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Next: Polarization of Plane Waves Up: The Free Space Wave Previous: The Wave Equation   Contents

Plane Waves

Plane waves can propagate in any direction. Any superposition of these waves, for all possible $\omega,\mbox{\boldmath$k$}$, is also a solution to the wave equation. However, recall that $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$B$}$ are not independent, which restricts the solution in electrodynamics somewhat.

To get a feel for the interdependence of $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$B$}$, let's pick $\mbox{\boldmath$k$} = \pm k\hat{\mbox{\boldmath$x$}}$ so that e.g.:

$\displaystyle \mbox{\boldmath$E$}(x,t)$ $\textstyle =$ $\displaystyle \mbox{\boldmath$E$}_+ e^{i(kx - \omega t)} +
\mbox{\boldmath$E$}_- e^{i(-kx - \omega t)}$ (9.22)
$\displaystyle \mbox{\boldmath$B$}(x,t)$ $\textstyle =$ $\displaystyle \mbox{\boldmath$B$}_+ e^{i(kx - \omega t)} +
\mbox{\boldmath$B$}_- e^{i(-kx - \omega t)}$ (9.23)

which are plane waves travelling to the right or left along the $x$-axis for any complex $\mbox{\boldmath$E$}_+, \mbox{\boldmath$E$}_-$, $\mbox{\boldmath$B$}_+, \mbox{\boldmath$B$}_-$. In one dimension, at least, if there is no dispersion we can construct a fourier series of these solutions for various $k$ that converges to any well-behaved function of a single variable.

[Note in passing that:

\begin{displaymath}
u(x,t) = f(x - vt) + g(x + vt)
\end{displaymath} (9.24)

for arbitrary smooth $f(z)$ and $g(z)$ is the most general solution of the 1-dimensional wave equation. Any waveform that preserves its shape and travels along the $x$-axis at speed $v$ is a solution to the one dimensional wave equation (as can be verified directly, of course). How boring! These particular harmonic solutions have this form (verify this).]

If there is dispersion (where the velocity of the waves is a function of the frequency) then the fourier superposition is no longer stable and the last equation no longer holds. Each fourier component is still an exponential, but all the velocities of the fourier components are different. As a consequence, any initially prepared wave packet spreads out as it propagates. We'll look at this shortly (in the homework) in some detail to see how this works for a very simple (gaussian) wave packet but for now we'll move on.

Note that $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$B$}$ are connected by having to satisfy Maxwell's equations even if the wave is travelling in just one direction (say, in the direction of a unit vector $\hat{\bf n}$); we cannot choose the wave amplitudes separately. Suppose

$\displaystyle \mbox{\boldmath$E$}(\mbox{\boldmath$x$},t)$ $\textstyle =$ $\displaystyle \mbox{\boldmath$\cal E$} e^{i(k\hat{\bf n} \cdot \mbox{\scriptsize\boldmath$x$} -
\omega t)}$  
$\displaystyle \mbox{\boldmath$B$}(\mbox{\boldmath$x$},t)$ $\textstyle =$ $\displaystyle \mbox{\boldmath$\cal B$} e^{i(k\hat{\bf n} \cdot \mbox{\scriptsize\boldmath$x$} -
\omega t)}$  

where $\mbox{\boldmath$\cal E$}$, $\mbox{\boldmath$\cal B$}$, and $\hat{\bf n}$ are constant vectors (which may be complex, at least for the moment).

Note that applying $(\nabla^2 + k^2)$ to these solutions in the HHE leads us to:

\begin{displaymath}
k^2 \hat{\bf n} \cdot \hat{\bf n} = \mu \epsilon \omega^2 =
\frac{\omega^2}{v^2}
\end{displaymath} (9.25)

as the condition for a solution. Then a real $\hat{\bf n} \cdot
\hat{\bf n} = 1$ leads to the plane wave solution indicated above, with $k = \frac{\omega}{v}$, which is the most familiar form of the solution (but not the only one)!

This has mostly been ``mathematics'', following more or less directly from the wave equation. The same reasoning might have been applied to sound waves, water waves, waves on a string, or ``waves'' $u(x,t)$ of nothing in particular. Now let's use some physics and see what it tells us about the particular electromagnetic waves that follow from Maxwell's equations turned into the wave equation. These waves all satisfy each of Maxwell's equations separately.

