12. Ch. 5, part 2

Lecture 12 - Dipoles and Magnetization

We have seen that the electrostatic field of a localized charge distribution can be expanded in electric multipoles. Similarly, the magnetic field of a stationary current distribution can be expanded in magnetic multipoles when the currents are localized. The general expansion involves vector spherical harmonics, which are left for later (Jackson Ch. 16). The multipole expansion is valid in the region external to a sphere that contains the currents.

12.1. Magnetic dipole fields

The leading term, the magnetic dipole, can be obtained directly with the use of certain differential identities for tensors and of the condition ∇ · J = 0.

12.1.1. Tensor identities for stationary currents

The identities we need all follow from the simple fact that the integral of a perfect differential is zero and are valid for Cartesian components only, or for vectors and tensors considered as abstract geometric quantities. We use the convention that a sum over repeated indices is implied.

The integral of J vanishes

We start from

∂_{i}(J_{i}x_{j}) = (∂_{i}J_{i}) x_{j} + J_{i}(∂_{i}x_{j}) = J_{j}  (12.1)

which is obtained by using ∂_{i}J_{i} ≡ ∇ · J = 0

∂_{i}x_{j} ≡ \frac{∂x_{j}}{∂x_{i}} = δ_{ij}

Integrating (12.1) we obtain the useful identity

∫ J_{j}(x) d^{3}x = 0  (12.2)

The first moment of J gives the magnetic moment

We want to prove that

\frac{1}{c} ∫ J_{i}x_{j} d^{3}x = −ε_{ijk} m_{k}  (12.3)

where ε_{ijk} is the totally antisymmetric tensor with ε_{123} = 1 (for a right-handed Cartesian system) and m is by definition the magnetic dipole moment

m =\frac{1}{2c} ∫ x × J(x) d^{3}x  (12.4)

Proceeding as in the derivation of (12.2), we integrate

∂_{i}(J_{i}x_{j}x_{k}) = (∂_{i}J_{i}) x_{j} x_{k} + J_{i}(∂_{i}x_{j})x_{k} + J_{i}x_{j}(∂_{i}x_{k}) = J_{j}x_{k} + J_{k}x_{j}

and obtain

∫ (J_{j}x_{k} + J_{k}x_{j}) d^{3}x = 0  (12.5)

We recognize that the integrand is twice the symmetric part of the first moment tensor J_{j}x_{k}. We can then decompose this tensor into symmetric and antisymmetric parts

J_{j}x_{k} = \frac{1}{2} (J_{j}x_{k} + J_{k}x_{j}) + \frac{1}{2} (J_{j}x_{k} − J_{k}x_{j})

and obtain the equation

∫ J_{j}x_{k} d^{3}x = \frac{1}{2} ∫ (J_{j}x_{k} − J_{k}x_{j}) d^{3}x  (12.6)

This does not look like progress at first, but it reduces the 9 components of J_{j}x_{k} to just 3, and these 3 can be identified with the components of J × x according to

(J × x)_{i} ≡ ε_{ijk}J_{j}x_{k} = \frac{1}{2}ε_{ijk} (J_{j}x_{k} − J_{k}x_{j})

J_{j}x_{k} − J_{k}x_{j} = ε_{jkl} (J × x)_{l}  (12.7)

These equations can easily be verified directly, because it is enough to verify the _{12} component and invoke antisymmetry. Collecting the results (12.6), (12.7) and the definition (12.4):

\frac{1}{c} ∫ J_{j}x_{k} d^{3}x = −ε_{jkl} m_{l}

which is the same as (12.3), completing the proof. (The names of tensor indices can be changed with impunity, and must often be changed when making substitutions).

