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Next: Green's functions Up: Time-dependent Maxwell's equations Previous: Potential formulation

Electromagnetic waves

This is an appropriate point at which to demonstrate that Maxwell's equations possess propagating wave-like solutions. Let us start from Maxwell's equations in free space (i.e., with no charges and no currents):
$\displaystyle \nabla\cdot{\bf E}$ $\textstyle =$ $\displaystyle 0,$ (430)
$\displaystyle \nabla\cdot{\bf B}$ $\textstyle =$ $\displaystyle 0,$ (431)
$\displaystyle \nabla\times{\bf E}$ $\textstyle =$ $\displaystyle - \frac{\partial {\bf B}}{\partial t} ,$ (432)
$\displaystyle \nabla\times{\bf B}$ $\textstyle =$ $\displaystyle \epsilon_0\mu_0 \frac{\partial {\bf E}}{\partial t}.$ (433)

Note that these equations exhibit a nice symmetry between the electric and magnetic fields.

There is an easy way to show that the above equations possess wave-like solutions, and a hard way. The easy way is to assume that the solutions are going to be wave-like beforehand. Specifically, let us search for plane-wave solutions of the form:

$\displaystyle {\bf E}({\bf r}, t)$ $\textstyle =$ $\displaystyle {\bf E}_0 \cos ({\bf k}\cdot {\bf r} - \omega  t),$ (434)
$\displaystyle {\bf B}({\bf r}, t)$ $\textstyle =$ $\displaystyle {\bf B}_0 \cos ({\bf k}\cdot {\bf r} - \omega  t+\phi).$ (435)

Here, ${\bf E}_0$ and ${\bf B}_0$ are constant vectors, ${\bf k}$ is called the wave-vector, and $\omega$ is the angular frequency. The frequency in hertz, $f$, is related to the angular frequency via $\omega = 2\pi f$. The frequency is conventionally defined to be positive. The quantity $\phi$ is a phase difference between the electric and magnetic fields. Actually, it is more convenient to write
$\displaystyle {\bf E}$ $\textstyle =$ $\displaystyle {\bf E}_0  {\rm e}^{ {\rm i} ({\bf k}\cdot{\bf r} - \omega  t)},$ (436)
$\displaystyle {\bf B}$ $\textstyle =$ $\displaystyle {\bf B}_0  {\rm e}^{ {\rm i} ({\bf k}\cdot{\bf r} - \omega  t)},$ (437)

where, by convention, the physical solution is the real part of the above equations. The phase difference $\phi$ is absorbed into the constant vector ${\bf B}_0$ by allowing it to become complex. Thus, ${\bf B}_0 \rightarrow {\bf B}_0  {\rm e}^{ {\rm i} \phi}$. In general, the vector ${\bf E}_0$ is also complex.

A wave maximum of the electric field satisfies

{\bf k}\cdot {\bf r} = \omega  t + n 2\pi+\phi,
\end{displaymath} (438)

where $n$ is an integer and $\phi$ is some phase angle. The solution to this equation is a set of equally spaced parallel planes (one plane for each possible value of $n$), whose normals lie in the direction of the wave-vector ${\bf k}$, and which propagate in this direction with phase-velocity
v = \frac{\omega}{k}.
\end{displaymath} (439)

The spacing between adjacent planes (i.e., the wave-length) is given by
\lambda = \frac{2\pi}{k}
\end{displaymath} (440)

(see Fig. 35).
Figure 35:
\epsfysize =2.5in

Consider a general plane-wave vector field

{\bf A} = {\bf A}_0  {\rm e}^{ {\rm i} ({\bf k}\cdot{\bf r} - \omega  t)}.
\end{displaymath} (441)

What is the divergence of ${\bf A}$? This is easy to evaluate. We have
$\displaystyle \nabla\cdot {\bf A}$ $\textstyle =$ $\displaystyle \frac{\partial A_x}{\partial x}+
\frac{\partial A_y}{\partial y}+...
...A_{0z}  {\rm i}  k_z)
 {\rm e}^{ {\rm i} ({\bf k}\cdot{\bf r} - \omega t)}$  
  $\textstyle =$ $\displaystyle {\rm i}  {\bf k}\cdot {\bf A}.$ (442)

