18.303 Problem Set 7 Solutions
Problem 1: ((10+10)+2)
(a) We solve for g(r) in 3d as follows:
(i) For r > 0, −∇2 g + κ2 g = 0 and hence κ2 g = ∇2 g = 1r (rg)00 =⇒ h00 = κ2 h where h(r) = rg(r). The
solution to this is h(r) = ce−κr + de+κr for some constants c and d. However, since g(r) → 0 as r → ∞, we
must have d = 0 as otherwise the eκr /r term diverges. Hence we can write:
(
0
r≤
g(r) = lim ce−κr
→0+
r>
r
to solve the PDE for all r > 0.
(ii) Given an arbitrary test function q(x), we now evaluate
ˆ
(Âg){q} = g{Âq} = g Âq


ˆ ∞
ˆ 2π
ˆ π
( 
 2
g ∂2
((
(θ,(
φ(
derivatives
=
r2 dr
dφ
sin θ dθ 
(((( of q) 
κ gq − r ∂r2 (rq) + (
|
{z
} 
0
0
integrate to 0, from class


 
ˆ ∞ 
 
∂2
 


c κ2 rqe−κr − e−κr 2 [rq]  dr
sin θ dθ dφ 
∂r

 0
|
{z
} 
int. by parts


 
¨
ˆ

 
∞
∞
∂
∂
 2 −κr

 
− κe−κr [rq]  dr
=
sin θ dθ dφ  −ce−κr [rq] + c
κ rqe
∂r


0
0
|
{z∂r }  
int. by parts
¨
ˆ ∞
∞
(((
2
−κr((2(−κr
−κr
κ rqe
−κ e
[rq] dr
=
sin θ dθ dφ cq(0) − cκe
[rq] 0 + c
((((
¨
=
0
= 4πcq(0),
and hence c = 1/4π . Thus
0
G(x, x0 ) =
e−κ|x−x |
4π|x − x0 |
(with the value at x = x0 being irrelevant in the distribution sense, e.g. we can assign it to zero, since this
is a regular distribution with a finite integral, similar to what we showed in class).
(b) The κ → 0+ limit gives 1/4π|x − x0 | as in class, by inspection.
(I was thinking of assigning the 2d case, but the solution involves a modified Bessel function and seemed a bit much
for homework. However, the 2d case with κ = 0 is quite doable — I assigned it in last year’s pset 6, which you can
look at online.)
Problem 2: (10+10)
(a) The first derivative gives the regular distribution:

