Derivation

Derivation

This page documents the important derivation to help people better understand the relationship between formulas and variables.

Magnetic field normalization

Reason to normalize

If we apply linear, source-free, monochromatic assumptions to the maxwell equation, we get the following equation:

$$\nabla\times\vec E=i\omega\mu\vec H$$

$$\nabla\times\vec H=-i\omega\varepsilon\vec E$$

Where the value of $\mu$ and $\varepsilon$ in free-space is:

• $\varepsilon_0 = 8.8541878128(13)×10^{−12}$ F⋅m⁻¹
• $\mu_0 = 1.25663706212(19)×10^{−6}$ N⋅A⁻²

The angular frequency of the visible light is typically $2.7×10^{15}$ rad/s to $4.7×10^{15}$ rad/s, which is hard to compute, and leads to larger numerical error.

How is normalization term derived

If we introduce a normalization term as $\tilde{\vec H}=\alpha\vec H$, the equation becomes:

$$\nabla\times\vec E=({i\omega\mu\over\alpha})\tilde{\vec H}$$

$$\nabla\times\tilde{\vec H}=(-i\omega\varepsilon\alpha)\vec E$$

We can balance the coefficient in free-space condition as:

$${i\omega\mu_0\over\alpha}=-i\omega\varepsilon_0\alpha$$

Result in:

$$\alpha^2=-{\mu_0\over\varepsilon_0}$$

$$\alpha=i\sqrt{\mu_0\over\varepsilon_0}$$

Normalized coefficient

Therefore, the coefficient becomes:

$${i\omega\mu\over\alpha}=\omega\mu_r\mu_0\sqrt{\varepsilon_0\over\mu_0}=\omega\mu_r\sqrt{\mu_0\varepsilon_0}={\omega\over c}\mu_r=k_0\mu_r$$

$$-i\omega\varepsilon\alpha=\omega\varepsilon_r\varepsilon_0\sqrt{\mu_0\over\varepsilon_0}=\omega\varepsilon_r\sqrt{\mu_0\varepsilon_0}={\omega\over c}\varepsilon_r=k_0\varepsilon_r$$

• $c$ : Speed of light in free-space ${1\over\sqrt{\mu_0\varepsilon_0}}={\omega\over k_0}$
• $k_0$ : Wave number in free-space $2\pi\over\lambda_0$
• $\lambda_0$ : Free-space wavelength

Normalized maxwell equation

$$\nabla\times\vec E=k_0\mu_r\tilde{\vec H}$$

$$\nabla\times\tilde{\vec H}=k_0\varepsilon_r\vec E$$

Fourier convolution

For any field $U$ (ex. $E$ or $H$) interact with material coefficient $u$ (ex. $\varepsilon_r$ or $\mu_r$):

$$U(x,y;z)=\sum_{mn}U(m,n;z)e^{i(k_x(m,n)x+k_y(m,n)y)}$$

$$u(x, y)=\sum_{mn}u(m,n)e^{i(G_x(m, n)x+G_y(m, n)y)}$$

The effect of $u$ acts on $U$ is:

$$u(x,y;z)\cdot U(x,y;z)$$

$$=\sum_{m_2n_2}u(m_2,n_2)e^{i(G_x(m_2, n_2)x+G_y(m_2, n_2)y)}\cdot\sum_{m_1n_1}U(m_1,n_1;z)e^{i(k_x(m_1,n_1)x+k_y(m_1,n_1)y)}$$

$$=\sum_{m_2n_2m_1n_1}u(m_2,n_2)U(m_1,n_1;z)\cdot e^{i(G_x(m_2, n_2)x+G_y(m_2, n_2)y+k_x(m_1,n_1)x+k_y(m_1,n_1)y)}$$

$$=\sum_{m_2n_2m_1n_1}u(m_2,n_2)U(m_1,n_1;z)\cdot e^{i(k_x(m_1+m_2,n_1+n_2)x+k_y(m_1+m_2,n_1+n_2)y)}$$

By substitution:

$$m=m_1+m_2$$

$$n=n_1+n_2$$

$$=\sum_{mn}\sum_{m_1n_1}u(m-m_1,n-n_1)U(m_1,n_1;z)\cdot e^{i(k_x(m,n)x+k_y(m,n)y)}$$

$$=\sum_{mn}U'(m,n;z)e^{i(k_x(m,n)x+k_y(m,n)y)}$$

Where

$$U'(m,n;z)=\sum_{m_1n_1}U(m_1,n_1;z)u(m-m_1,n-n_1)$$

Matrix equation

Continue from the normalized maxwell equation:

$$\nabla\times\vec E=k_0\mu_r\tilde{\vec H}$$

$$\nabla\times\tilde{\vec H}=k_0\varepsilon_r\vec E$$

Expanding the curl operation, in matrix form:

$$ \nabla\times\vec E= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \partial\over\partial x & \partial\over\partial y & \partial\over\partial z \\ E_x & E_y & E_z \end{vmatrix}= \begin{bmatrix} 0 & -{\partial\over\partial z} & {\partial\over\partial y} \\ {\partial\over\partial z} & 0 & -{\partial\over\partial x} \\ -{\partial\over\partial y} & {\partial\over\partial x} & 0 \\ \end{bmatrix} \begin{bmatrix} E_x \\ E_y \\ E_z \end{bmatrix} $$

