Multipole moments. November 9, 2015

Transcription

1 Multipole moments November 9, 5 The far field expansion Suppose we have a localized charge distribution, confined to a region near the origin with r < R. Then for values of r > R, the electric field must be described by solutions to the Laplace euation with vanishing boundary condition at infinity. This reuires A l for an expansion in Legendre polynomials and the potential for any azimuthally symmetric source is V r, θ l r l+ B lp l cos θ For large values of r, the lowest nonvanishing term will dominate this approximation, giving an excellent approximation for the potential Now consider the full solution for any localized source near the origin. Let the charge density be ρ x, and suppose we with to know the potential at a point x far from the source. The general solution for the potential is V x ρ x d 3 x x x Let the angle between x r r and x rr be θ, so the denominator is x x x + x xx cos θ r + r rr cos θ r Since r r, we could expand the suare root in a Taylor series and collect the resulting powers of cos θ into Legendre polynomials. Fortunately, there is an easier way. We would like the Legendre series for x x, and in order to vanish at infinity it must take the form x x l r l+ B l P l cos θ Consider what happens at θ. The Legendre polynomials all give sincep l cos P l, so when x and x are parallel the series reduces to r r l r l+ B l But r r r rr

2 and we know the Taylor series for x, f x x f x x f x x 3. f n x n! x n Evaluating these at x gives dn f n x n! so the Taylor series is x n n n n! f n x n n! n!xn x n Letting x r r and substituting into our expansion, this means that l Comparing like powers of r shows that r l+ B l r r r r r r r B l r l n This gives all of the coefficients in the original series! Therefore, x x l r l r l+ r n P l cos θ Now, returning to the potential, V x x x ρ x d3 x r l r l+ P l cos θ ρ x d 3 x l r l+ r l ρ x P l cos θ d 3 x l where the angle θ is the angle between x and x.

3 Multipole moments The integrals Q l r l ρ x P l cos θ d 3 x still depend on the angle between the charge increment at x and the observation point x. For the first few values of l, we now separate Q l into a part depending on the observation direction and another part which depends only on the charge distribution. These latter expressions are the multipole moments of the charge distribution. Once we know all of the multipole moments that characterize a charge distribution, we have the complete potential: V x l Q l r l+ It is important to be able to recognize and understand the properties of the first few multipole moments. They are extremely useful for describing arbitrary charge distributions. 3 Total charge The zeroth moment is proportional to the total charge, Q r ρ x P cos α d 3 x ρ x d 3 x Q total Notice that the Q total is independent of the observation point x. It characterizes only the source. The potential of a pure monopole is given by Coulomb s law for a single point charge, and the resulting electric field is V x Q r E V r Q r r Q r r as we know. For any source, at large distances r r the dominant term in the potential is the Coulomb potential of the total charge, Higher multipole moments give corrections to this. V x Q total r 3

4 4 Dipole moment The next term depends on the dipole moment, Q rρ x P cos θ d 3 x rρ x cos θ d 3 x rρ x cos θd 3 x r xρ x d 3 x We define the dipole moment to be p xρ x d 3 x A dipole moment may be thought of as a pair of eual and opposite charges, slightly displaced. If we take charges ± displaced by a vector dk, then the charge density is ρ r 3 r d k 3 r + d k so the dipole moment is p xρ x d 3 x x 3 x d k 3 x + d k d 3 x x x 3 d k d 3 r x x 3 + d k d 3 x d k + d k dk that is, the vector displacement between the two charges times one charge. The dipole moment p depends only on the charge distribution, not on the observation point. To see if this is a pure dipole, we check the higher moments of the distribution, Q l r l ρ x P l cos θ d 3 x r l x 3 d k 3 x + d k P l cos θ d 3 x r l x 3 d k 3 x + d k P l cos θ r dr sin θdθdϕ x 3 d k P l cos θ r l+ dr sin θdθdϕ x 3 + d k Letting x cos θ, the delta functions may be written as x 3 d k 4 πd r d x x 3 + d k 4 πd r d x + 4 P l cos θ r l+ dr sin θdθdϕ

5 so that Q l 4 πd πɛ d πɛ d { π r d r d x P l x r l+ drdϕ r l+ dr d l+ P l l d l d l l odd l even d x P l x l+ P l π r d x + P l x r l+ drdϕ r d r l+ dr x + P l x This means that the distribution is not a pure dipole, but contains higher moments as well. To obtain a pure dipole, we take the limit as the separation of charges vanishes, d, in such a way that the product p d remains constant, p ρ x lim 3 x d d d k 3 x + d k In this case, the odd moments become Q k+ lim d lim d p d d p d d k+ k k p lim d { p k k > and we have a pure dipole with dipole moment p pk For an arbitrary pure dipole with dipole moment p, the potential is therefore V r r p r The electric field of a dipole lying along the z-axis, p pk, is found by taking the gradient. For simplicity, we drop the subscript on the observation point. Then E V x r p r r r + θ r r r p cos θ r θ + ϕ p cos θ r sin θ ϕ r + θ p cos θ r θ r 5