12.1.2. The expansion of A and B

Now we have the tools to compute the vector potential

A_{i}(x) = \frac{1}{c} ∫ \frac{J_{i}(x′)}{|x − x′|} d^{3}x′

when |x| >> |x′| for all points x′ where J is non-negligible. We expand as usual

\frac{1}{|x − x′|} = \frac{1}{|x|} + \frac{x_{j}x′_{j}}{|x|^{3}} + …

and we find

A_{i}(x) = \frac{1}{c |x|} ∫ J_{i}(x′) d^{3}x′ + \frac{x_{j}}{c |x|^{3}} ∫ J_{i}(x′)x′_{j} d^{3}x′ + …

Using (12.2) and (12.3) this gives the desired result

A_{i}(x) = −\frac{x_{j}}{|x|^{3}} ε_{ijk}m_{k} + …

or

A(x) = \frac{m × x}{|x|^{3}} + …

The magnetic field is B = ∇ × A, or B_{i} = ε_{ijk}∂_{j}A_{k} with

A_{k} = −ε_{kpq} \frac{x_{p}}{|x|^{3}} m_{q} = ε_{kpq} (∂_{p} \frac{1}{|x|}) m_{q}

Thus, using ε_{ijk}ε_{kpq} = δ_{ip}δ_{jq} − δ_{iq}δ_{jp},

B_{i} = ε_{ijk}ε_{kpq} (∂_{j}∂_{p} \frac{1}{|x|}) m_{q} = (∂_{j}∂_{i} \frac{1}{|x|}) m_{j} − (∇^{2} \frac{1}{|x|}) m_{i}

The second term on the r.h.s. vanishes, except at x = 0 where the expansion is not to be used anyhow. The first term gives

B_{i} = ∂_{i} (m_{j}∂_{j} \frac{1}{|x|}) = − ∂_{i} (\frac{x_{j}m_{j}}{|x|^{3}} ) = \frac{3x_{i}x_{j}m_{j} − |x|^{2} m_{i}}{|x|^{5}}

or

B = ∇ (m · ∇\frac{1}{|x|}) = −∇ (\frac{x · m}{|x|^{3}}) = \frac{3x (x · m) − |x|^{2} m}{|x|^{5}}

in complete analogy to the E field for an electric dipole p.

12.1.3. Magnetic moments of a plane loop and an orbit

For a current I flowing in a loop eq. (12.4) becomes

m = \frac{I}{2c} ∮ x × dl

For a plane loop, m is perpendicular to the plane of the loop and has magnitude

m = \frac{IA}{c}

where A is the area of the loop (see Jackson Fig. 5.7).

For a stream of particles of charge q and mass m_{q}, the current density is

J(x) = q ∑_{n} v_{n}δ(x − x_{n})

where v_{n} = dx_{n}/dt. If the particles are in a closed orbit, or somehow confined, the magnetic moment is

m = \frac{q}{2c} ∑_{n} x_{n}×v_{n}

On the other hand, the angular momentum is

L = m_{q} ∑_{n} x_{n}×v_{n}

so that

m = \frac{q}{2m_{q}c} L

The ratio q/2m_{q}c is called the gyromagnetic ratio. In a quantum system, any component of L is quantized in units of h̄. For electrons in atoms then, the unit of magnetic moment is the Bohr magneton μ_{B} = |e|h̄/2m_{e}c = 0.927 · 10^{−20} gauss · cm^{3}.

As already discussed at the beginning of Lecture 11, electrons also have a magnetic moment μ_{B} (actually, 1.00116 μ_{B}) associated with their spin angular momentum \frac{1}{2}h̄. Thus the gyromagnetic ratio for intrinsic spin is (essentially) twice the orbital value.

12.2. Forces and torques

12.2.1. Force on a dipole

Consider a current distribution J localized near the point x. There is the question: how to choose x? Sometimes there is a natural choice, for instance the center of the circle for a circular current loop. But for an elliptical current loop either the center of the ellipse, or one of the foci could be the “natural” choice for x. Let us leave the question aside for the moment and write the current at the point x′ in the form J(x′ − x) to stress that J vanishes for large values of |x′ − x| . According to Ampère’s force law, this current in a field B is subject to the force F of components

F_{i}(x) = \frac{1}{c} ε_{ijk} ∫ J_{j}(x′ − x)B_{k}(x′) d^{3}x′

Expanding

B_{k}(x′) = B_{k}(x) + (x′_{p} − x_{p}) ∂_{p}B_{k}(x) + …

and using from (12.2)

