Arch. Rational Mech. Anal. 192 (2009) 589–611
Digital Object Identifier (DOI) 10.1007/s00205-008-0134-4
Discrete-to-Continuum Limit of Magnetic
Forces: Dependence on the Distance Between
Bodies
Anja Schlömerkemper & Bernd Schmidt
Communicated by A. Mielke
Abstract
We investigate force formulae for two rigid magnetic bodies in dependence on
their mutual distance. These formulae are derived as continuum limits of atomistic
dipole–dipole interactions. For bodies that are far apart in terms of the typical
lattice spacing we recover a classical formula for magnetic forces. For bodies whose
distance is comparable to the atomistic lattice spacing, however, we discover a new
term that explicitly depends on the distance, measured in atomic units, and the
underlying crystal lattice structure. This new term links the classical force formula
and a limiting force formula obtained earlier in the case of two bodies being in
contact on the atomistic scale.
1. Introduction
Multiscale models of materials have recently attracted a lot of interest in the
engineering and mathematical literature. Many interesting phenomena in materials
science can be understood only when taking into account effects that are due to an
intricate interplay between various models at different length scales. To understand
such behavior better one relates the models at different scales. In particular, bridging
the scales from atomistic models to continuum theory is an active area of research.
In this article we consider discrete-to-continuum limits for magnetic forces.
Formulae for the magnetic force between rigid magnetic bodies in contact have
been under discussion for quite some time, cf. [2,4,5,7] and in particular [1]. All
those formulae are obtained from models in a continuum setting, that is, from a
macroscopic point of view. This neglects contributions to the force from dipoles
close to the interface of the bodies as was already pointed out by Brown [2, p. 53]
and mathematically studied in [9–11,13]. In these studies a lattice of magnetic
dipole moments is considered and the discrete-to-continuum limit of the force
between two connected parts of the lattice is calculated, which yields an additional
lattice-dependent force term, called Fshort .
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Anja Schlömerkemper & Bernd Schmidt
Whereas the discrete-to-continuum results in [10,13] deal with the discussion
of magnetic forces between bodies being in contact on the lattice as well as on the
macroscopic scale, the main focus of the present paper is to study the discrete-tocontinuum limit of the magnetic force formula in various distance regimes between
two bodies being separated on the scale of the lattice, but being in contact on the
macroscopic scale. It turns out that the previous results have to be extended by an
additional term that depends on the microscopic distance and reflects the discrete
nature of the underlying atomic lattice, cf. Section 3, in particular Theorem 3.4.
More precisely, suppose the lattice spacing is equal to −1 , ∈ N, and consider
two bodies (sets of lattice points) whose mutual distance is given by ε = a , where
a is a measure for the microscopic distance in terms of atomistic units. We study
the discrete-to-continuum limit for magnetic forces for fixed a ∈ N in detail in
Section 3. As → ∞, ε tends to 0 and thus the macroscopic distance between the
two bodies vanishes, that is, the bodies are macroscopically in contact.
The discrete-to-continuum limit of the force which is exerted by a magnetic
rigid body B on a magnetic rigid body A is (Theorem 3.4)
Flim (A, B, a) = Flong (A, B) + Fshort (A, B) + G(a),
where Flim (A, B) = Flong (A, B) + Fshort (A, B) is the limiting formula which
was derived in [13,10] in the case a = 0, see also Equations (9) and (11). In
particular we have G(0) = 0. For a > 0 we obtain an additional a-dependent term
G(a), which has not been included in any force formula so far and which might
be particularly interesting with regard to nanoscale experiments. For the definition
of G(a) see Equations (11) and (13). In Example 4.4 we prove that this additional
term decreases exponentially in a and we give some numerical values.
In Section 4 we discuss properties of the additional a-dependent force term. In
particular we study the limit as a → ∞, which corresponds to two bodies that are
in contact on the macroscale but are infinitely far apart on the microscale. It turns
out (Theorem 4.3) that
lim Flim (A, B, a) = FBr (A, B),
a→∞
where FBr is a force formula which was extensively analyzed by Brown [2] in the
continuum setting and which is called Brown’s formula for short in the following,
cf. (25). In other words, we here give a discrete-to-continuum derivation for FBr .
Thereby we flesh out a claim by Brown [2, p. 53] who expected—phrased in
the notions of our multiscale setting—that the force contribution from dipole–
dipole interactions close to the (macroscopic) interface decreases rapidly as the
(microscopic) distance between the bodies increases.
Furthermore we study the case a ∼ , which corresponds to two bodies being
microscopically as well as macroscopically separated. This case yields a classical
well-known force formula for separated bodies as → ∞, see Proposition 5.1.
To give a complete multiscale picture of all the possible distance regimes, we also
include a study in which the micro-distance a = a() scales with such that
1 a() (Section 5.2).
Discrete-to-Continuum Limit of Magnetic Forces
591
In Section 6 we summarize the force formulae obtained for the different scaling
regimes in a table. The question arises which formula one should work with in
applications if the distance between the bodies is small. To answer this question,
one needs to know which scaling regime models the physical situation considered
best. The numerical experiments in [11] as well as related real-life experiments [6]
will hopefully yield further insight into this issue.
2. Preliminaries
Our discrete-to-continuum calculations start from an underlying Bravais lattice
L of magnetic dipole moments. We sometimes regard the lattice
d points as atoms.
µi ei , µi ∈ Z},
A precise definition of L is as follows: L = {x ∈ Rd |x = i=1
where e1 , . . . , ed is a basis of Rd , d 2. For definiteness we suppose that the unit
d
cell {x ∈ Rd |x = i=1
λi ei , λi ∈ [0, 1)} has volume one. For instance, L = Zd .
In order to pass from the discrete model to the continuum, we consider the scaled
Bravais lattice 1 L = {z ∈ Rd |z ∈ L}, for ∈ N.
In Fig. 1 we give an example for the domains considered in the following
assumption.
Assumption 1. A. Let d ∈ N be fixed.
1. A and B are bounded Lipschitz domains in Rd . A and B have polygonal boundaries and finitely many corners or edges such that A ∩ B = ∅. Moreover, A and
B are in contact, that is, the surface measure of ∂ A ∩ ∂ B ⊂ ∂ A is positive. The
set
Bε = B + εν,
(1)
where ν ∈ L is fixed and ε = a with a ∈ N, satisfies A ∩ Bε = ∅ for all ε > 0.
2. The corresponding magnetizations m A : A → Rd and m B : B → Rd are
Lipschitz continuous and are supported on A and B, respectively, that is, there
holds m A ∈ W 1,∞ (A) and m B ∈ W 1,∞ (B). Moreover, the magnetization
m Bε : Bε → Rd satisfies
m Bε (x) = m B (x − εν) for all x ∈ Bε .
All magnetization fields are extended by zero to the entire space Rd .
