Calculus 101 to a Nobel Prize in under 50 minutes William J. Parnell Faculty Research Fellow School of Mathematics, University of Manchester, UK [email protected]http://www.maths.man.ac.uk/∼wparnell. W.J.Parnell, School of Mathematics, University of Manchester. – p.1/51
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Calculus 101 to a Nobel Prize in under 50 minutes
William J. Parnell
Faculty Research Fellow
School of Mathematics, University of Manchester, UK
W.J.Parnell, School of Mathematics, University of Manchester. – p.12/51
Scattering
Now consider boundary conditions!
Uinc
r = a
What happens when an incidentwave
Uinc = exp(ikx) = exp(ikr cos θ)
is incident on a circular (cylindri-cal) cavity where
∂U
∂r= 0 on r = a
W.J.Parnell, School of Mathematics, University of Manchester. – p.13/51
First, we can use the so-called generating function
exp
(
kr
2(t − 1/t)
)
=∞∑
n=−∞
tnJn(kr)
with t = i exp(iθ) to give (Jacobi-Anger identity)
exp(ikr cos θ) =∞∑
n=∞
inJn(kr)einθ
Now write U = Us + Uinc with
Us =∞∑
n=∞
in(AnH(1)n (kr) + BnH(2)
n (kr))einθ
but obviously the scattered field will be outgoing so we choose
Bn = 0
W.J.Parnell, School of Mathematics, University of Manchester. – p.14/51
SolutionThe boundary condition
∂U
∂r= 0 on r = a
translates into
∞∑
n=−∞
in (J ′n(kr) + AnH ′
n(kr))∣
∣
∣
r=aeinθ = 0
so that we can satisfy this with An = −J ′n(ka)/H ′
n(ka), i.e.
Us =∞∑
n=∞
−inJ ′
n(ka)
H ′n(ka)
Hn(kr)einθ
so the total field is
U = eikx −∞∑
n=∞
inJ ′
n(ka)
H ′n(ka)
Hn(kr)einθ
W.J.Parnell, School of Mathematics, University of Manchester. – p.15/51
Low frequency scattering
Note that is ǫ = ak ≪ 1, we have
Us ∼ A0H0(kr) + A1H1(kr) cos θ
where
A0 =ǫ2π
4iA1 = −
ǫ2π
4i
W.J.Parnell, School of Mathematics, University of Manchester. – p.16/51
Multiple Scattering
W.J.Parnell, School of Mathematics, University of Manchester. – p.17/51
2D Composite Slab
x1
x2
h−h
D
θinc
O
pspj
rs
rj
θs
θj
θjs
Rjs
W.J.Parnell, School of Mathematics, University of Manchester. – p.18/51
Multiple scattering problem
Scale lengthscales on host wavenumber k0.So ǫ = ak0 is radius of fibres..
(∇2 + 1)U = 0
and boundary conditions
∂U
∂r= 0,
on rj = ǫ, j = 1, 2, ..., N .For ease of exposition let’s consider normal incidence, θinc = 0.
