Magnetic moments. S.M.Lea. January 2017

Transcription

1 Magnetic moments S.M.Lea January 17 1 The magnetic moment tensor Our goal here is to develop a set of moments to describe the magnetic field as we did for the electric field in Ch 4. Because there are no magnetic monopoles, the dominant contribution to B at a great distance from a current distribution is a dipole, so we start by looking at the dipole moment. For a planar loop, m is definedtobe(seeleaandburkech 9): m = IAˆn = I x d (1) where the loop is traversed counter-clockwise around ˆn according to the right hand rule. (See figure) (Note: the cross product gives the area of the parallelogram formed by x and d, but we need the area of the triangle, or half the area of the parallelogram. The cross product also conveniently gives a direction normal to the area element. ) Cross products are pseudo-vectors, so we often prefer to use an antisymmetric tensor to describe such quantities. The magnetic moment tensor is defined by: M ij I x i dx j () for a current loop, or more generally, M ij x i J j dv (3) if the current density is not confined to wire loops. Each component of M in equation represents the area of the projection of the loop onto the i j plane. Thus it should be related to the magnetic moment vector (1). Thus we consider the vector dual to M ij, defined loop 1

2 by (see Lea Optional topic A eqn A.7): Then m p = 1 ε prsi m p = 1 ε prsm rs (4) x r dx s = 1 I ε prs x r dx s = 1 x I d l p in agreement with equation (1), showing that the magnetic moment vector is dual to the tensor M ij. The dual relation has an inverse (Lea eqn A.8): ε jkp m p = ε jkp 1 ε prsm rs = 1 (δ jrδ ks δ js δ kr ) M rs But M ij is antisymmetric: M ij = I = 1 (M jk M kj ) x i dx j = I x i x j PP x j dx i = M ji Thus: ε jkp m p = M jk (5) Alternative proof of antisymmetry for general, localized, current density J: Firstnote that k (x i x j J k ) = δ ik x j J k + δ jk x i J k + x i x j k J k ρ = x j J i + x i J j x i x j t where we used the charge conservation relation in the last step. In a steady state the last term is zero. Then k (x i x j J k ) dv = V x i x j J k n k da = S (x j J i + x i J j ) dv = V since J =on the surface at infinity, and so x j J i dv = V V x i J j dv M ji = M ij (6)

3 Magnetic field due to a current loop Here we will find the multipole expansion of the magnetic field due to a current loop. We start with the vector potential (Notes 1 eqn.16). As we did in the electric case, we use a Taylor series expansion of 1/R (multipole moment notes section ) to get A = µ I dx x x = µ I dx 1 x x x + x 3 + x q x + where the tensor q has components q ij = 1 x i x j x x = xj x j x = x i x x 3 δ ij = x x 3 +3(x i x i ) xj xj x x 5 x = x = = δ ij x 3 +3x ix j x 5 (7) We may integrate term by term because the Taylor series converges uniformly. Then, moving the functions of unprimed coordinates out of the integrals, we get A i = µ I + x j x 3 x jdx i + q jk x jx kdx i = µ x j x 3 M ji + A i = µ x j x 3 ε jipm p + (8) where we used the dual inverse (5) in the last step. The leading term is the dipole: A dipole = µ m x x 3 (9) (Compare "magloop" notes eqn 5.) The next term in the expansion is the quadrupole: A i,quad = µ I q jk x jx kdx i = µ 8π q jkm jki where M jki = I x jx k dx i (1) is a rank three tensor. (More on this below.) Compare with the quadrupole term in eqn (8) of the "sphermult" notes. We can get B from A. The leading (dipole) term is: B = A = µ m x x 3 = µ = µ m( 1 x )+ 1 x m 1 x m 1 x m m 1 x But m is a constant vector, so all its derivatives are zero, and temporarily putting the z axis 3

