Problems
Application 3.2 The Ionization Potential of a Metal Cluster
4.2 The Electric Dipole
The second term in (4.7) describes the distant electrostatic potential produced by most electrically neutral (Q=0) objects like neutrons, atoms, molecules, plasmas, and ordinary matter. This is the electric dipolepotential,
ϕ(r)= 1 4π 0
p·r
r3 rR. (4.8)
The integral that defines the electric dipole momentpin (4.5) is non-zero for any spatially extended object where the “centers” of positive charge and negative charge do not coincide. We say that a system has apermanentelectric dipole moment whenp=0 is a property of its ground state charge density. A polar molecule like water is a good example (Figure 4.2). We speak of aninduceddipole moment whenp=0 is produced by an external electric field. This is the situation with essentially all conductors and dielectrics.
Whatever the physical origin of the dipole moment, (4.8)approximatesthe true potential far from the distribution. This statement is well defined because the numerical value ofpis independent of the choice of origin whenQ=0.To see this, letpdenote the new dipole moment when we shift the origin by a vectordsor=r+d. Then, becaused3r=d3randρ(r)=ρ(r),
p=
d3rρ(r)r=
d3r ρ(r)(r−d)=p−Qd. (4.9) Hence,p=p, ifQ=0.Conversely, the electric dipole moment is not uniquely defined for any system with a net charge.
Since E= −∇ϕ, the moment p also completely characterizes the asymptotic dipole electric field,
E(r)= 1 4π 0
3ˆr(ˆr·p)−p
r3 rR. (4.10)
This formula takes a simple form in polar coordinates because E(r) must be unchanged when the vector protates about its own axis. Thus, the choicep=pˆzand the identity ˆz=rˆcosθ−θˆsinθ reduce (4.10) to the azimuthally symmetric form
E(r, θ)= p 4π 0r3
,
2 cosθˆr+sinθθˆ-
. (4.11)
4.2 The Electric Dipole 93
p
Figure 4.3: Lines ofE(r) for an electric dipolep=pzˆlocated at the center of the diagram. The localized charge distribution responsible for the field is too small to be seen on the scale used in the diagram. The black dots are two points on a circular orbit around thez-axis. See text for details.
Uracyl Alkalai halides Thymine
CH2CN
CH2CHO
Dipole moment (D)
Electron binding (meV)
10−2 10−1 100 101 102 103
2 4 6 8 10
Figure 4.4: The binding energy of an electron captured by a polar molecule. One debye (D) is=3.36× 10−30C-m. Figure adapted from Desfranc¸oiset al.(1994).
An equation for the electric field lines follows from the polar analog of the Cartesian formula (3.27):
dr Er
= rdθ Eθ
⇒ 1 r
dr
dθ =2 cotθ ⇒ r=ksin2θ. (4.12)
The integration constant kparameterizes the family of electric field lines (solid curves) shown in Figure 4.3.
The characteristic 1/r3dependence of the dipole field (4.11) differs from the 1/r2behavior of the point charge electric field. A more important difference is that (4.11) is angle-dependent. This has many consequences. Not least of these is that the forceF=qE that an electric dipole exerts on a point chargeqis non-central and leads to non-trivial orbital dynamics. For example, the black dots in Figure 4.3 are two points on a circular orbit around thez-axis in a plane perpendicular to that axis.
This can be a stable orbit for an electron because the electric force is always centripetal (the vertical component ofEis zero). Figure 4.3 implies that an entire family of such circular orbits exists, each one requiring a particular orbital speed to satisfy Newton’s second law.
Formal classical mechanics confirms this qualitative prediction. Moreover, the main result of such an analysis—stable bound states exist if the electric dipole moment is large enough—survives the transition to quantum mechanics. Figure 4.3 thus provides a qualitative rationalization for the exper- imental fact that a neutral polar molecule can capture an incoming electron to form a stable negative ion. The data plotted in Figure 4.4 indicate that the ionic bound state does indeed disappear when the dipole moment of the host molecule is too small.
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Example 4.1 LetE(r) be the electric field produced by a charge densityρ(r) which lies entirely inside a spherical volumeV of radiusR. Show that the electric dipole moment of the distribution is given by
p= −30
V
d3rE(r).
Solution: Assume first thatρ(r) doesnotlie entirely insideV. If we place the origin of coordinates at the center ofV, Coulomb’s law gives
1 V
V
d3rE(r)= 1 V
V
d3r 1 4π 0
d3s ρ(s) r−s
|r−s|3.
Reversing the order of integration and extracting a minus sign produces 1
V
V
d3rE(r)= −1 V
d3s
⎡
⎣ 1 4π 0
V
d3r ρ(s) s−r
|s−r|3
⎤
⎦.