W.J.Parnell, School of Mathematics, University of Manchester. – p.19/51
Solutions
U = Uinc +N
∑
j=1
∞∑
n=−∞
AjnZnH(1)
n (rj)einθj , in the host phase,
Near jth inclusion, incident wave is (Jacobi-Anger identity)
Uinc(x, y) = eix =∞∑
n=−∞
inJn(rj)einθj
where (rj , θj) are local polar coordinates with origin (pj , qj).Hence solution in vicinity of sth inclusion is
W.J.Parnell, School of Mathematics, University of Manchester. – p.20/51
Solutions
U = Uinc +N
∑
j=1
∞∑
n=−∞
AjnZnH(1)
n (rj)einθj , in the host phase,
Near jth inclusion, incident wave is (Jacobi-Anger identity)
Uinc(x, y) = eix =∞∑
n=−∞
inJn(rj)einθj
where (rj , θj) are local polar coordinates with origin (pj , qj).Hence solution in vicinity of sth inclusion is
U(rs, θs) =∞∑
n=−∞
(
inJn(rs)einθs + As
nZnH(1)n (rs)e
inθs
)
+N
∑
j=1,j 6=s
∞∑
n=−∞
AjnZnH(1)
n (rj)einθj
W.J.Parnell, School of Mathematics, University of Manchester. – p.20/51
Graf’s addition theorem
Graf’s addition theorem says that in vicinity of jth inclusion:
H(1)n (rj)e
inθj =∞∑
m=−∞
H(1)n−m(|rj − rs|)e
i(n−m)θjsJm(rs)eimθs
and thus
W.J.Parnell, School of Mathematics, University of Manchester. – p.21/51
Graf’s addition theorem
Graf’s addition theorem says that in vicinity of jth inclusion:
H(1)n (rj)e
inθj =∞∑
m=−∞
H(1)n−m(|rj − rs|)e
i(n−m)θjsJm(rs)eimθs
and thus
U(rs, θs) =∞∑
n=−∞
(
inJn(rs)einθs + As
nZnH(1)n (rs)e
inθs
)
+
N∑
j=1,j 6=s
∞∑
n=−∞
AjnZn
∞∑
m=−∞
H(1)n−m(Rjs)e
i(n−m)θjsJm(rs)eimθs
W.J.Parnell, School of Mathematics, University of Manchester. – p.21/51
Application of boundary conditions and orthogonality thus gives
Asm +
N∑
j=1,j 6=s
∞∑
n=−∞
AjnZnHn−m(Rjs)e
i(n−m)θjs = −im
for m ∈ Z.
Various techniques have been developed in order to speed up thesummation of these series in order to calculate An.
W.J.Parnell, School of Mathematics, University of Manchester. – p.22/51
"Effective" SH Waves
When ǫ ≪ 1 an effective or homogenized wave propagates through theinhomogeneous medium and it can be shown that this wave is of theform
U = exp(ik∗x)
where
k2∗ =
ρ∗µ∗
ω2
and thus
ρ∗ = (1 − φ)ρ,
µ∗ =(1 − φ)
(1 + φ)µ.
W.J.Parnell, School of Mathematics, University of Manchester. – p.23/51
Wave propagation throughinhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.24/51
Wave Propagation through inhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.25/51
Wave Propagation through inhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.26/51
Wave Propagation through inhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.27/51
Wave Propagation through inhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.28/51
Wave Propagation through inhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.29/51
Wave Propagation through inhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.30/51
Wave Propagation through inhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.31/51
Wave Propagation through inhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.32/51
Wave Propagation through inhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.33/51
Anderson Localization - the Nobel Prize
W.J.Parnell, School of Mathematics, University of Manchester. – p.34/51
Anderson Localization
• Anderson was awarded the Nobel Prize in 1977.• This work was related to transport of electrons.• Change in behaviour from diffusive regime (in which Ohm’s law
holds) to a localized state where the material behaves as aninsulator. Effect due to interaction of electrons which haveundergone multiple scattering.
• Explained Metal-Insulator transitions.• 1980’s: what about classical wave physics (light, ultrasound, etc.)
W.J.Parnell, School of Mathematics, University of Manchester. – p.35/51
I still don’t really understand all of this........
For classical systems, essentially the argument is:• Propagation of light through inhomogeneous media can be
described as a diffusion process• For diffusion of light through a disordered medium, the same
Ohm’s law holds as for diffusion of electrons through a resistor:transmission decreases linearly with system length
• Anderson localization brings diffusion to a halt: transmissiondecreases exponentially rather than linearly with system length
• This transition requires very strong scattering
W.J.Parnell, School of Mathematics, University of Manchester. – p.36/51
Localization of light
λ = 1064nm, three powder types.
• Electromagnetic propagation through very strongly scatteringsemiconductor powders.
• Powder is Gallium Arsenide (GaAs).• Very little absorption.
W.J.Parnell, School of Mathematics, University of Manchester. – p.37/51
localization of light
W.J.Parnell, School of Mathematics, University of Manchester. – p.38/51
Localization of ultrasound
W.J.Parnell, School of Mathematics, University of Manchester. – p.39/51
Link with multiple scattering theory
• Clearly there must be one!• However it is still unclear how this effect links in with the theory
described here, it particular the lengthscale effects described.• In particular the sample length effect.
