# Waves in composites and metamaterials/Backus formula for laminates and rank-1 laminates

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.

## Recap

Recall the material laminated in the $x_{1}$ direction as shown in Figure 1.
* Figure 1. A laminate with direction of lamination $x_{1}$.
We showed that we could write
    ${\begin{bmatrix}-E_{1}\\\mathbf {D} _{t}\end{bmatrix}}={\boldsymbol {L}}(x_{1})\cdot {\begin{bmatrix}-D_{1}\\\mathbf {E} _{t}\end{bmatrix}}$
or,
    ${\begin{bmatrix}-\langle E_{1}\rangle \\\langle \mathbf {D} _{t}\rangle \end{bmatrix}}=\langle {\boldsymbol {L}}(x_{1})\rangle \cdot {\begin{bmatrix}-D_{1}\\\mathbf {E} _{t}\end{bmatrix}}$
where
    ${\boldsymbol {L}}(x_{1}):={\cfrac {1}{\epsilon _{11}}}~{\begin{bmatrix}-1&{\boldsymbol {\epsilon }}_{1t}\\{\boldsymbol {\epsilon }}_{t1}&~~\epsilon _{11}~{\boldsymbol {\epsilon }}_{tt}-{\boldsymbol {\epsilon }}_{t1}\cdot {\boldsymbol {\epsilon }}_{1t}\end{bmatrix}}~.$
We also showed that
    ${\begin{bmatrix}-\langle E_{1}\rangle \\\langle \mathbf {D} _{t}\rangle \end{bmatrix}}={\boldsymbol {L}}_{\text{eff}}\cdot {\begin{bmatrix}-D_{1}\\\mathbf {E} _{t}\end{bmatrix}}$
where
    ${\boldsymbol {L}}_{\text{eff}}:={\cfrac {1}{\epsilon _{11}^{\text{eff}}}}~{\begin{bmatrix}-1&{\boldsymbol {\epsilon }}_{1t}^{\text{eff}}\\{\boldsymbol {\epsilon }}_{t1}^{\text{eff}}&~~\epsilon _{11}~{\boldsymbol {\epsilon }}_{tt}^{\text{eff}}-{\boldsymbol {\epsilon }}_{t1}^{\text{eff}}\cdot {\boldsymbol {\epsilon }}_{1t}\end{bmatrix}}~.$
Therefore, we got the relation
    ${\boldsymbol {L}}_{\text{eff}}=\langle {\boldsymbol {L}}(x_{1})\rangle ~.$
This provides the relations
    ${\begin{aligned}\epsilon _{11}^{\text{eff}}&=\langle {\cfrac {1}{\epsilon _{11}}}\rangle ^{-1}\\\epsilon _{1j}^{\text{eff}}&=\langle {\cfrac {1}{\epsilon _{11}}}\rangle ^{-1}\langle {\cfrac {\epsilon _{1j}}{\epsilon _{11}}}\rangle \\\epsilon _{ij}^{\text{eff}}&=\langle \epsilon _{ij}-{\cfrac {\epsilon _{i1}~\epsilon _{1j}}{\epsilon _{11}}}\rangle +\langle {\cfrac {1}{\epsilon _{11}}}\rangle ^{-1}\langle {\cfrac {\epsilon _{i1}}{\epsilon _{11}}}\rangle \langle {\cfrac {\epsilon _{1j}}{\epsilon _{11}}}\rangle ~~,i,j\neq 1\end{aligned}}$
where
    $\langle f(x_{1})\rangle ={\cfrac {1}{V}}~\int _{0}^{l}f(x_{1})~{\text{d}}x_{1}~.$
When the off diagonal elements vanish, we get
    ${\begin{aligned}\epsilon _{11}^{\text{eff}}&=\langle {\cfrac {1}{\epsilon _{11}}}\rangle ^{-1}&&\qquad {\text{(Harmonic average)}}\\\epsilon _{jj}^{\text{eff}}&=\langle \epsilon _{jj\rangle }&&\qquad {\text{(Arithmetic average)}}\\\epsilon _{ab}^{\text{eff}}&=0&&\qquad {\text{if}}~a\neq b~.\end{aligned}}$
The harmonic average corresponds to a situation in which each layer may be thought of as a capacitor in series while the arithmetic average corresponds to a situation where the capacitors are in parallel.

