140 17 Electrostatics and Magnetostatics one obtains divB ≡ 0 , or better ∂V d 2 A(B ·n) ≡ 0 , ∀V (Maxwell II) ; (17.49) i.e., there are no (“true”) magnetic charges at all, but only magnetic dipoles. The above relation is the second Maxwell equation, which is again essen- tially based on experimental experience. Using these relations, the magnetic field H in the absence of electric currents, can be derived from a magnetic potential φ m , which is calculated from equations analogous to (17.46) and (17.48). (The reader – if a student undergoing examinations – should write down the equations as preparation for a possible question.) Solution: H(r)=−gradφ m (r) , with the two equivalent formulae a) dipole representation for magnetized bodies φ m (r)= G dV J(r ) ·(r −r ) 4πμ 0 |r −r | 3 ; (17.50) b) representation in terms of effective magnetic charges φ m (r)= G dV −(divJ)(r ) 4πμ 0 |r −r | + ∂G d 2 A J(r ) ·n(r ) 4πμ 0 |r −r | . (17.51) 17.2.6 Forces and Torques on Electric and Magnetic Dipoles The force on an electric dipole in an electric field may be calculated using the dumbbell approximation of two opposite charges at slightly different po- sitions: F = q · E r + a 2 − E r − a 2 ∼ = q 3 i=1 a i ∂ ∂x i E(r) → (p ·∇) E(r) . (17.52) In the dipole limit (q →∞, a → 0, but q·a → p(= 0), whereas qa i a j → 0) one obtains the result F =(p · grad)E(r) . A similar calculation yields the formula for the torque D: D = q r + a 2 × E r + a 2 − r − a 2 × E r − a 2 → p × E(r) . (17.53) For the magnetic case one only needs to replace p by m and E by H. 17.2 Electrostatic and Magnetostatic Fields in Polarizable Matter 141 17.2.7 The Field Energy We shall now introduce a number of different, but equivalent, expressions for the energy associated with an electric field. The expressions are fundamental and will be used later. Firstly, we shall start with a capacitor with dielectric material of dielectric constant ε between the plates. Let us transport an infinitesimal amount of charge 14 , δQ,fromthemetal plate at lower potential to that with the higher potential, where a capacitor voltage U(Q)= Q C has already been built up. The (infinitesimal) energy or work done in trans- porting the charge is δE = δQ ·U(Q) . By transporting the total charge in this way we finally obtain E = Q 0 d ˜ Q · U( ˜ Q) , with U( ˜ Q)= ˜ Q C , i.e., 1.) E = 1 2 U · Q = Q 2 2C = C 2 U 2 . (17.54) This is our first expression for the electric field energy. The following (seemingly more general) expression is equivalent: 2.) E = 1 2 φ(r)(r)dV. (17.55) Since the charge density vanishes in the space between the metal plates, ≡ 0, whereas at the lower plate the potential is φ ≡ 0 and at the upper plate φ ≡ U , and since dV can be replaced by σd 2 A (with σ = D ·n), we obtain the same result as before. Now, substituting the result φ(r)= dV (r ) 4πεε 0 |r −r | into equation (17.55), we obtain a third expression for the field energy, which is the energy of mutual repulsion of the charges at r and r: 3.) E = 1 2 r r dV dV (r )(r) 4πεε 0 |r −r | . (17.56) 14 “true charge” 142 17 Electrostatics and Magnetostatics (The factor “ 1 2 ” in all these expressions is most obvious here, since when calculating the mutual repulsion each pair of charges should only be counted once.) Finally, perhaps the most important expression is the following result, which is obtained from (17.55) by integrating by parts: In equation (17.55) we substitute = ∇·D and again shift the ∇-differentiation to the left. The result, with −∇φ = E , is: 4.) E = 1 2 S dV E · D. (17.57) (It is obvious that one only has to integrate over the space S between the metal plates 1 and 2, since in the interior of the plates the electric field E vanishes.) The energy density of the electric field, w e , e.g., to build up the field between the plates of a capacitor, is thus given by w e (r):= ε · ε 0 2 E(r) 2 . Similarly (without proof) in the magnetic case we have the following result for the energy density of the field, w m , e.g., to build up the magnetic field in the interior of a solenoid filled with material of relative permeability μ: w m (r)= μ · μ 0 2 H(r) 2 . 