Fig. 1. Sketch of the sets A and Bε
(2)
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Anja Schlömerkemper & Bernd Schmidt
Later we will use the following implication of the above assumptions. Let n A
denote the outer normal to ∂ A. Then n A (x) · ν > 0 for all x ∈ ∂ A ∩ ∂ B (if n A
exists).
Assumption A is natural in view of applications. However, in Remark 3.6, we
briefly indicate how the assumptions on the domains can be relaxed.
The magnetization is related to the magnetic dipole moments on the lattice
points of 1 L through the scaling law
1
1
m A (x) if x ∈ A ∩ L
d
1
1
()
m Bε (x) := d m Bε (x) if x ∈ Bε ∩ L.
m()
A (x) :=
(3)
The kth component of the magnetic force that all dipole moments in Bε exert
on all dipole moments in A is given by (see, for example, [13])
()
()
()
Fk (A, Bε ) := γ
∂i ∂ j ∂k N (x − y) m A (x) m Bε (y),
i
x∈A∩ 1 L y∈Bε ∩ 1 L
j
(4)
where γ is a constant which only depends on the choice of physical units. For
instance, γ = 4π in Gaussian units, cf. [2, p. 6]. Note that we use the Einstein
summation convention, that is, the above expression contains sums over
i, j = 1, . . . , d as these indices occur twice. The function N denotes the fundamental solution of Laplace’s equation, which is given by
1
− 2π log |x|
for d = 2,
N (x) :=
Γ ( d2 )
2−d
|x|
for d 3,
(d−2)2π d/2
where Γ (·) denotes the Gamma-function. In particular, N (x) =
Let ϕ be a smooth, radially symmetric function such that
1
if |z| < 21 ,
ϕ(z) =
0
if |z| > 1
and, for ϕ (δ) (z) := ϕ(z/δ), set
(ϕ (δ) ∂k N )(x − y)
for d
(δ)
Pk (x − y) :=
(δ)
for d
∂k (ϕ N )(x − y)
(1 − ϕ (δ) )∂k N (x − y)
(δ)
Rk (x − y) :=
∂k (1 − ϕ (δ) )N (x − y)
(δ)
1 1
4π |x|
= 2,
3,
(δ)
for d = 3.
(5)
for d = 2,
for d 3.
Note that Pk is zero if |x − y| is greater than δ, whereas Rk is zero if |x − y| < δ/2.
(δ)
(δ)
Since Pk (x − y) + Rk (x − y) = ∂k N (x − y), we can use these functions to split
F() into a so-called long range part and a so-called short range part:
Discrete-to-Continuum Limit of Magnetic Forces
∂i ∂ j Rk(δ) (x − y) m()
(x) m()
(y)
A
Bε
Fk() (A, Bε ) = γ
x∈A∩ 1 L y∈Bε ∩ 1 L
+γ
i
long(,δ)
j
(δ)
()
()
∂i ∂ j Pk (x − y) m A (x) m Bε (y)
i
x∈A∩ 1 L y∈Bε ∩ 1 L
=: Fk
593
short(,δ)
(A, Bε ) + Fk
j
(A, Bε ).
(6)
This allows us to compute the continuum limit of F() as follows:
Flim (A, B, a) := lim F() (A, B a )
→∞
= lim lim Flong(,δ) (A, B a ) + lim lim Fshort(,δ) (A, B a )
δ→0 →∞
δ→0 →∞
=: Flong (A, B, a) + Fshort (A, B, a),
(7)
given that the limits of Flong(,δ) (A, B a ) and Fshort(,δ) (A, B a ) exist, see below.
Note ε = a → 0 in the continuum limit as → ∞. However, we expect/obtain
that the continuum limit of the force depends on a.
For later reference we quote a formula for the magnetic field, H B . The magnetic field is a solution of the magnetostatic Maxwell equations curl H B = 0 and
div B B = 0 with B B = H B + γ m B . By setting H B = −∇φ B , φ B : Rd → R, these
equations can be written in the form of a Poisson equation div(−∇φ B + γ m B ) = 0
with transition conditions [∇φ B · n B ] = −γ m B · n B and [φ B ] = 0 on ∂ B, where
n B denotes the outer normal to ∂ B. If x ∈
/ ∂ B, an integral representation of H B
reads (cf., for example, [8, p. 73])
H B (x) = −γ
−γ
B
(−∇ · m B )(y)∇ N (x − y) dy
∂B
(m B · n B )(y)∇ N (x − y) ds y ,
(8)
where s y denotes the d − 1 dimensional surface measure on ∂ B. The magnetic field
H A∪B generated by the magnetization in A and B is defined and can be represented
similarly.
3. The discrete-to-continuum limit of the magnetic force
In this section we derive a formula for the limiting magnetic force Flim (A, B, a)
in Equation (7) for fixed a ∈ N. Although the resulting expression makes sense for
all a > 0, for technical reasons we restrict our analysis to a ∈ N which basically
allows us to work with some fixed underlying atomic lattice.
The following proposition on the long range part of the force is proven in [10]
in the case a = 0, to which the case a > 0 can be reduced, cf. the proof below. The
long range part of the magnetic interaction turns out to be independent of a.
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Anja Schlömerkemper & Bernd Schmidt
Proposition 3.1. Let Assumption A hold. Then the limit lim lim Flong(,δ) (A, Bε )
δ→0 →∞
exists, is independent of a and satisfies
γ
long
(m A · n A ) ((m A − m B ) · n A ) n A dsx .
F (A, B) = (m A · ∇)H A∪B dx +
2 ∂A
A
(9)
Here, m B denotes the outer trace on ∂ A with respect to A, that is, m B is equal to
0 on ∂ A\(∂ A ∩ ∂ B) and equals the inner trace of m B on ∂ A ∩ ∂ B with respect to
B.
Proof. By definition (6) and (2),
Flong(,δ) (A, Bε )
=γ
x∈A∩ 1 L y∈Bε ∩ 1 L
=γ
(δ)
()
()
∂i ∂ j Rk (x − y) m A (x) m Bε (y)
i
j
(δ)
()
∂i ∂ j Rk (x − y) m A (x)
i
x∈A∩ 1 L y∈Bε ∩ 1 L
()
m()
(y
−
εν)
−
m
(y)
B j
B j
(δ)
()
()
∂i ∂ j Rk (x − y) m A (x) m B
(y).
+γ
×
i
x∈A∩ 1 L y∈Bε ∩ 1 L
(10)
j
Since m is bounded and Lipschitz continuous on B, the first term on the right hand
side can be estimated as follows:
(δ)
()
()
()
∂i ∂ j Rk (x − y) m A (x) m B (y − εν)− m B (y) i
j
j
x∈A∩ 1 L y∈Bε ∩ 1 L
() ()
m
C(δ)
−d
B j (y − εν) − m B j (y)
x∈A∩ 1 L
⎛
⎜
C(δ) ⎝
y∈Bε ∩ 1 L
y∈Bε ∩B∩ 1 L
−d ε +
⎞
⎟
−d ⎠
y∈(Bε\B)∩ 1 L
1
−d
C(δ) ε + # (Bε \ B) ∩ L .