## Effective Elastic Properties of a Layered Medium

In this section we use the approach of arranging constant fields to find the effective elastic properties of a layered medium in which each layer is anisotropic. The approach used is that of Backus (Backus62).
Consider the layered medium shown in Figure 2. In this case the displacement field is continuous across the interfaces between the layers.
* Figure 2. An elastic layered medium with direction of lamination $x_{1}$.
The strain field (${\boldsymbol {\varepsilon }}$) is given by
    ${\boldsymbol {\varepsilon }}(\mathbf {x} )={\frac {1}{2}}\left[{\boldsymbol {\nabla }}\mathbf {u} +({\boldsymbol {\nabla }}\mathbf {u} )^{T}\right]~.$
In a Cartesian basis ($\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}$) with coordinates ($x_{1},x_{2},x_{3}$) we can write
    $\varepsilon _{ij}={\frac {1}{2}}\left({\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}\right)~.$
Therefore the strain components $\varepsilon _{22},\varepsilon _{33},\varepsilon _{23}$ are also continuous across the interfaces. Moreover, these components of strain are also constant in each layer. Since a piecewise constant field that is also continuous must be constant, the strain components $\varepsilon _{22},\varepsilon _{33},\varepsilon _{23}$ must be constant throughout the laminate.
The tractions (normal components of the stress) at each interface are given by
    $\mathbf {t} ={\boldsymbol {\sigma }}\cdot \mathbf {n} \qquad \implies \qquad t_{i}=\sigma _{ij}~n_{j}$
where ${\boldsymbol {\sigma }}$ is the stress and $\mathbf {n}$ is the normal at the interface. Now the tractions must be continuous at the interfaces. Since the normal components of the stress are piecewise constant in each layer, this implies that the normal components of the stress must also be constant throughout the laminate.
We have chosen $\mathbf {n}$ such that $\mathbf {n} =(1,0,0)$. Therefore the stress components $\sigma _{11},\sigma _{12},\sigma _{13}$ must be constant.
Recall that the constitutive relation for an anisotropic elastic material is given by
    ${\boldsymbol {\sigma }}={\boldsymbol {\mathsf {C}}}:{\boldsymbol {\varepsilon }}~.$
Following the approach that we used for the permittivity, we now write the constitutive relation in the form
    ${\begin{bmatrix}{\boldsymbol {\sigma }}_{n}\\{\boldsymbol {\sigma }}_{t}\end{bmatrix}}={\begin{bmatrix}{\boldsymbol {C}}_{nn}&{\boldsymbol {C}}_{nt}\\{\boldsymbol {C}}_{tn}&{\boldsymbol {C}}_{tt}\end{bmatrix}}{\begin{bmatrix}{\boldsymbol {\varepsilon }}_{n}\\{\boldsymbol {\varepsilon }}_{t}\end{bmatrix}}$
where
    ${\boldsymbol {\sigma }}_{n}={\begin{bmatrix}\sigma _{11}\\{\sqrt {2}}~\sigma _{12}\\{\sqrt {2}}~\sigma _{13}\end{bmatrix}}~;~~{\boldsymbol {\sigma }}_{t}={\begin{bmatrix}\sigma _{22}\\\sigma _{33}\\{\sqrt {2}}~\sigma _{23}\end{bmatrix}}~;~~{\boldsymbol {\varepsilon }}_{n}={\begin{bmatrix}\varepsilon _{11}\\{\sqrt {2}}~\varepsilon _{12}\\{\sqrt {2}}~\varepsilon _{13}\end{bmatrix}}~;~~{\boldsymbol {\varepsilon }}_{t}={\begin{bmatrix}\varepsilon _{22}\\\varepsilon _{33}\\{\sqrt {2}}~\varepsilon _{23}\end{bmatrix}}$
and
    ${\boldsymbol {C}}_{nn}={\begin{bmatrix}C_{1111}&{\sqrt {2}}~C_{1112}&{\sqrt {2}}~C_{1113}\\{\sqrt {2}}~C_{1211}&2~C_{1212}&2~C_{1213}\\{\sqrt {2}}~C_{1311}&2~C_{1312}&2~C_{1313}\end{bmatrix}}~;~~{\boldsymbol {C}}_{nt}={\begin{bmatrix}C_{1122}&C_{1133}&{\sqrt {2}}~C_{1123}\\{\sqrt {2}}~C_{1222}&{\sqrt {2}}~C_{1233}&2~C_{1223}\\{\sqrt {2}}~C_{1322}&{\sqrt {2}}~C_{1333}&2~C_{1323}\end{bmatrix}}~;~~$
    ${\boldsymbol {C}}_{tn}={\begin{bmatrix}C_{2211}&{\sqrt {2}}~C_{2212}&{\sqrt {2}}~C_{2213}\\C_{3311}&{\sqrt {2}}~C_{3312}&{\sqrt {2}}~C_{3313}\\{\sqrt {2}}~C_{2311}&2~C_{2312}&2~C_{2313}\end{bmatrix}}~;~~{\boldsymbol {C}}_{tt}={\begin{bmatrix}C_{2222}&C_{2233}&{\sqrt {2}}~C_{2223}\\C_{3322}&C_{3333}&{\sqrt {2}}~C_{3323}\\{\sqrt {2}}~C_{2322}&{\sqrt {2}}~C_{2333}&2~C_{2323}\end{bmatrix}}~.$
From the major symmetry of ${\boldsymbol {\mathsf {C}}}$, we see that ${\boldsymbol {C}}_{nt}={\boldsymbol {C}}_{tn}^{T}$. Also, ${\boldsymbol {C}}_{nn}$ and ${\boldsymbol {C}}_{tt}$ are symmetric.
Writing the first row out, we get
