The content of these notes is based on the lectures by Prof. Graeme W. Milton (University of Utah) given in a course on metamaterials in Spring 2007.
Recall from the previous lecture that the average rate of work done in a cycle
of oscillation of material with frequency dependent mass is
P
=
ω
2
π
∫
0
2
π
/
ω
F
(
t
)
⋅
V
(
t
)
d
t
=
Re
(
F
^
)
⋅
Re
(
V
^
)
+
Im
(
F
^
)
⋅
Im
(
V
^
)
2
=
ω
[
Re
(
V
^
)
⋅
Im
[
M
(
ω
)
]
⋅
Re
(
V
^
)
+
Im
(
V
^
)
⋅
Im
[
M
(
ω
)
]
⋅
Im
(
V
^
)
]
.
{\displaystyle {\begin{aligned}{\mathcal {P}}&={\cfrac {\omega }{2~\pi }}\int _{0}^{2\pi /\omega }~\mathbf {F} (t)\cdot \mathbf {V} (t)~{\text{d}}t\\&={\cfrac {{\text{Re}}({\widehat {\mathbf {F} }})\cdot {\text{Re}}({\widehat {\mathbf {V} }})+{\text{Im}}({\widehat {\mathbf {F} }})\cdot {\text{Im}}({\widehat {\mathbf {V} }})}{2}}\\&=\omega ~[{\text{Re}}({\widehat {\mathbf {V} }})\cdot {\text{Im}}[{\boldsymbol {M}}(\omega )]\cdot {\text{Re}}({\widehat {\mathbf {V} }})+{\text{Im}}({\widehat {\mathbf {V} }})\cdot {\text{Im}}[{\boldsymbol {M}}(\omega )]\cdot {\text{Im}}({\widehat {\mathbf {V} }})].\end{aligned}}}
This quadratic form will be non-negative for all choices of
V
^
{\displaystyle {\widehat {\mathbf {V} }}}
if
and only if
Im
(
M
(
ω
)
)
{\displaystyle {\text{Im}}({\boldsymbol {M}}(\omega ))}
is positive semidefinite for all real
ω
>
0
{\displaystyle \omega >0}
. Therefore, a restriction on the behavior of such materials
is that
Im
[
M
(
ω
)
]
≥
0
.
{\displaystyle {{\text{Im}}[{\boldsymbol {M}}(\omega )]\geq 0~.}}
Similarly, for electrodynamics, the average power dissipated into heat
is given by
(1)
P
=
ω
2
π
∫
0
2
π
/
ω
[
E
(
t
)
⋅
∂
D
(
t
)
∂
t
+
H
(
t
)
⋅
∂
B
(
t
)
∂
t
]
d
t
.
{\displaystyle {\text{(1)}}\qquad {\mathcal {P}}={\cfrac {\omega }{2~\pi }}\int _{0}^{2\pi /\omega }~\left[\mathbf {E} (t)\cdot {\frac {\partial \mathbf {D} (t)}{\partial t}}+\mathbf {H} (t)\cdot {\frac {\partial \mathbf {B} (t)}{\partial t}}\right]~dt~.}
In this case, the quantity
E
{\displaystyle \mathbf {E} }
is equivalent to the voltage and the
quantity rate of change of electrical displacement
∂
D
/
∂
t
{\displaystyle \partial \mathbf {D} /\partial t}
is equivalent to the current (recall that in electrostatics the power is
given by
P
=
V
I
{\displaystyle {\mathcal {P}}=V~I}
). In addition, we also have a contribution due to
magnetic induction.
Let us assume that the fields can be expressed in harmonic form, i.e.,
E
(
t
)
=
Re
[
E
^
e
−
i
ω
t
]
;
H
(
t
)
=
Re
[
H
^
e
−
i
ω
t
]
{\displaystyle \mathbf {E} (t)={\text{Re}}[{\widehat {\mathbf {E} }}~e^{-i\omega t}]~;~~\mathbf {H} (t)={\text{Re}}[{\widehat {\mathbf {H} }}~e^{-i\omega t}]}
or equivalently as
E
(
t
)
=
Re
(
E
^
)
cos
(
ω
t
)
+
Im
(
E
^
)
sin
(
ω
t
)
;
H
(
t
)
=
Re
(
H
^
)
cos
(
ω
t
)
+
Im
(
H
^
)
sin
(
ω
t
)
.