For example, from Gauss' Laws we see e.g. that:

$\displaystyle \mbox{\boldmath$\nabla$}\cdot {\mbox{\boldmath$E$}}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle \mbox{\boldmath$\nabla$}\cdot {\mbox{\boldmath$\cal E$} e^{i(k\hat{\bf n} \cdot \mbox{\scriptsize\boldmath$x$} - \omega t)} }$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle \mbox{\boldmath$\cal E$} \cdot \mbox{\boldmath$\nabla$}e^{i(k\hat{\bf n} \cdot \mbox{\scriptsize\boldmath$x$} - \omega t)}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle ik \mbox{\boldmath$\cal E$} \cdot \mbox{\boldmath$n$} e^{i(k\hat{\bf n} \cdot \mbox{\scriptsize\boldmath$x$} - \omega t)}$ $\textstyle =$ $\displaystyle 0$ (9.26)

or (dividing out nonzero terms and then repeating the reasoning for $\mbox{\boldmath$B$}$):
\begin{displaymath}
\hat{\bf n} \cdot \mbox{\boldmath$\cal E$} = 0 \quad \hbox{and} \quad \hat{\bf
n} \cdot \mbox{\boldmath$\cal B$} = 0.
\end{displaymath} (9.27)

Which basically means for a real unit vector $\hat{\bf n}$ that $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$B$}$ are perpendicular to $\hat{\bf n}$, the direction of propagation! A plane electromagnetic wave is therefore a transverse wave. This seems like it is an important thing to know, and is not at all a mathematical conclusion of the wave equation per se.

Repeating this sort of thing using one of the the curl eqns (say, Faraday's law) one gets:

\begin{displaymath}
\mbox{\boldmath$\cal B$} = \sqrt{\mu \epsilon} \left( \hat{\bf n} \times \mbox{\boldmath$\cal
E$}\right)
\end{displaymath} (9.28)

(the $i$ cancels, $k/\omega = 1/v = \sqrt{\epsilon\mu}$). This means that $E$ and $B$ have the same phase if $\hat{\bf n}$ is real9.4

If $\hat{\bf n}$ is a real unit vector in 3-space, then we can introduce three real, mutually orthogonal unit vectors $(\hat{\mbox{\boldmath$\epsilon_1$}}, \hat{\mbox{\boldmath$\epsilon_2$}}, \hat{\mbox{\boldmath$n$}})$ such that $\hat{\mbox{\boldmath$\epsilon_1$}}\times\hat{\mbox{\boldmath$\epsilon_2$}} = \hat{\mbox{\boldmath$n$}}$ and use them to express the field strengths:

\begin{displaymath}
\mbox{\boldmath$\cal E$}_1 = \hat{\mbox{\boldmath$\epsilon$...
... \hat{\mbox{\boldmath$\epsilon$}}_2
\sqrt{\mu \epsilon} E_0
\end{displaymath} (9.29)

and
\begin{displaymath}
\mbox{\boldmath$\cal E$}_2 = \hat{\mbox{\boldmath$\epsilon$...
...
\hat{\mbox{\boldmath$\epsilon$}}_1 \sqrt{\mu \epsilon} E_0'
\end{displaymath} (9.30)

where $E_0$ and $E_0'$ are constants that may be complex. It is worth noting that
\begin{displaymath}
\vert E\vert = v\vert B\vert
\end{displaymath} (9.31)

have the same dimensions and that the magnitude of the electric field is greater than that of the magnetic field to which it is coupled via Maxwell's Equations by a factor of the speed of light in the medium, as this will be used a lot in electrodynamics.