∫ J_{j}(x′ − x) d^{3}x′ = ∫ J_{j}(x′′) d^{3}x′′ = 0

we obtain

F_{i}(x) = \frac{1}{c} ε_{ijk} (∂_{p}B_{k}(x)) ∫ J_{j}(x′ − x) (x′_{p} − x_{p}) d^{3}x′ + …

Further, using (12.3) in the form

\frac{1}{c} ∫ J_{j}(x′′)x′′_{p} d^{3}x′′ = −ε_{jpq}m_{q}

and −ε_{ijk}ε_{jpq} = ε_{jik}ε_{jpq} = δ_{ip}δ_{kq} − δ_{iq}δ_{kp}, we arrive at

F_{i}(x) = (δ_{ip}δ_{kq} − δ_{iq}δ_{kp}) m_{q}∂_{p}B_{k}(x) = m_{k}∂_{i}B_{k}(x) − m_{i}∂_{k}B_{k}(x)

or, noting that ∂_{k}B_{k} ≡ ∇ · B = 0,

F_{i}(x) = ∂_{i} (m_{k}B_{k}(x)) F(x) = ∇ (m · B(x))

in complete analogy with the electrostatic case. This formula is exact for any current distribution if B varies linearly. In particular, there is no net force on a current loop in a uniform field, even if the loop is not small. Returning to the question of the choice of x, we note that m is actually independent of this choice, but ∂_{i}B_{k} is not.

12.2.2. Torque on a dipole

To compute the leading term of the torque N, it is enough to keep the leading term in the expansion B(x′) = B(x) + · · ·. The torque about x is

N = \frac{1}{c} ∫ (x′ − x) × [J(x′ − x) × B(x)] d^{3}x′

N_{i} = \frac{1}{c} ∫ x′′_{j} [B_{j}(x)J_{i}(x′′) − J_{j}(x′′)B_{i}(x)] d^{3}x′′

Using again (12.3) and the trace of (12.5) we obtain

N_{i} = −ε_{ijk}B_{j}(x)m_{k} N = −B (x) × m = m × B (x)

This formula is exact for a uniform field, and also for a field that is uniform over a current loop, but varies perpendicularly to it.

12.2.3. Energy of a permanent dipole

Both the force and the torque can be derived from the potential energy

U(x, θ) = −m · B (x) = −mB(x) cos θ

as follows:

F = −∇U(x)

and, with the angle θ oriented from B to m,

N = −∂U/∂θ

The dipole tends to orient itself parallel to the field and to be attracted to regions of high flux density. In cyclotron motion, the orbit is a spiral with an effective dipole moment that is antiparallel to the field. Thus the particles are repelled from regions of high flux density.

12.3. Hyperfine interaction in atoms

A nuclear magnetic moment μ^{(N)} interacts with the magnetic field eL/mcr^{3} that is caused by the orbiting motion of the electrons in a state with angular momentum L. The interaction energy is described by

H_{HF}^{(orbit)} = −\frac{e}{mc} L · μ^{(N)} ∫ \frac{ρ(r)}{r^{3}} d^{3}r

This “orbital” contribution to the hyperfine interaction vanishes for s states (L = 0).

An electron also carries a magnetic moment μ_{e} and associated with the electron density ρ there is a (spin) magnetization density M = ρμ_{e} and a vector potential

A(x) = ∫ \frac{M(x′)× (x − x′)}{|x − x′|^{3}} d^{3}x′ = ∫ M(x′) × ∇′ \frac{1}{|x − x′|} d^{3}x′

Integrating by parts, this becomes

A(x) = ∫ \frac{∇′ × M(x′)}{|x − x′|} d^{3}x′

showing that there is an effective magnetization current

J_{M} = c∇ × M = c∇ρ × μ_{e}

which causes a magnetic field

B(x) = ∫ (∇′ρ(x′) × μ^{(e)}) × \begin{matrix}x − x & ′ \\ |x − x′| & 3 \end{matrix} d^{3}x′