{\displaystyle \mathbf {E} (t)={\text{Re}}({\widehat {\mathbf {E} }})~\cos(\omega t)+{\text{Im}}({\widehat {\mathbf {E} }})~\sin(\omega t)~;~~\mathbf {H} (t)={\text{Re}}({\widehat {\mathbf {H} }})~\cos(\omega t)+{\text{Im}}({\widehat {\mathbf {H} }})~\sin(\omega t)~.}
Also, recall that,
∂
D
∂
t
=
∇
×
H
=
−
i
ω
ϵ
⋅
E
(
t
)
and
∂
B
∂
t
=
−
∇
×
E
=
−
i
ω
μ
⋅
H
(
t
)
.
{\displaystyle {\frac {\partial \mathbf {D} }{\partial t}}={\boldsymbol {\nabla }}\times \mathbf {H} =-i\omega ~{\boldsymbol {\epsilon }}\cdot \mathbf {E} (t)\qquad {\text{and}}\qquad {\frac {\partial \mathbf {B} }{\partial t}}=-{\boldsymbol {\nabla }}\times \mathbf {E} =-i\omega ~{\boldsymbol {\mu }}\cdot \mathbf {H} (t)~.}
Therefore, for real
ω
{\displaystyle \omega }
and real
P
{\displaystyle {\mathcal {P}}}
, we can write equation
(1) as (with the substitution
z
=
ω
t
{\displaystyle z=\omega t}
),
P
=
ω
2
π
∫
0
2
π
[
Re
(
E
^
)
cos
z
+
Im
(
E
^
)
sin
z
]
⋅
{
Im
(
ϵ
)
⋅
[
Re
(
E
^
)
cos
z
+
Im
(
E
^
)
sin
z
]
}
+
[
Re
(
H
^
)
cos
z
+
Im
(
H
^
)
sin
z
]
⋅
{
Im
(
μ
)
⋅
[
Re
(
H
^
)
cos
z
+
Im
(
H
^
)
sin
z
]
}
d
z
{\displaystyle {\begin{aligned}{\mathcal {P}}={\cfrac {\omega }{2~\pi }}\int _{0}^{2\pi }~&\left[{\text{Re}}({\widehat {\mathbf {E} }})~\cos z+{\text{Im}}({\widehat {\mathbf {E} }})~\sin z\right]\cdot \left\{{\text{Im}}({\boldsymbol {\epsilon }})\cdot \left[{\text{Re}}({\widehat {\mathbf {E} }})~\cos z+{\text{Im}}({\widehat {\mathbf {E} }})~\sin z\right]\right\}+\\&\left[{\text{Re}}({\widehat {\mathbf {H} }})~\cos z+{\text{Im}}({\widehat {\mathbf {H} }})~\sin z\right]\cdot \left\{{\text{Im}}({\boldsymbol {\mu }})\cdot \left[{\text{Re}}({\widehat {\mathbf {H} }})~\cos z+{\text{Im}}({\widehat {\mathbf {H} }})~\sin z\right]\right\}~{\text{d}}z\end{aligned}}}
Expanding out, and using the fact that
∫
0
2
π
cos
2
z
d
z
=
∫
0
2
π
sin
2
z
d
z
=
π
and
∫
0
2
π
sin
(
2
z
)
d
z
=
0
{\displaystyle \int _{0}^{2\pi }\cos ^{2}z~{\text{d}}z=\int _{0}^{2\pi }\sin ^{2}z~{\text{d}}z=\pi \quad {\text{and}}\quad \int _{0}^{2\pi }\sin(2z)~{\text{d}}z=0}
we have,
(2)
P
=
ω
2
[
Re
(
E
^
)
⋅
Im
(
ϵ
)
⋅
Re
(
E
^
)
+
Im
(
E
^
)
⋅
Im
(
ϵ
)
⋅
Im
(
E
^
)
+
Re
(
H
^
)
⋅
Im
(
μ
)
⋅
Re
(
H
^
)
+
Im
(
H
^
)
⋅
Im
(
μ
)
⋅
Im
(
H
^
)
]
.
{\displaystyle {\text{(2)}}\qquad {\mathcal {P}}={\cfrac {\omega }{2}}\left[{\text{Re}}({\widehat {\mathbf {E} }})\cdot {\text{Im}}({\boldsymbol {\epsilon }})\cdot {\text{Re}}({\widehat {\mathbf {E} }})+{\text{Im}}({\widehat {\mathbf {E} }})\cdot {\text{Im}}({\boldsymbol {\epsilon }})\cdot {\text{Im}}({\widehat {\mathbf {E} }})+{\text{Re}}({\widehat {\mathbf {H} }})\cdot {\text{Im}}({\boldsymbol {\mu }})\cdot {\text{Re}}({\widehat {\mathbf {H} }})+{\text{Im}}({\widehat {\mathbf {H} }})\cdot {\text{Im}}({\boldsymbol {\mu }})\cdot {\text{Im}}({\widehat {\mathbf {H} }})\right]~.}
Since
ω
≥
0
{\displaystyle \omega \geq 0}
and the power
P
≥
0
{\displaystyle {\mathcal {P}}\geq 0}
, the quadratic forms in
equation (2) require that
Im
(
ϵ
)
>
0
and
Im
(
μ
)
>
0
.
