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\documentclass[letterpaper]{article}
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\title{Using the One Dimensional Wave Equation to Represent Electromagnetic Waves in a Vacuum}
\author{Eric Minor}
\affil{Department of Physics, University of Colorado Boulder}
\date{4/23/2016}
\begin{document}
\maketitle
\begin{abstract}
The differential wave equation can be used to describe electromagnetic waves in a vacuum. In the one dimensional case, this takes the form $\frac{\partial^2\phi}{\partial x^2}-\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2} = 0$. A general function $f(x,t) = x \pm ct$ will propagate with speed c. To represent the properties of electromagnetic waves, however, the function $\phi(x,t) = \phi _0 sin(kx-\omega t)$ must be used. This gives the Electric and Magnetic field equations to be $E (z,t) = \hat{x} E _0 sin(kz-\omega t)$ and $B (z,t) = \hat{y} B _0 sin(kz-\omega t)$. Using this solution as well as Maxwell's equations the relation $\frac{E_0}{B_0} = c$ can be derived. In addition, the average rate of energy transfer can be found to be $\bar{S} = \frac{E_0 ^2}{2 c \mu _0} \hat{z}$ using the poynting vector of the fields.
\end{abstract}
\section{Introduction}
In 1861 James Maxwell published a set of equations to describe electromagnetism, which henceforth became known as Maxwell's equations [1]. They are as follows
\begin{enumerate}
\item $\nabla \cdot E = \frac{\rho}{\epsilon_0}$
\item $\nabla \cdot B = 0$
\item $\nabla \times B - \mu_0 \epsilon_0 \frac{\partial E}{\partial t} = \mu_0 J$
\item $\nabla \times E + \frac{\partial B}{\partial t} = 0$
\end{enumerate}
These equations describe the generation and properties of Electric and Magnetic fields. Applying these equations to the one dimensional wave equation reveals that Maxwell's equations can generate planes waves consisting of Electric and Magnetic Fields.
\section{Solving the Wave Equation}
\subsection{Propagation Speed}
The speed of light has been experimentally measured to a high degree of accuracy, and thus any solution for the propagation of an electromagnetic wave in a vacuum must propagate with speed c. If $\phi (x,t)$ is our waveform, then the parameters x and t must sum in the form $\phi (x,t) = f(x \pm ct) $ in order to generate a propagation speed of c. In order to track a point on f, the input value must be held constant. Calling our point $f(k)$ means that $k=x \pm ct$, so the x value must change with speed c in order for the quantity k to remain constant. The direction of the wave's propagation is determined by the sign of ct, with negative resulting in a wave moving in the positive x direction and positive resulting in a wave moving in the negative x direction
\subsection{Wave Equation}
[2] The function that satisfies both Maxwell's equations and the one dimensional wave equation is $\phi (x,t) = \phi _0 sin(kx-\omega t)$. To prove this, it is only necessary to differentiate $\phi (x,t)$ twice with respect to x and t.
\begin{description}
\item[First Derivatives] $\frac{\partial \phi}{\partial x} = k \phi_0 \cos(kx- \omega t)$, $\frac{\partial \phi}{\partial t} = - \omega \phi_0 \cos(kx- \omega t)$
\item[Second Derivatives] $\frac{\partial ^2 \phi}{\partial x^2} = -k^2 \phi_0 \sin(kx- \omega t)$, $\frac{\partial ^2 \phi}{\partial t^2} = - \omega ^2 \phi_0 \sin(kx- \omega t)$
Using these in the wave equation results in the following:
$$k^2 \phi \sin (kx - \omega t) = \frac{\omega ^2}{c^2} \phi_0 \sin (kx - \omega t)$$
\end{description}
Eliminating terms that appear on both sides results in:
$$k^2 = \frac{\omega ^2}{c^2}$$
Since the speed of a wave is given by $v = \frac{\omega}{k}$ and we know $v=c$, the equation can be rewritten as $k= \frac{\omega}{c}$. $k^2$ is therefore $\frac{\omega^2}{c^2}$, so $\phi (x,t) = \phi _0 sin(kx-\omega t)$ solves the wave equation.