4 along m, we get B = µ m( 1 x ) m z 1 x Now we get the last term from eqn 1 of the multipole moments notes: B = µ ẑ m [( δ (x)] + m r 3 3 x r 5 ẑ x + 3 δ (x) ẑ = µ 3 xr 5 m x mr 3 + 8π3 δ (x) m (11) which is Jackson s eqn As with the electric dipole, there is a delta-function at the origin, but in the magnetic case it is parallel to (not opposite) m. The direction and magnitude of the delta-function term may be understood by looking at a tiny current loop model for the magnetic dipole. The magnetic fieldatthecenterofaloop of radius a is (magloop notes page 4) I B = µ a ˆn Iπa = µ πa 3 ˆn = 3 µ m (a 3 /3) As a, the magnetic dipole density mδ (x) and we get B 3 µ mδ (x) as in (11). Note that the electric dipole field delta function term (multipole notes eqn 19) differs from this result by a factor of as well as the sign. See J pg 19 for applications of this to the energy of the hyperfine states of atomic systems. 3 Force and torque 3.1 Force The force exerted on a steady current distribution J by an external magnetic field B ext is: F = J B ext dv As we did in the electric case, we expand the external field in a Taylor series: F i = ε ijk J j B ext,k d 3 x B ext,k = ε ijk J j B ext,k () + x n x n + 1 x B ext,k nx p x n x p d 3 x (1) 4

5 We may use the time-independent Maxwell s equations to rewrite the first term: ε ijk B ext, k () J j d 3 x = ε ijkb ext, k () B µ j d3 x = ε ijkb ext, k () µ S ˆn B j d x = since B due to J is proportional to 1/r 3, as proved above. (For a current loop we have the more obvious result Idx j =.) Notice here that B (due to J) and B ext are distinct fields. Thus the first non-zero term in the force (1) is the second term: B ext, k F i = ε ijk J j x n x n d 3 x B ext, k = ε ijk M nj x n We may express this result in terms of the magnetic moment vector using (5): B ext, k F i = ε ijk ε njp m p x n B ext, k = (δ in δ kp δ ip δ kn ) m p x n B ext, k = m k B ext, k x i m i x k But B ext =, and we can bring the components m k through the differential operator because they are constants. We also need the fact that B ext =. Then F = m B ext (13) This is the first non-zero term in a Taylor expansion of F. Note the relation between this expression for F and the energy U = m B ext of a dipole in an external field. F = U as expected. (Note the discussion in J 5.16, pg 14, however, which shows the limitations of this interpretation.) Thenexttermis 1 B ext,k F i = ε ijk x n J j x m x n d 3 1 B ext,k x = ε ijk x n M mnj Compare with the electric field result in Jackson Problem Torque The torque exerted on a current distribution by the external field is τ = x df = x J Bext d 3 x = J x B ext B ext x J d 3 x 5

6 Usingthesameexpansionasbefore,anddroppingthesubscript"ext"on B = B ext for clarity, we have: B k τ i = J i x k B k () + J i x k x m + B i B i () x k J k x m x k J k + d 3 x Let s look at the terms one at a time. The firsttermis τ i,1 = J i x k B k () d 3 x = B k () M ki = B k () ε kip m p τ 1 = m B () (14) The third term is B i () x k J k d 3 x = B i () M kk = since M ij is antisymmetric, and thus its trace is zero. Thus (14) is the total torque if B is uniform. The other two terms are higher order corrections to the basic result (14) τ i (correction terms) = B k x k x m J i d 3 x B i x m x k J k d 3 x They involve the third rank tensor M ijk = x i x j J k d 3 x which also appeared in the expansion of A (1). This tensor is not easily expressed in terms of the vector m. It does have some symmetry: M ijk = M jik The correction terms are: τ i (correction terms) = B k M kmi B i M mkk (15) 3.3 An example Consider a current loop made of of two rectangles: One in the x y plane with dimensions a by b, and one in the y z plane with dimensions b by a. Imagine forming this thing by bending a rectangle a by b through 9. Then the magnetic moment tensor has components: b M 1 = I xdy = I ady = Iab M 13 = I xdz = a M 3 = I ydz = I bdz = Iab 6

7 Thus the tensor is: M = Iab The magnetic moment vector has components: m 1 = 1 ε 1jkMjk = 1 (M 3 M 3 )=Iab m = 1 ε jkmjk = 1 (M 31 M 13 )= m 3 = 1 ε 3jkMjk = 1 (M 1 M 1 )=Iab corresponding to the two planar parts of the loop. (Notice we can make up the bent loop from two planar loops stuck together along the y axis.) Thus the magnetic field produced by this loop, at a large distance from the loop, is (eqn 11): B = µ 1 3 (3ˆr (m ˆr) m) x = µ Iab x 3 3x x + z x ˆx ẑ Now we introduce the external magnetic field B ext = B (1 + αx)ˆx + B (1 αy)ŷ (Notice that B ext =and B ext =.) The force on the loop in this magnetic field is (eqn 13): F = m B ext = B Iab (1 + αx) =B Iabα ˆx Notice that this result is exact since the external magnetic field has no higher order derivatives. Check the dimensions! The leading term in the torque is (eqn14): τ = m B () = B Iab(ˆx +ẑ) (ˆx +ŷ) = B Iab(ẑ +ŷ ˆx) 7