    ${\text{(1)}}\qquad {\boldsymbol {\sigma }}_{n}={\boldsymbol {C}}_{nn}\cdot {\boldsymbol {\varepsilon }}_{n}+{\boldsymbol {C}}_{nt}\cdot {\boldsymbol {\varepsilon }}_{t}\qquad \implies \qquad {\boldsymbol {\varepsilon }}_{n}={\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {\sigma }}_{n}-\left[{\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {C}}_{nt}\right]\cdot {\boldsymbol {\varepsilon }}_{t}~.$
From the second row we get
    ${\boldsymbol {\sigma }}_{t}={\boldsymbol {C}}_{nt}^{T}\cdot {\boldsymbol {\varepsilon }}_{n}+{\boldsymbol {C}}_{tt}\cdot {\boldsymbol {\varepsilon }}_{t}~.$
Substituting the expression for ${\boldsymbol {\varepsilon }}_{n}$ from the first row, we get
    ${\text{(2)}}\qquad {\boldsymbol {\sigma }}_{t}={\boldsymbol {C}}_{nt}^{T}\cdot \left[{\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {\sigma }}_{n}-{\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {C}}_{nt}\cdot {\boldsymbol {\varepsilon }}_{t}\right]+{\boldsymbol {C}}_{tt}\cdot {\boldsymbol {\varepsilon }}_{t}=\left[{\boldsymbol {C}}_{nt}^{T}\cdot {\boldsymbol {C}}_{nn}^{-1}\right]\cdot {\boldsymbol {\sigma }}_{n}+\left[{\boldsymbol {C}}_{tt}-{\boldsymbol {C}}_{nt}^{T}\cdot {\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {C}}_{nt}\right]\cdot {\boldsymbol {\varepsilon }}_{t}~.$
Collecting (1) and (2) we get
    ${\begin{bmatrix}-{\boldsymbol {\varepsilon }}_{n}\\\\{\boldsymbol {\sigma }}_{t}\end{bmatrix}}={\begin{bmatrix}-{\boldsymbol {C}}_{nn}^{-1}&{\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {C}}_{nt}\\\\{\boldsymbol {C}}_{nt}^{T}\cdot {\boldsymbol {C}}_{nn}^{-1}&~~~{\boldsymbol {C}}_{tt}-{\boldsymbol {C}}_{nt}^{T}\cdot {\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {C}}_{nt}\end{bmatrix}}{\begin{bmatrix}-{\boldsymbol {\sigma }}_{n}\\\\{\boldsymbol {\varepsilon }}_{t}\end{bmatrix}}~.$
Taking a volume average gives
    ${\text{(3)}}\qquad {\begin{bmatrix}-\langle {\boldsymbol {\varepsilon }}_{n}\rangle \\\\\langle {\boldsymbol {\sigma }}_{t}\rangle \end{bmatrix}}={\begin{bmatrix}-\langle {\boldsymbol {C}}_{nn}^{-1}\rangle &\langle {\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {C}}_{nt}\rangle \\\\\langle {\boldsymbol {C}}_{nt}^{T}\cdot {\boldsymbol {C}}_{nn}^{-1}\rangle &~~~\langle {\boldsymbol {C}}_{tt}-{\boldsymbol {C}}_{nt}^{T}\cdot {\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {C}}_{nt}\rangle \end{bmatrix}}{\begin{bmatrix}-{\boldsymbol {\sigma }}_{n}\\\\{\boldsymbol {\varepsilon }}_{t}\end{bmatrix}}~.$
If the effective stiffness of the material is given by
    $\langle {\boldsymbol {\sigma }}\rangle ={\boldsymbol {\mathsf {C}}}_{\text{eff}}:\langle {\boldsymbol {\varepsilon }}\rangle$
we can also show that
    ${\text{(4)}}\qquad {\begin{bmatrix}-\langle {\boldsymbol {\varepsilon }}_{n}\rangle \\\\\langle {\boldsymbol {\sigma }}_{t}\rangle \end{bmatrix}}={\begin{bmatrix}-({\boldsymbol {C}}_{nn}^{\text{eff}})^{-1}&({\boldsymbol {C}}_{nn}^{\text{eff}})^{-1}\cdot {\boldsymbol {C}}_{nt}^{\text{eff}}\\\\({\boldsymbol {C}}_{nt}^{\text{eff}})^{T}\cdot ({\boldsymbol {C}}_{nn}^{\text{eff}})^{-1}&~~~{\boldsymbol {C}}_{tt}^{\text{eff}}-({\boldsymbol {C}}_{nt}^{\text{eff}})^{T}\cdot ({\boldsymbol {C}}_{nn}^{\text{eff}})^{-1}\cdot {\boldsymbol {C}}_{nt}^{\text{eff}}\end{bmatrix}}{\begin{bmatrix}-{\boldsymbol {\sigma }}_{n}\\\\{\boldsymbol {\varepsilon }}_{t}\end{bmatrix}}~.$
Comparing (3) and (4) we can show that
    ${\begin{aligned}{\boldsymbol {C}}_{nn}^{\text{eff}}&=\left\langle {\boldsymbol {C}}_{nn}^{-1}\right\rangle ^{-1}\\{\boldsymbol {C}}_{nt}^{\text{eff}}&=\left\langle {\boldsymbol {C}}_{nn}^{-1}\right\rangle ^{-1}\cdot \left\langle {\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {C}}_{nt}\right\rangle \\{\boldsymbol {C}}_{tt}^{\text{eff}}&=\left\langle {\boldsymbol {C}}_{tt}-{\boldsymbol {C}}_{tn}\cdot {\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {C}}_{nt}\right\rangle +\left\langle {\boldsymbol {C}}_{tn}\cdot {\boldsymbol {C}}_{nn}^{-1}\right\rangle \cdot \left\langle {\boldsymbol {C}}_{nn}^{-1}\right\rangle ^{-1}\cdot \left\langle {\boldsymbol {C}}_{nn}^{-1}\cdot {\boldsymbol {C}}_{nt}\right\rangle ~.\end{aligned}}$