\subsection{Wave Properties}
Having the solution be a sine wave entails certain properties. A sine wave takes completes a cycle in $2\pi$ radians, so the angular frequency is related to the frequency of the wave by $f=\frac{\omega}{2\pi}$. Since the frequency is the inverse of the period $T=\frac{2\pi}{\omega}$. The sine wave will also complete a cycle by holding time constant and moving one full wavelength. Since this requires $2\pi$ radians, k must equal $\frac{2\pi}{\lambda}$, where lambda is the wavelength.
\subsection{Energy Conservation}
Since we are considering an electromagnetic wave propagating in a perfect vacuum, energy must be conserved. The energy density of an electric field is given by [3] $\frac{1}{2} \epsilon_0 E^2$. The energy density of a magnetic field is given by [3] $\frac{1}{2} \frac{B^2}{\mu_0}$. Summing these gives $E_{den} = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$. Substituting in our equation for both the Electric and Magnetic fields yields
$$E_{den} = \frac{1}{2} \epsilon_0 E _0 ^2 \sin(kx-\omega t)^2 + \frac{1}{2} \frac{B _0 ^2 \sin(kx-\omega t)^2}{\mu_0}$$
Factoring out the sine portion gives
$$E_{den} = (\frac{1}{2} \epsilon_0 E _0 ^2 + \frac{1}{2} \frac{B _0 ^2 }{\mu_0}) \sin(kx-\omega t)^2$$
Integrating over one wavelength (respect to x) gives
$$E_{\lambda} = (\frac{1}{2} \epsilon_0 E _0 ^2 + \frac{1}{2} \frac{B _0 ^2 }{\mu_0}) \frac{\lambda}{2} $$
This result is not dependent on t, and thus energy is not being gained nor lost over time.
\subsection{Mathematical Basis}
In order to prove that the wave equation satisfies Maxwell's Equations, a proof of the divergence of the curl and the curl of curl vector identities must be established [4].
\subsubsection{Divergence of the Curl}
Divergence of the curl states that
\begin{equation}
\nabla \cdot (\nabla \times F) = 0.
\end{equation}
To prove this, we need only to expand the terms.
$$\nabla \times F = \hat{i} (\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}) - \hat{j} (\frac{\partial F_z}{\partial x} - \frac{\partial F_x}{\partial z}) + \hat{k} (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y})$$
The dot product yields
$$\nabla \cdot (\nabla \times F) = \frac{\partial F_z}{\partial x \partial y} - \frac{\partial F_y}{\partial x \partial z} - \frac{\partial F_z}{\partial y \partial x} + \frac{\partial F_x}{\partial y \partial z} + \frac{\partial F_y}{\partial z \partial x} - \frac{\partial F_x}{\partial z \partial y}$$
The terms can be rearranged to form
$$\nabla \cdot (\nabla \times F) = (\frac{\partial F_z}{\partial x \partial y}- \frac{\partial F_z}{\partial y \partial x}) + (\frac{\partial F_y}{\partial z \partial x} - \frac{\partial F_y}{\partial x \partial z}) + (\frac{\partial F_x}{\partial y \partial z} - \frac{\partial F_x}{\partial z \partial y})$$
as long as all partials are continuous and differentiable, all terms on the right side cancel leaving
$$\nabla \cdot (\nabla \times F) = 0$$
\subsubsection{Curl of the Curl}
[5] The Curl of the Curl is
\begin{equation}
\nabla \times (\nabla \times F) = - \nabla ^2 F + \nabla (\nabla \cdot F)
\end{equation}
