MULTIPOLE EXPANSIONS IN THE PLANE

Transcription

1 MULTIPOLE EXPANSIONS IN THE PLANE TSOGTGEREL GANTUMUR Contents 1. Electrostatics and gravitation 1 2. The Laplace equation 2 3. Multipole expansions 5 1. Electrostatics and gravitation Newton s law of universal gravitation, first published in his Principia in 1687, asserts that the force exerted on a point mass at x R 3 by the system of finitely many point masses q i at y i R 3, i = 1,..., m, is equal to F = m i=1 Cq i x y i 2 x y i x y i, 1 with a constant C < like masses attract. Here and q i are understood as real numbers that measure how much mass the corresponding points have, and a = a a2 2 + a2 3 is the Euclidean length of the vector a R 3. The same law of interaction between point charges was discovered experimentally by Charles Augustin de Coulomb and announced in 1785, now with C > like charges repel. Note that the numerical value of the constant C depends on the unit system one is using to measure force, mass or charge, and distance. It is convenient to view the force F = F x as a vector function of x, that is, a vector field. This means that we fix the configuration of the point masses {q i }, and think of as a test mass, that can be placed at any point in space to probe the field. The vector field m Cq i x y i Ex = x y i 2 x y i, 2 i=1 does not depend on the test mass, and given any test mass at x R 3, the force can be recovered as F = Ex. Therefore E can be thought of as a preexisting entity that characterizes the gravitational or electric field generated by the point masses {q i }. In fact, we call E the gravitational field or the electric field. In this short note, we give an introduction to multipole expansions, which is a powerful method to get a handle on the field E, when it is generated by a system of large number of particles, or by a body with continuous distribution of mass or charge. The method was originally developed to calculate the gravitational field of an irregular shaped object, such as the Earth, and is largely due to the three great mathematicians Joseph-Louis Lagrange, Pierre- Simon Laplace, and Adrien-Marie Legendre. It was later extended to treat wave propagation problems in electromagnetism and general relativity. Date: April 4,

2 2 TSOGTGEREL GANTUMUR For continuous distribution of mass, the sum in 2 must be replaced by an integral, as ρy x y Ex = C x y 2 dy, 3 x y Ω where ρ is the mass or charge density, and Ω R 3 is the region occupied by the body. By defining ρ = outside Ω, in 3 we may integrate over R 3. Now, the method of multipole expansions depends on a few crucial observations. Firstly, in 1773, Lagrange showed that the field E is minus the gradient of some scalar function u, called the potential, that is, there is a scalar function u such that E = grad u x 1, x 2, x 3. 4 Secondly, in 1782, Laplace observed that E is divergence-free in empty space, that is, dive E 1 x 1 + E 2 x 2 + E 3 x 3 = in free space, 5 where free space means a place where there is no mass or charge. From 4 and 5 we see that the potential satisfies u 2 u x u 1 x u 2 x 2 = in free space, 6 3 where is called the Laplace operator. The equation 6 is the Laplace equation, and its solutions are called harmonic functions. It should however be noted that the same equation had been considered by Lagrange in 176 in connection with his study of fluid flow problems. Laplace s result 6 was completed by his student Siméon Denis Poisson in 1813, when Poisson showed that u = 4πCρ in R 3, 7 for ρ vanishing outside some bounded set 1. This equation is called the Poisson equation, and is valid everywhere, as opposed to 6, which is only valid in free space. Note that in terms of the field E, the Poisson equation 7 is simply dive = 4πCρ. Finally, around 1785, Legendre and Laplace devised a method to expand u in a series of the form ux = u n x, 8 n= where u n behaves like u n x n for x R 3 far away from the origin. Then by summing only the first few terms in the right hand side of 8, we can approximate ux with high accuracy, especially for large. This expansion was the first instance of what came to be known as multipole expansions. In the next section, we will confirm the existence of a potential 4, divergence-free property of the field 5, and the Poisson equation 7. Then in the following section, we will derive the multipole expansion 8 in two dimensions. 2. The Laplace equation In order to confirm the existence of a potential 4, we simply define m Cq i ux = x y i, 9 i=1 1 In Gaussian type unit systems, one sets up the units so that C = ±1, and hence the Newtonian/Coulomb formulas 2 and 3 have simple expressions. In other systems such as SI, one has C = ± 1, meaning that 4π the Poisson equation 7 has a simple expression.