{\displaystyle {\text{Im}}({\boldsymbol {\epsilon }})>0\qquad {\text{and}}\qquad {\text{Im}}({\boldsymbol {\mu }})>0~.}
Note that if the permittivity is expressed as
ϵ
=
ϵ
0
+
i
σ
ω
{\displaystyle {\boldsymbol {\epsilon }}={\boldsymbol {\epsilon }}_{0}+i{\cfrac {\boldsymbol {\sigma }}{\omega }}}
the requirement
Im
(
ϵ
)
>
0
{\displaystyle {\text{Im}}({\boldsymbol {\epsilon }})>0}
implies that the conductivity
σ
>
0
{\displaystyle {\boldsymbol {\sigma }}>0}
.
Therefore, if the conductivity is greater than zero, there will be
dissipation.
Brief introduction to elastodynamics
edit
A concise introduction to the theory of elasticity can be found in
Atkin80 . In this section, we consider the linear theory of
elasticity for infinitesimal strains and small displacements.
Consider the body (
Ω
{\displaystyle \Omega }
) shown in Figure~1. Let
Γ
{\displaystyle \Gamma }
be a subpart of the body (in the interior of
Ω
{\displaystyle \Omega }
or sharing a
part of the surface of
Ω
{\displaystyle \Omega }
). Postulate the existence of a force
t
(
x
,
n
)
{\displaystyle \mathbf {t} (\mathbf {x} ,\mathbf {n} )}
per unit area on the surface of
Γ
{\displaystyle \Gamma }
where
n
{\displaystyle \mathbf {n} }
is
the outward unit normal to the surface of
Γ
{\displaystyle \Gamma }
. Then
t
{\displaystyle \mathbf {t} }
is the
force exerted on
Γ
{\displaystyle \Gamma }
by the material outside
Γ
{\displaystyle \Gamma }
or by surface
tractions.
Figure 1. Illustration of the concept of stress.
From the balance of forces on a small tetrahedron (
Γ
{\displaystyle \Gamma }
), we can show
that
t
(
x
,
n
)
{\displaystyle \mathbf {t} (\mathbf {x} ,\mathbf {n} )}
is linear in
n
{\displaystyle \mathbf {n} }
. Therefore,
t
=
σ
⋅
n
{\displaystyle \mathbf {t} ={\boldsymbol {\sigma }}\cdot \mathbf {n} }
where
σ
{\displaystyle {\boldsymbol {\sigma }}}
is a second-order tensor called the stress tensor .
Since the tetrahedron cannot rotate at infinite velocity as its size goes
to zero (conservation of angular momentum), we can show that the stress
tensor is symmetric, i.e.,
σ
=
σ
T
.
{\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{T}~.}
In particular, for a fluid,
σ
=
−
p
1
{\displaystyle {\boldsymbol {\sigma }}=-p~{\boldsymbol {\mathit {1}}}}
where
p
{\displaystyle p}
is the pressure.
Let us assume that the stress depends only on the strain (and not on strain
gradients or strain rates), where the strain is defined as
(3)
ϵ
=
1
2
[
∇
u
+
(
∇
u
)
T
]
.
{\displaystyle {\text{(3)}}\qquad {\boldsymbol {\epsilon }}={\frac {1}{2}}\left[{\boldsymbol {\nabla }}\mathbf {u} +({\boldsymbol {\nabla }}\mathbf {u} )^{T}\right]~.}
Here
u
(
x
,
t
)
{\displaystyle \mathbf {u} (\mathbf {x} ,t)}
is the displacement field. Note that a gradient of the
displacement field is used to define the strain because rigid body motions
should not affect
σ
{\displaystyle {\boldsymbol {\sigma }}}
and a rigid body rotation gives zero strains
(for small displacements).
Assume that
σ
{\displaystyle {\boldsymbol {\sigma }}}
depends linearly on
ϵ
{\displaystyle {\boldsymbol {\epsilon }}}
so that
σ
(
x
,
t
)
=
∫
d
x
′
[
∫
K
ε
(
x
,
x
′
,
t
′
−
t
)
:
ϵ
(
x
,
t
′
)
d
t
′
]
.
