,)),|o(x,)|3) again. (6.252) Let's say the string is moved:  The strings must come together again before or at , otherwise one can show there is an effect of moving the string (namely, infinite radiation!) We have o(A,)L2 - o(A,)L1 = gi( L2,, )do(x,)' x f((o(x,)-o(x,)'),|o(x,)-o(x,)'|3) - gi(L1,, )do(x,)' x f((o(x,)-o(x,)'),|o(x,)-o(x,)'|3), (6.253) i( 0,, ) = -i( ,0, ), (6.254) => o(A,)L2 - o(A,)L1 = -g o(,o(o,l(L2-L1))) do(x,)' x f((o(x,)-o(x,)'),|o(x,)-o(x,)'|3). (6.255) Easy to evaluate using our previous mathematical identity: (same as in HW problem for o(B,)) o(,o(o, )) do(l,)' x o(A,) = i(,, )isu(i,, )ds'i o(,)'Ai - i(,, )do(s,)' (o(,)'.o(A,)), (6.256) "membrane" => o(A,)L2 - o(A,)L1 = -g o(,)W(o(x,)) - 4pgi(,, )do(s,)'d(o(x,)-o(x,)'). (6.257) We see our membrane d-function again. It is present here for the same reason it was present for o(B,): it makes o(A,)L2 - o(A,)L1 continuous across the surface. The fields are single-valued but the difference between o(A,)'s is not just a divergence here, as in Jacksons discussion. Another point: can just regard this as a generalized gauge transformation. Usually, (time indep. L) o(A,) . o(A,) + o(,)L. (6.258) Make sure this expression for o(A,)L2 - o(A,)L1 is consistent with earlier expression for o(B,). Need identity: o(,) x i(,, )do(s,)'d(o(x,)-o(x,)') = i(,, )b(o(,) d(o(x,)-o(x,)')) x do(s,)' = i(,, )da o(n,^)' x o(,)'d(o(x,)-o(x,)') = o(,o(o,l(L2-L1))) do(x,)'d(o(x,)-o(x,)'). (6.259) Therefore o(,) xb(o(A,)L2 - o(A,)L1) = -4pg o(,o(o,l(L2-L1))) do(x,)'d(o(x,)-o(x,)'). (6.260) But from earlier expression: o(B,)1,2(o(x,)) = f(go(x,),|o(x,)|3) - 4pg i(l(L1,2),, ) do(x,)'d(o(x,)-o(x,)'), (6.261) => o(B,)2 - o(B,)1 = -4pgi(l(L2),, )do(x,)'d(o(x,)-o(x,)') - 4pgi(l(L1),, )(-do(x,)')d(o(x,)-o(x,)'), (6.262) = -4pg o(,o(o,l(L2-L1))) do(x,)'d(o(x,)-o(x,)'). (6.263) Same as above. Thus, the eqns o(,).o(D,) = 4p rs(free,e), (6.264) o(,) x o(H,) - f(1,c) f(o(D,),t) = f(4p,c) o(J,)s(free,e), (6.265) o(,) x o(E,) + f(1,c) f(o(B,),t) = 0, (6.266) o(,).o(B,) = 0. (6.267) if properly interpreted, can replace the earlier eqns with explicit rm, o(J,)m sources. (However, this involves a leap to the dynamical case, which I haven't justified.) There is an important consequence of all this for electric charge. Will have to go beyond the physics in this class for the argument, however. A charged particle in an em field picks up a path-dependent phase:  If we transport the particle in a closed loop, then  Let us imagine transporting our electrically charged particle in a closed loop when the string is at L1, then again when the string is at L2:  Uniqueness of the wave function requires (this gives the magnetic flux through C'; "e" is the charge) eo(,o(o, c'))do(x,)'.[o(A,)L2 - o(A,)L1 ] = 2pn, n = 0,1,2, ... . (6.268) The lhs gives eo(,o(o, c'))do(x,)'.{-g o(,)'W(o(x,)') - 4pgi(,, )do(s,)d(o(x,)-o(x,)')}, (6.269) = -4pgeo(,o(o, c'))i(,, )do(s,).do(x,)'d(o(x,)-o(x,)') = 4peg. (6.270)  Dirac: eagb = f(1,2) n. (f(eagb,hc) = f(n,2), gaussian units) (6.271) Explains quantitization of electric charge if only 1 magnetic monopole in the universe. (Im putting a,b subscripts on the charges to emphasize that the charges are on separate particles.) Actually, requires either e or g to be quantitized. In particular since in natural units es(2,electron) o(-,~) f(1,137), b(f(e2,hc) = f(1,137) a in Gaussian ), (6.272) if we let es(s,a) = magn of smallest nonzero electric charge o(-,~) f(1,3r(,137)) (down quark) (6.273) gs(s,b) = magn of smallest nonzero magnetic charge => b(gs(s,b))2 = f(9,4) x 137 !! HUGE (6.274) The above was the "Dirac string":  The "Schwinger string" looks likes:  Magnetization: o(M,)(o(x,)) = f(1,2) gi( L,, )do(x,)' [d(o(x,)-o(x,)') - d(o(x,)+o(x,)')]. (6.275) Now, for separate electric and magnetic charges:  We still require eo(,o(o, c'))do(x,)'.