An application of the following Lemma 3.2 with z = εν = a ν shows that this term
converges to 0 as → ∞.
Lemma 3.2. For z ∈ Rd with |z| c̃−1 there is a constant C only depending on
c̃ and B such that
Discrete-to-Continuum Limit of Magnetic Forces
595
1
# ((B + z)B) ∩ L C|z|d ,
where (B + z)B = ((B + z)\ B) ∪ (B \(B + z)).
Proof. This is an immediate consequence of Assumption A since all the elements
of the set on the left hand side of the above inequality lie in the z-neighborhood of
∂ B.
Since replacing Bε by B in the second term of the right hand side of (10)
leads to an error term of order −d #{(B \ Bε ) ∩ 1 L} because m()
B vanishes on the
complement of B, another application of Lemma 3.2 with z = εν = a ν shows that
in fact
lim Flong(,δ) (A, Bε ) = lim Flong(,δ) (A, B).
→∞
→∞
In particular this is independent of a. An application of Theorem 3.3 in [10], which
is based on [12, Theorem 13] hence yields the assertion.
Contrary to the long range part, the short range contribution Fshort to the limit
force does depend on a.
Proposition 3.3. Let Assumption A hold. Then the limit Fshort (A, B, a) :=
lim lim Fshort(,δ) (A, Bε ) exists and is given by
δ→0 →∞
Fkshort (A, B, a)
d
1 Si jkp
(m A )i (x)(m B ) j (x)(n A ) p (x) dsx
2
∂ A∩∂ B
i, j, p=1
−γ
(m A )i (x)(m B ) j (x)
∂i ∂ j ∂k N (z)
=
∂ A∩∂ B
(11)
z∈L\{0}
× (n A (x) · (z − aν))+ − (n A (x) · z)+ dsx
for k = 1, . . . , d, where (·)+ := max{0, ·}, m B is the outer trace of m B on ∂ A with
respect to A and
1
(δ)
∂i ∂ j Pk (z) z p d .
(12)
Si jkp := −γ lim lim
δ→0 →∞
1
z∈Bδ (0)∩ L\{0}
The first term in (11) equals Fkshort (A, B), the short range formula that was
obtained in [10, Theorem 3.4] in the case of ε = 0, see also [13, Theroem 2].
Numerical values of Si jkp are given for the square/cubic lattice L = Zd in d = 2
[10, Appendix A] and d = 3 [13, Section 6.1] dimensions. We denote the second,
new term in (11), which is also a surface integral about the interface of A and B,
by Gk (a) and thus we have
Fshort (A, B, a) = Fshort (A, B) + G(a),
(13)
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Anja Schlömerkemper & Bernd Schmidt
where G(a) = (G1 (a), . . . , Gd (a)). Moreover, we observe that G(0) = 0, that is,
Fshort (A, B, 0) = Fshort (A, B). For more properties of the additional term G(a),
see Section 4.
Propositions 3.1 and 3.3 imply
Theorem 3.4. Let Assumption A hold. Then
Flim (A, B, a) = Flong (A, B) + Fshort (A, B) + G(a).
short(,δ)
(A, Bε ) in (6). SimiProof of Proposition 3.3. Recall the definition of Fk
larly to [13, p. 233], we reorganize the sums in order to get rid of the hypersingular
(δ)
order of the kernel ∂i ∂ j Pk (x − y); set z = y − x and note that z ∈ 1 L by our
(δ)
(δ)
(δ)
assumptions, Pk (−z) = −Pk (z) and Pk (z) is supported on Bδ (0). We thus
obtain
Fkshort(,δ) (A, Bε )
=γ
(δ)
()
()
∂i ∂ j Pk (x − y) m A (x) m Bε (y)
x∈A∩ 1 L y∈Bε ∩ 1 L
=γ
i
(δ)
()
()
∂i ∂ j Pk (−z) m A (x) m Bε (x + z)
i
x∈A∩ 1 L z∈Bδ (0)∩ 1 L\{0}
x+z∈Bε
= −γ
j
(δ)
∂i ∂ j Pk (z)
x∈Aεz ∩ 1 L
z∈Bδ (0)∩ 1 L\{0}
j
()
()
m A (x) m Bε (x + z),
i
j
(14)
where Aεz := x ∈ A x + z ∈ Bε , which satisfies, by (1),
Aεz = A ∩ (Bε − z) = A ∩ (B − (z − εν)) =: A z−εν .
()
()
Since m Bε (x + z) = m B
(x + z − εν), cf. (2), we can apply a slightly
j
j
adapted version of Proposition 1 in [14], which is based on an idea of Cauchy [3]
developed for the discrete-to-continuum limit of elastic forces:
Lemma 3.5. Let A and B satisfy Assumption A and assume f : A z−εν → R to be
Lipschitz continuous. Let 0 < δ 1 and z ∈ Bδ (0) ∩ 1 L\{0}. Then there exists
an 0 ∈ N such that for all 0
1
4
f (x) −
f (x) (n A (x) · (z − εν))+ dsx C|z| 3 .
d
∂ A∩∂ B
x∈Az−εν ∩ 1 L\{0}
(15)
Here, the constants C and 0 only depend on sup( f ), on the Lipschitz constant of
f , on a, the dimension d and on the geometries of A and B.
Discrete-to-Continuum Limit of Magnetic Forces
597
Proof. Since |z| c̃ for some c̃ > 0 by assumption and |εν| = a |ν|, we have
|z −εν| c|z|. Moreover, for 0 large enough we obtain z −εν ∈ B2δ (0)∩ 1 L\{0}.
We can thus apply [14, Proposition 1] with z replaced with z − εν, which yields
1
d
f (x) −
∂ A∩∂ B
x∈A z−εν ∩ 1 L\{0}
4
f (x) (n A (x) · (z − εν))+ dsx C|z − εν| 3
and hence (15).
With f (x) = (m A )i (x)(m Bε ) j (x + z) = (m A )i (x)(m B ) j (x + z − εν), which
is Lipschitz continuous on A z−εν by assumption, we obtain
Fkshort(,δ) (A, Bε ) = −γ
∂i ∂ j Pk(δ) (z)
z∈Bδ (0)∩ 1 L\{0}
1
d
×
∂ A∩∂ B
(m A )i (x)(m B ) j (x) (n A (x) · (z − εν))+ dsx
(16)
up to terms of higher order. The terms of higher order converge to zero as → ∞
and δ → 0, cf., for example, [10, Proof of Theorem 3.4].