To prove this, an expansion of terms is sufficient
$$\nabla \times F = \hat{i} (\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}) - \hat{j} (\frac{\partial F_z}{\partial x} - \frac{\partial F_x}{\partial z}) + \hat{k} (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y})$$
\begin{multline*} \nabla \times (\nabla \times F) =
\\ \hat{i}(\frac{\partial ^2 F_y}{\partial y \partial x} - \frac{\partial ^2 F_x}{\partial y ^2} + \frac{\partial ^2 F_z}{\partial z \partial x} + \frac{\partial ^2 F_x}{\partial z^2})
\\ -\hat{j}(\frac{\partial ^2 F_y}{\partial x^2} - \frac{\partial ^2 F_x}{\partial x \partial y} - \frac{\partial ^2 F_z}{\partial z \partial y} - \frac{\partial ^2 F_y}{\partial z^2})
\\ +\hat{k}(- \frac{\partial ^2 F_z}{\partial x^2} + \frac{\partial ^2 F_x}{\partial x \partial z} - \frac{\partial ^2 F_z}{\partial y^2} - \frac{\partial ^2 F_y}{\partial y \partial z}) \\
\end{multline*}
Next week need to expand
$- \nabla ^2 F + \nabla (\nabla \cdot F)$
Starting with the term on the right we get
$$ \nabla \cdot F = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}$$
Taking the gradient gives
\begin{multline*}
\nabla (\nabla \cdot F) =
\\ \hat{i}(\frac{\partial ^2 F_x}{\partial x^2} + \frac{\partial ^2 F_y}{\partial x \partial y} + \frac{\partial ^2 F_z}{\partial x \partial z}) +
\\ \hat{i}(\frac{\partial ^2 F_x}{\partial y \partial x} + \frac{\partial ^2 F_y}{\partial y^2} + \frac{\partial ^2 F_z}{\partial y \partial z}) +
\\ \hat{i}(\frac{\partial ^2 F_x}{\partial z \partial x} + \frac{\partial ^2 F_y}{\partial z \partial y} + \frac{\partial ^2 F_z}{\partial z^2})\\
\end{multline*}
[6] The Vector Laplacian $\nabla ^2 F$ is applied to each component of a vector, yielding
\begin{multline*}
\nabla ^2 F =
\\ \hat{i}(\frac{\partial ^2 F_x}{\partial x^2} + \frac{\partial ^2 F_x}{\partial y^2} + \frac{\partial ^2 F_x}{\partial z^2}) +
\\ \hat{j}(\frac{\partial ^2 F_y}{\partial x^2} + \frac{\partial ^2 F_y}{\partial y^2} + \frac{\partial ^2 F_y}{\partial z^2}) +
\\ \hat{k}(\frac{\partial ^2 F_z}{\partial x^2} + \frac{\partial ^2 F_z}{\partial y^2} + \frac{\partial ^2 F_z}{\partial z^2})\\
\end{multline*}
Subtracting these two equations gives the final result of
\begin{multline*}
\nabla (\nabla \cdot F) - \nabla ^2 F =
\\ \hat{i}(\frac{\partial ^2 F_y}{\partial y \partial x} - \frac{\partial ^2 F_x}{\partial y ^2} + \frac{\partial ^2 F_z}{\partial z \partial x} - \frac{\partial ^2 F_x}{\partial z^2})
\\ +\hat{j}(-\frac{\partial ^2 F_y}{\partial x^2} + \frac{\partial ^2 F_x}{\partial x \partial y} + \frac{\partial ^2 F_z}{\partial z \partial y} - \frac{\partial ^2 F_y}{\partial z^2})
\\ +\hat{k}(- \frac{\partial ^2 F_z}{\partial x^2} + \frac{\partial ^2 F_x}{\partial x \partial z} - \frac{\partial ^2 F_z}{\partial y^2} - \frac{\partial ^2 F_y}{\partial y \partial z}) \\
\end{multline*}
This is equivalent to the expression given for the cross product of the cross product, so
$$ \nabla \times (\nabla \times F) = - \nabla ^2 F + \nabla (\nabla \cdot F) $$
\subsection{Satisfying Maxwell's Equations}
With the two previously proven identities, we can now show that the Wave Equation satisfies Maxwell's Equations.
