Faraday’s Cage
Acoustic and Electric Faraday Cages
P. A. Martin
Colorado School of Mines
“Everybody knows Faraday’s cage or metal-lined room, into
which he went with electroscopes, birds, frogs, and other
instruments, arranged that the outside should receive
violent flashes, and detected nothing inside.”
Oliver Lodge, 1890
Acknowledgment: D. P. Hewett (Oxford)
Faraday’s Cage: how does it work?
A quick
gives
A Faraday cage’s operation depends on the fact that
an external static electrical field will cause the
electrical charges within the cage’s conducting
material to redistribute themselves so as to cancel the
field’s effects in the cage’s interior.
Faraday’s Cage: how does it work?
A quick
gives
A Faraday cage’s operation depends on the fact that
an external static electrical field will cause the
electrical charges within the cage’s conducting
material to redistribute themselves so as to cancel the
field’s effects in the cage’s interior.
This is the classic
explanation for a closed
conductor – but what is
the effect of the holes,
the gaps between the
wires in the cage?
We consider such problems for electrostatics (∇2 V = 0) and
for acoustics (∇2 u + k 2 u = 0).
First, we need a model . . .
Hertz’s “Mouse-Mill”: a Ring of Parallel Wires
Hertz’s “Mouse-Mill”: a Ring of Parallel Wires
b
Terminology and figure taken
from 1890 paper by O. Lodge.
Also reprinted in his 1892 book:
Terminology and figure taken
from 1890 paper by O. Lodge.
Also reprinted in his 1892 book:
O
N wires equally spaced around
a circle of radius b, the “ring”.
Hertz’s “Mouse-Mill”: a Ring of Parallel Wires
Solution for Complete Ring: Acoustics
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b
Terminology and figure taken
from 1890 paper by O. Lodge.
Also reprinted in his 1892 book:
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x = r cos θ,
O
N wires equally spaced around
a circle of radius b, the “ring”.
Does the solution for this
problem converge to solution
for the ring as N → ∞?
If so, what is the rate of
convergence?
Solve Helmholtz equation, ∇2 u + k 2 u = 0, for u(r , θ)
outside a circle (the “ring”) of radius b
∞
X
ikx
Solution: u(r , θ) = e +
cn Hn (kr )einθ
n=−∞
Hn = Hankel function
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First term is incident wave
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Summation gives effect of the ring: for “soft” ring
u=0
on circle r = b
cn = −i n [Jn (kb)]/[Hn (kb)], Jn = Bessel function
Does the solution for the cage, with N wires
equally spaced around the ring, approach
the solution for a single ring as N → ∞?
b
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Maxwell’s Infinite Grating: ∇2 V = 0
Electrostatic Solution for Complete Ring
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Solve Laplace’s equation, ∇2 V = 0, for V (x, y) outside a
circle (the “ring”) of radius b
p
b2 y
Solution: V (x, y) = y − 2 , r = x 2 + y 2
r
First term is uniform field at infinity
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Second term is effect of the conducting ring,
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V =0
Note exponential decay.
on circle r = b
W = AV + C is another solution.
Adjust constants A and C so that
Does the solution for the cage, with N wires
equally spaced around the ring, approach
the solution for a single ring as N → ∞?
b
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W has the desired behaviour as
Y →∞
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W ' 0 on small circles (wires) at
Y = 0, X = 2nπ.
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There are classical results for an infinite straight row of wires . . .
Acoustic Scattering by N Scatterers
e−iωt ,
354
Phys. perspect.
This is example of an inverse method.
Philip L. Taylor and William J. Fickinger
Leslie L. Foldy (1919–2001)
(∇2
+
k 2 )u
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N sound-soft scatterers in 2D,
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For N circles, this problem can be solved exactly using
multipole methods, but we are interested in N 1.
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1
log 1 − 2eY cos X + e 2Y
2
Y − e−Y cos X , Y → +∞
V ∼
−eY cos X ,
Y → −∞
V =
=0
Born László Földi in
Czechoslovakia.
