Difference between revisions of "Maxwell Equations"

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* [http://en.wikipedia.org/wiki/Maxwell's_equations Maxwell's Equations (Wikipedia)]
 
* [http://en.wikipedia.org/wiki/Maxwell's_equations Maxwell's Equations (Wikipedia)]
 
* [http://web.mit.edu/sahughes/www/8.022/lec20.pdf Wave equation & electromagnetic radiation (Scott Hughes, MIT)]
 
* [http://web.mit.edu/sahughes/www/8.022/lec20.pdf Wave equation & electromagnetic radiation (Scott Hughes, MIT)]
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===Related Topics===
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* [[Electromagnetic Plane Waves]]
  
 
<latex>
 
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\end{eqnarray}
 
\end{eqnarray}
 
If you plug these into the wave equation at the top of this section, you'll see that it solves the equation.  In the general 3D case for $E$ and $B$, this means that any electromagnetic waveform that translates at speed $c$ solves these equations.  And I probably needn't point out that an electromagnetic waveform traveling at the speed of light is, well... light!
 
If you plug these into the wave equation at the top of this section, you'll see that it solves the equation.  In the general 3D case for $E$ and $B$, this means that any electromagnetic waveform that translates at speed $c$ solves these equations.  And I probably needn't point out that an electromagnetic waveform traveling at the speed of light is, well... light!
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* \emph{To see hoe this generalizes to plane waves, see [[Electromagnetic Plane Waves]].}
  
 
\subsection{The Relative Orientation of $E$ and $B$}
 
\subsection{The Relative Orientation of $E$ and $B$}

Revision as of 11:37, 4 December 2017

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Reference Material

Related Topics

<latex> \documentclass[]{article} \usepackage[top=1in,bottom=1in,left=1in,right=1in]{geometry} \usepackage{amsmath} \usepackage{graphicx} \usepackage{natbib}

\begin{document} \title{Maxwell's Equations for Electromagnetic Waves}

First, a word of caution. We're going to work in CGS units, and some of the reference material above use SI units. In CGS units, there is no $\epsilon_0$ (it's incorporated into the definition of charge) and $\mu_0$ becomes $1/c^2$. In these units, Maxwell's equations are written: \begin{eqnarray} \nabla\cdot E &= 4\pi\rho\\ \nabla\cdot B &= 0\\ \nabla\times E &= -\frac1c\frac{\partial B}{\partial t}\\ \nabla\times B &= \frac{4\pi}{c}J + \frac1c\frac{\partial E}{\partial t}\\ \end{eqnarray} where $E$ is the electric field, $B$ is the magnetic field, $J$ is the current (charge per time), and $\rho$ is the charge density. In words, these equations state: \begin{enumerate} \item Electrical charge creates a divergence of in the electric field (which gives rise to a radial potential). \item Magnetic fields do not diverge. There are no magnetic ``charges" (that we know of). \item A magnetic field that changes with time creates a circular electric potential that drives charges around loops. \item A moving electrical charge (a ``current") or an electric field that changes with time creates a circular magnetic potential. \end{enumerate}

\section{Maxwell's Equations in Free Space}

In free-space, where $\rho=0$ (no charge) and $J=0$ (no current), Maxwell's equations say \begin{eqnarray} \nabla\cdot E &= 0,\\ \nabla\cdot B &= 0,\\ \nabla\times E &= -\frac1c\frac{\partial B}{\partial t},\\ \nabla\times B &= \frac1c\frac{\partial E}{\partial t}.\\ \end{eqnarray} Using the vector identity \begin{equation} \nabla\times(\nabla\times F) = \nabla(\nabla\cdot F)-\nabla^2F, \end{equation} the curl ($\nabla\times$) of the last 2 Maxwell's equations reduces to the following wave equations (using that the divergence of each field is 0): \begin{eqnarray} \nabla^2E -\frac1{c^2}\frac{\partial^2 E}{\partial t^2} &= 0,\\ \nabla^2B - \frac1{c^2}\frac{\partial^2 B}{\partial t^2} &=0.\\ \end{eqnarray} Why are these called wave equations? To make it easier to see why, let's make them one dimensional.