\subsubsection{Electric Field}
[7] Starting with the Fourth equation we have
$$\nabla \times E = - \frac{\partial B}{\partial t}$$
Using $\nabla \times$ on both sides gives
$$\nabla \times (\nabla \times E) = - \frac{\partial \nabla \times B}{\partial t}$$
From identity (2) we know $\nabla \times (\nabla \times E) = - \nabla ^2 E + \nabla (\nabla \cdot E)$
Since there are no sources, $\rho = 0$ so the term $\nabla \cdot E = 0$ from the first of Maxwell's equations, making the term $\nabla(\nabla \cdot E) =0$. We are then left with $$\nabla \times (\nabla \times E) = -\nabla ^2 E$$. We can rearrange Maxwell's third equation to be
$$\nabla \times B = \mu _0 J + \mu _0 \epsilon _0 \frac{\partial E}{\partial t}$$
since J is 0 in a vacuum, the equation becomes
$$\nabla \times B = \mu _0 \epsilon _0 \frac{\partial E}{\partial t}$$
Substituting both of these terms into the equation
$$\nabla \times (\nabla \times E) = - \frac{\partial \nabla \times B}{\partial t}$$
and canceling the negative signs gives
$$ \nabla ^2 E = \mu _0 \epsilon _0 \frac{\partial ^2 E}{\partial ^2 t}$$
Since this is the one dimensional case $\nabla ^2 E = \frac{\partial ^2 E}{\partial x^2}$ which gives
$$ \frac{\partial ^2 E}{\partial x^2} = \mu _0 \epsilon _0 \frac{\partial ^2 E}{\partial ^2 t}$$
which is the form of the wave equation. This also implies that $\mu _0 \epsilon _0 = \frac{1}{c^2}$, which can be rearranged to give $c = \frac{1}{\sqrt{\mu _0 \epsilon _0}}$.
\subsubsection{Magnetic Field}
Starting with the third of Maxwell's equations, we have
$$\nabla \times B - \mu _0 \epsilon _0 \frac{\partial E}{\partial t} = \mu _0 J$$
Since $J=0$ the equation becomes
$$\nabla \times B - \mu _0 \epsilon _0 \frac{\partial E}{\partial t} = 0$$
Rearranging and applying $\nabla \times $ to the equation yields
$$\nabla \times (\nabla \times B) = \mu _0 \epsilon _0 \frac{\partial \nabla \times E}{\partial t}$$
Identity (2) gives $\nabla \times (\nabla \times B) = - \nabla ^2 B + \nabla (\nabla \cdot B)$. Maxwell's second equation says $ \nabla \cdot B = 0$. Applying this to the identity gives $\nabla \times (\nabla \times B) = - \nabla ^2 B$. \\
From Maxwell's fourth equation, we get $\nabla \times E = - \frac{\partial B}{\partial t}$.
Substituting these two equations into our modified form of Maxwell's third equation gives
$$ \nabla ^2 B = \mu _0 \epsilon _0 \frac{\partial ^2 B}{\partial t^2}$$
Since this is the one dimensional case $\nabla ^2 B$ is just $\frac{\partial ^2 B}{\partial x^2}$ which gives us
$$ \frac{\partial ^2 B}{\partial x^2} = \mu _0 \epsilon _0 \frac{\partial ^2 B}{\partial t^2}$$
This is the form of the wave equation and, like the electric field version, implies that $c=\frac{1}{\sqrt{\mu _0 \epsilon _0}}$
\subsection{Magnetic and Electric Field Waves}
In the Wave Equation section we found $\frac{\partial^2\phi}{\partial x^2}-\frac{1}{c^2}\frac{\partial^2\phi}{\partial x^2} = 0$ was solved by $\phi (x,t) = \phi _0 sin(kx-\omega t)$. Likewise, the Magnetic and Electric fields were shown to take the form $ \frac{\partial ^2 B}{\partial x^2} = \mu _0 \epsilon _0 \frac{\partial ^2 B}{\partial t^2}$ and $ \frac{\partial ^2 E}{\partial x^2} = \mu _0 \epsilon _0 \frac{\partial ^2 E}{\partial ^2 t}$ Which yields solutions $B (z,t) = \hat{y} B _0 sin(kz-\omega t)$ and $E (z,t) = \hat{x} E _0 sin(kz-\omega t)$. We know the directions of the E and B fields must be perpendicular in an electromagnetic wave, so the directions $\hat{x}$ and $\hat{y}$ were assigned to give an orientation.
