16.2 Plane electromagnetic waves  (Page 40/5)

Mechanical waves travel through a medium such as a string, water, or air. Perhaps the most significant prediction of Maxwell’s equations is the existence of combined electric and magnetic (or electromagnetic) fields that propagate through space as electromagnetic waves. Because Maxwell’s equations hold in free space, the predicted electromagnetic waves, unlike mechanical waves, do not require a medium for their propagation.

A general treatment of the physics of electromagnetic waves is beyond the scope of this textbook. We can, however, investigate the special case of an electromagnetic wave that propagates through free space along the x -axis of a given coordinate system.

Electromagnetic waves in one direction

An electromagnetic wave consists of an electric field, defined as usual in terms of the force per charge on a stationary charge, and a magnetic field, defined in terms of the force per charge on a moving charge. The electromagnetic field is assumed to be a function of only the x -coordinate and time. The y -component of the electric field is then written as $E_y\left(x,t\right),$ the z -component of the magnetic field as $B_z\left(x,t\right)$ , etc. Because we are assuming free space, there are no free charges or currents, so we can set $Q_{\text{in}}=0$ and $I=0$ in Maxwell’s equations.

The transverse nature of electromagnetic waves

We examine first what Gauss’s law for electric fields implies about the relative directions of the electric field and the propagation direction in an electromagnetic wave. Assume the Gaussian surface to be the surface of a rectangular box whose cross-section is a square of side l and whose third side has length $\text{Δ}x$ , as shown in [link] . Because the electric field is a function only of x and t , the y -component of the electric field is the same on both the top (labeled Side 2) and bottom (labeled Side 1) of the box, so that these two contributions to the flux cancel. The corresponding argument also holds for the net flux from the z -component of the electric field through Sides 3 and 4. Any net flux through the surface therefore comes entirely from the x -component of the electric field. Because the electric field has no y - or z -dependence, $E_x\left(x,t\right)$ is constant over the face of the box with area A and has a possibly different value $E_x\left(x+\text{Δ}x,t\right)$ that is constant over the opposite face of the box. Applying Gauss’s law gives

where $A=l\times l$ is the area of the front and back faces of the rectangular surface. But the charge enclosed is $Q_{\text{in}}=0$ , so this component’s net flux is also zero, and [link] implies $E_x\left(x,t\right)=E_x\left(x+\text{Δ}x,t\right)$ for any $\text{Δ}x$ . Therefore, if there is an x -component of the electric field, it cannot vary with x . A uniform field of that kind would merely be superposed artificially on the traveling wave, for example, by having a pair of parallel-charged plates. Such a component $E_x(x,t)$ would not be part of an electromagnetic wave propagating along the x -axis; so $E_x(x,t)=0$ for this wave. Therefore, the only nonzero components of the electric field are $E_y(x,t)$ and $E_z(x,t),$ perpendicular to the direction of propagation of the wave.