The factorρ(s) is a constant as far as the integration overris concerned. Therefore, the quantity in square brackets is the electric fieldE(s) produced by a sphere of volumeV =4π R3/3 with constant and uniform charge densityρ(s). From Gauss’ law, the latter is
E(s)=
⎧⎪
⎪⎨
⎪⎪
⎩ ρ(s)
30
s s < R, V
4π 0
ρ(s)
s3 s s > R.
Therefore, ifpinis the dipole moment due to the part ofρwhich lies insideV andEout(0) is the electric field at the origin due to the part ofρwhich lies outsideV,
1 V
V
d3rE(r)= − 1 30V
s<R
d3s ρ(s)s− 1 4π 0
s>R
d3s ρ(s)s s3 = − 1
30
pin
V +Eout(0).
We get the stated result if all ofρ(r) is contained inV soEout(0)=0 andpin=p. On the other hand, ifnoneof the charge is contained inV,
E(0)= 1 V
d3rE(r).
Parity and The Dipole Moment
A well-known theorem of quantum mechanics states that the electric dipole moment is zero for any microscopic system described by a wave function with definite parity:
p=
d3r ρ(r)r =< ψ|r|ψ >=0.
The theorem applies to isolated electrons, atoms, and molecules because the Hamiltonian for each of these is invariant under the parity operation. How, then, do we understand the dipole moment of the water molecule indicated in Figure 4.2 and the table of permanent electric dipole moments found in every handbook of molecular properties?
4.2 The Electric Dipole 95
O P
r r0
sb r r0 sb 0
r r
+q/s
–q/s
Figure 4.5: A dipole with total chargeQ=0 and dipole momentp=qb. The potential is measured at the observation pointP.Ois the origin of coordinates.
Handbooks writep=pn+pe, wherepnis a contribution from the nuclei treated as point charges andpe is computed from the integral above using electron wave functions computed in a body- fixed frame of reference. The latter arenoteigenstates of the parity operator applied to the electron coordinates alone. The quantum mechanical theorem applies to the total molecular wave function in the laboratory frame, including the nuclear coordinates. We getp=0 in that case because all orientations of the body frame dipole moment are equally probable.
4.2.1 The Point Electric Dipole
The word “dipole” comes from a geometrical construction due to Maxwell. It begins with the placement of two (“di”) equal and opposite charges (“poles”)±q/sat opposite ends of a vectorsb(Figure 4.5).
Now lets→0. Ifp=qb, the electrostatic potential of the resulting “point dipole” located atr0is
ϕ(r)=lim
s→0
1 4π 0
q/s
|r−r0−sb|− q/s
|r−r0|
= − 1 4π 0
p· ∇ 1
|r−r0|. (4.13) This formula is exact because the higher-order terms in the Taylor expansion of|r−r0−sb|−1all vanish in thes→0 limit. The potential (4.13) is identical to (4.8) except that (4.13) is valid at every point in space (exceptr0). This is reasonable becauserR is always valid when the source size R→0.
We can derive an analytic formula for the source charge density that produces the point dipole potential (4.13) by applying Maxwell’s limiting process directly to the charge density of two point charges. An alternative method exploits (4.13), Poisson’s equation (3.14), and the delta function identity (1.121) to get
ρD(r)= −0∇2ϕ(r)= p
4π · ∇∇2 1
|r−r0|= −p· ∇δ(r−r0). (4.14) As far as we know, point electric dipoles with the singular charge density (4.14) do not exist in Nature.
Nevertheless, we will findρD(r) a useful tool for computation when the size of a charge distribution is very small compared to all other characteristic lengths in an electrostatic problem.
4.2.2 The Singularity at the Origin
The electric field of a point electric dipole must be given by (4.10) whenr=r0. But what happens precisely atr=r0? Given that the point dipole was constructed from two delta function point charges,
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it is reasonable to guess that a delta function lies at the heart of the electric field of the dipole as well.
To check this, it is simplest to use Example 4.1 and a spherical integration volumeV centered atr0. In that case,
V
d3rE(r)= − p 30
. (4.15)
On the other hand, it is clear from the symmetry of Figure 4.3 that (4.11) integrates tozeroover the volume of any origin-centered sphere. Therefore, if this behavior persists asV →0, (4.15) will be true if the total electric field of a point dipole is1
E(r)= 1 4π 0
3 ˆn( ˆn·p)−p
|r−r0|3 −4π
3 pδ(r−r0)
. (4.16)
The unit vector ˆn=(r−r0)/|r−r0|. A useful way to think about the delta function in (4.16) will emerge in Application 6.1 of Chapter 6, devoted to dielectric matter.