### Isotropic layers

If the material in each layer is isotropic, then the constitutive relation is
    ${\boldsymbol {\sigma }}={\boldsymbol {\mathsf {C}}}:{\boldsymbol {\epsilon }}=\lambda (x_{1})~{\text{tr}}({\boldsymbol {\varepsilon }})~{\boldsymbol {\mathit {1}}}+2~\mu (x_{1})~{\boldsymbol {\varepsilon }}$
where $\lambda$ is the Lame modulus and $\mu$ is the shear modulus. In that case the effective properties of the laminate are
    ${\begin{aligned}C_{1111}^{\text{eff}}&=\langle {\cfrac {1}{\lambda +2~\mu }}\rangle ^{-1}~;~~C_{1122}^{\text{eff}}=C_{1133}^{\text{eff}}=\langle {\cfrac {\lambda }{\lambda +2~\mu }}\rangle ~\langle {\cfrac {1}{\lambda +2~\mu }}\rangle ^{-1}\\C_{1212}^{\text{eff}}&=C_{1313}^{\text{eff}}=\langle {\cfrac {1}{\mu }}\rangle ^{-1}~;~~C_{2323}^{\text{eff}}=\langle \mu \rangle ~;~~C_{1112}^{\text{eff}}=C_{1113}^{\text{eff}}=C_{1123}^{\text{eff}}=0\\C_{2222}^{\text{eff}}&=C_{3333}^{\text{eff}}=\langle {\cfrac {4~\mu ~\lambda ~(\lambda +\mu )}{(\lambda +2~\mu )^{2}}}\rangle +\langle {\cfrac {\lambda }{\lambda +2~\mu }}\rangle ~\langle {\cfrac {1}{\lambda +2~\mu }}\rangle ^{-1}\\C_{2233}^{\text{eff}}&=\langle {\cfrac {2~\mu ~\lambda }{\lambda +2~\mu }}\rangle +\langle {\cfrac {\lambda }{\lambda +2~\mu }}\rangle ^{2}~\langle {\cfrac {1}{\lambda +2~\mu }}\rangle ^{-1}\end{aligned}}$

