User:Rutikangimpuruza/sandbox

INTRODUCTION

make a note about SI units

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The inhomogeneous electromagnetic wave equation for the electric field can be written in terms of the electric Hertz Potential, $$ \boldsymbol{\pi}_{\mathrm{e}} $$, in the Lorenz gauge as

$$ \nabla^{2} \boldsymbol{\pi}_{\mathrm{e}}-\frac{1}{c^{2}} \frac{\partial^{2} \boldsymbol{\pi}_{\mathrm{e}}}{\partial t^{2}}=-\frac{\mathbf{P} }{\epsilon_{0}} $$.

The electric field in terms of the Hertz vectors is given as

$$\mathbf{E}=\nabla \times \nabla \times \boldsymbol\pi_{\mathrm{e}}-\nabla \times \frac{\partial \boldsymbol{\pi}_{\mathrm{m}}}{\partial t}-\frac{\mathbf{P}}{\epsilon_{0}}$$,

but the magnetic Hertz vector $$ \boldsymbol{\pi}_{\mathrm{m}} $$is 0 since the the material is assumed to be non-magnetizable and there is no external magnetic field. Therefore the electric field simplifies to

$$ \mathbf{E}=\nabla \times \nabla \times \boldsymbol\pi_{\mathrm{e}}-\frac{\mathbf{P}}{\epsilon_{0}} $$.

In order to calculate the electric field we must first solve the inhomogeneous wave equation for $$ \boldsymbol{\pi}_{\mathrm{e}} $$. To do this, split $$ \boldsymbol{\pi}_{\mathrm{e}} $$in the homogeneous and particular solutions

$$ \boldsymbol{\pi}_{\mathrm{e}}(\mathbf{r}, t)=\boldsymbol{\pi}_{e,h}(\mathbf{r}, t)+\boldsymbol{\pi}_{\mathrm{e,p}}(\mathbf{r}, t) $$.

Linearity then allows us to write

$$ \mathbf{E}(\mathbf{r}, t)=\mathbf{E}_{h}(\mathbf{r}, t)+\mathbf{E}_{\mathrm{p}}(\mathbf{r}, t) $$.

The homogeneous solution, $$ \mathbf{E}_{h}(\mathbf{r}, t) $$, is the initial plane wave traveling with wave vector $$ k_{0}=\omega/c

$$ in the positive $$z$$ direction

$$ \mathbf{E}_{h}(\mathbf{r}, t) = \mathbf{E}_{0}

e^{i\left(k_{0} z-\omega t\right)} . $$

We do not need to explicitly find $$ \boldsymbol{\pi}_{e,h}(\mathbf{r}, t) $$since we are only interested in finding the field.

The particular solution, $$ \boldsymbol{\pi}_{\mathrm{e,p}}(\mathbf{r}, t) $$and therefore, $$ \mathbf{E}_{\mathrm{p}}(\mathbf{r}, t) $$, is found using a time dependent Green's function method on the inhomogeneous wave equation for $$ \boldsymbol{\pi}_{\mathrm{e,p}} $$which produces the retarded integral

$$ \boldsymbol{\pi}_{\mathrm{e,p}}(\mathbf{r}, t)=\frac{1}{4 \pi \epsilon_{0}} \int d^{3} r^{\prime} \frac{\mathbf{P}\left(\mathbf{r}^{\prime}, t-\left|\mathbf{r}-\mathbf{r}^{\prime}\right| / c\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} $$.

Since the initial electric field is polarizing the material, the polarization vector must have the same space and time dependence $$\mathbf{P}(\mathbf{r}, t)=\mathbf{P}_{0} e^{ik z-\omega t}.$$Plugging this into the integral and expressing in terms of Cartesian coordinates produces

$$ \boldsymbol{\pi}_{\mathrm{e,p}}(\mathbf{r}, t)=\frac{\mathbf{P}_{0} e^{ i(k z-\omega t)}} {4 \pi \epsilon_{0}} \int_{0}^{\infty} d z^{\prime} e^{ i k\left(z^{\prime}-z\right)} \int_{-\infty}^{\infty} d x^{\prime} \int_{-\infty}^{\infty} d y^{\prime} \frac{e^{ i k_{0}\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}. $$

First, consider only the integration over $$x^{\prime} $$and $$y^{\prime} $$and convert this to cylindrical coordinates $$(x,y,z)\rightarrow(\rho,\varphi,z) $$and call $$ {\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=R $$

$$ I:=\int_{-\infty}^{\infty} d x^{\prime} \int_{-\infty}^{\infty} d y^{\prime} \frac{e^{ i k_{0}\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} = \int_{0}^{2\pi} d \varphi^{\prime} \int_{0}^{\infty} d \rho^{\prime} \frac{e^{ i k_{0}R}}{R} = 2\pi \int_{0}^{\infty} d \rho^{\prime} \frac{e^{ i k_{0}R}}{R}. $$