Emigrated to USA, 1920
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BS in Physics, Case Western
Reserve University,
Cleveland, Ohio, 1941
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MS, Wisconsin, with Léon
Brillouin, 1942
We use Foldy’s method. For one scatterer at the origin,
basic assumption is that we can represent the total field as
u(r) = uinc (r) + guinc (0)H0 (k |r|)
(1)
(1)
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1942–1945: New York
uinc = incident field
H0 = H0 = Hankel function.
g = scattering coefficient. For circular scatterer, radius a,
can take g = −[J0 (ka)]/[H0 (ka)].
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PhD, Princeton, 1948, with
Oppenheimer
The assumption (1) says the scattering is isotropic. It is
justified for small soft (u = 0 on boundary) scatterers.
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Then, back to Case for entire
career
Fig. 5. Leslie Foldy (1919–2001) in 1974. Credit: Courtesy of Frank Cverna.
P.L. Taylor & W.J. Fickinger, Physics in Perspective 9 (2007) 346–356
lege to be his colleagues at Case Western Reserve University, where he remained for
Foldy in New York (1942–1945)
Foldy’s Method (1945)
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Underwater Sound Laboratory,
Division of War Research,
Columbia University
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64th floor, Empire State Building
[J. B. Keller also there]
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Problem: understand
transmission of sound through
submarine wakes, and echoes
received from wakes
Scattering by N identical small objects (cylinders, wires).
Assume each scatters isotropically:
u(r) = uinc (r) +
Cylinders centred at r = rj . The “exciting field” for n th cylinder
is incident field + field scattered by all the other cylinders:
Scattering by a Ring of Identical Small Scatterers
n th scatterer at θ = nh, where
h = 2π/N is angular spacing
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b
Foldy linear system is
N
X
Cn−j Aj = fn ,
j=1
C0 = −g −1 ,
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θ
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Aj H0 (k|r − rj |),
Solution
Recall system:
N
X
Cj = H0 (2kb| sin (jπ/N)|),
j 6= 0 mod N
The system matrix is a circulant matrix: we can solve for Aj
using discrete Fourier transforms . . .
n = 1, 2, . . . , N
j=1
N
X
Aj φ
mj
j=1
[Cj+mN = Cj , all m],
Cn−j Aj = fn ,
Discrete Fourier transform:
Ãm =
n = 1, 2, . . . , N
where fn = −uinc (rn ), Cj is N-periodic
N
X
(3)
Evaluate (3) at r = rn . As An = gun (rn ) for self-consistency, we
get linear N × N system for Aj .
Solve. Then put these numbers in (2) to get u(r).
We apply this method to a ring of scatterers . . .
Saturday 28 July 1945
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(2)
j=1
j6=n
Bubbly
liquids
behind
◦◦ ◦◦ ◦
Scatterers at rn , n = 1, 2, . . . , N.
Aj H0 (k|r − rj |)
j=1
un (r) = uinc (r) +
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N
X
Solution:
N
1X
Am =
Ãj φ−mj
N
φ = e2πi/N
j=1
Ãn = f̃n /C̃n
At centre of the ring:
u(0) = uinc (0) + Ã0 H0 (kb)
The far-field pattern is also expressible in terms of Ãn
Dependence on N: f̃n
Recall:
Ãn = f̃n /C̃n .
Trapezoidal Rule
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Start with f̃n .
Basic formula.
Z
Incident plane wave, uinc (r) = eikx = eikr cos θ . As fj = −uinc (rj ),
f̃n =
N
X
nj
fj φ = −
j=1
N
X
2π
0
e
ikb cos (jh)
e
Let h = 2π/N.
N−1
X
h
h
F (θ) dθ ' F (0) + h
F (jh) + F (2π)
2
2
j=1
injh
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j=1
When
F (0) = F (2π),
Z
where h = 2π/N and φ = eih .