17.2.8 The Demagnetization Tensor With the aid of the above equations (17.46) and (17.48) one can always calculate the electrostatic field E(r) and the magnetostatic field H(r)of an electrically or magnetically polarized system, at every sampling point r; however, in general the calculation is difficult and the results are complicated, except for the case of an ellipsoid. It can be shown that the field outside an ellipsoid has exactly the same form as the field due to a dipole at the center, with dipole moment p = P ΔV , where ΔV is the volume of the ellipsoid and P is the electric polarization. This field is of course inhomogeneous, but rather simple. In the interior of the ellipsoid one has an even simpler result, a homogeneous field with the 17.2 Electrostatic and Magnetostatic Fields in Polarizable Matter 143 three components E i = − 1 ε 0 3 k=1 N ik P k . (The proof is omitted here). N i,k (= N k,i ) is a tensor, which is diagonal w.r.t. the principal axes of the ellipsoid and has the property that the eigenvalues depend only on the ratios of these axes. In the magnetic case, this same tensor is called the demagnetization tensor. It always satisfies the identity 3 i=1 N ii ≡ 1 . The three eigenvalues of the demagnetization tensor are called demagneti- zation factors; i.e., for a sphere they have the value 1 3 ; for an infinitely-long circular cylinder two of the eigenvalues (the “transverse” ones) are 1 2 ,whereas the “longitudinal” one is zero; and finally for a very thin extended plane the two “in-plane” eigenvalues are zero, whereas the “out-of-plane” eigenvalue is 1. For other geometries the eigenvalues can be found in tables. 17.2.9 Discontinuities at Interfaces; “Interface Divergence” and “Interface Curl” At interfaces between systems with different material properties the fields are usually discontinuous, but the integral formulations of Maxwell’s equations are valid, whenever the integrals can be performed, e.g., for piecewise con- tinuous functions, which are non-differentiable. From Gauss’s law (“Maxwell I”) (17.32), by applying it to a so-called Gauss interface box (which is a box running parallel to the 2d-interface that is aligned, e.g., horizontally,such that the top of the box is contained in the region above the interface and the bottom is below the interface, whereas the height of the side surfaces is negligibly small) the following equation can be derived n · D (+) − D (−) = σ. (17.58) The D (±) are the fields at the outer and inner sides of the interface, respectively, and σ is the (2d) interface charge density. The operation n · (v + − v − ) corresponding to the l.h.s. of equation (17.58) is called the interface diver- gence of the vector field v(r). This quantity is obtained from divv by formally replacing the (vectorial) differential operation ∇·v by the difference operation n · (v + − v − ) appearing in (17.58). 144 17 Electrostatics and Magnetostatics One can proceed similarly with the curl operator: Calculating the circula- tion of the electric field E(r)alongaStokes interface loop (i.e., a small closed loop running in one direction on the upper side of the horizonal interface and in the opposite direction on the lower side but with negligible vertical height), one obtains from curlE = ∇×E =0: n × E (+) − E (−) =0. (17.59) From (17.58) and (17.59) one can derive a law of refraction the electric field lines at the interface between two different dielectric materials. This law follows from the fact that the tangential components of E are continuous, whereas the normal components n · E (i) with i =1, 2 (i.e., corresponding to the two different materials) are inversely proportional to the respective ε i . It then follows that tan α 2 tan α 1 ≡ ε 2 ε 1 , with angles α i to the normal. For ε 2 /ε 1 →∞one obtains conditions such as those for a metal surface in vacuo, α 2 → 90 ◦ , α 1 → 0 (a sketch is recommended). 