Then using the substitution

$$R^{2}=\rho^{2}+\left|z^{\prime}-z\right|^{2} \Rightarrow \rho^{2}=R^{2}-\left|z^{\prime}-z\right|^{2} \Rightarrow \rho d \rho =RdR $$

and

$$\rho=\sqrt{R^{2}-\left|z^{\prime}-z\right|^{2}} $$

so the limits become

$$\rho=0=\sqrt{R^{2}-\left|z^{\prime}-z\right|^{2}} \Rightarrow R=\left|z^{\prime}-z\right| $$

and

$$\rho = \infty=\sqrt{R^{2}-\left|z^{\prime}-z\right|^{2}} \Rightarrow R=\infty. $$

Then introduce a convergence factor $$e^{-\epsilon R} $$with $$\epsilon \in \R $$ into the integrand since it does not change the value of the integral,

$$ \begin{align} I &= 2\pi \int_{\left|z^{\prime}-z\right|}^{\infty} d R e^{ i k_{0}R} \\&= 2\pi \lim_{\epsilon \to 0}\int_{\left|z^{\prime}-z\right|}^{\infty} d R e^{(i k_{0}-\epsilon)R} \\&= \left. 2\pi \lim_{\epsilon \to 0} \left[\frac{e^{(i k_{0}-\epsilon)R}}{i k_{0}-\epsilon}\right] \right|_{\left|z^{\prime}-z\right|}^\infty \\&= 2\pi \lim_{\epsilon \to 0} \left[\frac{e^{(i k_{0}-\epsilon)\infty}}{i k_{0}-\epsilon}-\frac{e^{(i k_{0}-\epsilon){\left|z^{\prime}-z\right|}}}{i k_{0}-\epsilon}\right]

\end{align} $$

Then $$\epsilon \in \R $$implies $$\lim_{\epsilon \to 0} e^{-\epsilon\infty}=0 $$, hence $$\lim_{\epsilon \to 0}e^{(i k_{0}-\epsilon)\infty} = \lim_{\epsilon \to 0}e^{i k_{0}\infty} e^{-\epsilon\infty}=0 $$. Therefore,

$$ \begin{align} I &= 2\pi \lim_{\epsilon \to 0} \left[0-\frac{e^{(i k_{0}-\epsilon){\left|z^{\prime}-z\right|}}}{i k_{0}-\epsilon}\right] \\&= -2\pi \frac{e^{i k_{0}{\left|z^{\prime}-z\right|}}}{i k_{0}} \\&= 2\pi i \frac{e^{i k_{0}{\left|z^{\prime}-z\right|}}}{k_{0}}

\end{align} $$

Now, plugging this result back into the z-integral yields

$$ \boldsymbol{\pi}_{\mathrm{e,p}}(z, t)= \frac{i\mathbf{P}_{0}

e^{ i(k z-\omega t)}

}{2 k_{0}\epsilon_{0}} \int_{0}^{\infty} d z^{\prime}

e^{ i k\left(z^{\prime}-z\right) }

e^{ i k_{0}\left|z-z^{\prime}\right|} $$

Notice that $$ \boldsymbol{\pi}_{\mathrm{e,p}} $$is now only a function of $$ z $$and not $$ \mathbf{r},

$$which was expected for the given symmetry.

This integration must be split into two due to the absolute value $$\left|z-z^{\prime}\right|$$inside the integrand. The regions are $$z<0$$and $$z>0.$$ Again, a convergence factor must be introduced to evaluate both integrals and the result is

$$\boldsymbol{\pi}_{\mathrm{e,p}}(z, t)=-\frac{\mathbf{P}e^{ -i \omega t}} {2 \epsilon_{0} k_{0}}

\left\{\begin{array}{l} & {z<0} \\ {\frac{2 k_{0}}{k_{0}^{2}-k^{2}} e^{ i k z} + \frac{1}{k-k_{0}} e^{ i k_{0} z}} & {z>0.} \end{array}\right.$$

Instead of plugging $$ \boldsymbol{\pi}_{\mathrm{e,p}} $$directly into the expression for the electric field, several simplifications can be made. Begin with the curl of the curl vector identity,

$$ \nabla \times(\nabla \times \mathbf{\boldsymbol{\pi}_{\mathrm{e,p}}})=\nabla(\nabla \cdot \mathbf{\boldsymbol{\pi}_{\mathrm{e,p}}})-\nabla^{2} \mathbf{\boldsymbol{\pi}_{\mathrm{e,p}}} $$,

therefore,

$$ \begin{align} \mathbf{E}_{\mathrm{p}} &= \nabla \times \nabla \times \boldsymbol\pi_{\mathrm{e}}-\frac{\mathbf{P}}{\epsilon_{0}} \\&= \nabla(\nabla \cdot \boldsymbol{\pi}_{\mathrm{e,p}})-\nabla^{2} \boldsymbol{\pi}_{\mathrm{e,p}}-\frac{\mathbf{P}}{\epsilon_{0}}.