\subsection{A One-Dimensional Wave Equation}

Along one dimension (let's say, $\hat x$), the wave equation for the electric field (derived above as a solution of Maxwell's Equations in free space) is written \begin{equation} \frac{\partial^2 E}{\partial x^2} - \frac1{c^2}\frac{\partial^2 E}{\partial t^2} = 0. \end{equation} Let's suppose that $E(x)=E(x\pm ct)$, and we define a new variable $u\equiv x\pm ct$. The Chain Rule tells us that a spatial derivative of $E$ gives us \begin{equation} \frac{\partial E(u)}{\partial x}=\frac{\partial E}{\partial u}\frac{\partial u}{\partial x}=\frac{\partial E}{\partial u}. \end{equation} Similarly, for the time derivative, we have \begin{equation} \frac{\partial E(u)}{\partial t}=\frac{\partial E}{\partial u}\frac{\partial u}{\partial t}=\pm\frac1c\frac{\partial E}{\partial u}. \end{equation} Based on these identities, we have \begin{eqnarray} \frac{\partial^2 E}{\partial x^2}&=\frac{\partial^2 E}{\partial u^2},\\ \frac{\partial^2 E}{\partial t^2}&=c^2\frac{\partial^2 E}{\partial u^2}.\\ \end{eqnarray} If you plug these into the wave equation at the top of this section, you'll see that it solves the equation. In the general 3D case for $E$ and $B$, this means that any electromagnetic waveform that translates at speed $c$ solves these equations. And I probably needn't point out that an electromagnetic waveform traveling at the speed of light is, well... light!

\subsection{The Relative Orientation of $E$ and $B$}

From the previous section, we see that in free space, the Maxwell equations give us 2 wave equations. One propagates the $E$-field at the speed of light, and the other propagates the $B$-field at the speed of light, but we haven't related the $E$ and $B$ fields to one another. To do this, we need to re-examine the Maxwell equations for free space.

We'll defer a rigorous derivation of this until we've examined {\it plane waves} as a solution to these wave equations. In the meantime, here's a sketch of the arguments for the relative phases of $E$ and $B$. Let's suppose that $E$ and $B$ are both described by sine waves of the same frequency, but with some arbitrary phase relative to one another. Let's say that $\vec E=E_0\sin(\vec x+ct)$, so that $\vec E=0$ at $x=0,t=0$. Since a sine wave is anti-symmetric around 0, one can draw a loop encircling $x=0$, and since $\vec E$ is positive on one side of that loop, and negative on the other side, $\nabla\times E$ is not zero, and in fact, is maximal at $x=0$.

\begin{figure} \includegraphics[width=2in]{curlE.png} \caption{The curl of a sine wave (in 2 or more dimensions) is maximized around $x=0$.} \end{figure}

Since we know that \begin{equation} \nabla\times E = -\frac1c\frac{\partial B}{\partial t}, \end{equation} It must be true that $\frac{\partial B}{\partial t}$ is maximized at $x=0,t=0$. For $\sin(x+ct)$, this happens when $x=0,t=0$. So by this hand-wavy argument (which we can can make more rigorous later), $E$ and $B$ must be in-phase with one another (i.e. they hit 0 at the same time). Moreover, because $\nabla\times \vec E$ is perpendicular to $\vec E$ and perpendicular to the spatial gradient of $\vec E$ (which we've set to be in the $\hat x$ direction, the direction the wave moves with time), $\vec B$ must be perpendicular to $\vec E$ and perpendicular to the direction the wave propagates. Finally, if you work out the spatial and temporal derivatives at $x=0,t=0$, you'll see that $E_0=B_0$.