3 MULTIPOLE EXPANSIONS IN THE PLANE 3 for a system of point charges, and ρydy ux = C R 3 x y, 1 for a continuous distribution of charge, and check that E = grad u indeed gives 2 and 3, respectively. To this end, we compute 1 x 1 = 2 /2 1 = 2 3/2 2x1 = 1 x x1, 11 and so grad 1 = 1 2 x, or grad 1 x y = 1 x y 2 for a fixed y R 3. This means that if u is given by 9, then m grad ux = grad Cq i m x y i = Cq i x y i 2 i=1 i=1 x y x y, 12 x y i = Ex, 13 x y i confirming 4. The case 1 can be treated similarly. Next, we show that E is divergence-free in free space. When E is generated by a single particle at the origin, we have 2 3/2 x1 = 3 2 5/2 2x1 x /2 2x 2 = 1 + x2 2 + x2 3 x 1 2 5, 14 and so div x 3 = x 1 2 3/2 x1 + x 2 2 3/2 x2 + x 3 2 3/2 x3 = 2x2 1 + x2 2 + x x2 2 + x2 1 + x x2 3 + x2 1 + x2 2 5 =, for x. Therefore, for either 2 or 3, we have dive = in free space. 15 If we combine this with E = grad u, we get the Laplace equation u = in free space. 16 To proceed further, let us recall the divergence theorem, which asserts that divv = V n, 17 U where U R 3 is a bounded domain with smooth boundary, V : U R 3 is a 3-dimensional vector field, and n is the outward pointing unit normal to the boundary. This theorem first appeared in Lagrange s 176 work, and was proved in a special case by Gauss in The general 3-dimensional case was treated by Mikhail Vasilievich Ostrogradsky in U n n Figure 1. The setting of the divergence theorem.

4 4 TSOGTGEREL GANTUMUR If V = grad u then divv = u, and V n = n grad u = n u, 18 where n u is the normal derivative of u at, and so the divergence theorem yields u = n u 19 Now consider the potential generated by a point mass q at some point y U: U ux = Cq x y. 2 Let B ε = {x R 3 : x y < ε}, and let us apply 19 to the domain U ε = U \ B ε. Since u is harmonic in U ε, we infer = u = n u = n u U ε ε B ε r, 21 where r = x y is the radial variable centred at y. We can compute B ε r = Cq {r=ε} r r = Cq Cq {r=ε} r 2 = 4πε2 = 4πCq, 22 ε2 and thus Ω n u = 4πCq. 23 This is for the potential generated by a charge at y U. On the other hand, if y U, then u = in U, and hence we have n u = u =. 24 Therefore, for the potential generated by a set of point charges 9, we conclude that n u = 4πC q i, 25 U {i:y i U} where the sum is taken over the indices i such that y i U, meaning that the sum {i:y i U} q i is simply the total charge contained in the domain U. Similarly, for the potential generated by a continuous distribution of charge 1, we get n u = 4πC ρ. 26 Finally, let B ε = {y R 3 : y x < ε} for ε > small, and note that for any continuous function f, one has 1 fx = lim f, 27 ε B ε B ε where B ε is the volume of the ball B ε. By applying this result to u, with u given by 1, we infer ux 1 u = 1 n u = 4πC ρ 4πCρx, 28 B ε B ε B ε B ε B ε B ε where we used the identity 19 in the second step, and 26 in the third step. Upon taking the limit ε, we get ux = 4πCρx, 29 which is Poisson s equation. U

5 MULTIPOLE EXPANSIONS IN THE PLANE 5 3. Multipole expansions In this section, we want to restrict ourselves to the two dimensional situation, which can be thought of as a stepping stone to understanding the full three dimensional setting. The main equations of two dimensional electrostatics and gravitation can be derived by assuming that the field potential ux = ux 1, x 2, x 3 does not depend on x 3. Since 2 u/ x 2 3 =, the three dimensional Laplacian reduces to the two dimensional Laplacian u = 2 u x u y 2, 3 where we have used x, y instead of x 1, x 2. Thus in free space, we have u =. 31 The potential generated by a point charge at, in 2 dimensions or equivalently, the potential of a uniformly charged line coinciding with the z-axis in three dimensions must clearly be radial, meaning that it must depend only on r = x 2 + y 2. We know that the only radial solutions of u = in 2 dimensions is ur, θ = A + B log r. 32 Since the constant A has no effect on the field strength E = grad u, for simplicity, we pick A =. Then the potential generated by a point charge q at the origin is given by ur, θ = Cq log r = Cq log 1 r, 33 where the sign convention for the constant C is chosen so that if Cq >, then E = grad u points away from the origin. This is the Newton/Coulomb law in 2 dimensions. r a In 2 dimensions, the potential generated by a point charge at the origin is unbounded both as, and as x. b In 3 and higher dimensions, the potential generated by a point charge at the origin is unbounded as x, but tends to as. r Figure 2. The potential generated by a point charge in different dimensions. Now by invoking the identity 19 on the disk D ε = {x, y : x 2 + y 2 < ε 2 }, we get u = D ε D ε r = Cq {r=ε} r log 1 r = Cq {r=ε} r = 2πεCq = 2πCq, 34 ε and proceeding as in the 3 dimensional case, we derive the Poisson equation where µ is the charge or mass density in 2 dimensions. u = 2πCµ 35