\end{align}

$$

Notice that $$\nabla \cdot \boldsymbol{\pi}_{\mathrm{e,p}}=0$$because $$\mathbf{P}$$has no $${\mathbf{z}}$$dependence and is always perpendicular to $$\hat{\mathbf{z}}.$$Also, notice that the second and third terms are equivalent to the inhomogeneous wave equation, therefore,

$$ \begin{align} \mathbf{E}_{\mathrm{p}} &= -\frac{1}{c^{2}} \frac{\partial^{2} \boldsymbol{\pi}_{\mathrm{e,p}}}{\partial t^{2}} \\&= -\frac{1}{c^{2}} (-i\omega)^2 \boldsymbol{\pi}_{\mathrm{e,p}} \\&= k_{0}^2 \boldsymbol{\pi}_{\mathrm{e,p}}

\end{align} $$

Therefore, the total field is

$$ \mathbf{E}(z, t)=\mathbf{E}_{0} e^{i\left(k_{0} z-\omega t\right)} +k_{0}^{2} \boldsymbol\pi_{\mathrm{e,p}}(z, t) $$

which becomes,

$$\mathbf{E}(z, t) = \left\{\begin{array}{l} {\mathbf{E}_{0} e^{i\left(k_{0} z-\omega t\right)}-\frac{\mathbf{P}}{2 \epsilon_{0}} \frac{k_{0}}{k+k_{0}} e^{-i\left(k_{0} z+\omega t\right)}} & {z<0} \\ {\mathbf{E}_{0} e^{i\left(k_{0} z-\omega t\right)}-\frac{\mathbf{P}}{2 \epsilon_{0}} \frac{k_{0}}{k-k_{0}} e^{i\left(k_{0} z-\omega t\right)}-\frac{\mathbf{P}}{\epsilon_{0}} \frac{k_{0}^{2}}{k_{0}^{2}-k^{2}} e^{i(k z-\omega t)}} & {z>0}\end{array}\right.$$

Now focus on the field inside the dielectric. Using the fact that $$ \mathbf{E}(z, t) $$is complex, we may immediately write

$$ \mathbf{E}(z>0, t)=\mathbf{E} e^{i\left(k_{0} z-\omega t\right)} $$

recall also that inside the dielectric we have $$\mathbf{P}=\epsilon_{0} \chi \mathbf{E}$$.

Then by coefficient matching we find,

$$ e^{i\left(k z-\omega t\right)} \Rightarrow 1=-\chi \frac{k_{0}^{2}}{k_{0}^{2}-k^{2}} $$

and

$$ e^{i\left(k_{0} z-\omega t\right)} \Rightarrow

0=\mathbf{E}_{0}-\frac{\chi}{2} \frac{k_{0}}{k-k_{0}} \mathbf{E} $$.

The first relation quickly yields the wave vector in the dielectric in terms of the incident wave as

$$ \begin{align} k &=\sqrt{1+\chi} k_{0}

\\&=n k_{0}.

\end{align} $$

Using this result and the definition of $$ \mathbf{P}

$$in the second expression yields the polarization vector in terms of the incident electric field as

$$ \mathbf{P} = 2 \epsilon_{0}(n-1) \mathbf{E}_{0}. $$

These results can be immediately used substituted into the the expression for the electric field

$$\mathbf{E}(z, t) = \left\{\begin{array}{l} {\mathbf{E}_{0} e^{i\left(k_{0} z-\omega t\right)}

-\left(\frac{n-1}{n+1}\right) \mathbf{E}_{0} e^{-i\left(k_{0} z+\omega t\right)}} & {z<0} \\ {\left(\frac{2}{n+1}\right) \mathbf{E}_{0} e^{i\left(n k_{0} z-\omega t\right)}} & {z>0}\end{array}\right.$$

$$\boldsymbol{\pi}_{\mathrm{p}}(z, t)=-\frac{\mathbf{P}} {2 \epsilon_{0} k_{0}}

\left\{\begin{array}{l} {} & {z<0} \\ {} & {z>0}\end{array}\right.$$