Next we add and subtract (n A · z)+ in (16). This yields another term discussed
below
1
(δ)
∂i ∂ j Pk (z) d
(m A )i (x)(m B ) j (x)(n A (x) · z)+ dsx ,
−γ
∂ A∩∂ B
1
z∈Bδ (0)∩ L\{0}
which is the same as in [10, (3.25)] and thus can be estimated accordingly. It
converges to
Fkshort (A,
1
B) := Si jkp
2
∂ A∩∂ B
(m A )i (m B ) j (n A ) p dsx ,
(17)
where Si jkp is defined as in (12).
It remains to estimate
−γ
z∈Bδ (0)∩ 1 L\{0}
∂i ∂ j Pk(δ) (z)
1
×
d
∂ A∩∂ B
(m A )i (x)(m B ) j (x)
× (n A (x) · (z − εν))+ − (n A (x) · z)+ dsx
= −γ
(m A )i (x)(m B ) j (x) ×
∂ A∩∂ B
z∈Bδ (0)∩ 1 L\{0}
× (n A (x) · (z − εν))+ − (n A (x) · z)+ dsx .
(δ)
∂i ∂ j Pk (z)
1
d
(18)
598
Anja Schlömerkemper & Bernd Schmidt
Recalling that ε = a , we can write the sum in (18) as
y∈L\{0}
(δ)
∂i ∂ j Pk
y
1
d+1
(n A (x) · (y − aν))+ − (n A (x) · y)+ χ Bδ (0) (y).
(19)
Firstly we show that the terms in this sum are dominated by a function which is
summable over Bδ (0) ∩ 1 L\{0}. Clearly
(n A (x) · (y − aν))+ − (n A (x) · y)+ |n A (x) · (y − aν) − n A (x) · y|
= an A (x) · ν c,
(20)
so an upper bound for the summands in (19) is given by c|y|−d−1 , which is summable over L\{0}. As an aside, note that the above c depends on a by the estimate
in (20), that is, we have only proved absolute convergence of the sum in (18) for
fixed a ∈ R.
Using the following identity, which can be immediately obtained with the chain
rule, cf. [14, (27)],
(δ)
∂i ∂ j Pk
y
1
(δ)
= ∂i ∂ j Pk (y),
d+1
(21)
we
see next that the terms in (19) converge pointwise to ∂i ∂ j ∂k N (y)
(n A (x) · (y − aν))+ − (n A (x) · y)+ as → ∞, because if → ∞, ϕ (δ) (y)
as well as χ Bδ (0) (y) converge to 1 pointwise. Note that the limit does not depend
on δ. Thus we finally obtain by (5) and the theorem on dominated convergence that
(18) converges to
(m A )i (x)(m B ) j (x)
∂i ∂ j ∂k N (y)
−γ
∂ A∩∂ B
y∈L\{0}
× (n A (x) · (y − aν))+ − (n A (x) · y)+ dsx
as → ∞ (and δ → 0), which finishes the proof of Proposition 3.3.
Finally, we comment on how Assumption A can be modified to also include
more general domains A and B, cf. [10, Remark 3.7].
Remark 3.6. Assumption A is primarily supposed for an application of results in
[12,14], cf. Proposition 3.1 and Lemma 3.5. However, in those articles, only weaker
assumptions on A and B are imposed, which read:
A and B are Lipschitz domains with piecewise C 1,1 boundaries. See [14, Definition 2] for a precise definition. Furthermore, rather technical conditions are
assumed, which we outline only briefly (see [12,14] for details): (i) A ∪ B satisfies an outer cone property [14, Assumption A1 ]; (ii) ∂ A, ∂ B and ∂ A ∪ ∂ B satisy
a so-called non-degeneracy condition (S) that controls the number of isolated points
which have the same tangent vector, cf. [14, Definition 3]; (iii) ∂ A ∩ ∂ B satisfies
Discrete-to-Continuum Limit of Magnetic Forces
599
a so-called neighborhood estimate which allows one to bound volumes of tubes
about relative boundaries of portions of ∂ A ∩ ∂ B, see [14, Definition 4].
These weaker assumptions are for instance satisfied for the domains in Assumptions A. We remark that all lemmas, propositions and theorems in this article also
hold under the above weaker assumptions. (In particular, the non-degeneracy condition (S) guarantees that {x ∈ ∂ A : n A · ν = 0} has Hd−1 -measure zero. With
this observation also the necessary modifications in the proof of Proposition 4.2
become straightforward.)
4. Properties of the additional, a-dependent contribution to the limit force
Here we study properties of the term
(m A )i (x)(m B ) j (x)
∂i ∂ j ∂k N (z)
Gk (a) = −γ
∂ A∩∂ B
z∈L\{0}
× (n A (x) · (z − aν))+ − (n A (x) · z)+ dsx .
(22)
Recall that n A (x) · ν > 0 for all x ∈ ∂ A ∩ ∂ B by Assumption A. See Fig. 2 for
sketches of the sets defined in the following lemma.
Lemma 4.1. Let x ∈ ∂ A ∩ ∂ B.
(i) On Ca,x := z ∈ Rd \{0} (n A (x) · z) a(n A (x) · ν) we have
(n A (x) · (z − aν))+ − (n A (x) · z)+ = −an A (x) · ν.
Fig. 2. Sketch of the sets Ca,x and Da,x with a = 2
600
Anja Schlömerkemper & Bernd Schmidt
(ii) On Da,x := z ∈ Rd \{0} 0 (n A (x) · z) < a(n A (x) · ν) we have
(n A (x) · (z − aν))+ − (n A (x) · z)+ = −n A (x) · z.
(iii) On Rd \(Ca,x ∪ Da,x ) we have
(n A (x) · (z − aν))+ − (n A (x) · z)+ = 0.
In the spirit of Lemma 4.1 we rewrite the additional term G(a) as
Gk (a) = a γ
(m A )i (x)(m B ) j (x)(n A (x) · ν)
∂i ∂ j ∂k N (z) dsx
∂ A∩∂ B
z∈Ca,x ∩L
(23)
+γ
∂ A∩∂ B
(m A )i (x)(m B ) j (x)
∂i ∂ j ∂k N (z)(n A (x) · z) dsx .
z∈Da,x ∩L
(24)
Next we study the limit of G(a) as a tends to ∞, which corresponds to two
bodies which are infinitely far apart on the lattice scale (lim→∞ lima→∞ ε = ∞)
but are in contact on the continuum scale (lima→∞ lim→∞ ε = 0).
Proposition 4.2. Let Assumption A be satisfied. Then
γ
lim G(a) = −Fshort (A, B) +
(m A · n A )(x)(m B · n A )(x)n A (x) dsx .
a→∞
2 ∂ A∩∂ B
Brown’s formula [2, p. 57], see also [10, Section 3.1], is given by
γ
FBr (A, B) = (m A · ∇)H A∪B dx +
(m A · n A )2 n A dsx .