18 Magnetic Field of Steady Electric Currents 18.1 Amp`ere’s Law For centuries it had been assumed that electricity and magnetism were com- pletely separate phenomena. Therefore it was quite a scientific sensation when in 1818 the Danish physicist Hans Christian Ørsted proved experimentally that magnetic fields were not only generated by permanent magnetic dipoles, but also by electric currents, and when slightly later Andr´eMarieAmp`ere showed quantitatively that the circulation of the magnetic field H along a closed loop followed the simple relation: ∂F H(r) · dr = I(F ) (Ampere’s law) . (18.1) Here, I(F )istheflux of electric current through a surface F inserted into the closed loop Γ = ∂F 1 I(F ):= F j ·nd 2 A. (18.2) j := (r)v(r) is the vector of the current density (dimensionality: A/cm 2 =C/(cm 2 s)). With Stokes’s integral theorem it follows that the differential form of Amp`ere’s law (18.1) is given by: curlH = j . (18.3) For the special case of a thin wire aligned along the z-axis from (−∞)to (+∞), in which a steady electric current I flows, using cylindrical coordinates one obtains H ‘z−wire = e ϕ I 2πr ⊥ . (18.4) Just as the electrostatic field of a point charge possesses a (three- dimensional) δ-divergence, 1 The surface F is not uniquely defined by Γ , since different surfaces can be in- serted into the same closed loop. This is the topological reason underlying gauge freedom of the vector potential, which is discussed below. 146 18 Magnetic Field of Steady Electric Currents div qr 4πε 0 r 3 = qδ(x, y, z) , an analogous relation is also valid for the curl of the magnetic field of the above “z-wire”: (curlH “z−wire )(x, y, z)=Iδ(x, y)e z . We can formulate these ideas in a general way: The effective electric charges are the sources of the electrostatic field E(r) (whereas the vortices of E vanish); in contrast the vortices of the magneto- static field B(r) correspond to effective electric currents (whereas the sources of B vanish). Generally, a vector field v(r) is determined by its sources and vortices. Note that we have written B,notH, and “effective” quantities, not “true” ones (see above). In particular, the relations between E and D as well as B and H are not quite simple, and not all magnetic fields are produced by electric currents (Sect. 18.5 → spin magnetism). 18.1.1 An Application: 2d Boundary Currents for Superconductors; The Meissner Effect As already detailed in Sect. 17.2.9, at an interface Amp`ere’s equation curlH = j must be generalized to n × (H + − H − )=j s , where j s is an interface-current density (dimensionality: A/cm, not A/cm 2 ; and we have j s ≡ σv, analogously to j ≡ v). As we shall see, this formulation yields a simple explanation of the so- called Meissner effect of superconductivity. This effect amounts to “expelling” the magnetic field from the interior of a superconducting material, by loss- free interface (super)currents that flow tangentially at the interface between a superconducting region “1” (e.g., the r.h.s. of a plane) and a normally conducting region “2” (e.g., vacuum on the l.h.s.). For example, if the in- terface normal (from “1” to “2”) is in the (−x)-direction and the external magnetic field (in the normal conducting region “2”) is (as usual) in the +z- direction, then in “1” (at the interface towards “2”) supercurrents flow in the y-direction, producing in “1” a field −Be z , which is different from zero only in a very thin layer of typical width Δx = λ ≈ 10 nm . 18.2 The Vector Potential; Gauge Transformations 147 For energy reasons (the magnetic field energy in region “1” can be saved) the supercurrents flow with such a strength that in the interior of region “1”, outside the above-mentioned interface zone of width Δx = λ, the external magnetic field is exactly compensated. Further details cannot be given here. 18.2 The Vector Potential; Gauge Transformations Since curlH = j(=0), the magnetic field can no longer be calculated from a scalar potential: With H(r)=−gradφ m (r) one would derive curlH ≡ 0 , since curl gradφ m (r) ≡ 0 for arbitrary scalar functions φ m (r). (∇×(∇φ m ) is formally a cross-product of two identical vectors and thus ≡ 0.) Fortunately we have divB(r) ≡ 0 , so that one can try: B =curlA(r) , because div curlv(r) ≡ 0 for all vector fields v(r), as can easily be shown. (Formally div curlv is a so- called spate product, the determinant of a 3 × 3-matrix, i.e., of the form u · [v × w], with two identical vectors, ∇·[∇×v], and therefore it also vanishes identically.) In fact an important mathematical theorem, Poincar´e’s lemma, states the following: For source-free vector fields B, i.e., if ∂G d 2 AB ·n ≡ 0 , in a convex open region G (e.g., in the interior of a sphere) with a sufficiently well-behaved connected boundary ∂G, one can write vector potentials A with B =curlA . 