W.J.Parnell, School of Mathematics, University of Manchester. – p.40/51
Pre-stressed inhomogeneous media
W.J.Parnell, School of Mathematics, University of Manchester. – p.41/51
Problem description
aA
Undeformed Deformed
p∞
p∞
p∞
p∞
p∞
p∞
p∞p∞
W.J.Parnell, School of Mathematics, University of Manchester. – p.42/51
Problem description
a
p∞
p∞
p∞
p∞
p∞
p∞
p∞p∞
exp iΓx1
W.J.Parnell, School of Mathematics, University of Manchester. – p.43/51
Radial stressOnly 1 equilbrium equation:
dΣrr
dr+
1
r(Σrr − Σθθ) = 0
which can be integrated to give Σrr:
2 4 6 8 10
-5
-4
-3
-2
-1
rΣrr
p∞µ
= 1
p∞µ
= 2
p∞µ
= 4
W.J.Parnell, School of Mathematics, University of Manchester. – p.44/51
Incremental deformationSuperpose small amplitude waves on the finite deformation:
u = u + ηu′
where u is the finite displacement, η ≪ 1 and
u′ = (0, 0, w(r, θ)) exp(iωt).
W.J.Parnell, School of Mathematics, University of Manchester. – p.45/51
Incremental deformationSuperpose small amplitude waves on the finite deformation:
u = u + ηu′
where u is the finite displacement, η ≪ 1 and
u′ = (0, 0, w(r, θ)) exp(iωt).
Transpires that the modified wave equation is:
(
1 +k
r2
)
∂2w
∂r2+
1
r
(
1 −k
r2
)
∂w
∂r+
1
r2
(
1 −k
r2 + K
)
∂2w
∂θ2+ γ2w = 0.
Can think of this as:
(∇2 + Lk + γ2)w = 0
where Lk = 0 for no pre-stress.W.J.Parnell, School of Mathematics, University of Manchester. – p.45/51
Low Frequency Scattering
Once again assume a low frequency plane incident wave incomingfrom infinity:
winc = eiγr cos θ
so that the scattered wave is described by:
(∇2 + Lk + γ2)ws + I(r, θ) = 0
W.J.Parnell, School of Mathematics, University of Manchester. – p.46/51
Matched Asymptotics - Schematic
Scaling lengthscales on either a or γ gives inner and outer problems.
W.J.Parnell, School of Mathematics, University of Manchester. – p.47/51
Matching
Outer solution:F0(ro) = ǫ2
(
B0H(1)0 (ro) + E0(ro)
)
+ o(r2o)
Inner solution:
f0(ri) = ǫ2(1
4(1 + k) log(r2
i + k) + C0 log ǫ + D0
)
+ o(ǫ2)
Van Dyke’s matching principle says that f(1,1)0 = F
(1,1)0 which gives the
Modified Monopole Scattering Coefficient due to the pre-stress:
B0 =πǫ204i
[
a2L
A2(1 + (L − 1)S1)
(
1 +(A2 − a2)LS1
a2(1 + (L − 1)S1)
)]
W.J.Parnell, School of Mathematics, University of Manchester. – p.48/51
Monopole Scattering CoefficientB0
-3 -2 -1 1 2 3
1.0
1.5
2.0
p∞/µ
B0/Γ2
S1 = 1
S1 = 0.9
S1 = 0.8
W.J.Parnell, School of Mathematics, University of Manchester. – p.49/51
Dipole Scattering Coefficient:B1:
-3 -2 -1 0 1 2 3
0.5
1.0
1.5
2.0
p∞/µ
|B1|/Γ2
S1 = 1
S1 = 0.9
S1 = 0.8
W.J.Parnell, School of Mathematics, University of Manchester. – p.50/51
Effective shear propertiesWe can use this canonical problem in multiple scattering theory inorder to deduce the effective response of a pre-stressedinhomogeneous medium:
0.1 0.2 0.3 0.4
0.5
0.6
0.7
0.8
0.9
1.0
µ∗
φ0
p∞/µ = 1
p∞/µ = 0
p∞/µ = −1
W.J.Parnell, School of Mathematics, University of Manchester. – p.51/51