{\displaystyle {\boldsymbol {\sigma }}(\mathbf {x} ,t)=\int {\text{d}}\mathbf {x} '~\left[\int {\boldsymbol {\mathsf {K}}}_{\varepsilon }(\mathbf {x} ,\mathbf {x} ',t'-t):{\boldsymbol {\epsilon }}(\mathbf {x} ,t')~{\text{d}}t'\right]~.}
Note that this assumption ignores preexisting internal stresses such as
those found in prestressed concrete. If the material can be approximated
as being local, then
(4)
σ
(
x
,
t
)
=
∫
K
ε
(
x
,
t
′
−
t
)
:
ϵ
(
x
,
t
′
)
d
t
′
.
{\displaystyle {\text{(4)}}\qquad {\boldsymbol {\sigma }}(\mathbf {x} ,t)=\int {\boldsymbol {\mathsf {K}}}_{\varepsilon }(\mathbf {x} ,t'-t):{\boldsymbol {\epsilon }}(\mathbf {x} ,t')~{\text{d}}t'~.}
Taking the Fourier transforms of equation (4), we get
(5)
σ
^
(
x
,
ω
)
=
C
(
x
,
ω
)
:
ϵ
^
(
x
,
ω
)
{\displaystyle {\text{(5)}}\qquad {\widehat {\boldsymbol {\sigma }}}(\mathbf {x} ,\omega )={\boldsymbol {\mathsf {C}}}(\mathbf {x} ,\omega ):{\widehat {\boldsymbol {\epsilon }}}(\mathbf {x} ,\omega )}
where
C
(
x
,
ω
)
=
∫
K
ε
(
x
,
τ
)
e
−
i
ω
τ
d
τ
;
τ
=
t
′
−
t
.
{\displaystyle {\boldsymbol {\mathsf {C}}}(\mathbf {x} ,\omega )=\int {\boldsymbol {\mathsf {K}}}_{\varepsilon }(\mathbf {x} ,\tau )~e^{-i\omega \tau }~{\text{d}}\tau ~;~~\tau =t'-t~.}
In index notation, equation (5) can be written as
σ
^
i
j
=
C
i
j
k
l
ϵ
^
k
l
.
{\displaystyle {\widehat {\sigma }}_{ij}=C_{ijkl}~{\widehat {\epsilon }}_{kl}~.}
Causality implies that stresses at time
t
{\displaystyle t}
can only depend on strains of
previous times, i.e., if
t
′
≤
t
{\displaystyle t'\leq t}
or
τ
≤
0
{\displaystyle \tau \leq 0}
. Therefore,
K
ε
(
x
,
τ
)
=
0
if
τ
>
0
.
{\displaystyle {\boldsymbol {\mathsf {K}}}_{\varepsilon }(\mathbf {x} ,\tau )={\boldsymbol {\mathsf {0}}}\qquad {\text{if}}\qquad \tau >0~.}
This in turn implies that the integral converges only if
Im
(
ω
)
>
0
{\displaystyle {\text{Im}}(\omega )>0}
,
i.e.,
C
(
x
,
ω
)
{\displaystyle {\boldsymbol {\mathsf {C}}}(\mathbf {x} ,\omega )}
is analytic when
Im
(
ω
)
>
0
{\displaystyle {\text{Im}}(\omega )>0}
.
In the absence of body forces, the equation of motion of the body can be
written as
(6)
∇
⋅
σ
=
ρ
(
x
)
∂
2
u
∂
t
2
{\displaystyle {\text{(6)}}\qquad {\boldsymbol {\nabla }}\cdot {\boldsymbol {\sigma }}=\rho (\mathbf {x} )~{\frac {\partial ^{2}\mathbf {u} }{\partial t^{2}}}}
where
ρ
{\displaystyle \rho }
is the mass density,
∇
⋅
σ
{\displaystyle {\boldsymbol {\nabla }}\cdot {\boldsymbol {\sigma }}}
is the internal force
per unit volume, and
∂
2
u
/
∂
t
2
{\displaystyle \partial ^{2}\mathbf {u} /\partial t^{2}}
is the acceleration.
Hence, this is just the expression of Newton's second law for continuous
systems.
For a material which has a frequency dependent mass, equation
(6) may be written as
(7)
∇
⋅
σ
=
∫
−
∞
∞
ρ
(
x
,
t
−
t
′
)
∂
2
u
∂
t
2
d
t
′
{\displaystyle {\text{(7)}}\qquad {\boldsymbol {\nabla }}\cdot {\boldsymbol {\sigma }}=\int _{-\infty }^{\infty }\rho (\mathbf {x} ,t-t')~{\frac {\partial ^{2}\mathbf {u} }{\partial t^{2}}}~{\text{d}}t'}
where causality implies that if
t
′
>
t
{\displaystyle t'>t}
then
ρ
=
0
{\displaystyle \rho =0}
.
Taking the Fourier transform of equation (7), we get
(8)
∇
⋅
σ
^
(
x
,
ω
)
=
−
ω
2
ρ
^
(
x
,
ω
)
u
^
(
x
,
ω
)
.