2
A
∂A
(25)
By Propositions 3.1 and 4.2 and Theorem 3.4 we thus obtain
Theorem 4.3. Let Assumption A be satisfied. Then
lim Flim (A, B, a) = FBr (A, B).
a→∞
This proves a claim of Brown [2, p. 53], who expected additional contributions
to the magnetic force from dipole–dipole interactions close to the (macroscopic)
interface between A and B to be of short range character, that is, he expected those
contributions to converge to zero as the (microscopic) distance becomes large.
Proof of Proposition 4.2. We will in fact prove that the integrands in (23) and (24)
converge uniformly in x. So without loss of generality we may assume that x is
fixed with n A (x) = e1 , see Step 6 for a verification.
Step 1 We consider the term on the right hand side in (23) as a → ∞ and show
that it converges to zero. For this we replace the sum with a Riemann integral. Set
z = ay and apply an analog of (21). Then
∂i ∂ j ∂k N (z) =
∂i ∂ j ∂k N (y)a −d ,
a
z∈Ca,x ∩L
y∈C1,x ∩ a1 L
Discrete-to-Continuum Limit of Magnetic Forces
601
which converges to C1,x ∂i ∂ j ∂k N (y) dy as a → ∞ since |∂i ∂ j ∂k N (y)| c|y|−d−1
implies that both
−d
|∂i ∂ j ∂k N (y)|a
and
|∂i ∂ j ∂k N (y)| dy
C1,x\B R (0)
y∈(C1,x\B R (0))∩ a1 L
are bounded by c R −1 for arbitrary R > 0. The claim then follows by choosing R
large and applying a Riemann sum argument for the domain C1,x ∩ B R (0).
Now, let x ∗ be the projection of 0 on ∂C1,x , cf. Fig. 2. Since |∂ j ∂k N (y)| c(R − |x ∗ |)−d for all y ∈ C1,x ∩ ∂ B R (x ∗ ), an integration by parts yields
∂i ∂ j ∂k N (y) dy = lim
−(n A )i (x)∂ j ∂k N (y) ds y ,
R→∞ ∂ C1,x ∩B R (x ∗ )
C1,x
which is zero for i = 1 as n A = e1 by assumption. If j = 1 we can integrate by
parts once more and obtain
y j
∂ N (y) dl y ,
lim (n A )i (x)
k
R→∞
∂ C1,x ∩∂ B R (x ∗ ) |y |
where l y denotes the d − 2 dimensional surface measure on ∂C1,x ∩ B R (x ∗ ) and
y := (y2 , . . . , yd ), that is, |yy | is the outward normal to ∂C1,x ∩ ∂ B R (x ∗ ) within the
hyperplane ∂C1,x . The integral is now easily seen to be bounded by c R −d+1 R d−2 .
2
This is still true if both i and
j are2 equal to 1 because of ∆N = 0, that is, ∂1 N =
2
2
−(∂2 N + · · · + ∂d N ), and C1,x ∂i ∂k N (y) dy = 0 for i = 1. Sending R to infinity
shows that C1,x ∂i ∂ j ∂k N (y) dy = 0 for all i, j, k = 1, . . . , d.
In the remaining steps of the proof we will show that the term in (24) converges
to
γ
− Fkshort (A, B) +
(m A · n A )(x)(m B · n A )(x)n A (x) dsx
2 ∂ A∩∂ B
as a → ∞. To this end we prove in Steps 2 to 4 that
∂i ∂ j ∂k N (z)z 1
z∈Da,x ∩L
−→ δi1 δ j1
∂C1,x
∂ j ∂k N (y)y1 ds y − δi1 δ j1
∂C1,x
∂k N (y) ds y −
1
Si jk1 (26)
2γ
as a → ∞.
Step 2 Again we replace z with ay and obtain for the sum in (24)
∂i ∂ j ∂k N (z)(n A (x) · z) =
∂i ∂ j ∂k N (y)(n A (x) · y)a −d
z∈Da,x ∩L
y∈D1,x ∩ a1 L
=
y∈D1,x ∩ a1 L
∂i ∂ j ∂k N (y)y1 a −d .
(27)
602
Anja Schlömerkemper & Bernd Schmidt
Note that the terms in the sum are symmetric in y with respect to the origin. Thus
∂i ∂ j ∂k N (y)y1 a −d =
y∈D1,x ∩ a1 L
1
2
∂i ∂ j ∂k N (y)y1 a −d ,
1,x ∩ 1 L
y∈D
a
a,x := z ∈ Rd \{0} − aν1 < z 1 < aν1 , which is invariant under the
where D
transformation z → −z. Set r = ν1 . It will turn out to be useful to split the above
sum as follows:
1
1
∂i ∂ j ∂k N (y)y1 a −d =
∂i ∂ j ∂k N (y)y1 a −d
2
2
1
1
1,x ∩ L
y∈D
a
1,x ∩ L∩Br (0)
y∈D
a
+
1
2
∂i ∂ j ∂k N (y)y1 a −d .
1,x ∩ 1 L)\Br (0)
y∈(D
a
(28)
We consider the first term in (28) further below
(see Step 4). The second term in
(28) converges to the Riemann integral 21 D
1,x\Br (0) ∂i ∂ j ∂k N (y)y1 dy as a → ∞
by a Riemann sum argument. To prove this, we split the sum into one part where
|y| R and another part where |y| > R for some R > r . The first part is a bounded
Riemann sum on a bounded domain. The error terms, that is, the sum, respectively.,
integral, over y with |y| R are seen to converge to 0 as R → ∞ similarly as in
1,x .
Step 1, since |y1 | is bounded on D
Step 3 Integrations by parts of 21 D
1,x\Br (0) ∂i ∂ j ∂k N (y)y1 dy yield
1
2
yi
1
∂ j ∂k N (y)y1 ds y +
µi ∂ j ∂k N (y)y1 ds y
1,x
2 ∂D
∂ Br (0) |y|
yj
1
1
∂k N (y) ds y − δi1
+ δi1
µ j ∂k N (y) ds y ,
1,x
2
2
∂ Br (0) |y|
∂D
−
(29)
1,x . (Note that the fourth integral does
where µ denotes the outer normal to ∂ D
not converge
absolutely. This integral is understood in the principal value sense
lim R→∞ ∂ D
1,x ∩B R (0) (. . .) ds y . The aforementioned integrations by parts are jus1,x ∩ ∂ B R (0) is bounded by c R d−2 while
tified since the surface measure of D
|∂ j ∂k N (y)y1 | and |∂k N (y)| are bounded by c R −d+1 and the corresponding integrals thus tend to zero.)