We have carefully chosen the polarization directions so that the (time-averaged) Poynting vector for any particular component pair points in the direction of propagation, $\hat{\mbox{\boldmath$n$}}$:

$\displaystyle \mbox{\boldmath$S$}$ $\textstyle =$ $\displaystyle \frac{1}{2} \left( \mbox{\boldmath$E$} \times \mbox{\boldmath$H$}^\ast
\right )$ (9.32)
  $\textstyle =$ $\displaystyle \frac{1}{2\mu} \left( \mbox{\boldmath$E$} \times \mbox{\boldmath$B$}^\ast
\right )$ (9.33)
  $\textstyle =$ $\displaystyle \frac{\sqrt{\epsilon\mu}}{2\mu} \left( \mbox{\boldmath$E$}
\times v \mbox{\boldmath$B$}^\ast \right )$ (9.34)
  $\textstyle =$ $\displaystyle \frac{1}{2} \sqrt{\frac{\epsilon}{\mu}} \mid E_0 \mid^2
\hat{\mbox{\boldmath$n$}}$ (9.35)

Note well the combination $\sqrt{\frac{\epsilon}{\mu}}$, as it will occur rather frequently in our algebra below, so much so that we will give it a name of its own later. So much for the ``simple'' monochromatic plane wave propagating coherently in a dispersionless medium.

Now, kinky as it may seem, there is no real9.5 reason that $\mbox{\boldmath$k$}= k\hat{\bf n}$ cannot be complex (while $k$ remains real!) As an exercise, figure out the complex vector of your choice such that

\begin{displaymath}
\hat{\bf n} \cdot \hat{\bf n} = 1.
\end{displaymath} (9.36)

Did you get that? What, you didn't actually try? Seriously, you're going to have to at least try the little mini-exercises I suggest along the way to get the most out of this book.

Of course, I didn't really expect for you to work it out on such a sparse hint, and besides, you gotta save your strength for the real problems later because you'll need it then. So this time, I'll work it out for you. The hint was, pretend that $\hat{\bf n}$ is complex. Then it can be written as:

\begin{displaymath}
\hat{\bf n} = \hat{\bf n}_R + i\hat{\bf n}_I
\end{displaymath} (9.37)


\begin{displaymath}
n_R^2 - n_I^2 = 1
\end{displaymath} (9.38)


\begin{displaymath}
\hat{\bf n}_R \cdot \hat{\bf n}_I = 0.
\end{displaymath} (9.39)

So, $\hat{\bf n}_R$ must be orthogonal to $\hat{\bf n}_I$ and the difference of their squares must be one. For example:
\begin{displaymath}
\hat{\bf n}_R = \sqrt{2} \, \hat{\mbox{\boldmath$\bf i$}} \quad \hat{\bf n}_I = 1 \, \hat{\mbox{\boldmath$\bf j$}}
\end{displaymath} (9.40)

works, as do infinitely more More generally (recalling the properties of hyberbolics functions):
\begin{displaymath}
\hat{\bf n} = \hat{\mbox{\boldmath$e$}}_1 \cosh \theta + i \hat{\mbox{\boldmath$e$}}_2 \sinh \theta
\end{displaymath} (9.41)

where the unit vectors are orthogonal should work for any $\theta$.

Thus the most general $\mbox{\boldmath$\cal E$}$ such that ${\bf n} \cdot
\mbox{\boldmath$\cal E$} = 0$ is

\begin{displaymath}
\mbox{\boldmath$\cal E$} = (i \hat{\mbox{\boldmath$e$}}_1 \...
...ldmath$e$}}_2 \cosh \theta) A +
\hat{\mbox{\boldmath$e$}}_3 B
\end{displaymath} (9.42)

where (sigh) $A$ and $B$ are again, arbitrary complex constants. Note that if $\hat{\bf n}$ is complex, the exponential part of the fields becomes:
\begin{displaymath}
e^{i(k\hat{\bf n} \cdot \mbox{\scriptsize\boldmath$x$} - \o...
...{\bf n}_R \cdot \mbox{\scriptsize\boldmath$x$} - \omega t)}.
\end{displaymath} (9.43)

This inhomogeneous plave wave exponentially grows or decays in some direction while remaining a ``plane wave'' in the other (perpendicular) direction.

Fortunately, nature provides us with few sources and associated media that produce this kind of behavior (imaginary $\hat{\bf n}$? Just imagine!) in electrodynamics. So let's forget it for the moment, but remember that it is there for when you run into it in field theory, or mathematics, or catastrophe theory.

We therefore return to a more mundane and natural discussion of the possible polarizations of a plane wave when $\hat{\bf n}$ is a real unit vector, continuing the reasoning above before our little imaginary interlude.


next up previous contents
Next: Polarization of Plane Waves Up: The Free Space Wave Previous: The Wave Equation   Contents
Robert G. Brown 2013-01-04