6 6 TSOGTGEREL GANTUMUR The main purpose of this section is to compute the potential generated by a compact body located near the origin, in the form of a series. Suppose that µ = outside the disk D R of radius R centred at the origin. Then u = outside D R, and hence u can be written as ur, θ = a log 1 r + n=1 a n cosnθ + b n sinnθ r n 36 There are no terms proportional to r n in the expansion, because the term log 1 r must dominate for large r. This expansion is known as the multipole expansion of harmonic potential functions in 2 dimensions. The question is now how to compute the coefficients a, a 1,... and b 1, b 2,... in terms of the density µ. To get some initial insight, let us compute the coefficient a first. For large r, the expansion 36 becomes a log 1 r, so we expect that a must be equal to the total charge of the body. To confirm this expectation, we start with the Poisson equation 35 and invoke the identity 19, yielding 2πC µ = 2πC µ = u = R 2 D R D R r. 37 Then from 36 we compute r = a and integrate it over, to get r = 2π r n=1 n a n cosnθ + b n sinnθ r n+1, 38 r R, θrdθ = a R 2πR = 2πa. 39 By comparing this with 37, we conclude a = C µ R 2 4 which is the total charge/mass, up to the constant C. In order to derive formulas for the remaining coefficients, we need some preparations. Let v be a sufficiently smooth scalar function. Then by applying the divergence theorem to the vector field V = u grad v, we get Green s first identity grad u grad v + D R u v = D R u v r. 41 Interchanging the roles of u and v in this identity, and subtracting the resulting identity from 41, we infer Green s second identity u v v u = u v D R D R r v r. 42 Note that 19 follows from 41 by putting v 1. The identities 41 and 42 can be considered as instances of, and are often called, integration by parts in higher dimensions. Now the idea is to put a suitably chosen function v with v = into 42. Then the left hand side becomes essentially an integral of vµ, and the right hand side can be computed from 36. In other words, we use 42 as an extension of 37. A general harmonic function in D R can be written as vr, θ = A + r n A n cosnθ + B n sinnθ. 43 n=1

7 MULTIPOLE EXPANSIONS IN THE PLANE 7 We consider the simple choices vr, θ = r n cosnθ and vr, θ = r n sinnθ, and compute the terms in the right hand side of 42, as follows. For vr, θ = r n cosnθ, we have u v 2π r = ur, θ v R, θrdθ r 2π = a log 1 R + a n cosnθ + b n sinnθ R n nr n cosnθ Rdθ and = πna n, v r = = 2π 2π = πna n. n=1 vr, θ R, θrdθ r R n cosnθ a R n=1 n a n cosnθ + b n sinnθ R n+1 Rdθ Plugging this into 42, we get 2πna n = u v r v r = v u = 2πC vµ 44 D R R 2 implying that a n = C n R 2 vµ = C n R 2 r n cosnθµ 45 Similarly, for vr, θ = r n sinnθ, we can compute u v r = πnb n, and v r = πnb n, 46 leading to the formula b n = C n R 2 vµ = C n R 2 r n sinnθµ 47 The formulas 4, 45, and 47 provide a complete prescription to compute the coefficients of the multipole expansion 36 in terms of the density µ. Definition 1. The number a in the multipole expansion 36 is called the monopole moment. The vectors a 1, b 1 and a 2, b 2 are called the dipole moment and the quadrupole moment, respectively. Figure 3 illustrates the first 3 terms of the multipole expansion ur, θ = a log 1 + a 1 cos θ + b 1 sin θ }{{ r }}{{ r } monopole term dipole term + a 2 cos 2θ + b 2 sin 2θ r 2 }{{} quadrupole term We depicted the dipole and quadrupole potentials with a 1, b 1 = 1, and a 2, b 2 = 1,, respectively. A dipole or quadrupole potential with an arbitrary moment a 1, b 1 or a 2, b 2 is simply a rotation and scaling of the case depicted. In the following examples, we are going to set C = 1, that is, we choose the normalization that the potential of a unit charge at the origin is given by log 1 r, cf. 33.