By construction, µ equals the constant vectors n A (x) = e1 and −n A (x) =
−e1 , respectively.
By antisymmetry of ∂ j ∂k N (y)y1 , the second term in (29) is
equal to δi1 ∂ C1,x ∂ j ∂k N (y)y1 ds y . Similarly, the fourth term in (29) becomes
−δi1 δ j1 ∂ C1,x ∂k N (y) ds y .
Hence, if j = 1, the fourth term in (29) vanishes and the second term can be
y
integrated by parts once more and becomes lim R→∞ δi1 ∂ C1,x ∩∂ B R (x ∗ ) |y j | ∂k N (y)
y1 dl y , which is zero. Indeed,
∂ C1,x ∩∂ B R (x ∗ )
y j
|y |
∂k N (y)y1 dl y
Discrete-to-Continuum Limit of Magnetic Forces
=
=
=
R −1 (∂ C1,x ∩∂ B R (x ∗ ))
603
(Ry ) j
∂k N (Ry) Ry1 R d−2 dl y
|Ry |
y j
∂ N (y)y1 dl y
k
{y1 =R −1 ν1 ,|y |2 =1} |y |
y j
∂ N (y)y1 dl y + O(R −1 ),
k
{y1 =0,|y |2 } |y |
since ∂k N (y)y1 is uniformly Lipschitz continuous in y1 on y ∈ Rd |y |2 = 1 .
By antisymmetry of the integrand, the last integral vanishes. Hence we obtain
y
indeed that ∂ C1,x ∩∂ B R (x ∗ ) |y j | ∂k N (y)y1 dl y → 0 as R → ∞. We study the remaining terms in (29) and the case j = 1 further below.
Step 4 Next we consider the first term in (28) and show that
∂i ∂ j ∂k N (y)y1 a −d
1,x ∩ 1 L∩Br (0)
y∈D
a
yj
yi
− ∂ j ∂k N (y)y1 + δi1 ∂k N (y) ds y
+
|y|
|y|
∂ Br (0)
1
−→ − Si jk1 as a → ∞,
γ
(30)
which implies (26). To prove (30), recall [14, Theorem 13] and observe that
1
− Si jk1 = lim lim
δ→0 →∞
γ
n→∞
1
d
(n)
∂i ∂ j Pk (z) z 1
z∈Bn (0)∩L\{0}
a→∞
(δ)
∂i ∂ j Pk (z) z 1
z∈Bδ (0)∩ 1 L\{0}
= lim
= lim
(2ar )
∂i ∂ j Pk
(z) z 1
z∈B2ar (0)∩L\{0}
for some r = ν1 > 0 as above. We split the latter sum into a sum over Bar (0) and
1,x ∩ L ∩ Bar (0) and
a sum over B2ar (0) \ Bar (0). Since Bar (0) ∩ L \ {0} = a D
(2ar )
(z) = ∂k N (z) on Bar (0) by definition, we have
Pk
(2ar )
∂i ∂ j Pk
(z) z 1 =
∂i ∂ j ∂k N (y)y1 a −d .
(31)
z∈Bar (0)∩L\{0}
1,x ∩ 1 L∩Br (0)
y∈D
a
Hence, it remains to study the sum over B2ar (0)\ Bar (0). A scaling of the lattice
with a1 yields
(2r )
∂i ∂ j Pk (y) y1 a −d
y∈(B2r (0)\Br (0))∩ a1 L\{0}
−→
(2r )
B2r (0)\Br (0)
∂i ∂ j Pk
(y) y1 dy
604
Anja Schlömerkemper & Bernd Schmidt
(2r )
as a → ∞. Finally, by the definition of Pk in (5) and integrations by parts we
obtain
(2r )
∂i ∂ j Pk (y) y1 dy
B2r (0)\Br (0)
yj
yi
=
− ∂ j ∂k N (y)y1 ds y + δi p
∂k N (y) ds y
∂ Br (0) |y|
∂ Br (0) |y|
and thus (30).
Step 5 Collecting all terms, we have shown that
aν1
∂i ∂ j ∂k N (z) +
z∈Ca,x ∩L
∂i ∂ j ∂k N (z)z 1 +
z∈Da,x ∩L
1
Si jk1
2γ
(32)
converges to 0 as a → ∞ for j = 1. Indeed, the first term converges to zero by
Step 1 and the convergence of the second term follows by (26). By symmetry in i
and j, this is also true for j = 1, i = 1. For i = j = 1, (32) converges to
∂1 ∂k N (y)y1 ds y −
∂k N (y) ds y
∂ C1,x
= δ1k
∂ C1,x
∂ C1,x
(33)
∂12 N (y)y1 − ∂1 N (y) ds y
as a → ∞ by a symmetry argument. Now using that
Γ d2 y1
Γ d2
dy12
1
2
∂1 N (y) = − d/2 d and ∂1 N (y) =
− d
2π
|y|
2π d/2 |y|d+2
|y|
we find that
∂12 N (y)y1 − ∂1 N (y) =
Γ ( d2 ) dy13
,
2π d/2 |y|d+2
and (33) becomes
dΓ d2
y13
δ1k
ds y
2π d/2 ∂ C1,y |y|d+2
dΓ d2
ν13
= δ1k
dy2 . . . dyd
2π d/2 Rd−1 (ν12 + y22 + · · · + yd2 )d/2+1
∞
dΓ d2
r d−2
(d−1)
3
= δ1k
|∂
B
(0)|ν
dr
1
1
2π d/2
(ν12 + r 2 )d/2+1
0
dΓ d2 2π (d−1)/2 3 ∞
r d−2
ν
dr
= δ1k
1
2
d−1
d/2
2π
(ν1 + r 2 )d/2+1
Γ 2
0
∞
dΓ ( d2 )
r d−2
dr.
= δ1k
d−1 (1 + r 2 )d/2+1
π 1/2 Γ 2
0
(34)
Discrete-to-Continuum Limit of Magnetic Forces
605
1
2
1
To calculate the latter integral, we change variables via r → 1−t
−1
and
obtain
∞
1
r d−2
1 1 d−3
dr
=
t 2 (1 − t) 2 dt.
2
d/2+1
(1 + r )
2 0
0
This can be written in terms of the Euler beta function B(a, b) :=
(b)
t)b−1 dt = ΓΓ(a)Γ
(a+b) :
1
2
1
t
0
d−3
2
1
(1 − t) dt = B
2
1
2
d −1 3
,
2
2
1
0
t a−1 (1 −
3
Γ 2
1 Γ d−1
2
=
d
.
2 Γ 2 +1
1
With d2 Γ ( d2 ) = Γ ( d2 + 1) and Γ ( 23 ) = 21 π 2 we obtain that (34) and hence (33) is
equal to 21 δ1k .