[o(A,)L2 - o(A,)L1] = 2pn but now the lhs gives eo(,o(o, c'))do(x,)'.{- g o(,)'W(o(x,)) - 2pg i(,, )do(s,)[d(o(x,)-o(x,)') - d(o(x,)+o(x,)')] }, (6.276)  pick up one or the other contribution, but not both in general = 2peg, (6.277) => eagb = n. (6.278) Actually, if one assumes each particle can have both magnetic and electric charge, then Schwinger's condition is eagb - ebga = n, "Dyon" (6.279) Notice it is invariant under our original transformation, ea,b = e'a,b cosx + g'a,b sinx, (6.280) ga,b = -e'a,b sinx + g'a,b cosx, (6.281) whereas Dirac's condition, which makes a real invariant distinction between electric and magnetic charge, does not. Using Schwinger's condition, the magnitude of the smallest magnetic charge (with no accompanying electric charge) is (gs)2 o(-,~) 9 x 137. The binding energy (in hydrogen-like systems) is given by the Bohr formula: Es(b,n) = - f((gs)4m,2n2). (6.282) m = reduced mass. It is clear that for gs o(~,>) 1 no positronium (electron-positron bound state) like system can exist in electrodynamics. In Schwinger's original application to strong interactions, this was dealt with by assuming that the observed states have n 137. However, the strong interactions are now understood to arise from a confining theory of colored quarks, Quantum Chromodynamics (QCD). The color of the quarks comes in three types which are separately conserved; this is a generalization of the single type of electric charge. Unfortunately, magnetic monopoles are a theory in search of an application. In Eq.(6.242) there is a reference to an integration that extends to infinity, where the end of the string ends. This is not entirely satusfactory. We can remove reference to integrations at infinity if we only allow an equal number of magnetic and anti-magnetic charges. The strings then extend from one type of charge to the opposite and there is no need to involve integrations to spatial infinity. I emphasize again that there are no observable effects of the strings themselves; they are necessary mathematical artifices which only serve to define the magnetic charges. The above is a purely classical approach to the subject of magnetic charge. As far as I know, no clearly recognized consistent quantum theory of point magnetic charge exists. A consistent basis for a quantum theory of magnetic charge exists, however. In modern "grand unified" theories, magnetic monopoles emerge as soliton solutions and have an electromagnetic size determined by the Compton wavelength of the associated gauge bosons of the theory. These solitons obey the Dirac quantization condition and have masses on the order of the early universe symmetry breaking scale, ~ 1016 GeV/c2. The role of the strings is taken over by a scalar (zero spin) field generated by the so-called Higgs particle. Such objects are indeed connected by strings which, unlike the strings described above, is a real, observable object. Problems 1. Show, to first order in o(v,) only, that the Maxwell equations o(,).o(E,) = 4pr, o(,)Xo(B,) - f(1,c) f(o(E,),t) = f(4p,c) o(J,), o(,)Xo(E,) + f(1,c) f(o(B,),t) = 0, o(,).o(B,) = 0, are form invariant under the transformations BLC{(A(o(,) . o(,) + f(o(v,),c2) f(,t),f(,t) . f(,t) + o(v,) .o(,))), BLC{(A(o(E,) . o(E,) + f(o(v,),c) x o(B,), o(B,) . o(B,) - f(o(v,),c) x o(E,))) , BLC{(A(r . r - f(o(v,),c2) . o(J,), o(J,) . o(J,) - o(v,)r)). 