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The Transfer Matrix for a Boundary

1. Introduction

Let’s apply the material from Making Friends with Electromagnetic Boundary Conditions to calculate a transfer matrix. We will consider a TE wave travelling between materials with parameters {\mu_1,\epsilon_1} and {\mu_2,\epsilon_2}:

TEDiagram

There are a few things to note:

  • This is a TE wave, so the electric field {\mathbf{E}} is transverse to the plane of incidence. In this case we chose to let {\mathbf{E}} point into the page, which then determines the direction of the magnetic field since {(\mathbf{k},\mathbf{E},\mathbf{B})} form a right-handed coordinate system.
  • The total wavevector {\mathbf{k}} is made up of a component {k_z} in the {z}-direction and {k_x} in the {x} direction.
  • The field on each side is made up of a forward-propagating wave ({a_1,c_1}) and a backward propagating wave ({b_1,d_1}). On the left we have

    \displaystyle E=\left(a_1e^{i(k_{1z}z-\omega t)}+b_1e^{i(-k_{1z}z-\omega t)}\right)e^{ik_xx}, \ \ \ \ \ (1)

    and similarly for the right.

We want {(a_1,b_1)} to represent the electric field in the first medium, but do we want them to represent {\mathbf{E}} or {\mathbf{D}}? To decide, let’s take a look at the boundary conditions:

\displaystyle \begin{gathered} \hat{\mathbf{n}}\cdot\left(\mathbf{B}_2-\mathbf{B}_1\right)=0;\; \hat{\mathbf{n}}\times\left(\mathbf{E}_1-\mathbf{E}_2\right)=0, \\ \hat{\mathbf{n}}\cdot\left(\mathbf{D}_2-\mathbf{D}_1\right)=0;\;\hat{\mathbf{n}}\times\left(\mathbf{H}_2-\mathbf{H}_1\right)=0. \end{gathered} \ \ \ \ \ (2)

The electric field is perpendicular to the boundary, so {\hat{\mathbf{n}}\cdot(\mathbf{D}_2-\mathbf{D}_1)=0}. On the other hand since {\mathbf{E}} is perpendicular to {\hat{\mathbf{n}}}, you can convince yourself that the other boundary condition for the electric field gives {\mathbf{E}_1=\mathbf{E}_2}. Thus the electric field is continuous across the boundary, so we want {(a_1,b_1)} to represent the electric field. In this case if {a_1,b_1} are the fields just before the boundary and {c_1,d_1} the fields just after, we have

\displaystyle \begin{aligned} \mathbf{E}_1 &= \mathbf{E}_2, \\ a_1+b_1&=c_1+d_1. \end{aligned} \ \ \ \ \ (3)

Note that we have chosen coordinates such that the exponentials in Eq. (1) are one right at the boundary.

Aside: If we had a TM wave we would choose {(a_1,b_1)} to represent {\mathbf{H}} rather than {\mathbf{B}}, since {\mathbf{H}} is what would be continuous across the boundary and we would again have {a_1+b_1=c_1+d_1}. Note that this is just a convenience; you could use {\mathbf{B}} if you really wanted to, and using {\mathbf{B}=\mu \mathbf{H}} the boundary condition would then become {(a_1+b_1)/\mu_1=(c_1+d_1)/\mu_2}.

We need one more relation between the {a_1,b_1,c_1,d_1}, and we have two two boundary conditions we haven’t used so far:

\displaystyle \hat{\mathbf{n}}\cdot(\mathbf{B}_2-\mathbf{B}_1)=0;\;\hat{\mathbf{n}}\times(\mathbf{H}_2-\mathbf{H}_1)=0. \ \ \ \ \ (4)

These are in terms of the magnetic field, which we want to relate to the electric field. This comes from Maxwell’s equation:

\displaystyle \nabla\times\mathbf{E}=-\partial_t\mathbf{B}=i\omega\mathbf{B}. \ \ \ \ \ (5)

What would happen if we took the {\mathbf{B}} equation? Since {\mathbf{B}} is being dotted with {\hat{\mathbf{n}}}, we see that the {z}-component of {\mathbf{B}} is continuous across the boundary. This tells us that {(\nabla\times\mathbf{E})_z} is continuous, which is:

\displaystyle (\nabla\times\mathbf{E})_z=\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}=\frac{\partial E_y}{\partial x}, \ \ \ \ \ (6)

since only {E_y} is nonzero. Meanwhile the {\mathbf{H}} equation gives us a relation for the {x}-component of {\mathbf{H}}, and

\displaystyle (\nabla\times\mathbf{E})_x=\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}=-\frac{\partial E_y}{\partial z}. \ \ \ \ \ (7)

The wave is propagating in the {z} direction, so this is what we want.

Using {\mathbf{B}=\mu\mathbf{H}}, Eq. (7) tells us that

\displaystyle \left.\frac{1}{\mu_1}\frac{\partial E_y}{\partial z}\right\rvert_{Left}=\left.\frac{1}{\mu_2}\frac{\partial E_y}{\partial z}\right\rvert_{Right}. \ \ \ \ \ (8)

Looking at \cref{eq:TotalEField} the left hand side is

\displaystyle \frac{ik_{1z}}{\mu_1}\left(a_1-b_1\right), \ \ \ \ \ (9)

while the right hand side will be

\displaystyle \frac{ik_{2z}}{\mu_2}\left(c_1-d_1\right). \ \ \ \ \ (10)

Thus this boundary condition is:

\displaystyle a_1-b_1=\frac{\mu_1}{\mu_2}\frac{k_{2z}}{k_{1z}}(c_1-d_1). \ \ \ \ \ (11)

The last thing is to write {k_{2z}/k_{1z}} in terms of the parameters of the material. We know that {|k|=\omega\sqrt{\mu\epsilon}}, and {k_x^2+k_z^2=k^2}. Thus

\displaystyle \begin{aligned} k_{1z} &= \sqrt{\omega^2\mu_1\epsilon_1-k_x^2}, \\ k_{2z} &= \sqrt{\omega^2\mu_2\epsilon_2-k_x^2}. \\ \end{aligned} \ \ \ \ \ (12)

Eq. (11) thus gives us our second boundary equation:

\displaystyle a_1-b_1=\frac{\mu_1}{\mu_2}\sqrt{\frac{\omega^2\mu_2\epsilon_2-k_x^2}{\omega^2\mu_1\epsilon_1-k_x^2}}(c_1-d_1). \ \ \ \ \ (13)

This is in terms of the frequency of the wave {\omega}, the parameters {\mu_i,\epsilon_i} of the material, as well as {k_x}, which is a parameter telling us the angle of the input wave. For normal incidence we have {k_x=0}.

2. References

[1] The LaTeX was written using the excellent tool LaTeX to WordPress:

LaTeX to WordPress

 

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