## Laminates with Arbitrary Direction of Lamination n

So far we have dealt with laminates with a single direction of lamination that was oriented with the $x_{1}$ axis. In this section we generalize our approach to deal with laminates with a normal $\mathbf {n}$ which is not parallel to the $x_{1}$ axis.
Recall that the normal component of $\mathbf {D}$, i.e., $\mathbf {D} \cdot \mathbf {n}$, is constant and the tangential components of $\mathbf {E}$ are constant throughout the entire laminate (if there is only one direction of lamination).
Let us introduce the second-order tensor basis
    ${\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )=\mathbf {n} \otimes \mathbf {n} ~;~~{\boldsymbol {\mathit {\Gamma }}}_{2}(\mathbf {n} )={\boldsymbol {\mathit {1}}}-\mathbf {n} \otimes \mathbf {n} ={\boldsymbol {\mathit {1}}}-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )~.$
These are useful because
    ${\text{(5)}}\qquad {\begin{aligned}{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \mathbf {D} (\mathbf {x} )&={\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \langle \mathbf {D} \rangle &&\quad {\text{normal component}}\\{\boldsymbol {\mathit {\Gamma }}}_{2}(\mathbf {n} )\cdot \mathbf {E} (\mathbf {x} )&={\boldsymbol {\mathit {\Gamma }}}_{2}(\mathbf {n} )\cdot \langle \mathbf {E} \rangle &&\quad {\text{tangential components.}}\end{aligned}}$
Therefore,
    ${\text{(6)}}\qquad {\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \mathbf {E} (\mathbf {x} )=\mathbf {E} (\mathbf {x} )-\langle \mathbf {E} \rangle +{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \langle \mathbf {E} \rangle ~.$
Let us now introduce a polarization field
    ${\text{(7)}}\qquad \mathbf {P} (\mathbf {x} )=[{\boldsymbol {\epsilon }}(\mathbf {x} )-\epsilon _{0}~{\boldsymbol {\mathit {1}}}]\cdot \mathbf {E} (\mathbf {x} )=\mathbf {D} (\mathbf {x} )-\epsilon _{0}~\mathbf {E} (\mathbf {x} )$
where $\epsilon _{0}$ is an arbitrary constant.
The volume averaged polarization field is given by
    ${\text{(8)}}\qquad \langle \mathbf {P} \rangle ={\boldsymbol {\epsilon }}_{\text{eff}}\cdot \langle \mathbf {E} \rangle -\epsilon _{0}~\langle \mathbf {E} \rangle =({\boldsymbol {\epsilon }}_{\text{eff}}-\epsilon _{0}~{\boldsymbol {\mathit {1}}})\cdot \langle \mathbf {E} \rangle ~.$
Define
    ${\begin{aligned}{\boldsymbol {S}}(\mathbf {x} )&:=\epsilon _{0}~[\epsilon _{0}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}(\mathbf {x} )]^{-1}\\{\boldsymbol {S}}_{\text{eff}}&:=\epsilon _{0}~[\epsilon _{0}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}_{\text{eff}}]^{-1}~.\end{aligned}}$
Then,
    ${\text{(9)}}\qquad {\begin{aligned}{\boldsymbol {S}}(\mathbf {x} )\cdot \mathbf {P} (\mathbf {x} )&=-\left(\epsilon _{0}~[\epsilon _{0}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}(\mathbf {x} )]^{-1}\right)\cdot \left([\epsilon _{0}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}(\mathbf {x} )]\cdot \mathbf {E} (\mathbf {x} )\right)=-\epsilon _{0}~\mathbf {E} (\mathbf {x} )\\{\boldsymbol {S}}_{\text{eff}}\cdot \langle \mathbf {P} \rangle &=-\left(\epsilon _{0}~[\epsilon _{0}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}_{\text{eff}}]^{-1}\right)\cdot \left([\epsilon _{0}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}_{\text{eff}}]\cdot \langle \mathbf {E} \rangle \right)=-\epsilon _{0}~\langle \mathbf {E} \rangle ~.\end{aligned}}$
Applying the projection ${\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )$ to (7), we get