2. Two small, planar circuits with areas a1 and a2 are located a large distance, r, apart. (r>>sizes of the circuits.)  o(n,^)1 and o(n,^)2 are unit vectirs pointing in the direction of their magnetic moments. Find their mutual inductance (see Jackson section 5.17) and demonstrate explicitly that M12=M21. 3. A small planar circuit of arbitray shape and area A is located at the center of a much larger circular circuit of radius R. The planes of the two circuits coincide. The smaller circuit carries a steady current I1 and the larger one a steady current I2. Find the mutual inductance, M12, of this system. 4. A very long flat rectangular circuit of mass M, which has a source of emf, e0, and a resistance, R, produces a current by Ohms law. Gravity plays no role in this problem and the circuit moves frictionlessly. (a) Given that the end of the circuit is immersed in a uniform magnetic field, o(B,), directed out of the page, find the equation of the circuit that follows from the Lorentz force and Faradays law. That is, find the differential equation that determines x(t). Neglect the self-inductance of the circuit. Show the motion exponentially approaches a terminal velocity.  (b) Now the circuit is considered superconducting so that e0=0 (no source of emf present) and the resistance, R, is zero. The initial current in the circuit is I(0) and the self-inductnce of the circuit, L, can no longer be ignored. Find the new equation of motion in the x-direction. Show the motion is simple harmonic. 5. Assuming localized, time-dependent, non-overlapping current distributions o(J,)1 (giving rise to the field o(A,)1, treated here as "external") and o(J,)2, show that the interaction energy Wint = f(1,c) i ( , , )d3x o(A,)1 . o(J,)2 when o(A,)1(x) is expanded in a Taylor series about the origin gives rise to the usual force expression found already in the script: o(F,) = o(,)(o(m,).o(B,)1). [Hint: Think about the relationship between Wint and o(F,) in the light of the discussion of section 5.16, p.214 of Jackson.] 6. An object of permeability m is placed in a magnetic field whose current sources are fixed. Show that the change in energy, DW W - W0, where W = f(1,2c) i ( , , )d3x o(J,).o(A,), gives DW = f(1,2) i (v, , )d3x o(M,).o(B,)0, where V is the volume occupied by the object. (See Jackson, p.214). 7.(a) An approximately uniform magnetic field, o(B,)0, is produced in a superconducting solenoid of length L, producing a situation of constant magnetic flux within the solenoid. A thick rod of material of permeability m is inserted into the solenoid. Considering the BC at the rod's end and neglecting edge effects, find the force, Fx, on the rod.  (b) Consider the situation described in the script with a thin permeable rod inserted a distance x into a solenoid with an approximate uniform field. The current in the solenoid is fixed at I producing the initial field o(B,)0. The approbriate BC is now on the solenoid's side giving that the o(H,) is a constant everywhere. Find the approximate force on the rod, Fx. 8. A long straight wire with a circular cross-section carries a uniform current density flowing up along the z-axis, as shown. The total current is I, a constant in time.  