{\displaystyle {\text{(8)}}\qquad {\boldsymbol {\nabla }}\cdot {\widehat {\boldsymbol {\sigma }}}(\mathbf {x} ,\omega )=-\omega ^{2}~{\widehat {\rho }}(\mathbf {x} ,\omega )~{\widehat {\mathbf {u} }}(\mathbf {x} ,\omega )~.}
Substituting equation (5) into equation (8)
we get
(9)
∇
⋅
(
C
:
ϵ
^
)
+
ω
2
ρ
^
u
^
=
0
.
{\displaystyle {\text{(9)}}\qquad {\boldsymbol {\nabla }}\cdot ({\boldsymbol {\mathsf {C}}}:{\widehat {\boldsymbol {\epsilon }}})+\omega ^{2}~{\widehat {\rho }}~{\widehat {\mathbf {u} }}=0~.}
Also, taking the Fourier transform of equation (3), we have
(10)
ϵ
^
=
1
2
[
∇
u
^
+
(
∇
u
^
)
T
]
.
{\displaystyle {\text{(10)}}\qquad {\widehat {\boldsymbol {\epsilon }}}={\frac {1}{2}}\left[{\boldsymbol {\nabla }}{\widehat {\mathbf {u} }}+({\boldsymbol {\nabla }}{\widehat {\mathbf {u} }})^{T}\right]~.}
Since
σ
^
{\displaystyle {\widehat {\boldsymbol {\sigma }}}}
and
ϵ
^
{\displaystyle {\widehat {\boldsymbol {\epsilon }}}}
are symmetric, we must have
C
i
j
k
l
=
C
j
i
k
l
=
C
i
j
l
k
.
{\displaystyle C_{ijkl}=C_{jikl}=C_{ijlk}~.}
Because of this symmetry, we can replace
ϵ
^
{\displaystyle {\widehat {\boldsymbol {\epsilon }}}}
by
∇
u
^
{\displaystyle {\boldsymbol {\nabla }}{\widehat {\mathbf {u} }}}
in
equation (9) to get
∇
⋅
(
C
:
∇
u
^
)
+
ω
2
ρ
^
u
^
=
0
.
{\displaystyle {\boldsymbol {\nabla }}\cdot ({\boldsymbol {\mathsf {C}}}:{\boldsymbol {\nabla {\widehat {\mathbf {u} }})}}+\omega ^{2}~{\widehat {\rho }}~{\widehat {\mathbf {u} }}=0~.}
Dropping the hats, we then get the wave equation for elastodynamics
(11)
∇
⋅
(
C
:
∇
u
)
+
ω
2
ρ
u
=
0
.
{\displaystyle {\text{(11)}}\qquad {{\boldsymbol {\nabla }}\cdot ({\boldsymbol {\mathsf {C}}}:{\boldsymbol {\nabla \mathbf {u} )}}+\omega ^{2}~\rho ~\mathbf {u} =0~.}}
Let us now consider the case of antiplane shear. Assume that
C
{\displaystyle {\boldsymbol {\mathsf {C}}}}
is
isotropic, i.e.,
C
:
∇
u
=
μ
[
∇
u
+
(
∇
u
)
T
]
+
λ
tr
(
∇
u
)
1
{\displaystyle {\boldsymbol {\mathsf {C}}}:{\boldsymbol {\nabla }}\mathbf {u} =\mu ~[{\boldsymbol {\nabla }}\mathbf {u} +({\boldsymbol {\nabla }}\mathbf {u} )^{T}]+\lambda ~{\text{tr}}({\boldsymbol {\nabla }}\mathbf {u} )~{\boldsymbol {\mathit {1}}}}
where
μ
{\displaystyle \mu }
is the shear modulus and
λ
{\displaystyle \lambda }
is the Lame modulus.
Let us assume that
μ
{\displaystyle \mu }
and
λ
{\displaystyle \lambda }
are independent of
x
1
{\displaystyle x_{1}}
, i.e.,
μ
≡
μ
(
x
2
,
x
3
)
and
λ
≡
λ
(
x
2
,
x
3
)
.
{\displaystyle \mu \equiv \mu (x_{2},x_{3})\qquad {\text{and}}\qquad \lambda \equiv \lambda (x_{2},x_{3})~.}
Let us look for a solution with
u
2
=
u
3
=
0
{\displaystyle u_{2}=u_{3}=0}
and
u
1
{\displaystyle u_{1}}
independent of
x
1
{\displaystyle x_{1}}
, i.e.,
u
1
≡
u
1
(
x
2
,
x
3
)
{\displaystyle u_{1}\equiv u_{1}(x_{2},x_{3})}
. This is an out of plane mode of
deformation.