8 The next term involves the tensor M ijk = I x i x j dx k and the first derivatives of B. The only non-zero derivatives are B x / x = B α and B y / y = B α. Thecorrectionterms(15)arethus: τ 1 (correction terms) = B k M km1 B 1 M mkk = B α (M 111 M 1 M 111 M 1 M 133 ) = B α (M 1 + M 1 + M 133 ) τ (correction terms) = B k M km B M mkk = B α (M 11 M + M 11 + M + M 33 ) = B α (M 11 + M 11 + M 33 ) and τ 3 (correction terms) = B k M km3 B 3 M mkk = B α (M 113 M 3 ) Let s find the values of M that we need: b M 11 = I xxdy = I a dy = Ia b M 113 = I xxdz = Thus the correction terms are: M 1 = I M 133 = I M 11 = I M 1 = I M 3 = I M 33 = I τ 1 (correction terms) = B α xydy = I xzdz = yxdx = I yydx = I yydz = I yzdz = I b a a a a aydy = I ab bxdx = I a b b dx = Iab b dz = Iab bzdz = I a b Iab + I ab + = 1 B αiab 8

9 τ (correction terms) =B α Ia b I a b + I a b = B αia b and τ 3 (correction terms) =B α Iab = B αiab Thus: τ = B Iab ẑ (1 αb)+ŷ (1 + αa) ˆx 1 αb Again this result is exact as there are no higher derivatives of B. Check the dimensions of the result. If you need more terms, it is probably wise to choose a different approach. 4 Connection between magnetic moment and angular momentum If a current distribution is made up of N particles, where particle i has position x i, charge q i, mass µ i, and moves with velocity v i, then the current is due to the particles motion N j = q i v i δ (x x i ) and then the magnetic moment tensor components are N M pr = x p j r dv = q i x p v i,r δ (x x i ) dv i=1 i=1 and the vector components are (eqn 4) m k = 1 ε kprm pr = 1 N q i ε kpr x p v i,r δ (x x i ) dv i=1 m = 1 N N q i q i x i v i = L i µ i=1 i=1 i where L i = µ i x i v i is the angular momentum of particle i about the origin. If all the particles are electrons with mass µ e, for example, then the magnetic moment is m = e N L i = e L (16) µ e µ i=1 e where L is the total angular momentum of the collection of electrons. Relation (16) is very important, and holds even on the atomic scale. However, it needs modification when applied to the internal angular momentum of individual particles, when quantum mechanics plays an important role. We can take the QM effects into account by 9

10 introducing a "fudge factor" g.. For an electron, for example, m = g e µ e s where s is the electron spin and g. See Jackson page 565 for precise values of g. See Jackson problem 6.5 for the relation between m and the electromagnetic field momentum. 1

11 5 Finding B from the Biot-Savart Law Alternatively, we can find the field by starting from the Biot-Savart law: B (x) = µ d I (x x ) x x 3 = µ I d 1 x x Inserting the Taylor expansion of 1/R, weget B (x) = µ I d 1 x + x x x 3 + x q x + (17) Now 1 x =,sothefirst term in equation 17 is zero. In the second term, (x x )=x, and d =, so term = µ I d x x 3 = Thus the first non-zero term is the third: B i = µ 8π Iε ijk dx j k (x lq lm x m) and x x 3 = k (x lq lm x m) = δ kl q lm x m + x lq lm δ km = q km x m + x lq lk = q km x m since q lm (7) is symmetric. Thus the dominant term in B is: B i = µ Iε ijkq km dx jx m = µ ε ijkq km M mj = µ ε ijk δ km x 3 +3 x kx m x 5 M mj Then using equation (4), ε ijk δ km M mj = ε ijk M kj = ε ijk M jk = m i and using the inverse relation (5): ε ijk M mj = ε ijk ε mjp m p = ε jki ε jpm m p = (δ kp δ im δ km δ ip ) m p And thus: B i = µ = δ im m k δ km m i m i x 3 +3x kx i m k x k x k m i x 5 11

12 or: B = µ 1 x 3 m+3 x (m x) x = µ 1 3 (3ˆr (m ˆr) m) x which is Jackson equation This is a dipole field, as expected, but we do not get the delta function this way. Thus the result is valid for r>. 1