How about the curl of ${\bf A}$? This is slightly more difficult. We have
$\displaystyle (\nabla\times{\bf A})_x$ $\textstyle =$ $\displaystyle \frac{\partial A_z}{\partial y}
-\frac{\partial A_y}{\partial z} = ({\rm i} k_y A_z - {\rm i} 
k_z A_y)$  
  $\textstyle =$ $\displaystyle {\rm i} ({\bf k} \times {\bf A})_x.$ (443)

This is easily generalized to
\nabla\times{\bf A} = {\rm i}  {\bf k} \times{\bf A}.
\end{displaymath} (444)

We can see that vector field operations on a plane-wave simplify to replacing the $\nabla$ operator with ${\rm i} {\bf k}$.

The first Maxwell equation (430) reduces to

{\rm i}  {\bf k} \cdot {\bf E}_0 = 0,
\end{displaymath} (445)

using the assumed electric and magnetic fields (436) and (437), and Eq. (442). Thus, the electric field is perpendicular to the direction of propagation of the wave. Likewise, the second Maxwell equation gives
{\rm i} {\bf k} \cdot {\bf B}_0 = 0,
\end{displaymath} (446)

implying that the magnetic field is also perpendicular to the direction of propagation. Clearly, the wave-like solutions of Maxwell's equation are a type of transverse wave. The third Maxwell equation gives
{\rm i} {\bf k}\times {\bf E}_0 = {\rm i}  \omega  {\bf B}_0,
\end{displaymath} (447)

where use has been made of Eq. (444). Dotting this equation with ${\bf E}_0$ yields
{\bf E}_0 \cdot {\bf B}_0 = \frac{
{\bf E}_0 \cdot {\bf k} \times {\bf E}_0 }{ \omega }= 0.
\end{displaymath} (448)

Thus, the electric and magnetic fields are mutually perpendicular. Dotting equation (447) with ${\bf B}_0$ yields
{\bf B}_0 \cdot {\bf k} \times{\bf E}_0 = \omega B_0^{ 2} > 0.
\end{displaymath} (449)

Thus, the vectors ${\bf E}_0$, ${\bf B}_0$, and ${\bf k}$ are mutually perpendicular, and form a right-handed set. The final Maxwell equation gives
{\rm i}  {\bf k}\times{\bf B}_0 = -{\rm i} \epsilon_0\mu_0  \omega 
{\bf E}_0.
\end{displaymath} (450)

Combining this with Eq. (447) yields
{\bf k}\times ({\bf k} \times {\bf E}_0) =
({\bf k} \cdot {...
..._0 =k^2 {\bf E}_0
= - \epsilon_0 \mu_0 \omega^2  {\bf E}_0,
\end{displaymath} (451)

k^2 = \epsilon_0 \mu_0  \omega^2,
\end{displaymath} (452)

where use has been made of Eq. (445). However, we know from Eq. (439) that the phase-velocity $c$ is related to the magnitude of the wave-vector and the angular wave frequency via $c = \omega/k$. Thus, we obtain
c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}.
\end{displaymath} (453)

So, we have found transverse wave solutions of the free-space Maxwell equations, propagating at some phase-velocity $c$, which is given by a combination of $\epsilon_0$ and $\mu_0$. The constants $\epsilon_0$ and $\mu_0$ are easily measurable. The former is related to the force acting between stationary electric charges, and the latter to the force acting between steady electric currents. Both of these constants were fairly well-known in Maxwell's time. Maxwell, incidentally, was the first person to look for wave-like solutions of his equations, and, thus, to derive Eq. (453). The modern values of $\epsilon_0$ and $\mu_0$ are

$\displaystyle \epsilon_0$ $\textstyle =$ $\displaystyle 8.8542\times 10^{-12}  {\rm C}^2 {\rm N}^{-1} {\rm m}^{-2},$ (454)
$\displaystyle \mu_0$ $\textstyle =$ $\displaystyle 4\pi\times 10^{-7} {\rm N}  {\rm A}^{-2}.$ (455)

