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Classical light-waves

Consider a classical, monochromatic, linearly polarized, plane light-wave, propagating through a vacuum in the $x$-direction. It is convenient to characterize a light-wave (which is, of course, a type of electromagnetic wave) by giving its associated electric field. Suppose that the wave is polarized such that this electric field oscillates in the $y$-direction. (According to standard electromagnetic theory, the magnetic field oscillates in the $z$-direction, in phase with the electric field, with an amplitude which is that of the electric field divided by the velocity of light in vacuum.) Now, the electric field can be conveniently represented in terms of a complex wave-function:
\begin{displaymath}
\psi (x,t) = \bar{\psi}\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)}.
\end{displaymath} (27)

Here, ${\rm i} = \sqrt{-1}$, $k$ and $\omega$ are real parameters, and $\bar{\psi}$ is a complex wave amplitude. By convention, the physical electric field is the real part of the above expression. Suppose that
\begin{displaymath}
\bar{\psi} = \vert\bar{\psi}\vert\,{\rm e}^{-{\rm i}\,\varphi},
\end{displaymath} (28)

where $\varphi$ is real. It follows that the physical electric field takes the form
\begin{displaymath}
E_y(x,t) = {\rm Re}[\psi(x,t)] = \vert\bar{\psi}\vert\,\cos(k\,x-\omega\,t -\varphi),
\end{displaymath} (29)

since $\exp({\rm i}\,\theta)\equiv \cos\theta+{\rm i}\,\sin\theta$. As is well-known, the cosine function is a periodic function with period $2\pi$: i.e., $\cos(\theta+2\pi)\equiv \cos\theta$ for all $\theta $. Hence, the wave electric field is periodic in space, with period
\begin{displaymath}
\lambda = \frac{2\pi}{k},
\end{displaymath} (30)

and periodic in time, with period $T=2\pi/\omega$, and repetition frequency
\begin{displaymath}
\nu= T^{-1} = \frac{\omega}{2\pi}.
\end{displaymath} (31)

In other words, $E_y(x+\lambda,t+T)=E_y(x,t)$ for all $x$ and $t$. Here, $\lambda $ is called the wave-length, whereas $\nu $ is called the wave frequency. The parameters $k$ and $\omega$ are known as the wave-number, and the wave angular frequency, respectively. It is generally more convenient to represent a wave in terms of $k$ and $\omega$, rather than $\lambda $ and $\nu $. The parameter $\vert\bar{\psi}\vert$ is obviously the amplitude of the electric field oscillation (since $\cos\theta$ oscillates between $+1$ and $-1$ as $\theta $ varies). Finally, the parameter $\varphi$, which determines the positions and times of the wave peaks and troughs, is known as the phase-angle. Note that the complex wave amplitude $\bar{\psi}$ specifies both the amplitude and the phase-angle of the wave--see Eq. (28).

According to electromagnetic theory, light-waves propagate through a vacuum at the fixed velocity $c= 3\times 10^{\,8}\,{\rm m/s}$. So, from standard wave theory,

\begin{displaymath}
c = \nu\,\lambda,
\end{displaymath} (32)

or
\begin{displaymath}
\omega = k\,c.
\end{displaymath} (33)

Equations (29) and (33) yield
\begin{displaymath}
E_y(x,t) =\vert\bar{\psi}\vert\,\cos\left(k\,[x-(\omega/k)\,...
...)= \vert\bar{\psi}\vert\,\cos\left(k\,[x-c\,t]-\varphi\right).
\end{displaymath} (34)

Note that $E_y$ depends on $x$ and $t$ only via the combination $x-c\,t$. It follows that wave peaks and troughs satisfy
\begin{displaymath}
x - c\, t = {\rm constant}.
\end{displaymath} (35)

Thus, the wave peaks and troughs indeed propagate in the $x$-direction at the fixed velocity
\begin{displaymath}
\frac{dx}{dt} = c.
\end{displaymath} (36)

An expression, such as (33), which determines the wave angular frequency as a function of the wave-number, is generally termed a dispersion relation. Furthermore, it is clear, from Eq. (34), that a plane-wave propagates at the characteristic velocity

\begin{displaymath}
v_p = \frac{\omega}{k},
\end{displaymath} (37)

which is known as the phase-velocity. Hence, it follows from Eq. (33) that the phase-velocity of a plane light-wave is $c$.

Finally, from standard electromagnetic theory, the energy density (i.e., the energy per unit volume) of a light-wave is

\begin{displaymath}
U = \frac{E_y^{\,2}}{\epsilon_0},
\end{displaymath} (38)

where $\epsilon_0= 8.85\times 10^{-12}\,{\rm F/m}$ is the permittivity of free space. Hence, it follows from Eqs. (27) and (29) that
\begin{displaymath}
U \propto \vert\psi\vert^{\,2}.
\end{displaymath} (39)

Furthermore, a light-wave possesses linear momentum, as well as energy. This momentum is directed along the wave's direction of propagation, and is of density
\begin{displaymath}
G = \frac{U}{c}.
\end{displaymath} (40)


next up previous contents
Next: The photoelectric effect Up: Wave-particle duality Previous: Introduction   Contents
Richard Fitzpatrick 2006-12-12