0
RHS looks like (repeated) trapezoidal rule:
N
X
f̃n
=h
F (jh),
N
j=1
2π
1
F (θ) = − eikb cos θ einθ
2π
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F (θ) dθ ' h
N
X
F (jh)
j=1
When F is 2π-periodic and smooth, the error goes down
exponentially fast as N increases.
L.N. Trefethen and J.A.C. Weideman, The exponentially
convergent trapezoidal rule. SIAM Review 56 (2014) 385–458.
Dependence on N: f̃n
Dependence on N: f̃n
Incident plane wave, uinc (r) = eikx = eikr cos θ . As fj = −uinc (rj ),
f̃n =
N
X
nj
fj φ = −
j=1
N
X
e
ikb cos (jh)
e
injh
j=1
f̃n =
j=1
where h = 2π/N and φ = eih .
RHS looks like (repeated) trapezoidal rule:
N
X
f̃n
F (jh),
=h
N
Incident plane wave, uinc (r) = eikx = eikr cos θ . As fj = −uinc (rj ),
1
F (θ) = − eikb cos θ einθ
2π
N
X
nj
fj φ = −
j=1
N
X
eikb cos (jh) einjh
j=1
where h = 2π/N and φ = eih .
RHS looks like (repeated) trapezoidal rule:
N
X
f̃n
F (jh),
=h
N
j=1
F (θ) = −
1 ikb cos θ inθ
e
e
2π
As F (θ) is entire 2π-periodic function of θ, the trapezoidal rule
converges exponentially fast,
f̃n
∼
N
Z
0
2π
F (θ) dθ = −i n Jn (kb),
N → ∞.
Dependence on N: C̃n
Recall:
Ãn = f̃n /C̃n .
N−1
X
1
C̃n
=h
v (jh) −
,
N
gN
j=1
Dependence on N: C̃n
Next, C̃n :
Recall:
Next, C̃n :
N−1
X
1
C̃n
=h
v (jh) −
,
N
gN
einθ
v (θ) =
H0 (2kb | sin (θ/2)|)
2π
v (θ) is 2π-periodic but there are log singularities at θ = 0, 2π.
The sum looks like the trapezoidal rule in which the endpoint
contributions have been “ignored”.
Ãn = f̃n /C̃n .
j=1
v (θ) =
einθ
H0 (2kb | sin (θ/2)|)
2π
v (θ) is 2π-periodic but there are log singularities at θ = 0, 2π.
The sum looks like the trapezoidal rule in which the endpoint
contributions have been “ignored”. Fortunately (for me), the
convergence of such quadrature rules has been studied.
A theorem of A. Sidi (2004) shows that the sum converges to
Z
0
2π
v (θ) dθ = Jn (kb) Hn (kb),
but the error is O(N −1 log N): it’s a very poor quadrature rule!
In fact, Sidi gives an asymptotic expansion for the error: the
error estimate is sharp.
Dependence on N: Ãn
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For fixed n,
− in
log N
Ãn =
+O
,
Hn (kb)
N
Concluding Remarks
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The slow convergence with N is surprising (to me).
It contrasts with the exponential convergence with distance
seen with Maxwell’s grating.
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We have made similar calculations for potential problems
(Laplace’s equation). This required a (new?) static version
of Foldy’s method.
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Extension to sound-hard scatterers would require a
modified form of Foldy’s method, with dipole contributions.
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Direct extension to three dimensions is not possible
because there is no analogous circulant structure.
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Reference: Proc. Roy. Soc. A 470 (2014)
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Related work by Chapman, Hewett & Trefethen, using very
different methods, on potential problems.
Finally: Ãn = f̃n /C̃n .
N → ∞.
The leading-order term shows that we recover the
expected far-field pattern for scattering by a sound-soft
circle of radius b.
The total field at the centre of the ring vanishes slowly, as
N −1 log N.