Find the differential pressure, dP(r) = f(do(F,).o(r,^),da), exerted on a small volume of material by the current at a distance r from the center of the wire. Integrate your expression to find P(r). Is the pressure inward or outward? 9. Given the magnetization and polarization of a rigid, moving magnetic string (the magnetic charge, g, is located at o(R,)(t); the integration limits on o(x,)'' are 0 to along some given path) o(M,)(o(x,)) = -gi(,, )do(x,)d(o(x,)-o(R,)(t)-o(x,)''), o(P,)(o(x,)) = f(o(v,)(t),c) o(M,)(o(x,)), (o(v,)(t) = f(do(R,),dt)) show that the sourceless Maxwell equations, O(,).O(B,) = 0, O(,) O(E,) + F(1,c) F(O(B,),t) = 0, are transformed into (O(B,) = O(H,) + 4pO(M,), O(E,) = O(D,) - 4pO(P,)) O(,).O(H,) = 4p gd(o(x,)-o(R,)(t)), O(,) O(D,) + F(1,c) F(O(H,),t) = - f(4p,c) go(v,)(t)d(o(x,)-o(R,)(t)). [This shows that the O(D,), O(H,) fields are just the nonsingular fields of a moving magnetic point charge located at o(R,)(t); all reference to the string has disappeared! The general statement, f(,t) f(o(x,)-o(R,)(t))+ o(,).b(o(v,)(t)f(o(x,)-o(R,)(t))) = 0, where f is an arbitrary function of its argument, may be useful.] 10. Previously, we considered an arbitrary (truncated) line of magnetization; this turned out to be a description of a magnetic point charge. Consider now a uniform layer of magnetization, o(M,)s, on a flat surface, S, extending to infinity in one direction, as shown. The magnetization is pointed along the direction which extends to infinity. The sheet ends on an arbitrary line, L. (It's length is also L.) The sheet is bordered by lines L1 and L2, also extending to infinity.  Show that this describes a magnetic string. Where is the magnetic charge located? What is the total charge in the system? (Carefully explain the meanings of the symbols you use.) Other Problems 11. A plane interface has a magnetic suseptibility m>>1. A small magnetic dipole o(m,) is located a distance d from the plane. The direction of o(m,) is located with the azimuthal and polar angles f and q, respectively, measured with respect to the perpendicular to the plane (the z axis).  Find the force on the magnetic dipole. Is it attractive or repulsive? 12. We may write the Maxwell equations as (r and o(J,) are the total charge density, current) o(,).o(E,) = 4pr, (1) o(,) x o(E,) + f(1,c) f(o(B,),t) = 0, (2) o(,) x o(B,) - f(1,c) f(o(E,),t) = f(4p,c) o(J,), (3) o(,).o(B,) = 0. The formal solution to (3) is o(B,)(o(x,),t) = f(1,c) o(,o(o,)) d3o(x,)'(o(J,)+ f(1,4p) f(o(E,),t))x f((o(x,)-o(x,)'),|o(x,)-o(x,)'|3). (4) Show that the f(1,4p) f(o(E,),t) term in (4) leads, however, to a vanishing contribution in the quasi-magnetostatic limit f(1,c) f(o(B,),t) 0, which is equivalent to o(E,)(o(x,),t) -o(,)F(o(x,),t) from (2). (This limit is also equivalent to the statement that the energy radiated is small.) 13. A circular current loop of radius a and current I in vacuum has it's plane parallel to that of a flat interface of constant permeability m o(=,/) 1 as shown. Take the interface's plane to be given by z=0.  Using the field energy expression DW = f(1,2) i(,, )d3x o(M,).o(B,)0, show that the force of the current loop on the interface is given by Eq.(5.287) (which gives the force of the interface on the loop, so I supply a minus sign) of Ch.5: Fz = f(42I2a2,c2) (f(-1,+1))i( 0,, )dkke-2kdJ12(ka). (This is just Example 2 from Ch.6.) 