Then, noting that
tr
(
∇
u
)
=
∇
⋅
u
{\displaystyle {\text{tr}}({\boldsymbol {\nabla }}\mathbf {u} )={\boldsymbol {\nabla }}\cdot \mathbf {u} }
, we have
tr
(
∇
u
)
=
∂
u
i
∂
x
i
=
0
.
{\displaystyle {\text{tr}}({\boldsymbol {\nabla }}\mathbf {u} )={\frac {\partial u_{i}}{\partial x_{i}}}=0~.}
Therefore,
[
C
:
∇
u
]
i
j
=
μ
[
∂
u
i
∂
x
j
+
∂
u
j
∂
x
i
]
{\displaystyle [{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\nabla }}\mathbf {u} ]_{ij}=\mu ~\left[{\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}\right]}
or
C
:
∇
u
=
μ
[
0
∂
u
1
∂
x
2
∂
u
1
∂
x
3
∂
u
1
∂
x
2
0
0
∂
u
1
∂
x
3
0
0
]
.
{\displaystyle {\boldsymbol {\mathsf {C}}}:{\boldsymbol {\nabla }}\mathbf {u} =\mu {\begin{bmatrix}0&{\frac {\partial u_{1}}{\partial x_{2}}}&{\frac {\partial u_{1}}{\partial x_{3}}}\\\\{\frac {\partial u_{1}}{\partial x_{2}}}&0&0\\\\{\frac {\partial u_{1}}{\partial x_{3}}}&0&0\end{bmatrix}}~.}
Therefore
∇
⋅
(
C
:
∇
u
)
=
[
∂
∂
x
2
(
μ
∂
u
1
∂
x
2
)
+
∂
∂
x
3
(
μ
∂
u
1
∂
x
3
)
0
0
]
.
{\displaystyle {\boldsymbol {\nabla }}\cdot ({\boldsymbol {\mathsf {C}}}:{\boldsymbol {\nabla \mathbf {u} )}}={\begin{bmatrix}{\frac {\partial }{\partial x_{2}}}\left(\mu ~{\frac {\partial u_{1}}{\partial x_{2}}}\right)+{\frac {\partial }{\partial x_{3}}}\left(\mu ~{\frac {\partial u_{1}}{\partial x_{3}}}\right)\\0\\0\end{bmatrix}}~.}
Plugging into the wave equation (11) we get
∂
∂
x
2
(
μ
∂
u
1
∂
x
2
)
+
∂
∂
x
3
(
μ
∂
u
1
∂
x
3
)
+
ω
2
ρ
u
1
=
0
{\displaystyle {\frac {\partial }{\partial x_{2}}}\left(\mu ~{\frac {\partial u_{1}}{\partial x_{2}}}\right)+{\frac {\partial }{\partial x_{3}}}\left(\mu ~{\frac {\partial u_{1}}{\partial x_{3}}}\right)+\omega ^{2}~\rho ~u_{1}=0}
or (using the two-dimensional gradient operator
∇
¯
{\displaystyle {\overline {\boldsymbol {\nabla }}}}
)
(12)
∇
¯
⋅
(
μ
∇
u
1
)
¯
+
ω
2
ρ
u
1
=
0
.
{\displaystyle {\text{(12)}}\qquad {{\overline {\boldsymbol {\nabla }}}\cdot (\mu ~{\overline {{\boldsymbol {\nabla }}u_{1})}}+\omega ^{2}~\rho ~u_{1}=0~.}}
TM and TE modes in electromagnetism
edit
Let us now consider the TM (transverse magnetic field) and TE (transverse
electric field) modes in electromagnetism and look for parallels with
antiplane shear in elastodynamics.
Recall the Maxwell equations (with hats dropped)
(13)
∇
×
E
=
i
ω
μ
H
;
∇
×
H
=
−
i
ω
ϵ
E
.
{\displaystyle {\text{(13)}}\qquad {\boldsymbol {\nabla }}\times \mathbf {E} =i\omega \mu \mathbf {H} ~;~~{\boldsymbol {\nabla }}\times \mathbf {H} =-i\omega \epsilon \mathbf {E} ~.}
Assume that
μ
{\displaystyle \mu }
and
ϵ
{\displaystyle \epsilon }
are scalars which are independent of
x
1
{\displaystyle x_{1}}
,
i.e.,
μ
≡
μ
(
x
2
,
x
3
)
{\displaystyle \mu \equiv \mu (x_{2},x_{3})}
and
ϵ
≡
ϵ
(
x
2
,
x
3
)
{\displaystyle \epsilon \equiv \epsilon (x_{2},x_{3})}
.