    ${\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \mathbf {P} (\mathbf {x} )={\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \mathbf {D} (\mathbf {x} )-\epsilon _{0}~{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \mathbf {E} (\mathbf {x} )~.$
Using equations (5)$_{1}$ and (6), we have
    ${\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \mathbf {P} (\mathbf {x} )={\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \langle \mathbf {D} \rangle -\epsilon _{0}~\mathbf {E} (\mathbf {x} )+\epsilon _{0}~\langle \mathbf {E} \rangle -\epsilon _{0}~{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \langle \mathbf {E} \rangle ~.$
From the definitions (9) we can then write
    ${\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \mathbf {P} (\mathbf {x} )={\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \langle \mathbf {D} \rangle +{\boldsymbol {S}}(\mathbf {x} )\cdot \mathbf {P} (\mathbf {x} )-{\boldsymbol {S}}_{\text{eff}}\cdot \langle \mathbf {P} \rangle -\epsilon _{0}~{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \langle \mathbf {E} \rangle ~.$
Define
    ${\text{(10)}}\qquad \mathbf {V} (\mathbf {n} ,\mathbf {x} ):={\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \langle \mathbf {D} \rangle -{\boldsymbol {S}}_{\text{eff}}\cdot \langle \mathbf {P} \rangle -\epsilon _{0}~{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \langle \mathbf {E} \rangle ~.$
Then we have
    ${\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \mathbf {P} (\mathbf {x} )={\boldsymbol {S}}(\mathbf {x} )\cdot \mathbf {P} (\mathbf {x} )+\mathbf {V} (\mathbf {n} ,\mathbf {x} )$
or
    ${\text{(11)}}\qquad \mathbf {V} (\mathbf {n} ,\mathbf {x} )=-\left[{\boldsymbol {S}}(\mathbf {x} )-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]\cdot \mathbf {P} (\mathbf {x} )~.$
Also, form equations (10) and (8) we have
    $\mathbf {V} (\mathbf {n} ,\mathbf {x} )={\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \left[\langle \mathbf {D} \rangle -\epsilon _{0}~\cdot \langle \mathbf {E} \rangle \right]-{\boldsymbol {S}}_{\text{eff}}\cdot \langle \mathbf {P} \rangle ={\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\cdot \langle \mathbf {P} \rangle -{\boldsymbol {S}}_{\text{eff}}\cdot \langle \mathbf {P} \rangle$
or,
    ${\text{(12)}}\qquad \mathbf {V} (\mathbf {n} ,\mathbf {x} )=-\left[{\boldsymbol {S}}_{\text{eff}}-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]\cdot \langle \mathbf {P} \rangle ~.$
Inverting (11) and (12) we have
    ${\text{(13)}}\qquad {\begin{aligned}\mathbf {P} (\mathbf {x} )&=-\left[{\boldsymbol {S}}(\mathbf {x} )-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]^{-1}\cdot \mathbf {V} (\mathbf {n} ,\mathbf {x} )\\\langle \mathbf {P} \rangle &=-\left[{\boldsymbol {S}}_{\text{eff}}-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]^{-1}\cdot \mathbf {V} (\mathbf {n} ,\mathbf {x} )~.\end{aligned}}$
Also, taking the volume average of (13)$_{1}$, we have
    ${\text{(14)}}\qquad \langle \mathbf {P} \rangle =-\langle \left[{\boldsymbol {S}}(\mathbf {x} )-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]^{-1}\rangle \cdot \mathbf {V} (\mathbf {n} ,\mathbf {x} )~.$
Therefore, comparing (13)$_{2}$ and (14) and invoking the arbitrary nature $\epsilon _{0}$, we have
    ${\left[{\boldsymbol {S}}_{\text{eff}}-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]^{-1}=\langle \left[{\boldsymbol {S}}(\mathbf {x} )-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]^{-1}\rangle ~.}$
This relation provides us with a means of computing the effective permittivity of a layered medium oriented at a random angle (given by the normal $\mathbf {n}$) with respect to the coordinate basis.

