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\begin{document}

\section{Waves in Special Relativity}

\begin{figure}[h]
\psfig{figure=wave1.eps}
\caption{Wave fronts in spacetime are represented by the dashed lines.
The wavelength $\lambda$ and the period $T$ are shown for the
illustrated wave.}
\label{wave1}
\end{figure}

A spacetime picture of a wave is shown in figure \ref{wave1}.  The
wave fronts are represented by the dashed lines in this figure.  The
phase speed of the wave divided by the speed of light is $\lambda /
c T$, which is also the inverse of the slope of the wave fronts.  The
wavelength, $\lambda$, is the interval between successive wave fronts
in the space direction, while $c$ times the period, $T$, is the
interval in the time direction.

Since ``space direction'' and ``time direction'' are not
relativistically invariant concepts, it is clear that the period and
wavelength of a wave will be different in different coordinate
systems.  Let us in particular transform to the primed coordinate
system in which the wave fronts are horizontal.  This view of the wave
is shown in figure \ref{wave2}.  The velocity of the primed coordinate
system divided by the speed of light is $U/c = cT/ \lambda$.  This
follows from the earlier conclusion that the slope of the wave fronts
in the unprimed frame is $cT/ \lambda$.  In the primed frame of figure
\ref{wave2}, these wave fronts are horizontal, or parallel to the
lines of simultaneity of the primed frame.  Thus, they are parallel to
the wave fronts in the unprimed frame as well, and therefore also have
the slope $cT/ \lambda$ in this frame.  Finally, we note that the
slope of a line of simultaneity is equal to the velocity of the
associated reference frame divided by $c$, from which the above
assertion follows.

\begin{figure}
\psfig{figure=wave2.eps}
\caption{View of the wave of figure \ref{wave1} in a reference frame
in which the wave fronts are horizontal.}
\label{wave2}
\end{figure}

If the period of the wave in the primed frame is $T_0$, we can
relate the wavelength and period in the unprimed frame to $T_0$ and
$U$ using the Pythagorean theorem of spacetime.  In particular,
$A = c T_0 / (U/c) = c^2 T_0 / U$ and $B = c T_0 (U/c) = U T_0$.
Then, since $\lambda^2 = A^2 - c^2 T_0^2$ and $c^2 T^2 = c^2 T_0^2 -
B^2$, we find that $\lambda = c T_0 / \gamma$ and $T = T_0 / \gamma$,
where $\gamma = (1 - U^2 /c^2 )^{-1/2}$.

If we express wave parameters in terms of angular velocity, $\omega =
2 \pi /T$, and wavenumber, $k = 2 \pi / \lambda$, and further define
$\omega_0 = 2 \pi /T_0$, then we easily show that
\begin{equation}
	\omega = \omega_0 \gamma ~~~~~ k = \omega_0 U \gamma / c^2 .
\end{equation}
we further note that
\begin{equation}
	\omega^2 - k^2 c^2 = \omega_0^2 \gamma^2 - \omega_0^2 \gamma^2
	U^2 / c^2 = \omega_0^2 .
\label{dispreln}
\end{equation}
This is the dispersion relation for relativistic waves.  The phase and
group velocities of a wave can be obtained from the dispersion
relation.  Since $\omega_0$ is a constant, we immediately see that
$\omega d \omega = c^2 k d k$, which means that the group velocity,
$u_g \equiv d \omega / dk$ is proportional to the inverse of the phase
velocity: $u_{\phi} \equiv \omega / k = \lambda / T = c^2 / U$.  In
particular, we find that
\begin{equation}
	u_g = \frac{d \omega}{dk} = \frac{c^2 k}{\omega} = U .
\label{groupvel}
\end{equation}

What does all this mean?  First of all, if the idea of a wave front is
relativistically invariant, then the dispersion relation of the wave
must take the form given by equation (\ref{dispreln}).  Second, the
group velocity resulting from this dispersion relation behaves
sensibly when a Lorentz transformation is performed, in that it
transforms like other velocities.  Note in particular that the group
velocity is zero when the wave fronts are horizontal in spacetime.
This follows from equation (\ref{groupvel}) because $k = 0$ in this
case.  Furthermore, in a coordinate system moving with velocity $-U$
with respect to this special coordinate system, the group velocity is
$+U$.  This may seem like a trivial result, but in fact the {\em phase
velocity} does {\em not} behave this way at all!

Planck and Einstein established that light consisted of photons with
energy $E = \hbar \omega$.  Louis de Broglie extended this idea to
ordinary particles and further showed by arguments related to those
given above that the particle momentum had to be related to the
wavenumber by $p = \hbar k$.  The key point in this argument is to
note the relationship between the relativistic wave dispersion
relation, $\omega^2 - k^2 c^2 = \omega_0^2$ and the relativistic
relationship between energy and momentum, $E^2 - p^2 c^2 = m^2 c^4$.
In addition to the relationship between momentum and wavenumber, we
see that $\omega_0 = m c^2$.

\end{document}