For the TE case, we look for solutions with
E
2
=
E
3
=
0
{\displaystyle E_{2}=E_{3}=0}
and
E
1
{\displaystyle E_{1}}
independent of
x
1
{\displaystyle x_{1}}
, i.e.,
E
1
≡
E
1
(
x
2
,
x
3
)
{\displaystyle E_{1}\equiv E_{1}(x_{2},x_{3})}
.
Then,
∇
×
E
=
[
0
,
∂
E
1
∂
x
3
,
−
∂
E
1
∂
x
2
]
.
{\displaystyle {\boldsymbol {\nabla }}\times \mathbf {E} =\left[0,{\frac {\partial E_{1}}{\partial x_{3}}},-{\frac {\partial E_{1}}{\partial x_{2}}}\right]~.}
This implies that
H
=
[
0
,
1
i
ω
μ
∂
E
1
∂
x
3
,
−
1
i
ω
μ
∂
E
1
∂
x
2
]
.
{\displaystyle \mathbf {H} =\left[0,{\cfrac {1}{i\omega \mu }}~{\frac {\partial E_{1}}{\partial x_{3}}},-{\cfrac {1}{i\omega \mu }}~{\frac {\partial E_{1}}{\partial x_{2}}}\right]~.}
Therefore,
∇
×
H
=
[
−
∂
∂
x
3
(
1
i
ω
μ
∂
E
1
∂
x
3
)
−
∂
∂
x
2
(
1
i
ω
μ
∂
E
1
∂
x
2
)
,
0
,
0
]
.
{\displaystyle {\boldsymbol {\nabla }}\times \mathbf {H} =\left[-{\frac {\partial }{\partial x_{3}}}\left({\cfrac {1}{i\omega \mu }}~{\frac {\partial E_{1}}{\partial x_{3}}}\right)-{\frac {\partial }{\partial x_{2}}}\left({\cfrac {1}{i\omega \mu }}~{\frac {\partial E_{1}}{\partial x_{2}}}\right),0,0\right]~.}
or,
∇
×
H
=
i
ω
[
∂
∂
x
2
(
1
μ
∂
E
1
∂
x
2
)
+
∂
∂
x
3
(
1
μ
∂
E
1
∂
x
3
)
,
0
,
0
]
=
i
ω
[
∇
¯
⋅
(
1
μ
(
x
2
,
x
3
)
∇
¯
E
1
)
,
0
,
0
]
.
{\displaystyle {\boldsymbol {\nabla }}\times \mathbf {H} ={\cfrac {i}{\omega }}\left[{\frac {\partial }{\partial x_{2}}}\left({\cfrac {1}{\mu }}~{\frac {\partial E_{1}}{\partial x_{2}}}\right)+{\frac {\partial }{\partial x_{3}}}\left({\cfrac {1}{\mu }}~{\frac {\partial E_{1}}{\partial x_{3}}}\right),0,0\right]={\cfrac {i}{\omega }}\left[{\overline {\boldsymbol {\nabla }}}\cdot \left({\cfrac {1}{\mu (x_{2},x_{3})}}{\overline {\boldsymbol {\nabla }}}E_{1}\right),0,0\right]~.}
Plugging into equation (13) we get the TE equation
∇
¯
⋅
(
1
μ
(
x
2
,
x
3
)
∇
¯
E
1
)
+
ω
2
ϵ
(
x
2
,
x
3
)
E
1
=
0
.
{\displaystyle {{\overline {\boldsymbol {\nabla }}}\cdot \left({\cfrac {1}{\mu (x_{2},x_{3})}}{\overline {\boldsymbol {\nabla }}}E_{1}\right)+\omega ^{2}~\epsilon (x_{2},x_{3})~E_{1}=0~.}}
This equation has the same form as equation (12).