Recall the definition of fields under plane-wave decomposition, we are able to solve derivative in x, y direction:

$${\partial\over\partial x}\vec E=ik_x\vec E$$

$${\partial\over\partial y}\vec E=ik_y\vec E$$

Maxwell equation in matrix form:

$$ \begin{bmatrix} 0 & -{d\over dz} & ik_y \\ {d\over dz} & 0 & -ik_x \\ -ik_y & ik_x & 0 \\ \end{bmatrix} \begin{bmatrix} E_x \\ E_y \\ E_z \end{bmatrix}= k_0\mu_r \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \\ \tilde{H_z} \end{bmatrix} $$

$$ \begin{bmatrix} 0 & -{d\over dz} & ik_y \\ {d\over dz} & 0 & -ik_x \\ -ik_y & ik_x & 0 \\ \end{bmatrix} \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \\ \tilde{H_z} \end{bmatrix}= k_0\varepsilon_r \begin{bmatrix} E_x \\ E_y \\ E_z \end{bmatrix} $$

Normalize by the free-space wave number

The equation can be further normalized by dividing $k_0$:

$$ \begin{bmatrix} 0 & -{d\over d\bar z} & iK_y \\ {d\over d\bar z} & 0 & -iK_x \\ -iK_y & iK_x & 0 \\ \end{bmatrix} \begin{bmatrix} E_x \\ E_y \\ E_z \end{bmatrix}= \mu_r \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \\ \tilde{H_z} \end{bmatrix} $$

$$ \begin{bmatrix} 0 & -{d\over d\bar z} & iK_y \\ {d\over d\bar z} & 0 & -iK_x \\ -iK_y & iK_x & 0 \\ \end{bmatrix} \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \\ \tilde{H_z} \end{bmatrix}= \varepsilon_r \begin{bmatrix} E_x \\ E_y \\ E_z \end{bmatrix} $$

• $K_x={k_x\over k_0}$, $K_y={k_y\over k_0}$ : Normalized wave vector
• $\bar z=k_0 z$ : Normalized z coordinate

Solving matrix equation

Since we only expand in x, y dimensions but not z, we can solve the equation separately:

$$ \begin{bmatrix} 0 & -{d\over d\bar z}\\ {d\over d\bar z} & 0\\ \end{bmatrix} \begin{bmatrix} E_x \ E_y \end{bmatrix} + \begin{bmatrix} iK_y\\ -iK_x \end{bmatrix} E_z= \mu_r \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \end{bmatrix} $$

$$ \begin{bmatrix} -iK_y & iK_x \end{bmatrix} \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \end{bmatrix}= \varepsilon_r E_z $$

Where $E_z$ Can be solved:

$$ E_z= \varepsilon_r^{-1} \begin{bmatrix} -iK_y & iK_x \end{bmatrix} \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \end{bmatrix} $$

And substitude back:

$$ \begin{bmatrix} 0 & -{d\over d\bar z}\\ {d\over d\bar z} & 0\\ \end{bmatrix} \begin{bmatrix} E_x \ E_y \end{bmatrix}- \begin{bmatrix} K_y\\ -K_x \end{bmatrix} \varepsilon_r^{-1} \begin{bmatrix} -K_y & K_x \end{bmatrix} \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \end{bmatrix}= \mu_r \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \end{bmatrix} $$

To get:

$$ {d\over dz} \begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix} \begin{bmatrix} E_x \ E_y \end{bmatrix}= \left( \begin{bmatrix} K_y\\ -K_x \end{bmatrix} \varepsilon_r^{-1} \begin{bmatrix} -K_y & K_x \end{bmatrix} + \mu_r \right) \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \end{bmatrix} $$

Matrix differential equation

$$ {d\over dz} \begin{bmatrix} E_x \ E_y \end{bmatrix}= \left( \begin{bmatrix} K_x\\ K_y \end{bmatrix} \varepsilon_r^{-1} \begin{bmatrix} K_y & -K_x \end{bmatrix} + \begin{bmatrix} 0&\mu_r\\ -\mu_r&0 \end{bmatrix} \right) \begin{bmatrix} \tilde{H_x} \\ \tilde{H_y} \end{bmatrix} $$

$$ {d\over dz} \begin{bmatrix} \tilde{H_x} \ \tilde{H_y} \end{bmatrix}= \left( \begin{bmatrix} K_x\\ K_y \end{bmatrix} \mu_r^{-1} \begin{bmatrix} K_y & -K_x \end{bmatrix} + \begin{bmatrix} 0&\varepsilon_r\\ -\varepsilon_r&0 \end{bmatrix} \right) \begin{bmatrix} E_x \\ E_y \end{bmatrix} $$

(WIP)