148 18 Magnetic Field of Steady Electric Currents One should note that A is not at all unique, i.e., there is an infinity of different vector potentials A, but essentially they are all identical. If one adds an arbitrary gradient field to A, then curlA is not changed at all. A so-called gauge transformation: A → A := A +gradf (r) , (18.5) with arbitrary f(r), implies curlA ≡ curlA , since curl gradf ≡ 0 . Therefore, the physical quantity B is unchanged. 18.3 The Biot-Savart Equation In the following we consider, as usual, G = R 3 . a) Firstly, we shall use a gauge such that divA(r)=0(Landau gauge). b) Secondly, from Amp`ere’s law, curlH = j , with B = μ 0 H + J , we conclude that curlB = μ 0 j +curlJ =: μ 0 j B , with the effective current j B := j + M , where M := J μ 0 is the magnetization and J the magnetic polarization 2 . c) Thirdly, we now use the general identity curl curlA ≡ grad divA −∇ 2 A . (18.6) Hence, due to curl B =: μ 0 j B , the Cartesian components of A satisfy the Poisson equations −∇ 2 A i = μ 0 ·(j B ) i , for i = x, y, z . The solution of these equations is analogous to the electrostatic problem, viz A(r)= dV μ 0 j B (r ) 4π|r −r | . (18.7) 2 In the cgs system the corresponding quantities are M “ =(ΔV ) −1 P r i ∈ΔV m i ” and 4πM . 18.4 Amp`ere’s Current Loops and their Equivalent Magnetic Dipoles 149 One can easily show by partial integration that this result also satisfies the equation divA ≡ 0 , since divj B =0. Later, in the context of the so-called continuity equation, this relation will be discussed more generally. By applying the curl operator, equation (18.7) leads to the formula of Biot and Savart: B(r)= dV μ 0 4π j B (r ) × (r −r ) |r −r | 3 . (18.8) In the integrand one has the same dependence on distance as in Coulomb’s law for E, but complemented by the well-known right-hand rule connecting the directions of the effective current j B and the magnetic induction B, i.e., the product 1 ε 0 E (r ) (r − r ) |r −r | 3 is replaced by the cross-product μ 0 j B (r ) × (r − r ) |r −r | 3 . (It is no coincidence that the equation for A, (18.7), is easier to remember than its consequence, the Biot-Savart equation (18.8).) 18.4 Amp`ere’s Current Loops and their Equivalent Magnetic Dipoles This section is especially important, since it shows that the relationships be- tween electric currents and magnetic dipoles are very strong indeed. Firstly we state (without proof, but see the next footnote) that the magnetic induc- tion B(r) produced by a current loop Γ = ∂F (current I) is quantitatively identical to the magnetic field that would be produced by an infinitesimal film of magnetic dipoles inserted into the same loop, i.e., for the fictitious 2d-dipole density dm of that film the following formula would apply: dm ≡ μ 0 Ind 2 A. a) For a current loop, one obtains from Biot and Savart’s equation B(r)= μ 0 I 4π ∂F dr × r −r |r −r | 3 . (18.9) . arbitrary gradient field to A, then curlA is not changed at all. A so-called gauge transformation: A → A := A +gradf (r) , (18.5) with arbitrary f(r), implies curlA ≡ curlA , since curl gradf ≡. the plates. Let us transport an infinitesimal amount of charge 14 , δQ,fromthemetal plate at lower potential to that with the higher potential, where a capacitor voltage U(Q)= Q C has already been. . Therefore, the physical quantity B is unchanged. 18.3 The Biot-Savart Equation In the following we consider, as usual, G = R 3 . a) Firstly, we shall use a gauge such that divA(r)=0(Landau gauge). b) Secondly,