More generally, if
μ
=
μ
(
x
2
,
x
3
)
=
[
μ
11
0
0
0
μ
22
μ
23
0
μ
23
μ
33
]
=
[
μ
11
[
0
]
[
0
]
[
M
]
]
where
[
M
]
=
[
μ
22
μ
23
μ
23
μ
33
]
≡
M
{\displaystyle {\boldsymbol {\mu }}={\boldsymbol {\mu }}(x_{2},x_{3})={\begin{bmatrix}\mu _{11}&0&0\\0&\mu _{22}&\mu _{23}\\0&\mu _{23}&\mu _{33}\end{bmatrix}}={\begin{bmatrix}\mu _{11}&\left[{\mathsf {0}}\right]\\\left[{\mathsf {0}}\right]&\left[{\mathsf {M}}\right]\end{bmatrix}}\quad {\text{where}}\quad \left[{\mathsf {M}}\right]={\begin{bmatrix}\mu _{22}&\mu _{23}\\\mu _{23}&\mu _{33}\end{bmatrix}}\equiv {\boldsymbol {M}}}
and
ϵ
=
ϵ
(
x
2
,
x
3
)
=
[
ϵ
11
[
0
]
[
0
]
[
N
]
]
where
[
N
]
=
[
ϵ
22
ϵ
23
ϵ
23
ϵ
33
]
≡
N
{\displaystyle {\boldsymbol {\epsilon }}={\boldsymbol {\epsilon }}(x_{2},x_{3})={\begin{bmatrix}\epsilon _{11}&\left[{\mathsf {0}}\right]\\\left[{\mathsf {0}}\right]&\left[{\mathsf {N}}\right]\end{bmatrix}}\quad {\text{where}}\quad \left[{\mathsf {N}}\right]={\begin{bmatrix}\epsilon _{22}&\epsilon _{23}\\\epsilon _{23}&\epsilon _{33}\end{bmatrix}}\equiv {\boldsymbol {N}}}
we get the TE equation
∇
¯
⋅
[
(
R
⊥
⋅
M
−
1
⋅
R
⊥
T
)
⋅
∇
¯
E
1
]
+
ω
2
ϵ
11
E
1
1
=
0
where
R
⊥
≡
[
R
]
⊥
=
[
0
1
−
1
0
]
.
{\displaystyle {{\overline {\boldsymbol {\nabla }}}\cdot \left[\left({\boldsymbol {R}}_{\perp }\cdot {\boldsymbol {M}}^{-1}\cdot {\boldsymbol {R}}_{\perp }^{T}\right)\cdot {\overline {\boldsymbol {\nabla }}}E_{1}\right]+\omega ^{2}~\epsilon _{11}~E_{1}~\mathbf {1} =\mathbf {0} }\qquad {\text{where}}\qquad {\boldsymbol {R}}_{\perp }\equiv \left[{\mathsf {R}}\right]_{\perp }={\begin{bmatrix}0&1\\-1&0\end{bmatrix}}~.}
Similarly, there is a TM equation with
H
2
=
H
3
=
0
{\displaystyle H_{2}=H_{3}=0}
of the form
∇
¯
⋅
[
(
R
⊥
⋅
N
−
1
⋅
R
⊥
T
)
⋅
∇
¯
H
1
]
+
ω
2
μ
11
H
1
1
=
0
{\displaystyle {{\overline {\boldsymbol {\nabla }}}\cdot \left[\left({\boldsymbol {R}}_{\perp }\cdot {\boldsymbol {N}}^{-1}\cdot {\boldsymbol {R}}_{\perp }^{T}\right)\cdot {\overline {\boldsymbol {\nabla }}}H_{1}\right]+\omega ^{2}~\mu _{11}~H_{1}~\mathbf {1} =\mathbf {0} }}
which for the isotropic case reduces to
∇
¯
⋅
(
1
ϵ
(
x
2
,
x
3
)
∇
¯
H
1
)
+
ω
2
μ
(
x
2
,
x
3
)
H
1
=
0
.
{\displaystyle {{\overline {\boldsymbol {\nabla }}}\cdot \left({\cfrac {1}{\epsilon (x_{2},x_{3})}}{\overline {\boldsymbol {\nabla }}}H_{1}\right)+\omega ^{2}~\mu (x_{2},x_{3})~H_{1}=0~.}}
The general solution independent of
x
1
{\displaystyle x_{1}}
is a superposition of the TE and
TM solutions. This can be seen by observing that the Maxwell equations
decouple under these conditions and a general solution can be written as
(
E
1
,
E
2
,
E
3
)
=
(
E
1
,
0
,
0
)
+
(
0
,
E
2
,
E
3
)
{\displaystyle (E_{1},E_{2},E_{3})=(E_{1},0,0)+(0,E_{2},E_{3})}
where the first term represents the TE solution. We can show that the
second term represents the TM solution by observing that
∇
×
(
0
,
E
2
,
E
3
)
=
[
∂
E
3
∂
x
2
−
∂
E
2
∂
x
3
,
0
,
0
]
{\displaystyle {\boldsymbol {\nabla }}\times (0,E_{2},E_{3})=\left[{\frac {\partial E_{3}}{\partial x_{2}}}-{\frac {\partial E_{2}}{\partial x_{3}}},0,0\right]}
implying that
H
2
=
H
3
=
0
{\displaystyle H_{2}=H_{3}=0}
which is the TM solution.
A resonant structure
edit
R. J. Atkin and N. Fox. An introduction to the theory of elasticity . Longman, New York, 1980.
A. B. Movchan and S. Guenneau. Split-ring resonators and localized modes. Physical Review B , 70:125116, 2004.