# hw06a - Physics 505 Homework Assignment#6 Due Thursday...

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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 Φ or the energy of a dipole in an external ﬁeld, the dipole can be described by an eﬀective charge density ρ eﬀ ( ~x ) = - ~ p ·∇ δ ( ~x - ~x 0 ) We ﬁrst consider the potential Φ( ~x ) = 1 4 π± 0 Z ρ ( ~x 0 ) 1 | ~x - ~x 0 | d 3 x 0 = - 1 4 π± 0 Z [ ~ p · ~ x 0 δ 3 ( ~x 0 - ~x 0 )] 1 | ~x - ~x 0 | d 3 x 0 = 1 4 π± 0 Z δ 3 ( ~x 0 - ~x 0 ) ~ p · ~ x 0 ± 1 | ~x - ~x 0 | ² d 3 x 0 = 1 4 π± 0 ~ p · ~ x 0 ± 1 | ~x - ~x 0 | ²³ ³ ³ ³ ~x 0 = ~x 0 = 1 4 π± 0 ~ p · ( ~x - ~x 0 ) | ~x - ~x 0 | 3 This is the expected potential for a dipole. Similarly, we can work out the energy of the dipole in an external ﬁeld W = Z ρ ( ~x )Φ( ~x ) d 3 x = - Z [ ~ p · ~ δ 3 ( ~x - ~x 0 )]Φ( ~x ) d 3 x = Z δ 3 ( ~x - ~x 0 ) ~ p · ~ Φ( ~x ) d 3 x = ~ p · ~ Φ( ~x 0 ) = - ~ p · ~ E ( ~x 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 ( ∂E z /∂z ) 0 along the z axis at the position of the nucleus.

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a ) Show that the energy of quadrupole interaction is W = - e 4 Q ± ∂E z ∂z ² 0 The quadrupole interaction energy is W = - 1 6 Q ij ∂E i ∂x j 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 11 = Q 22 = - 1 2 Q 33 (Note that this is obtained by demanding symmetry and tracelessness.) Since the nuclear quadrupole moment Q (without indices) is (1 /e ) Q 33 we end up with Q 11 = Q 22 = - 1 2 eQ, Q 33 = eQ The energy is thus W = - eQ 6 ± - 1 2 ∂E x ∂x - 1 2 ∂E y ∂y + ∂E z ∂z ² (1) We now make use of the fact that the (external) electric ﬁeld is divergence free 0 = ~ ∇· ~ E = ∂E x ∂x + ∂E y ∂y + ∂E z ∂z and cylindrically symmetric (so that
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hw06a - Physics 505 Homework Assignment#6 Due Thursday...

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