Using tensor notation and r = |x − x′|^{−1}

B_{i} = ∫ ε_{ijk} ε_{jpq} (∂′_{p}ρ) μ_{q}^{(e)} ∂′_{k} \frac{1}{r} d^{3}x′ = ∫ ε_{ikj}ε_{jpq}ρ μ_{q}^{(e)}∂′_{k}∂′_{p} \frac{1}{r} d^{3}x′

where we have integrated by parts and used ε_{ikj} = −ε_{ijk}. Now we use

ε_{ikj}ε_{jpq} = δ_{ip}δ_{kq} − δ_{iq}δ_{kp}

and obtain

B_{i}(x) = ∫ ρ(x′) (μ_{k}^{(e)}∂′_{k}∂′_{i} \frac{1}{r} − μ_{i}^{(e)}∇^{′ 2} \frac{1}{r}) d^{3}x′

The tensor ∂′_{k}∂′_{i}(1/r) has a singularity at r = 0 in its isotropic part \frac{1}{3} δ_{ki}∇^{2} (1/r) (The traceless part averages to zero over a small sphere and its singular behavior does not affect the integral). Thus the parenthesis in the integrand is conveniently written

μ_{k}^{(e)} (∂′_{k}∂′_{i} − \frac{1}{3} δ_{ki}∇^{′ 2}) \frac{1}{r} − \frac{2}{3}μ_{i}^{(e)}∇^{′ 2} \frac{1}{r}

and, with no ambiguity at r = 0, it becomes

μ_{k}^{(e)} (\frac{3r_{i}r_{k}}{r^{5}} − \frac{δ_{ik}}{r^{3}} ) + \frac{8π}{3} μ_{i}^{(e)}δ(r)

where r = x − x′. Finally

B_{i}(x) = ∫ ρ(x′) [μ_{k}^{(e)} (\frac{3r_{i}r_{k}}{r^{5}} − \frac{δ_{ik}}{r^{3}} ) + \frac{8π}{3} μ_{i}^{(e)}δ(r)] d^{3}x′

The hyperfine interaction H_{HF} in an atom is −μ_{i}^{(N)}B_{i}(0), where μ_{i}^{(N)} is the magnetic moment of the nucleus. From the above we get

H_{HF}^{(spin)} = ∫ ρ(r) (\frac{μ^{(e)}· μ^{(N)}}{r^{3}} − \frac{3 (r · μ^{(e)}) (r · μ^{(e)})}{r^{5}} ) d^{3}r
−\frac{8π}{3} μ^{(e)}· μ^{(N)}ρ(0)  (12.8)

The first line of this equation is the standard external interaction between two dipoles; the second line must be added when the two dipoles are on top of each other. It is this “contact term” that is responsible for hyperfine splitting in l = 0 atomic states, including the splitting of the 1s state of atomic hydrogen that gives rise to radio absorption at 21 cm in interstellar space.

12.4. Magnetic potential of a loop

We have seen that an infinitesimal loop of area dA has the magnetic scalar potential of a dipole dm = (I/c)dA. A large loop can be built up from many small loops and has the magnetic potential of a dipole layer bounded by the loop. The shape of the dipole layer is arbitrary. For instance, for a circular loop, we can choose the fictitious dipole layer in the form of any spherical cap bounded by the loop, or the flat circle inside the loop. We know from electrostatics that Φ_{M} = IΩ/c, where Ω is the solid angle subtended by the loop. In general, to compute Ω is easier said than done. We turn to an example.

12.5. Circular current loop

The magnetic field of a circular loop of radius a carrying a current I can be computed in several ways. One can compute the vector potential, which has only the component A_{ϕ}, or one can compute the magnetic potential Φ_{M}, which is just the solid angle subtended by the loop. One can work directly with elliptic integrals, or use cylindrical coordinates and work with Bessel functions, or use spherical polar coordinates and spherical harmonics.