Let us use these values to find the phase-velocity of ``electromagnetic waves.'' We obtain
c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} = 2.998\times 10^8 {\rm m} {\rm s}^{-1}.
\end{displaymath} (456)

Of course, we immediately recognize this as the velocity of light. Maxwell also made this connection back in the 1870's. He conjectured that light, whose nature had previously been unknown, was a form of electromagnetic radiation. This was a remarkable prediction. After all, Maxwell's equations were derived from the results of benchtop laboratory experiments, involving charges, batteries, coils, and currents, which apparently had nothing whatsoever to do with light.

Maxwell was able to make another remarkable prediction. The wave-length of light was well-known in the late nineteenth century from studies of diffraction through slits, etc. Visible light actually occupies a surprisingly narrow wave-length range. The shortest wave-length blue light which is visible has $\lambda= 0.4$ microns (one micron is $10^{-6}$ meters). The longest wave-length red light which is visible has $\lambda= 0.76$ microns. However, there is nothing in our analysis which suggests that this particular range of wave-lengths is special. Electromagnetic waves can have any wave-length. Maxwell concluded that visible light was a small part of a vast spectrum of previously undiscovered types of electromagnetic radiation. Since Maxwell's time, virtually all of the non-visible parts of the electromagnetic spectrum have been observed. Table 1 gives a brief guide to the electromagnetic spectrum. Electromagnetic waves are of particular importance because they are our only source of information regarding the universe around us. Radio waves and microwaves (which are comparatively hard to scatter) have provided much of our knowledge about the centre of our own galaxy. This is completely unobservable in visible light, which is strongly scattered by interstellar gas and dust lying in the galactic plane. For the same reason, the spiral arms of our galaxy can only be mapped out using radio waves. Infrared radiation is useful for detecting proto-stars, which are not yet hot enough to emit visible radiation. Of course, visible radiation is still the mainstay of astronomy. Satellite based ultraviolet observations have yielded invaluable insights into the structure and distribution of distant galaxies. Finally, X-ray and $\gamma$-ray astronomy usually concentrates on exotic objects in the Galaxy, such as pulsars and supernova remnants.

Table 1: The electromagnetic spectrum
Radiation Type Wave-length Range ($m$)
Gamma Rays $<10^{-11}$
X-Rays $10^{-11}$-$10^{-9}$
Ultraviolet $10^{-9}$-$10^{-7}$
Visible $10^{-7}$-$10^{-6}$
Infrared $10^{-6}$-$10^{-4}$
Microwave $10^{-4}$-$10^{-1}$
TV-FM $10^{-1}$-$10^1$
Radio $>10^1$

Equations (445), (447), and the relation $c = \omega/k$, imply that

B_0= \frac{E_0}{c}.
\end{displaymath} (457)

Thus, the magnetic field associated with an electromagnetic wave is smaller in magnitude than the electric field by a factor $c$. Consider a free charge interacting with an electromagnetic wave. The force exerted on the charge is given by the Lorentz formula
{\bf f} = q  ({\bf E} +{\bf v}\times{\bf B}).
\end{displaymath} (458)

The ratio of the electric and magnetic forces is
\frac{f_{\rm magnetic}}{f_{\rm electric}} \sim \frac{v B_0}{E_0} \sim
\end{displaymath} (459)

So, unless the charge is relativistic, the electric force greatly exceeds the magnetic force. Clearly, in most terrestrial situations electromagnetic waves are an essentially electric phenomenon (as far as their interaction with matter goes). For this reason, electromagnetic waves are usually characterized by their wave-vector (which specifies the direction of propagation and the wave-length) and the plane of polarization (i.e., the plane of oscillation) of the associated electric field. For a given wave-vector ${\bf k}$, the electric field can have any direction in the plane normal to ${\bf k}$. However, there are only two independent directions in a plane (i.e., we can only define two linearly independent vectors in a plane). This implies that there are only two independent polarizations of an electromagnetic wave, once its direction of propagation is specified.