8 8 TSOGTGEREL GANTUMUR a Monopole potential b Dipole potential c uadrupole potential Figure 3. The first 3 terms of the multipole expansion up to rotation. Example 1. Suppose that the unit disk D = {x, y : x 2 + y 2 < 1} is uniformly charged with total charge 1 Case 1, Figure 4. Let us compute the electrostatic potential generated by the disk. It is immediate that µ = 1 π and a = 1. Furthermore, we have a 1 = 1 2π r cos θ 1 π rdθdr = 1 π 1 r 2 dr 2π cos θdθ =, 49 and 1 2π b 1 = r sin θ 1 π rdθdr = 1 1 2π r 2 dr sin θdθ =. 5 π In fact, basically the same computation shows that a n = b n = for all n = 1, 2,..., and hence ur, θ = log 1 r. 51 We conclude that the electric field generated by a uniformly charged disk, measured outside the disk, is identical to the field generated by a point charge Figure 4. Examples of charged bodies. Example 2. Now suppose that a half of the unit disk D carries a charge of + 1 2, and the other half carries a charge of 1 2. More precisely, suppose that the half disk {x > } D is uniformly charged with charge density + 1 π, and the half disk {x < } D is uniformly charged with charge density 1 π Case 2, Figure 4. Obviously, we have a =. Then for a 1, we have a 1 = 1 π = 1 π 1 π/2 1 r cos θ rdθdr 1 π/2 π sin θ π/2 r 2 dr 1 π/2 π 1 3π/2 1 π/2 sin θ 3π/2 π/2 r cos θ rdθdr r 2 dr = 4 3π.

9 Furthermore, we get b 1 = 1 π = 1 π 1 π/2 1 MULTIPOLE EXPANSIONS IN THE PLANE 9 r sin θ rdθdr 1 π/2 π cos θ π/2 r 2 dr 1 π/2 π 1 3π/2 π/2 1 cos θ r sin θ rdθdr 3π/2 π/2 In fact, it can be shown that a n = b n = for all n = 2, 3,..., and hence r 2 dr =. ur, θ = 4 cos θ 3πr. 52 Example 3. Let the union of the two squares =, 1 2 and =, 2 be uniformly charged with total charge +1 Case 3, Figure 4. We have µ = 1 2 and a = 1. Since r cos θ = x and r sin θ = y, we have the general formulas a 1 = C xµx, y dxdy R 2 and b 1 = C yµx, y dxdy R 2 53 Invoking these formulas, we infer a 1 = x 1 2 dxdy + x 1 2 dxdy = 1 2 and similarly, b 1 = = x dx x dx = y 1 2 dxdy + y 1 2 dxdy = 1 2 As for the quadrupole moment, noting that and 1 x dx x dx =, 1 y dy 1 1 dy dx x dx dy y dy dx =. r 2 cos 2θ = r 2 cos 2 θ sin 2 θ = x 2 y 2, 54 r 2 sin 2θ = 2r 2 sin θ cos θ = 2xy, 55 we derive the general formulas a 2 = C x 2 y 2 µx, y dxdy and b 2 = C xyµx, y dxdy 56 2 R 2 R 2 With the help of these formulas, we compute a 2 = 1 x 2 y dxdy + x 2 y dxdy = 1 4 and b 2 = 1 1 x 2 dxdy y 2 dxdy xy 1 2 dxdy + xy 1 2 dxdy = x dx x 2 dxdy y dy y 2 dxdy =, x dx y dy = 1 4. Finally, we conclude that the multipole expansion of the potential up to and including the quadrature term is ur, θ = log 1 r + sin 2θ 4r = log 1 r + sin θ cos θ 2r

10 1 TSOGTGEREL GANTUMUR Example 4. Consider the situation where =, 1 2 is uniformly charged with total charge + 1 2, and =, 2 is uniformly charged with total charge 1 2 Case 4, Figure 4. We have µ = 1 2 in, µ = 1 2 in, and a =. To compute the dipole moment, we use the formulas 53, and get a 1 = x 1 2 dxdy + x dxdy = x dx dy 1 x dx dy = x dx 1 2 x dx = 1 2, and similarly, b 1 = y 1 2 dxdy + y 1 1 dxdy = y dy 1 dx 1 2 As for the quadrupole moment, the formulas 56 yield a 2 = 1 x 2 y dxdy x 2 y 2 1 dxdy =, 2 and b 2 = xy 1 2 dxdy + xy 1 1 dxdy = x dx 1 y dy 1 2 y dy dx = 1 2. x dx y dy =. Therefore, the potential has the expansion cos θ + sin θ ur, θ = + O 1, 58 2r r 3 where the error term Or 3 is to indicate that there is no quadrupole term in the expansion.