The propagation of electromagnetic radiation through a dispersive medium

Let us now investigate the propagation of electromagnetic radiation through a dispersive medium by studying a simple one-dimensional problem. Suppose that our dispersive medium extends from $x=0$, where it interfaces with a vacuum, to $x=\infty$. Suppose further that a wave is incident normally on the medium, so that the field quantities only depend on $x$ and $t$. The wave is specified as a given function of $t$ at $x=0$. Since we are not interested in the reflected wave, let this function, $f(t)$, say, give the wave amplitude just inside the surface of the dispersive medium. Suppose that the wave arrives at this surface at $t=0$, and that

$$f(t) =\left\{\begin{array}{ccl} 0 &\mbox{\hspace{2cm}}& \mbo... ...2\pi t}{\tau}\right)&&\mbox{for $t\geq 0$}. \end{array}\right.$$ (704)

How does the wave subsequently develop in the region $x>0$? In order to answer this question we must first of all decompose $f(t)$ into harmonic components of the form $\exp(-{\rm i}\,\omega t)$ (i.e., Fourier harmonics). Unfortunately, if we attempt this using only real frequencies, $\omega$, we encounter convergence difficulties, since $f(t)$ does not vanish at $t=\infty$. For the moment, we can circumvent these difficulties by only considering finite (in time) wave forms. In other words, we now imagine that $f(t) =0$ for $t<0$ and $t>T$. Such a wave form can be thought of as the superposition of two infinite (in time) wave forms, the first beginning at $t=0$ and the second at $t=T$ with the opposite phase, so that the two cancel for all time $t>T$.

According to standard Fourier transform theory

$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} d\omega \int_{-\infty}^{\infty} f(t')\,{\rm e}^{-{\rm i}\,\omega(t-t')}\,dt'.$$ (705)

Since $f(t)$ is a real function of $t$ which is zero for $t<0$ and $t>T$, we can write

$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} d\omega \int_0^Tf(t')\,\cos[\omega(t-t')\,]\,dt'.$$ (706)

Finally, it follows from symmetry (in $\omega$) that

$$f(t) = \frac{1}{\pi}\int_0^{\infty} d\omega \int_0^Tf(t')\,\cos[\omega(t-t')\,]\,dt'.$$ (707)

Equation (4.69) yields

$$f(t) = \frac{1}{\pi}\int_0^{\infty} d\omega \int_0^T \sin\!\left(\frac{2\pi t'}{\tau}\right) \,\cos[\omega(t-t')\,]\,dt',$$ (708)

or

$$f(t) = \frac{1}{2\pi}\int_0^\infty d\omega\left\{ \frac{\cos[2\pi t'/\ta... ...ac{\cos[2\pi t'/\tau -\omega(t-t')\,]}{\omega+2\pi/\tau}\right\}_{t'=0}^{t'=T}.$$

Let us assume, for the sake of simplicity, that

$$T= N\tau,$$ (709)

where $N$ is a positive integer. This ensures that $f(t)$ is continuous at $t=T$. Equation (4.74) reduces to

$$f(t) = \frac{2}{\tau} \int_0^\infty \frac{d\omega}{\omega^2 ... .../\tau)^2} \left(\, \cos[\omega(t-T)\,] -\cos\omega t\,\right).$$ (710)

This expression can be written

$$f(t) = \frac{1}{\tau} \int_{-\infty}^{\infty} \frac{d\omega}... .../\tau)^2} \left(\, \cos[\omega(t-T)\,] -\cos\omega t\,\right),$$ (711)

or

$$f(t) = \frac{1}{2\pi}\, {\rm Re}\! \int_{-\infty}^{\infty} \... ...^{-{\rm i}\,\omega(t-T)} -{\rm e}^{-{\rm i}\,\omega t}\right).$$ (712)

It is not entirely obvious that Eq. (4.78) is equivalent to Eq. (4.77). However, we can easily prove that this is the case by taking Eq. (4.78) and using the standard definition of a real part (i.e., half the sum of the quantity in question and its complex conjugate) to give

$$f(t) = \frac{1}{4\pi} \int_{-\infty}^{\infty} \frac{d\omega}{\omega -2\pi/\tau}\left( {\rm e}^{-{\rm i}\,\omega(t-T)} -{\rm e}^{-{\rm i}\,\omega t}\right)$$

$$+\frac{1}{4\pi} \int_{-\infty}^{\infty} \frac{d\omega}{\omega -2\... ...au}\left( {\rm e}^{+{\rm i}\,\omega(t-T)} -{\rm e}^{+{\rm i}\,\omega t}\right).$$ (713)

Replacing the dummy integration variable $\omega$ by $-\omega$ in the second integral and then making use of symmetry, it is easily seen that the above expression reduces to Eq. (4.77).