Hence, by (32), (33) and the definition of Gk (a) in (23) and (24), we finally
obtain that
∂i ∂ j ∂k N (z) +
∂i ∂ j ∂k N (z)z 1
aν1
z∈Ca,x ∩L
z∈Da,x ∩L
δ1k 21
if i = j = 1
else
1
−→ − Si jk1 +
2γ
0
(35)
as a → ∞.
Step 6 Recall that we have assumed n A (x) = e1 so far. The general case is
obtained as follows. Let Q = (qi j ) = Q(x) be some orthogonal matrix mapping
n A (x) to e1 and transform coordinates according to Qz = z̃. Then
a(n A (x) · ν)
∂i ∂ j ∂k N (z)
z∈Ca,x ∩L
= a(e1 · Qν)
qri ∂z̃r qs j ∂z̃ s qtk ∂z̃ t N (z̃),
z̃∈Q Ca,x ∩Q L
where QCa,x = z̃ z̃ 1 a(Qν)1 . Similarly, for the sum in (24) we obtain
z∈Da,x ∩L
∂i ∂ j ∂k N (z)(n A (x) · z) =
qri ∂z̃r qs j ∂z̃ s qtk ∂z̃ t N (z̃)(e1 · z̃)
z̃∈Q Da,x ∩Q L
with QDa,x = z̃ 0 z̃ 1 < a(Qν)1 . The sum of these two terms converges to
1 Q
1
qri qs j qtk − Sr stu δ1u + δ1r δ1s δ1t
2γ
2
as a → ∞ by (35). Here, S Q denotes the tensor that arises when replacing the
lattice L by QL.
606
Anja Schlömerkemper & Bernd Schmidt
Q
Now note that Sr stu = qri qs j qtk qup Si jkp , which follows from (12) and the
(δ)
definition of Pk similarly as the previous transformation rules. Since r qri2 =
2
2
T
s qs j =
t qtk = 1 and Q e1 = n A (x), the above expression equals
−
1
1
qup δ1u Si jkp + qri δ1r qs j δ1s qtk δ1t
2γ
2
1
1
= − (n A ) p (x)Si jkp + (n A )i (x)(n A ) j (x)(n A )k (x).
2γ
2
Summarizing, we obtain
d
1 lim Gk (a) = −
Si jkp
(m A )i (x)(m B ) j (x)(n A ) p (x) dsx
a→∞
2
∂ A∩∂ B
i, j, p=1
γ
+
(m A · n A )(x)(m B · n A )(x)(n A )k (x) dsx ,
2 ∂ A∩∂ B
which proves Proposition 4.2. In order to calculate G(a) for finite values of a, we
write G(a) as
(m A )i (x)(m B ) j (x)Hi jk (a, x) dsx
(36)
Gk (a) = −γ
∂ A∩∂ B
with
Hi jk (a, x) =
∂i ∂ j ∂k N (z) (n A (x) · (z − aν))+ − (n A (x) · z)+ .
z∈L\{0}
Example 4.4. Suppose that γ = 1, n A ≡ e1 ∈ Rd on ∂ A ∩ ∂ B and that the
underlying lattice is the cubic lattice L = Zd . If ν = e1 , then for all x ∈ ∂ A ∩ ∂ B,
Hi jk (a, x) =
∞
((z 1 − a)+ − (z 1 )+ )
z 1 =1
∂i ∂ j ∂k N (z).
z 2 ,...z d ∈Z
Due to the symmetry of N , the inner sum is zero unless two of the indices i, j, k are
equal and the third one equals one. Moreover, if the two like indices are different
1
from one, then Hi jk (a) = − d−1
H111 (a) since ∆N = 0 on Rd \{0}. Observe that
3
the Fourier transform of ∂1 N (z) with respect to (z 2 , . . . , z d ) ∈ Rd−1 is given by
∂13 N (z 1 , . . . , z d )e−i(z 2 ζ2 +...+z d ζd ) dz 2 · · · dz d
Rd−1
1
−z
= − (ζ22 + · · · + ζd2 )e 1
2
ζ22 +···+ζd2
for z 1 > 0. (This can be seen, for example, by noting that the full d-dimensional
Fourier transform of N is −(ζ12 + · · · + ζd2 )−1 . An inverse Fourier transform of
this expression with respect to ζ1 and a subsequent calculation of the three partial
Discrete-to-Continuum Limit of Magnetic Forces
607
derivatives with respect to z 1 yields the asserted formula.) So we obtain by Poisson
summation that
H111 (a) = 2π
∞
2
((z 1 )+ − (z 1 − a)+ )
z 1 =1
= 2π
2
(ζ22
2
2
2 −2π z 1 ζ2 +...+ζd
+ . . . + ζd )e
ζ2 ,...,ζd ∈Z
(ζ22
+ . . . + ζd2 )
ζ2 ,...,ζd ∈Z
⎛
a−1
×⎝
z1e
−2π z 1
ζ22 +...+ζd2
ζ ∈Zd−1\{0}
⎞
ae
−2π z 1 ζ22 +...+ζd2 ⎠
z 1 =a
z 1 =1
= 2π 2
+
∞
|ζ |2
e−2π |ζ | (1 − e−2πa|ζ | )
.
(1 − e−2π |ζ | )2
That is, H111 (a) turns out to be monotonically increasing and converges to its
limit value for a → ∞ exponentially fast. But then, by (36), also G(a) converges
exponentially fast as a → ∞.
For the physically interesting cases d = 2 and 3 a numerical evaluation of
H111 (a) − H111 (a − 1) (which is more efficient than the evaluation of H111 ) yields:
a
a
a
a
a
=1
=2
=3
=4
H111 (a) − H111 (a − 1)
for d = 2
H111 (a) − H111 (a − 1)
for d = 3
0.07441. . .
1.379. . . · 10−4
2.575. . . · 10−7
4.810. . . · 10−10
0.1713. . .
2.788. . . · 10−4
5.155. . . · 10−7
9.620. . . · 10−10
If for example m A = m B = e1 , by (36) we can write the limiting force in Theorem 3.4 as
Flim (A, B, a) = Flim (A, B, a − 1) − (H111 (a) − H111 (a − 1)) |∂ A ∩ ∂ B|e1 ,
where we applied the relation G(a) = −H111 (a)|∂ A∩∂ B|e1 from (36). Let A and B
be for instance two cuboids of width 1, depth 8 and height 8 such that |∂ A∩∂ B| = 64
and such that A and B are two lattice spacings apart (which corresponds to a = 1).
As was computed in [11, Exp. 33D ], Flim (A, B, 0) = Flim (A, B) = 17.414. Hence
we obtain Flim (A, B, 1) = 6.451. For a comparison we note that Flim (A, B, ∞) =
FBr (A, B) = 6.431 [11, Exp. 33D ], which demonstrates the exponential decay of
G(a) with a strikingly.