\subsubsection{Relationship between $E_0$ and $B_0$}
By examining Maxwell's third equation, we can derive a relationship between the values of $E_0$ and $B_0$. Maxwell's third equation states
$$\nabla \times B - \mu_0 \epsilon_0 \frac{\partial E}{\partial t} = \mu_0 J$$
Taking the proper derivatives and curl we get
$$\nabla \times B = -k \hat{i} B_0 \cos (kz-\omega t)$$
and
$$\frac{\partial E}{\partial t} = -\omega E_0 \cos (kz-\omega t) \hat{k}$$
Substituting these into Maxwell's third equation and using J = 0 we obtain
$$\omega \mu _0 \epsilon _0 \cos (kz-\omega t) \hat{k}= kB_0 \cos(kz-\omega t)\hat{k}$$
Dividing out like terms yields
$$\omega \mu _0 \epsilon _0 E_0 = kB_0$$
which can be rearranged to obtain
$$\frac{E_0}{B_0} = \frac{k}{\omega \mu _0 \epsilon _0}$$
It has been shown that $\omega = ck$ and that $ c = \frac{1}{\sqrt{\mu _0 \epsilon _0}}$
so the equation can be rewritten as
$$\frac{E _0}{B _0} = c$$
\subsubsection{Poynting Vector}
The Poynting Vector of an Electromagnetic wave is defined as
$$S = \frac{1}{\mu_ 0 }E \times B$$
Carrying out the cross product yields
$$S = \frac{E_0 B_0 \sin ^2 (kz-\omega t)}{\mu _0} \hat{z}$$
Taking the average intensity of the wave over a period yields
$$\bar{S} = \frac{E_0 B_0}{2 \mu _0} \hat{z}$$ since we know the relationship between $E_0$ and $B_0$ to be $\frac{E_0}{B_0} = c$ the expression can also be written as \\
$\bar{S} = \frac{E_0 ^2}{2 c \mu _0} \hat{z}$ or $\bar{S} = \frac{B_0 ^2 c}{2 \mu _0} \hat{z}$
\section{Conclusion}
It has been shown that the one dimensional wave equation can be used to satisfy Maxwell's equations for both the Electric and Magnetic fields, and that this solution can be used to derive fundamental properties of an electromagnetic waves. The equations for the Electric and Magnetic fields were computed to be $E (z,t) = \hat{x} E _0 sin(kz-\omega t)$ and $B (z,t) = \hat{y} B _0 sin(kz-\omega t)$ respectively. This was proved using Maxwell's equations as well as the curl of the curl and divergence of the curl vector identities. Using Maxwell's third equation it was determined that $E_0$ and $B_0$ were related by the equation $\frac{E_0}{B_0} = c$. Furthermore, the poynting vector of the two fields was used to determine the average rate of energy transfer of the waves, which came out to be $\bar{S} = \frac{E_0 ^2}{2 c \mu _0} \hat{z}$. The one dimensional wave equation can thus be seen as a useful representation of electromagnetic waves propagating in a vacuum, and provides a basis for the properties of the wave.
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\end{document}
```