Equation (4.77) can be written

$$f(t) = \frac{2}{\tau}\int_{-\infty}^{\infty}d\omega\, \sin[\omega(t-T/2)]\, \frac{\sin(\omega T/2)}{\omega^2 -(2\pi/\tau)^2}.$$ (714)

Note that the integrand is finite at $\omega=2\pi/\tau$, since at this point the vanishing of the denominator is compensated for by the simultaneous vanishing of the numerator. It follows that the integrand in Eq. (4.78) is also not infinite at $\omega=2\pi/\tau$, as long as we do not separate the two exponentials. Thus, we can replace the integration along the real axis through this point by a small semi-circle in the upper half of the complex plane. Once this has been done, we can deform the path still further and can integrate the two exponentials in Eq. (4.78) separately:

$$f(t) = \frac{1}{2\pi}\,{\rm Re}\int_C {\rm e}^{-{\rm i}\,\om... ... e}^{-{\rm i}\,\omega (t-T)} \frac{d\omega}{\omega -2\pi/\tau}$$ (715)

The contour $C$ is sketched in Fig. 6. Note that it runs from $+\infty$ to $-\infty$, which accounts for the change of sign between Eqs. (4.78) and (4.81).

Figure 6: Sketch of the integration contours used to evaluate Eqs. (4.78) and (4.81)

We have already noted that a finite wave form which is zero for $t<0$ and $t>T$ can be through of as the superposition of two out of phase infinite wave forms, one starting at $t=0$ and the other at $t=T$. It is plausible, therefore, that the first term in the above expression corresponds to the infinite wave form starting at $t=0$, and the second to the infinite wave form starting at $t=T$. If this is the case then the signal (4.69), which starts at $t=0$ and ends at $t=\infty$, can be written in the form

$$f(t) = \frac{1}{2\pi}\,{\rm Re}\int_C {\rm e}^{-{\rm i}\,\omega t} \frac{d\omega}{\omega -2\pi/\tau}.$$ (716)

Let us test this proposition. In order to do this we must replace the original path of integration $C$ by two equivalent paths.

First, consider $t<0$. In this case, $-{\rm i}\,\omega t$ has a negative real part in the upper half plane which increases indefinitely with increasing distance from the axis. Thus, we can replace the original path of integration by the path $A$ (see Fig. 7). The integral clearly vanishes along this path if we let $A$ approach infinity in the upper half plane. Consequently,

$$f(t) =0$$ (717)

for $t<0$.

Next, consider $t>0$. Now, $-{\rm i}\,\omega t$ has a negative real part in the lower half plane, so that the exponential vanishes at infinity in this half plane. If we attempt to deform $C$ to infinity in the lower half plane, the path of integration ``catches'' on the singularity of the integrand at $\omega=2\pi/\tau$ (see Fig. 7). The path of integration $B$ therefore consists of three parts: the part at infinity, $B_1$, where the integral vanishes due to the exponential factor ${\rm e}^{-{\rm i}\,\omega t}$; $B_2$, the two parts leading to infinity which cancel each other and thus contribute nothing to the integral; the path $B_3$ around the singularity. This latter contribution can easily be evaluated using the Cauchy residue theorem:

$$B_3 = \frac{1}{2\pi}\, {\rm Re} \,(2\pi {\rm i}\,\,{\rm e}^{-2\pi {\rm i}\,t/\tau}) = \sin\left(\frac{2\pi t}{\tau}\right).$$ (718)

Thus, it is proven that the expression (4.82) actually describes a wave form beginning at $t=0$ whose subsequent motion is specified by Eq. (4.69).

Figure 7: Sketch of the integration contours used to evaluate Eq. (4.82)

Equation (4.82) can immediately be generalized to give the wave motion in the region $x>0$:

$$f(x,t) = \frac{1}{2\pi}\,{\rm Re}\int_C {\rm e}^{{\rm i}\,(k x- \omega t)} \frac{d\omega}{\omega -2\pi/\tau}.$$ (719)

This follows from standard wave theory, because we know that an unterminated wave motion at $x=0$ of the form ${\rm e}^{-{\rm i}\,\omega t}$ takes the form ${\rm e}^{\,{\rm i}\,(kx-\omega t)}$ after moving a distance $x$ in the dispersive medium, provided that $k$ and $\omega$ are related by the appropriate dispersion relation. For a medium consisting of a single resonant species this dispersion relation is written (see Eq. (4.17))

$$k^2 = \frac{\omega^2}{c^2}\left( 1 + \frac{(Ne^2/\epsilon_0 ... ...\omega_0^{~2} -\omega^2 -{\rm i}\,g\,\omega\,\omega_0}\right).$$ (720)