−1 0 < x < ∆x
1 
0
δ∆x
(x) =
+1 −∆x < x ≤ 0 ,
(∆x)2 

0
|x| > ∆x
1
(with the values at the jumps irrelevant in the distribution sense).1 The second derivative gives delta functions
at the jumps:
δ(x + ∆x) − 2δ(x) + δ(x − ∆x)
00
,
δ∆x
(x) =
(∆x)2
where the delta functions have amplitudes 1, −2, and 1 as the value jumps from 0 to +1 at x = −∆x, from +1
to −1 at x = 0, and from −1 to 0 at x = +∆x.
The derivative δ 00 (x) corresponds to δ 00 {φ} = δ 0 {−φ0 } = δ{φ00 } = φ00 (0) for a test function φ(x). Correspondingly,
00
δ∆x
{φ} =
φ(−∆x) − 2φ(0) + φ(+∆x)
,
(∆x)2
00
=
which is precisely a center-difference approximation that converges to φ00 (0) as ∆x → 0. Hence lim∆x→0+ δ∆x
00
δ as expected.
(b) In terms of p,
ρ(x) = p
δ(x − d/2) − δ(x + d/2)
,
d
and hence for any test function φ we obtain:
ρ{φ} = p
φ(d/2) − φ(−d/2)
→ pφ0 (0) = −pδ 0 {φ}
d
as d → 0 (this is just a center-difference approximation). Hence ρ → −pδ 0 : a point dipole is exactly a deltafunction derivative.
Problem 3: (10+5+10)
In the class notes on Green’s functions in inhomogeneous media, section 3.1, the solution of a “dipole” source p (e.g.
x−x0
the potential an electrostatic point dipole p) is given as Dp (x, x0 ) = p · ∇0 G0 (x, x0 ) = p · 4π|x−x
0 |3 . This is the solution
u0 (x) = Dp (x, x0 ) from a point dipole at x0 , i.e. −∇2 u0 = −p · ∇δ(x − x0 ).
(a) We know from the notes that the 1st-Born solution u1 to a delta-function source, −∇ · (c∇u1 ) = δ(x − x0 ), is
u1 (x) =
1
(x1 − x)
(x1 − x0 )
+ ln(1/c1 )
·
V.
3
4π|x − x0 |
4π|x1 − x| 4π|x1 − x0 |3
Now, just as in the notes for Dp , to find the first-Born solution u for −∇·(c∇u) = −p·∇δ(x−x0 ) = p·∇0 δ(x−x0 ),
we can just write u = p · ∇0 u1 , where ∇0 denotes the gradient with respect to x0 . We thereby obtain:
(x1 − x)
p
3p · (x1 − x0 )
u(x) = Dp (x, x0 ) −
· −
+ (x1 − x0 )
V ln(c1 ).
4π|x1 − x|3
4π|x1 − x0 |3
4π|x1 − x0 |5
This could be written as
u(x) = Dp (x, x0 ) + Dq (x, x1 )
where q is an effective dipole located at x1 with strength
p
3p · (x1 − x0 )
q= −
+
(x
−
x
)
V ln c1 = [−∇1 Dp (x1 , x0 )] V ln c1 ,
1
0
4π|x1 − x0 |3
4π|x1 − x0 |5
where the [· · · ] term, the gradient of Dp , could in electrostatics be interpreted as the “electric field” of the dipole
p at x1 , which “polarizes” the inhomogeneity V , where ∇1 denotes the gradient with respect to x1 .
0 {φ} =
we wanted to write this out more explicityly in terms of the distributional derivative definition, we would do δ∆x
h ´
i
h
i
´ ∆x
´0
´ ∆x
0
0
∆x
1
1
0
0
0
δ∆x {−φ } = (∆x)2 − −∆x (∆x + x)φ dx − 0 (∆x − x)φ dx = (∆x)2 (∆x + x)φ|−∆x + (∆x − x)φ|0 + −∆x φdx − 0 φdx =
h
(( ´ 0 φdx − ´ ∆x φdxi ,which is the same as the δ 0 distribution given above.
1
∆x φ(0)(
−(
∆x φ(0) + −∆x
∆x
0
(∆x)2 (((
1 If
2
(b) In the notation above, the change in u from V is the Dq term, so the interaction energy U is:
q
3q · (x0 − x1 )
U = p · ∇Dq |x=x0 = p ·
− (x0 − x1 )
,
4π|x0 − x1 |3
4π|x0 − x1 |5
where the derivative is exactly the same as the ∇1 Dp above (with p → q and x1 − x0 → x0 − x1 ). Note that
we are evaluating the electric field at the location of p, i.e. at x = x0 . We leave it in terms of q from above to
keep things simple for now, and substitute q below.
(c) First, we can simplify a bit. If we look at the products of p and q (proportional to p) in U , we will get a bunch
of terms pi pj for various components i and j of p. For a polarization rotating/fluctuating randomly in time
with zero mean, however, is is reasonable to suppose that the mean pi pj = 0 for i 6= j: the various components
of p must be uncorrelated if the orientation is random. That means we can focus on the p2i terms, which have
nonzero mean, one by one, thinking of them as dipoles oriented along one direction at a time.
Without loss of generality, we can take (x1 − x0 ) to lie along the x direction (this is just a choice of
coordinate systems). Then there are two cases: p2x terms corresponding to fluctuations when pk(x1 − x0 ), and
p2y or p2z terms corresponding to fluctuations p ⊥ (x1 −x0 ). Everything else is a linear combination of these terms.
When pk(x1 − x0 ), then 3(x1 − x0 )p · (x1 − x0 ) = 3p|x1 − x0 |2 and q simplifies to:
qk = 2
p
,
4π|x1 − x0 |3
and similarly for the qk(x1 − x0 ) terms in U . Hence we obtain
Uk = −
p2x
V ln(c1 ).
4π 2 |x1 − x0 |6
When p ⊥ (x1 − x0 ), the p · (x1 − x0 ) term vanishes and q simplifies to:
q⊥ = −
p
,
4π|x1 − x0 |3
and similalry for the q terims in U . Hence we obtain:
U⊥ = −
p2y + p2z
V ln(c1 ).
16π 2 |x1 − x0 |6
Therefore, the total interaction energy is
1
U =− 2
4π |x1 − x0 |6
p2x
p2y + p2z
+
4
!
V ln(c1 )
where the p2i terms represent the mean-square components of p.
Thus, when c1 > 1 and ln c1 > 0, we have U ∼ − d16 where d = |x1 − x0 |. Thus, U is smaller (more negative) for
smaller d and the force is attractive. Equivalently, the force in the radial (d) direction is Fd = −∂U/∂d ∼ −1/d7 ,
which is < 0 and hence attractive. This is the well-known 1/d7 scaling of the van der Waals force (more specifically, the London dispersion force).
3