12.2: Electromagnetic Waves

General Electromagnetic Plane Waves

12-1

We saw in chapters 8 and 9 that an electromagnetic plane wave traveling in the \(+z\) direction looks like this, \[E_{x}(z, t)=\varepsilon_{x} e^{i(k z-\omega t)}, \quad E_{y}(z, t)=\varepsilon_{y} e^{i(k z-\omega t)},\]

\[B_{x}(z, t)=\beta_{x} e^{i(k z-\omega t)}, \quad B_{y}(z, t)=\beta_{y} e^{i(k z-\omega t)},\]

\[E_{z}(z, t)=B_{z}(z, t)=0,\]

where the \(\beta\)s are determined by Maxwell’s equations to be \[\beta_{y}=\frac{n}{c} \varepsilon_{x}, \quad \beta_{x}=-\frac{n}{c} \varepsilon_{y} .\]

As usual, we have written the wave with the irreducible time dependence, \(e^{-i \omega t}\). To get the real electric and magnetic fields, we take the real part of (12.14)-(12.15). Note, in particular, that the constants \(\varepsilon_{j}\) and \(\beta_{j}\) for \(j = x\) and \(y\) may be complex.

The restriction to motion in the \(z\) direction is not important. Because the physics of Maxwell’s equations is invariant under rotations in three-dimensional space, we can write down the form of a plane wave moving with an arbitrary \(\vec{k}\) vector by extracting the features of (12.14)-(12.17) that do not depend on the direction. These are:

1. \(\vec{k}\), \(\vec{E}\) and \(\vec{B}\) are mutually orthogonal vectors, \[\vec{k} \cdot \vec{E}=\vec{k} \cdot \vec{B}=\vec{E} \cdot \vec{B}=0 ;\]
2. \(\vec{B}\) is determined by the cross product \[\vec{B}=\frac{n}{c} \hat{k} \times \vec{E}=\frac{1}{\omega} \vec{k} \times \vec{E} ,\]
   where \(\hat{k}\) is a unit vector in the direction of the \(\vec{k}\) vector, the direction of propagation of the wave.

These two conditions imply that a general real electromagnetic plane wave can be written as \[\begin{aligned}
&\vec{E}=\operatorname{Re}\left(\vec{e}(\vec{k}) e^{i \vec{k} \cdot \vec{r}-i \omega t}\right) \\
&\vec{B}=\operatorname{Re}\left(\vec{b}(\vec{k}) e^{i \vec{k} \cdot \vec{r}-i \omega t}\right)
\end{aligned}\]

where the vectors, \(\vec{e}\) and \(\vec{b}\), are complex, in general, satisfying \[\vec{b}(\vec{k})=\frac{1}{\omega} \vec{k} \times \vec{e}(\vec{k})=\frac{n}{c} \hat{k} \times \vec{e}(\vec{k}) \quad \text { and } \quad \hat{k} \cdot \vec{e}(\vec{k})=0 .\]

There are two things to note about the relations, (12.21).

1. It is enough to specify the direction of the electric field, \(\vec{e}(\vec{k})\). The magnetic field is then determined by (12.21). The vector, \(\vec{e}\) is called the “polarization” of the electromagnetic wave.
2. Because of (12.21), the polarization is perpendicular to \(\vec{k}\), and thus lives in a two-dimensional vector space.

In the two-dimensional space perpendicular to \(\vec{k}\), we can choose a basis of real vectors, \(\hat{e}_{1}\) and \(\hat{e}_{2}\), where \[\hat{e}_{1} \cdot \hat{k}=\hat{e}_{2} \cdot \hat{k}=\hat{e}_{1} \cdot \hat{e}_{2}=0, \quad \hat{e}_{1} \times \hat{e}_{2}=\hat{k}.\]

For example, for a plane wave traveling in the \(z\) direction, \(\hat{k}=\hat{z}\), we could take \(e_{1}=\hat{x}\) and \(e_{2}=\hat{y}\). Then we can write \[\vec{e}(\vec{k})=\psi_{1} \hat{e}_{1}+\psi_{2} \hat{e}_{2}.\]

The components, \(\psi_{1}\) and \(\psi_{2}\) go into the the two-dimensional vector, (12.3), that describes the polarization state of the electromagnetic wave, just as it describes the polarization state of the string.¹ We can always go back to the components of the electric field using (12.23) and (12.20) and then find the magnetic field using (12.21).

Now the entire discussion of transverse waves on a string from (12.4) to (12.13) can be taken over to describe polarized light. The direction of displacement of the string goes over directly into the direction of the electric field. Thus the animation in program 12-1 applies just as well to the electric field in a polarized wave as to polarization in a string.

Energy and Intensity

The energy density in an electromagnetic field is \[\mathcal{E}=\frac{1}{2}\left(\epsilon \vec{E}^{2}+\frac{1}{\mu} \vec{B}^{2}\right) .\]

Because the energy density is a nonlinear function of the field strengths, we must use real fields in (12.24). The momentum density is \[\overrightarrow{\mathcal{P}}=\epsilon \vec{E} \times \vec{B} .\]

The Poynting vector, a measure of energy flow, is \[\vec{S}=c^{2} \overrightarrow{\mathcal{P}} .\]

These quantities satisfy \[\frac{\partial}{\partial t} \mathcal{E}+\vec{\nabla} \cdot \vec{S}=0 .\]

The Poynting vector is useful because it measures the intensity of the wave, the energy per unit time per unit area carried by the electromagnetic wave. The relation, (12.27), then expresses conservation of energy. The sum of the change in the energy density at any point plus the energy flowing away from it is zero.