### Case 1: Simple or Rank-1 Laminate

Consider the Rank-1 laminate shown in Figure 3. The layers have permittivities alternating between ${\boldsymbol {\epsilon }}_{1}$ and $\epsilon _{2}~{\boldsymbol {\mathit {1}}}$. The volume fraction of phase $1$ is $f_{1}$ while that of phase $2$ is $f_{2}$ such that $f_{1}+f_{2}=1$.
* Figure 3. A Rank-1 laminate.
Recall that
    ${\boldsymbol {S}}(\mathbf {x} )=\epsilon _{0}~[\epsilon _{0}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}(\mathbf {x} )]^{-1}~.$
Let us take the limit as $\epsilon _{0}\rightarrow \epsilon _{2}$. Since $\epsilon (\mathbf {x} )=\epsilon _{2}~{\boldsymbol {\mathit {1}}}$ in phase $2$, we have
    ${\boldsymbol {S}}(\mathbf {x} )\rightarrow \infty \qquad {\text{in phase}}~2~.$
Therefore,
    $\left[{\boldsymbol {S}}(\mathbf {x} )-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]^{-1}\rightarrow 0\qquad {\text{in phase}}~2~.$
Hence, right hand side of
    $\left[{\boldsymbol {S}}_{\text{eff}}-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]^{-1}=\langle \left[{\boldsymbol {S}}(\mathbf {x} )-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]^{-1}\rangle$
reduces to an average only over phase $1$. If we define
    ${\boldsymbol {S}}_{1}:=\epsilon _{0}~[\epsilon _{0}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}_{1}]^{-1}\rightarrow \epsilon _{2}~[\epsilon _{2}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}_{1}]^{-1}$
we get
    ${\text{(15)}}\qquad \left[{\boldsymbol {S}}_{\text{eff}}-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]^{-1}=f_{1}~\left[{\boldsymbol {S}}_{1}-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]^{-1}~.$
Taking the inverse of both sides of (15) gives
    $f_{1}~\left[{\boldsymbol {S}}_{\text{eff}}-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )\right]={\boldsymbol {S}}_{1}-{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )$
or,
    $f_{1}~{\boldsymbol {S}}_{\text{eff}}={\boldsymbol {S}}_{1}-(1-f_{1})~{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )={\boldsymbol {S}}_{1}-f_{2}~{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )~.$
Since
    ${\boldsymbol {S}}_{\text{eff}}=\epsilon _{0}~[\epsilon _{0}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}_{\text{eff}}]^{-1}\rightarrow \epsilon _{2}~[\epsilon _{2}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}_{\text{eff}}]^{-1}$
we then have
    ${f_{1}~\epsilon _{2}~[\epsilon _{2}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}_{\text{eff}}]^{-1}=\epsilon _{2}~[\epsilon _{2}~{\boldsymbol {\mathit {1}}}-{\boldsymbol {\epsilon }}_{1}]^{-1}-f_{2}~{\boldsymbol {\mathit {\Gamma }}}_{1}(\mathbf {n} )~.}$
This is the formula of Tartar, Murat, Lurie, and Cherkaev and can be shown to be equivalent to the Backus formula.

## Footnotes

* ↑ The discussion in this lecture is based on Milton02. Please consult that book for more details and references. The method of Backus (Backus62) (see also Postma55 and Tartar76) has been used.