14. Show that the (second order) Maxwell equations, 2F + f(1,c) f(,t) (o(,).o(A,)) = 4(rfree - o(,).o(P,)), 2o(A,) - f(1,c2) f(2o(A,),t2) - o(,)(o(,).o(A,) + f(1,c) f(F,t)) = - f(4,c) (o(J,)free + co(,)xo(M,) + f(o(P,),t)), are invariant under the generalized gauge transformation, o(A,) -> o(A,) -go(,)Ws - 4pgi(,, )do(s,)'d(o(x,)-o(x,)'), o(M,) -> o(M,) - g o(,o(o,l(c))) do(x,)'d(o(x,)-o(x,)'), F -> F, o(P,) -> o(P,), where Ws is the solid angle subtended by a finite open surface, S, the contour of which is given by two possible locations of a magnetic string. 15. Show these equations are also invariant under the more general time dependent transformations, describing a moving magnetic charge, o(A,) -> o(A,) - go(,)Ws(o(x,),t) - 4pgi(,, )do(s,)" d(o(x,)-o(R,)(t)-o(x,)"), o(M,) -> o(M,) - g o(,o(o,l(c))) do(x,)" d(o(x,)-o(R,)(t)-o(x,)"), F -> F - f(4pg,c) f(do(R,),dt) .i(,, )do(s,)" d(o(x,)-o(R,)(t)-o(x,)") + f(g,c) f(Ws(o(x,),t),t), o(P,) -> o(P,) - f(g,c) f(do(R,),dt) x o(,o(o,l(c))) do(x,)" d(o(x,)-o(R,)(t)-o(x,)"), where o(R,)(t) describes the given path of the magnetically charged particle. 16. The middle of a long, thin cylinder of radius a and length L, with permeability m is located a large distance z from a point magnetic dipole, o(m,) (z>>L>>a). It is oriented with it's long lengthwise dimension oriented perpendicular to the direction to the dipole, but along o(m,), as shown.  The interaction field energy of the configuration is (cgs or SI) DW = f(1,2) i(,, )d3x o(M,).o(B,)0, where o(M,) is the magnetization and o(B,)0 is the field before the introduction of the cylinder. Find the force of the dipole on the cylinder. Are they attracted or repelled for m>1? 17. Let's return to the spinning uniformly charged (surface charge density s) spherical shell of radius R. Again, the magnetic fields inside and outside the sphere are: o(B,)(o(x,)) = blc{(aal(l(f(8pRsw,3c) o(z,^), r < R),l(f(4pR4sw,3cr3) b(3 f(z,r) o(r,^) - o(z,^)) , r > R .))) (o(z,^), and o(r,^) are unit vectors pointing in the o(z,) and o(r,) directions, respectively.) Using the explicit force expression (6.198) of the notes, show that the total force on the top hemisphere is the same as before (attractive), Fz = - f(p2,c2) R4w2s2. 18. A long, straight solenoid (circular cross section) carrying a current I in it's wires and having n turns per length, is situated such that it's end is touching the surface of a large, flat piece of magnetic material with magnetic permeability m. (a) Argue that the B field of the solenoid is the same as from a magnet with uniform lengthwise magnetization, M, where M = f(nI,c). (b) Using (a) and an image source, show that the solenoid sticks to the flat surface with a force of attraction given by F = 2p M M', where M' = b(f(m-1,m+1)) M (repulsive if m<1). [Hint: Eq.(6.213) and prob.19(a) of Ch.5.] 19. A very long flat rectangular circuit of mass M, which has a resistance, R, and a self-inductance, L, produces a current by Ohms law. Gravity plays no role in this problem and the circuit moves frictionlessly. There is no battery in the circuit.  Given that the end of the circuit is immersed in a uniform magnetic field, o(B,), directed out of the page, find the equation of the circuit that follows from the Lorentz force, Ohm's law, and Faradays law. 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