Because of cylindrical symmetry, computations can be restricted to the xz plane (ϕ = 0). The geometry is defined by

x = r (sin θ, 0, cos θ) = (ρ, 0, z) field point x′ = a (cos ϕ′, sin ϕ′, 0) source point |x − x′|^{2} = r^{2} + a^{2} − 2ra sin θ cos ϕ′ dl′ = a (− sin ϕ′, cos ϕ′, 0)dϕ′ wire element

Starting formulas are

A(x) = \frac{I}{c} ∮ \frac{dl′}{|x − x′|}

A_{ϕ}(ρ, z) = \frac{I}{c} ∫_{0}^{2π} \frac{a cos ϕ′ dϕ′}{\sqrt{r^{2} + a^{2} − 2aρ cos ϕ′}}  (12.9)
B(x) = − \frac{I}{c} ∮ \frac{(x − x′) × dl′}{|x − x′|^{3}}  (12.10)

12.5.1. Elementary results Directly from eq. (12.10) on the z axis (ρ = 0) B_{ρ} = 0 and

B_{z} = \frac{2πI}{ca} \frac{a^{3}}{(a^{2} + z^{2})^{3/2}}  (12.11)

From eq. (12.9) in the limit where r^{2} + a^{2} >> 2aρ

A_{ϕ}(ρ, z) ≃ \frac{I}{c} ∫_{0}^{2π} \frac{a cos ϕ′}{\sqrt{r^{2} + a^{2}}} (1 + \frac{aρ cos ϕ′}{r^{2} + a^{2}} ) dϕ′ = \frac{I}{c} \frac{πa^{2}ρ}{(r^{2} + a^{2})^{3/2}}  (12.12)

B_{ρ} = −\frac{∂A_{ϕ}}{∂z} ≃ \frac{πIa^{2}}{c} \frac{3ρz}{(r^{2} + a^{2})^{5/2}}

B_{z} = \frac{1}{ρ} \frac{∂ (ρA_{ϕ})}{∂ρ} ≃ \frac{πIa^{2}}{c} \frac{2 (r^{2} + a^{2}) − 3ρ^{2}}{(r^{2} + a^{2})^{5/2}}

In particular along the z axis (ρ = 0) we recover (12.11).

Far from the loop (r >> a) we recover the dipole field:

A_{ϕ}(r, θ) ≃ \frac{πIa^{2}}{c} \frac{sin θ}{r^{2}}

B_{θ} = −\frac{1}{r} \frac{∂ (rA_{ϕ})}{∂r} = \frac{πIa^{2}}{c} \frac{sin θ}{r^{3}}

B_{r} = \frac{1}{r sin θ} \frac{∂ (sin θ A_{ϕ})}{∂θ} = \frac{πIa^{2}}{c} \frac{2 cos θ}{r^{3}}

12.5.2. By elliptic integrals We work out this method in detail and leave the method of spherical harmonics for reading in Jackson.

The charged ring First we review the similar problem of computing the electrostatic potential of a uniformly charged ring, which is

Φ(ρ, z) = \frac{Q}{π} ∫_{0}^{π} \frac{dϕ′}{\sqrt{r^{2} + a^{2} − 2aρ cos ϕ′}}

The integral is transformed to standard form by the change of variable ϕ′ = π − 2φ, cos ϕ′ = 2 sin^{2} φ − 1, so that

\sqrt{r^{2} + a^{2} − 2aρ cos ϕ} = \sqrt{a^{2} + r^{2} + 2aρ}\sqrt{1 − \frac{4aρ}{a^{2} + r^{2} + 2aρ} sin^{2} φ}

The result is

Φ = \frac{2Q}{π\sqrt{a^{2} + r^{2} + 2aρ}}K (\frac{4aρ}{a^{2} + r^{2} + 2aρ})

where K(m) is the complete elliptic integral of the first kind

K(m) = ∫_{0}^{π/2} \frac{dφ}{\sqrt{1 − x sin^{2} φ}}  (12.13)

Vector potential of the loop Now, for the circular current loop we have, recalling eq. (12.9),

A_{ϕ}(ρ, z) = \frac{2I}{c} ∫_{0}^{π} \frac{a cos ϕ′ dϕ′}{\sqrt{r^{2} + a^{2} − 2aρ cos ϕ′}}