Let us now derive the velocity of light from Maxwell's equation the hard way. Suppose that we take the curl of the fourth Maxwell equation, Eq. (433). We obtain

\nabla\times\nabla\times{\bf B} = \nabla(\nabla\cdot{\bf B})...
...on_0\mu_0 \frac{\partial  \nabla\times {\bf E}}{\partial t}.
\end{displaymath} (460)

Here, we have used the fact that $\nabla\cdot {\bf B}=0$. The third Maxwell equation, Eq. (432), yields
\left( \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}\right) {\bf B} = {\bf0},
\end{displaymath} (461)

where use has been made of Eq. (456). A similar equation can obtained for the electric field by taking the curl of Eq. (432):
\left( \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}\right) {\bf E} = {\bf0},
\end{displaymath} (462)

We have found that electric and magnetic fields both satisfy equations of the form

\left( \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}\right) {\bf A} = {\bf0}
\end{displaymath} (463)

in free space. As is easily verified, the most general solution to this equation (with a positive frequency) is
$\displaystyle A_x$ $\textstyle =$ $\displaystyle F_x ({\bf k}\cdot{\bf r} - k c t),$ (464)
$\displaystyle A_y$ $\textstyle =$ $\displaystyle F_y ({\bf k}\cdot{\bf r} - k c t),$ (465)
$\displaystyle A_z$ $\textstyle =$ $\displaystyle F_z ({\bf k}\cdot{\bf r} - k c t),$ (466)

where $F_x(\phi)$, $F_y(\phi)$, and $F_z(\phi)$ are one-dimensional scalar functions. Looking along the direction of the wave-vector, so that ${\bf r} = ({\bf k} /k) r$, we find that
$\displaystyle A_x$ $\textstyle =$ $\displaystyle F_x [ k (r-c t) ],$ (467)
$\displaystyle A_y$ $\textstyle =$ $\displaystyle F_y [ k (r-c t) ],$ (468)
$\displaystyle A_z$ $\textstyle =$ $\displaystyle F_z [ k (r-c t) ].$ (469)

The $x$-component of this solution is shown schematically in Fig. 36. It clearly propagates in $r$ with velocity $c$. If we look along a direction which is perpendicular to ${\bf k}$ then ${\bf k}\cdot{\bf r} = 0$, and there is no propagation. Thus, the components of ${\bf A}$ are arbitrarily shaped pulses which propagate, without changing shape, along the direction of ${\bf k}$ with velocity $c$. These pulses can be related to the sinusoidal plane-wave solutions which we found earlier by Fourier transformation. Thus, any arbitrary shaped pulse propagating in the direction of ${\bf k}$ with velocity $c$ can be broken down into lots of sinusoidal oscillations propagating in the same direction with the same velocity.
Figure 36:
\epsfysize =2.5in

The operator

\nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}
\end{displaymath} (470)

is called the d'Alembertian. It is the four-dimensional equivalent of the Laplacian. Recall that the Laplacian is invariant under rotational transformation. The d'Alembertian goes one better than this, since it is both rotationally invariant and Lorentz invariant. The d'Alembertian is conventionally denoted $\Box^2$. Thus, electromagnetic waves in free space satisfy the wave equations
$\displaystyle \Box^2 {\bf E}$ $\textstyle =$ $\displaystyle {\bf0},$ (471)
$\displaystyle \Box^2 {\bf B}$ $\textstyle =$ $\displaystyle {\bf0}.$ (472)

When written in terms of the vector and scalar potentials, Maxwell's equations reduce to
$\displaystyle \Box^2 \phi$ $\textstyle =$ $\displaystyle - \frac{\rho}{\epsilon_0},$ (473)
$\displaystyle \Box^2 {\bf A}$ $\textstyle =$ $\displaystyle - \mu_0  {\bf j}.$ (474)

These are clearly driven wave equations. Our next task is to find the solutions to these equations.

next up previous
Next: Green's functions Up: Time-dependent Maxwell's equations Previous: Potential formulation
Richard Fitzpatrick 2006-02-02