4.2.3 The Dipole Force
The electric dipole moment emerges in a natural way when we calculate the force exerted on a neutral charge distributionρ(r) by an external electric fieldE(r) that changes slowly in space. By
“slowly” we mean that, over the entire spatial extent of the distribution,E(r) is well approximated by a two-term Taylor series expansion around a reference pointrlocated somewhere insideρ(r). Since
∇f(r)|r=r≡ ∇f(r), the expansion in question is E(r)=E(r)+
(r−r)· ∇
E(r)+ · · ·. (4.17) The total charge (4.4) is zero by assumption. Therefore, (4.17) gives the first non-zero contribution to the force onρ(r) as
F=
d3rρ(r)E(r)=
d3rρ(r)(r· ∇)E(r). (4.18) Using the definition ofpin (4.5), we learn that the force onρ(r) is entirely characterized by its dipole moment:
F=(p· ∇)E(r). (4.19)
Equation (4.19) is completely general. Whenpis a constant vector, the fact that∇ ×E=0 and the identity∇(p·E)=p×(∇ ×E)+E×(∇ ×p)+(p· ∇)E+(E· ∇)ptransform (4.19) to2
F= ∇(p·E). (4.20)
We emphasize that an electric fieldgradientis needed to generate a force.
An alternative derivation of (4.19) exploits the fact that the force depends only onp. The idea is to calculate the force thatEexerts on apointdipole density (4.14) with the same dipole momentpas ρ(r). After an integration by parts, the result of this second calculation for a constantpis
F=
d3rE(r)ρD(r)=p·
d3rδ(r−r)∇E(r). (4.21) Equation (4.21) reproduces (4.19) because the delta function changes∇to∇as well asE(r) toE(r).
1 Equation (4.16) also follows directly from (1.122).
2 Equation (4.20) isnotvalid for induced electric moments wherep∝Eand thus varies in space.
4.2 The Electric Dipole 97
4.2.4 The Dipole Torque
We can mimic the force calculation in (4.21) to find the Coulomb torque (3.4) exerted onρ(r) by a fieldEthat varies slowly in space:
N=
d3rr×E(r)ρD(r)=
d3rδ(r−r)(p· ∇)(r×E). (4.22) The delta function integration gives (p· ∇)(r×E). Therefore, because∇ir=δi, thekth component of the torque is3
Nk=pi∇ikmrEm=kmpi{δiEm+r∇iEm}. (4.23) Using the dipole forceFin (4.19), the vector form of (4.23) is
N=p×E+r×F. (4.24)
The first term in (4.24) is a torque which tends to rotateparound its center of mass into the direction ofE. The second term in (4.24) is a torque which tends to rotate the center of mass ofparound an origin defined by the tail of the position vectorr.
4.2.5 The Dipole Potential Energy
The potential energy of interaction betweenρ(r) and a slowly varying fieldE(r) can be calculated usingVEfrom (3.64) and the singular charge density (4.14). In detail,
VE(r)=
d3rϕ(r)ρD(r)= −
d3rϕ(r)p· ∇δ(r−r). (4.25) Whenpis constant, the (by now) familiar integration by parts yields
VE(r)= −p·E(r). (4.26)
A quick check uses (4.26) to compute the force and torque onp. We reproduce (4.20) immediately because
F= −∇VE= ∇(p·E). (4.27) To find the torque, we ask how (4.26) changes when the dipole momentp rotates rigidly by an infinitesimal amountδαabout its center.4The usual rules of mechanics give the change in the dipole moment as
δp=δα×p. (4.28)
Therefore, the change inVEis
δVE= −δp·E= −(δα×p)·E= −(p×E)·δα. (4.29) This confirms (4.24) becauseδVE = −N·δαdefines the torque.
3 Recall from (1.37) thatak=k mbcmis thekth Cartesian component ofa=b×c.
4 The right-hand rule determines the rotation sense by an angle|δα|with respect to an axis pointed in the direction ofδα.
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Figure 4.6: Fluorescence microscopic image (40µm×160µm) looking down onto a liquid mixture of cholesterol and a phospholipid trapped at an air-water interface parallel to the image. The black regions are long, thin “drops” of the liquid, with a thickness of exactly one molecule. Figure from Seul and Chen (1993).
4.2.6 The Dipole-Dipole Interaction
We can apply (4.26) to compute the total electrostatic energyUEof a collection ofNpre-existing point dipoles located at positionsr1,r2, . . . ,rN. As in the corresponding point charge problem (Section 3.6), it costs no energy to bring the first dipolep1into position. The work done byusto bringp2into position is exactly the interaction energy (4.25), with the electric field (4.10) ofp1playing the role ofEext:
W12= −p2·E1(r2)= 1 4π 0
p1·p2
|r2−r1|3 −3p1·(r2−r1)p2·(r2−r1)
|r2−r1|5
+
. (4.30)
Repeating the logic of the point charge example leads to the total energy:
UE =W= 1 4π 0
1 2
N i=1
N j=i
pi·pj
|ri−rj|3 −3pi·(ri−rj)pj·(ri−rj)
|ri−rj|5
+
. (4.31)
A more compact form of (4.31) makes use of the electric fieldE(ri) at the position of theithdipole produced by all theotherdipoles:
UE = −1 2
N i=1
pi·E(ri). (4.32)