5. Further distance regimes
In this section we investigate the limiting force for spatially separated bodies
and for a scaling with . We first consider the case that A and B are separated by
a macroscopic distance greater that 0.
608
Anja Schlömerkemper & Bernd Schmidt
5.1. A and B macroscopically separated
In the discrete-to-continuum setting, this case corresponds to a ∼ , that is,
ε = a = const. > 0.
Proposition 5.1. Suppose A and B are not in contact, that is, A ∩ B = ∅, but still
satisfy the remaining conditions of Assumption A. Then the discrete-to-continuum
lim (A, B) exists and satisfies
limit Fsep
lim
(37)
Fsep (A, B) = (m A (x) · ∇)H B (x) dx.
A
Proof. As in Section 3, we split the discrete sum into long and short range contribu(δ)
tions, cf. (6). It is easily seen that Fshort (A, B) = 0. Indeed, since Pk (x − y) = 0
short(,δ)
for |x − y| 0, we have F
(A, B) = 0 if δ dist(A, B).
The limit → ∞ of the long range part can be identified by a Riemann sum
argument as
long(,δ)
(A, B) = γ
∂i ∂ j Rk(δ) (x − y)(m A )i (x)(m B ) j (y) dydx.
lim Fk
→∞
A
B
Since this expression is in fact independent of δ for δ dist(A, B), it follows that
long
Fk (A, B) = γ
∂i ∂ j ∂k N (x − y)(m A )i (x)(m B ) j (y) dy dx
A B
= γ (m A )i (x)∂i
∂ j ∂k N (x − y)(m B ) j (y) dy dx
B
A
∂ j (m B ) j (y)∂k N (x − y) dy
= γ (m A )i (x)∂i
A
B
(m B ) j (y)(n B ) j (y)∂k N (x − y)ds y dx
−
∂B
and hence
Flong (A,
B) =
A (m A (x) · ∇)H B (x) d x
by (8).
lim (A, B) is equal to the well-known formula for the
Remark 5.2. The formula for Fsep
magnetic force, F(A, B), between two bodies being a macroscopic distance apart,
see, for example, [2]. As proven in [10], F converges to FBr as dist(A, B) → 0.
Furthermore, by Theorem 3.1 in [10], a formal application of the formulae for Flim
and FBr to the case of separated bodies yields (37) since then the trace of m B on
∂ A is zero, that is, formally, all forces coincide in this case.
5.2. -dependent a
In order to complete the picture of limiting forces in dependence of the distance
of two bodies, we finally analyze the regime in which ε = a/, where a = a() →
∞ such that a/ → 0 as → ∞. That is, we consider two bodies which are in
contact on the macroscale whose microscopic distance tends to infinity with the
lattice constant −1 converging to 0. By Theorem 4.3 and Remark 5.2 we expect
to recover Brown’s formula. In fact, this turns out to be true:
Discrete-to-Continuum Limit of Magnetic Forces
609
Theorem 5.3. Let ε() := a()/, a() ∈ N, such that ε() → 0 and a() → ∞
as → ∞ and let Assumption A hold. Then
lim Fk() (A, B) = FBr (A, B),
→∞
where Fk() (A, B) is as in (4) with ε = ε().
Proof. We split into long and short range parts as before. Literally the same argument as for fixed a (Proposition 3.1) shows that again
long
F (A, B) = (m A · ∇)H A∪B dx
A
(38)
γ
+
(m A · n A ) ((m A − m B ) · n A ) n A dsx .
2 ∂A
The short range term is more subtle. Note that Lemma 3.5 also holds if a
depends on such that a()/ → 0. To prove this, note that the left-hand side in
(15) vanishes unless |z| cε. Hence we obtain (16) and it remains to estimate
−γ
(m A )i (x)(m B ) j (x)
∂ A∩∂ B
×
(δ)
∂i ∂ j Pk (z)
z∈ 1 L\{0}
1 (n A (x) · (z − ε()ν))+ − (n A (x) · z)+ dsx . (39)
d
Observe that (n A (x) · (z − ε()ν))+ − (n A (x) · z)+ cε(). Moreover, since
(δ)
|Pk (z) − ∂i ∂ j ∂k N (z)| can be estimated by a constant c(δ) for |z| δ and
(δ)
by c|z|−d−1 for |z| δ, replacing the term ∂i ∂ j Pk (z) in the sum in (39) by
∂i ∂ j ∂k N (z) leads to an error term which is bounded by
z∈ 1 L,|z|δ
c(δ)
ε()
+
d
c|z|−d−1
z∈ 1 L,|z|δ
ε
c(δ)|Bδ (0)| d + cε
= c(δ)ε() → 0
∞
d
δ
ε()
d
r −d−1r d−1 dr
as → ∞. So (39) equals Gk (a) in (22) up to negligible errors. Now Theorem 5.3
follows by Proposition 4.2.
6. Conclusions
We have derived limiting force formulae for two magnetic bodies in dependence
on their mutual distance starting from atomistic dipole–dipole interactions. Brown’s
classical formula has been justified in the regime where the two bodies are far apart
from each other on a microscopic scale, that is, their distance is large compared to
the lattice spacing −1 . For distances of only few atomic lattice spacings, however,
610
Anja Schlömerkemper & Bernd Schmidt
our results show that the limiting forces have to be augmented by an additional,
distance-dependent term that reflects the discrete nature of the underlying atomic
lattice. In particular, for bodies in contact on the microscale, we recover the results
of [10,13]. The following table summarizes the regimes that we have investigated
and the corresponding limit forces:
ε
Micro
distance
Macro
distance
Force
ε=0
ε = a , a ∈ N const
a=0
a finite
a→∞
0
0
0
Flim = Flong + Fshort
Flim (a) = Flim + G(a)
lima→∞ Flim (a) = FBr
ε = a()
, a() → ∞ such
that ε → 0 as → ∞
ε > 0 const
a() → ∞
0
Flim = FBr
∞
>0
lim = F
Fsep
Acknowledgements. The largest part of this work was performed while Bernd Schmidt
was affiliated with the Max Planck Institute for Mathematics in the Sciences, Leipzig. Anja
Schlömerkemper was partly supported by the RTN network MRTN-CT2004-50522.
Open Access This article is distributed under the terms of the Creative Commons Attribution Noncommercial License which permits any noncommercial use, distribution, and
reproduction in any medium, provided the original author(s) and source are credited.
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Max Planck Institute for Mathematics in the Sciences,
Inselstraße 22-26,
04103 Leipzig,
Germany.
e-mail: [email protected]
and
Zentrum Mathematik,
Technische Universität München,
Boltzmannstraße 3,
85747 Garching,
Germany.
e-mail: [email protected]
(Received March 15, 2007 / Accepted February 16, 2008)
Published online July 25, 2008 – © The Authors(s) (2008)