Using the same change of variables as above

A_{ϕ}(ρ, z) = \frac{2I}{c} \frac{2a}{\sqrt{r^{2} + a^{2} + 2aρ}} ∫_{0}^{π/2} \frac{2 sin^{2} φ − 1}{\sqrt{1 − k^{2} sin^{2} φ}} dφ

where we have defined for convenience

k^{2} = \frac{4aρ}{a^{2} + r^{2} + 2aρ} = \frac{4aρ}{(a + ρ)^{2} + z^{2}}

Writing

sin^{2} φ = \frac{1 − (1 − k^{2} sin^{2} φ)}{k^{2}}

Lecture 12 (Ch 5 part 2) Dipoles and magnetization 11 the integral becomes

∫_{0}^{π/2} \frac{2 sin^{2} φ − 1}{\sqrt{1 − k^{2} sin^{2} φ}} dφ = (\frac{2}{k^{2}} − 1) K(k^{2}) − \frac{2}{k^{2}} E(k^{2})

where E(m) is the complete elliptic integral of the second kind

E(m) = ∫_{0}^{π/2} \sqrt{1 − m sin^{2} φ} dφ  (12.14)

The final result for the vector potential is

A_{ϕ}(ρ, z) = \frac{4Ia}{c\sqrt{(a + ρ)^{2} + z^{2}}} [\frac{(2 − k^{2})K(k^{2}) − 2E(k^{2})}{k^{2}}]  (12.15)

The elementary results can be recovered from this. Magnetic field We now compute

B_{ρ} = −\frac{∂A_{ϕ}}{∂z} B_{z} = \frac{1}{ρ} \frac{∂}{∂ρ} (ρA_{ϕ})

with the help of the formulas (Arfken, page 326)

\frac{dE}{dm} = \frac{E − K}{m} \frac{dK}{dm} = \frac{E}{m(1 − m)} − \frac{K}{m}

The result is

B_{ρ} = \frac{2Iaz}{cρ\sqrt{(a + ρ)^{2} + z^{2}}} [\frac{a^{2} + ρ^{2} + z^{2}}{(a − ρ)^{2} + z^{2}} E(k^{2}) − K(k^{2})]

B_{z} = \frac{2Ia}{c\sqrt{(a + ρ)^{2} + z^{2}}} [\frac{a^{2} − ρ^{2} − z^{2}}{(a − ρ)^{2} + z^{2}} E(k^{2}) + K(k^{2})]

The simple results on the z axis, for instance, are recovered using K(0) = E(0) = π/2.

12.6. Magnetization

Because the electrons carry a magnetic moment, there are actually two contributions to the magnetization, the orbital magnetization

M_{orbit}(x) = \frac{1}{2c} x × J(x)

that we have already discussed, and the spin magnetization

M_{spin}(x) = ∑_{i} ρ_{i}(x)μ_{i}^{(e)}

where μ_{i}^{(e)} is the spin magnetic moment of the i-th electron. The integral of M_{orbit} over the volume of an atom (or whatever) gives its orbital magnetic moment

m_{orbit} = \frac{1}{2c} ∫ x × J(x) d^{3}x

in agreement with our earlier definition (12.4) (but we use m_{orbit} to be more precise). It is also true that M_{orbit} is the orbital magnetic moment per unit volume, and M_{spin} is the spin magnetic moment per unit volume.

With an atom density N of almost 10^{23}/cm^{3}, which is typical of a solid, one Bohr magneton per atom corresponds to a magnetization M of about 1 kilogauss or to B ≃ 10 kgauss = 1 Tesla (we are anticipating B = 4πM).

The spontaneous magnetization of Fe is due to the alignment of the spins, and the same is generally true for ferromagnetic materials. The actual value of M for Fe at room temperature is 1.7 kgauss, in agreement with the order-of-magnitude arguments of the preceding paragraph. Orbital magnetism is usually quite weak. We discuss magnetization more in detail in the next lecture.