6 r We may rewrite the second term using together with p cos θ r 3 θ p sin θ r 3 r 3 r p r + p sin θθ r r p r p r p r r r p r p r r p r ϕp sin θ θp sin θ The electric field becomes 5 Quadrupole E E r p r + r p r p r3 3 r p r p r 3 For the uadrupole moment, we need to compute Q d 3 x r ρ x P cos θ where θ is the angle with the observation direction. Then substituting for P x, we have Q d 3 x ρ x r 3x 8πɛ We need to rewrite this to separate the direction r at which we wish to know the field from the integration variables. First, write the final terms in vector notation, 3x r 3r cos θ r Now, using index notation, we insert r r 3 r r r r ij r ir j into the second term and separate the observation direction from the integration variables, 3 r r r r 3 i,j r i x i i r i i j r j x j r j ij r ir j i,j r j 3xi x j r ij 6

7 Now we may write the uadrupole moment in a way that is independent of where we choose the direction of r: Q 8πɛ 8πɛ [ r ir j d 3 x ρ x 3x i x j r ] ij i,j r i Q ij r j i,j where we define the uadrupole moment tensor, Q ij d 3 x ρ x 3x i x j r ij Notice that Q ij describes only the charge distribution, and is independent of the observation point. We can develop a pure uadrupole moment tensor in the same way as we built a dipole moment vector, by placing four charges of alternating sign at the corners of a suare of side d in the xy plane. Then the charge density is ρ x lim 3 x d d î d ĵ 3 x d î + d ĵ + 3 x + d î + d ĵ 3 x + d î d ĵ and the uadrupole moment may be evaluated easily in Cartesian coordinates. The Dirac delta functions become, for example, x 3 d î d ĵ z x d y d and so on, changing signs appropriately. Then, noticing that the uadrupole tensor is a symmetric matrix, Q ij Q ji, we need to compute Q xx, Q xy, Q xz, Q yy, Q yz, Q zz. We can avoid one of thes by noticing that the trace of the matrix is tr Q ij i Q ii d 3 xρ x 3xi x i r ii i d 3 xρ x 3r 3r Therefore, Q zz Q xx Q yy. In Cartesian coordinates, it is easy to write the components of the matrix 3x i x j r ij. For i,, 3, we just replace x, x, x 3 x, y, z. For example if i, j, 3x x r 3xy since. For i j, 3x x r 3x r 3x x + y + z x y z 7

8 Now we compute one component at a time. For the first element, and for Q, Q Q + d + dzρ x x y z x d y d x d y + d x y x + d y + d x + d y d x y d d + d d dzρ x y x z x d y d x d y + d y x x + d y + d x + d y d y x d + and therefore Q 33 Q Q as well. For the off-diagonal components, Q The remaining two, 3 +3 d 3 + 3d d dzρ x 3xy d x d y d x d y + d xy x + d y + d x + d y d xy d Q 3 + d + 8 d dzρ x 3xz

9 both depend on the z-integral Q 3 dz z z dzρ x 3yz and therefore vanish. The limit as d in such a way that Q d remains constant gives a pure uadrupole. The uadrupole matrix is then simply Q ij 3Q The second moment is therefore Q 8πɛ i,j 3Q 8πɛ x x x i Q ij x j where we may write the components of the unit vector in the observation direction x x, y, z as The potential due for a uadrupole is therefore or, dropping the subscripts, Writing this in spherical coordinates, x i r x, y, z V x V x l Q r 3 Q r l+ 3Q 8πɛ r 3 x x 3Q 8πɛ r 5 x y V x 3Q xy 8πɛ r 5 3Q 8πɛ r 3 sin θ cos ϕ sin ϕ 3Q 6πɛ r 3 sin θ sin ϕ 6 Potentials of the first three multipoles Collecting our results, the potentials of the particular monopole, dipole and uadrupole we studied are V M x r 9

10 V D x p cos θ r 3Q V Q x 6πɛ r 3 sin θ sin ϕ where is the total charge, p lim d and Q lim d. Notice that the strength of each successive multipole falls off with a higher power of r. The dipole and uadrupole formulas above apply only to dipoles and uadrupoles oriented as in our examples. For a dipole oriented in a general direction, p pn, the potential is V x pn r r where the observation point is x rr. The dipole moment is given by p xρ x d 3 x The potential of a general uadrupole moment tensor has the form where Q Q ij 8πɛ r 5 Q ij x i x j i,j d 3 x ρ x 3x i x j r ij In general, an arbitrary charge distribution will have all multipole moments nozero. The potential is then the sum of potentials like those above, but at large enough distances it is only the lowest multipole that accounts for the bulk of the field. For example, a charge distribution with no net total charge, but all other multipoles nonzero is well approximated by 7 Exercises reuired V x p x r. Find the total charge and dipole moments of a charged ring of radius R lying in the xy plane, with constant charge density given by ρ x z ρ R πr. Find the total charge and dipole moments of a charged ring of radius R lying in the xy plane, with charge density given by ρ x sin ϕ z ρ R R 3. Three charges lie in the xy plane: + at the origin, on the x-axis at dî, and on the y-axis at dĵ. Without computing, what is the direction of the dipole moment? 4. Compute the uadrupole moment tensor of the three charges in the previous exercise.