Physics 505 Fall 2005 Homework Assignment #6 — Due Thursday, October 27
Textbook problems: Ch. 4: 4.2, 4.6 a) and b), 4.7 a) and b), 4.8
4.2 A point dipole with dipole moment p~ is located at the point ~x 0. From the properties of the derivative of a Dirac delta function, show that for calculation of the potential or the energy of a dipole in an external field, the dipole can be described by an e↵ective charge density ⇢ (~x )= p~ (~x ~x ) e↵ ·r 0 We first consider the potential
1 1 3 (~x )= ⇢(~x 0) d x0 4⇡✏ ~x ~x 0 Z | 0| 1 3 1 3 = [p~ ~ (~x 0 ~x )] d x0 4⇡✏ · rx0 0 ~x ~x 0 Z | 0| 1 3 1 3 = (~x 0 ~x )p~ ~ d x0 4⇡✏ 0 · rx0 ~x ~x 0 Z ✓| 0|◆ 1 ~ 1 1 p~ (~x ~x 0) = p~ x0 = · 3 4⇡✏0 · r ~x ~x 0 4⇡✏0 ~x ~x 0 ✓ ◆ ~x 0=~x 0 | | | | This is the expected potential for a dipole. Similarly, we can work out the energy of the dipole in an external field
W = ⇢(~x ) (~x ) d3x Z = [p~ ~ 3(~x ~x )] (~x ) d3x · r 0 Z = 3(~x ~x )p~ ~ (~x ) d3x 0 · r Z = p~ ~ (~x )= p~ E~ (~x ) · r 0 · 0 Essentially the derivative of a delta function serves to pick out derivatives of the function that it is multiplied against. Of course, a delta function itself is rather singular, and its derivatives are even more so. But as far as formal expressions (or distribution theory expressions) are concerned, these manipulations are in fact legitimate.
4.6 A nucleus with quadrupole moment Q finds itself in a cylindrically symmetric electric field with a gradient (@Ez/@z)0 along the z axis at the position of the nucleus. a) Show that the energy of quadrupole interaction is
e @E W = Q z 4 @z ✓ ◆0 The quadrupole interaction energy is
1 @Ei W = Qij 6 @xj
To work out this expression, we first note that the cylindrically symmetric charge distribution for the nucleus gives rise to non-vanishing quadrupole moments
Q = Q = 1 Q 11 22 2 33 (Note that this is obtained by demanding symmetry and tracelessness.) Since the nuclear quadrupole moment Q (without indices) is (1/e)Q33 we end up with
Q = Q = 1 eQ, Q = eQ 11 22 2 33 The energy is thus
eQ 1 @E 1 @E @E W = x y + z (1) 6 2 @x 2 @y @z ✓ ◆ We now make use of the fact that the (external) electric field is divergence free
@E @E @E 0=~ E~ = x + y + z r· @x @y @z
and cylindrically symmetric (so that @Ex/@x = @Ey/@y) to write
@E @E 1 @E x = y = z @x @y 2 @z
Making this substitution in (1) allows us to write W entirely in terms of @Ez/@z
eQ @E W = z 4 @z
28 2 b) If it is known that Q =2 10 m and that W/h is 10 MHz, where h is Planck’s ⇥ 3 2 2 constant, calculate (@Ez/@z)0 in units of e/4⇡✏0a0,wherea0 =4⇡✏0¯h /me = 10 0.529 10 m is the Bohr radius in hydrogen. ⇥ Note that @Ez 4 W 4h (W/h) = = 2 2 @z e Q ea0 (Q/a0) 3 In units of e/4⇡✏0a0, we have @E e 16⇡h✏ a (W/h) 8⇡ a (W/h) z = 0 0 = 0 @z 4⇡✏ a3 e2 (Q/a2) ↵ c (Q/a2) ✓ 0 0 ◆ 0 0 2 where ↵ = e /4⇡✏0¯hc =1/137.036 ... is the fine structure constant. Putting in numbers yields @E e z 0.085 @z 4⇡✏ a3 ⇡ ✓ 0 0 ◆ 4.7 A localized distribution of charge has a charge density
1 2 r 2 ⇢(~r )= r e sin ✓ 64⇡ a) Make a multipole expansion of the potential due to this charge density and deter- mine all the nonvanishing multipole moments. Write down the potential at large distances as a finite expansion in Legendre polynomials. This charge distribution is azimuthally symmetric. As a result, only m =0 moments will be nonvanishing. Furthermore, noting that sin2 ✓ =1 cos2 ✓ = 2 [P (cos ✓) P (cos ✓)] 3 0 2 we may write down the moments
l 2 ql0 = r Yl⇤0(✓, )⇢(r, ✓) r dr d d(cos ✓) Z 2l +1 =2⇡ rlP (cos ✓)⇢(r, ✓) r2 dr d(cos ✓) 4⇡ l r Z 1 2⇡ 2 2l +1 1 l+4 r = r e dr Pl(cos ✓)[P0(cos ✓) P2(cos ✓)] d(cos ✓) 64⇡ 3 4⇡ 0 1 r Z Z 1 2l +1 2 = (l + 5)[2 l,0 5 l,2] 48r 4⇡ As a result, we read o↵the only nonvanishing multipole moments 1 5 q00 = ,q20 = 6 r4⇡ r4⇡ The multipole expansion then yields the large distance potential
1 4⇡ Ylm(✓, ) = qlm l+1 4⇡✏0 2l +1 r Xl,m 1 4⇡ P (cos ✓) l (2) = ql0 l+1 4⇡✏0 2l +1 r l r X 1 1 6 = P (cos ✓) 4⇡✏ r r3 2 0