To see what these quantities look like in terms of the vector, \(Z\), let us compute the electric and magnetic fields explicitly using (12.20) and (12.21): \[\begin{aligned}
&\vec{E}=\operatorname{Re}\left(\vec{e}(\vec{k}) e^{i \vec{k} \cdot \vec{r}-i \omega t}\right) \\
&\vec{B}=\operatorname{Re}\left(\vec{b}(\vec{k}) e^{i \vec{k} \cdot \vec{r}-i \omega t}\right)
\end{aligned}\]
\[\vec{b}(\vec{k})=\frac{1}{\omega} \vec{k} \times \vec{e}(\vec{k})=\frac{n}{c} \hat{k} \times \vec{e}(\vec{k}) \quad \text { and } \quad \hat{k} \cdot \vec{e}(\vec{k})=0 .\]

The result is \[\begin{gathered}
\vec{E}=A_{1} \hat{e}_{1} \cos \left(\vec{k} \cdot \vec{r}-\omega t+\phi_{1}\right)+A_{2} \hat{e}_{2} \cos \left(\vec{k} \cdot \vec{r}-\omega t+\phi_{2}\right) , \\
\vec{B}=\sqrt{\mu \epsilon}\left(A_{1} \hat{e}_{2} \cos \left(\vec{k} \cdot \vec{r}-\omega t+\phi_{1}\right)-A_{2} \hat{e}_{1} \cos \left(\vec{k} \cdot \vec{r}-\omega t+\phi_{2}\right)\right) .
\end{gathered}\]

Putting this into (12.24) and (12.26) gives \[\mathcal{E}=\frac{\epsilon}{4 \pi}\left(A_{1}^{2} \cos ^{2}\left(\vec{k} \cdot \vec{r}-\omega t+\phi_{1}\right)+A_{2}^{2} \cos ^{2}\left(\vec{k} \cdot \vec{r}-\omega t+\phi_{2}\right)\right), \]
\[\vec{S}=\hat{k} \sqrt{\frac{\epsilon}{\mu}} \frac{c}{4 \pi}\left(A_{1}^{2} \cos ^{2}\left(\vec{k} \cdot \vec{r}-\omega t+\phi_{1}\right)+A_{2}^{2} \cos ^{2}\left(\vec{k} \cdot \vec{r}-\omega t+\phi_{2}\right)\right) .\]

You can check explicitly that (12.27) is satisfied. Because \(\omega\) is very large for interesting electromagnetic waves, we are almost always interested in only the time averaged values of \(\mathcal{E}\) and \(\vec{S}\). These are \[\langle\mathcal{E}\rangle=\frac{\epsilon}{8 \pi}\left(A_{1}^{2}+A_{2}^{2}\right),\]
\[\langle\vec{S}\rangle=\hat{k} \sqrt{\frac{\epsilon}{\mu}} \frac{c}{8 \pi}\left(A_{1}^{2}+A_{2}^{2}\right) .\]

Note that the time averaged values depend only on the quantity \[|Z|^{2} \equiv\left|\psi_{1}\right|^{2}+\left|\psi_{2}\right|^{2}=A_{1}^{2}+A_{2}^{2} .\]

The intensity of the light is proportional to \(|Z|^{2}\).

Circular Polarization and Spin

Although linear polarization is more familiar and perhaps easier to understand, there is a sense in which circular polarization is the more fundamental. The plane electromagnetic wave in the \(\hat{k}\) direction can be rotated around the \(\hat{k}\) axis without changing anything but its polarization state. The rotation symmetry of the physics suggests that we ought to be able to find states that behave simply under such a rotation, and just get multiplied by a phase factor. These states are, in fact, the circular polarization states. The action of a rotation by an angle \(\theta\) about the \(\hat{k}\) axis on the polarization vector, \(Z\), is represented by the matrix \[R_{\theta}=\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right) .\]

For example, \(R_{\theta}\) acting on \(u_{1}\), (12.4), gives \(u_{\theta}\), (12.6): \[R_{\theta}\left(\begin{array}{l}
1 \\
0
\end{array}\right)=\left(\begin{array}{l}
\cos \theta \\
\sin \theta
\end{array}\right) .\]

But on the left- and right-circularly polarized states, \[R_{\theta}\left(\begin{array}{l}
1 \\
i
\end{array}\right)=e^{-i \theta}\left(\begin{array}{l}
1 \\
i
\end{array}\right), \quad R_{\theta}\left(\begin{array}{c}
1 \\
-i
\end{array}\right)=e^{i \theta}\left(\begin{array}{c}
1 \\
-i
\end{array}\right) .\]

This is related to the fact that the circularly polarized states carry the maximum angular momentum possible, which in turn is related to the quantum mechanical property of the spin of the photon.

__________________________

¹This is sometimes called the Jones vector. See Hecht, page 323.