# 8.2.1 Monochromatic Waves in Vacuum

The simplest case for electromagnetic waves is to examine a single frequency propagating as a plane.

Hi, I'm Jonathan Gardner, and I'm covering section 8.2.1 of Griffith's Introduction to Electrodynamics, 2nd Edition. We've just talked about the Wave Equation, and now we're going to see how electromagnetic waves, specifically, behave in a vacuum. First, we're going to look at what an EM plan wave will look like.

Before we begin, let's talk about what "monochromatic" means. "mono" - one. "chromatos" - color. So one-color. What does color have to do with EM waves? Well, it turns out that light of every color is simply an EM wave. This is hardly surprising to us today, but remember how exciting it was to unify the field of optics to the field of electromagnetism back at the turn of last century.

Not only visible light, but all EM waves come in different frequencies. Remember that wavelength and frequency are related by $\nu = c / \lambda\;$ so as the frequencies get faster, the wavelengths get shorter.

At the very highest frequencies, the tightest wavelengths, we have gamma rays, rays that come from space or from decaying radioactive material. Gamma rays, of course, are very dangerous. At lower frequencies, longer wavelengths, we get X-Rays, which we know can be used to see inside of people's bodies. Ultraviolet light comes next. You might know it as "dark light", dark since you can't see it but many materials will glow when exposed to it. What's happening is the EM waves are disturbing the electrons in orbit around the nucleus, and when the electrons settle, they emit visible light.

Visible light are EM waves of about 400 to 760 nanometers in length. A nanometer is one millionth of a millimeter, so you can pack about 1,500 to 2,500 of these into a single millimeter, depending on the frequency and wavelength.

Beyond visible light is infrared radiation, which we feel from warm sources. Then we move into microwave frequencies, then TV and FM signals. These wavelengths are about 1 meter long. Longer than that, we get AM signals, and then we get into lower and lower frequencies and longer and longer wavelengths.

The nice thing is that two different waves of different frequencies don't interfere with each other. Likewise, two waves of the same frequency traveling in different directions don't interact. We need only think about waves of a particular frequency at a particular direction to consider everything that is relevant.

So, we discovered back in section 1 that in a vacuum, Maxwell's Laws can be modified into this form by taking the curl of Faraday's Law and Ampere's Law with Maxwell's Correction and ignoring the charge density and current since they are zero.

The equations are:

\begin{align} \nabla^2 \vec{E} &= {1 \over c^2} {\partial^2 \vec{E} \over \partial t^2} \\ \nabla^2 \vec{B} &= {1 \over c^2} {\partial^2 \vec{B} \over \partial t^2} \\ \end{align}

There are several interesting shapes for waves. One of which is the spherical wave, which there is a problem on that Griffith's encourages you to try and tackle. Here, we'll discuss the plane wave since it is the simplest and because it doesn't give away too many secrets, but gives us a good sense of what is really going on.

Plane waves are waves which travel in a single direction. We'll align that direction with our x axis. The y-z planes that travel with this wave all have the same E and B fields. There is no y or z dependence in either the E or B fields, in other words.

We can write out our E and B fields as thus:

$\tilde{\vec{E}}(x,t) = \tilde{\vec{E}}_0e^{i(\kappa x - \omega t}, \qquad \tilde{\vec{B}}(x,t) = \tilde{\vec{B}}_0e^{i(\kappa x - \omega t}$

Don't get confused. These are complex E and B fields, but we're only interested in the real parts of them, which represent the actual E and B fields. Just like we did with the waves along a string, we note that the complex notation is much easier to work with, and the real part of the complex number will give us what we're looking for.

Complex vectors are just vectors where each component is a complex number. I don't want you to think very hard about what that means. Basically, you have to image that each of the axis are now doubled, since each comes with a real and an imaginary part. In essence, we're working in six dimensions with a time dependence, so we're doing some very complex things but just ignore that. It's irrelevant bookkeeping.

Anyways, the phase constant delta is embedded in these complex E_0 and B_0 vector constants.

Now, these complex fields obey Maxwell's Laws. It's easy to see why this is: The imaginary part is different from the real part only by replacing cosine by sin. This means whatever rules we used to constrain the real part also ends up constraining the imaginary part as well.

Maxwell's Laws says that the divergence of E and B must be zero: no charge and no monopoles, right? Let's take the divergence and see what it says about E_0 and B_0.

\begin{align} \nabla \cdot \tilde{\vec{E}} &= 0 \\ \nabla \cdot \tilde{\vec{E}}_0e^{i(\kappa x - \omega t} &= 0 \\ {\partial \over \partial x} \tilde{\vec{E}}_0e^{i(\kappa x - \omega t} &= 0 \\ {\partial \over \partial y} \tilde{\vec{E}}_0e^{i(\kappa x - \omega t} &= 0 \\ {\partial \over \partial z} \tilde{\vec{E}}_0e^{i(\kappa x - \omega t} &= 0 \\ \end{align}

For the y and z directions, it's trivial to see that E_0 could have a magnitude in those directions, and the constraint is satisfied. Why? Because the vector is a constant with no dependence on x, y, z, or t; and because the exponential part has only a dependence on x and t.

For the x direction, however, unless E_0x is 0, the law will not be satisfied. So we see that there can be no x component of the E_0 vector. Similar reasoning is used for B, so B_0x is also 0. That is, the E and B fields point perpendicular to the direction of travel in plane waves.

The next two laws, Faraday's and Ampere's, describe how the time dependence of the E field is related to the curl of the B field, and so on.

For Faraday's Law, minus the time derivative of the B field must be equal to the curl of the E field. Let's calculate the time derivative of the B field first.

$-{\partial \over \partial t} \tilde{\vec{B}} = -i\omega\tilde{\vec{B_0}}e^{i(\kappa x - \omega t)}$

Remember, we already showed that the x component of the B field is zero.

And now the curl of the E field:

$\nabla \times \tilde{\vec{E}} = \det \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ {\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ \tilde{E}_x & \tilde{E}_y & \tilde{E}_z \\ \end{vmatrix}$

E_x is zero. We already saw that.

The x component of the curl must be zero. That's because the B field has no x component.

The y component of the curl is:

$-{\partial \over \partial x} \tilde{E_z} = -i\kappa(\tilde{E_0})_ze^{i(\kappa x - \omega t)}$

The z component of the curl is:

${\partial \over \partial x} \tilde{E_y} = i\kappa(\tilde{E_0})_ye^{i(\kappa x - \omega t)}$

Bringing it together:

$-\kappa (\tilde{E_0})_z = \omega (\tilde{B_0}_y), \qquad \kappa (\tilde{E_0})_y = \omega (\tilde{B_0}_z)$

This looks like they are going to be perpendicular to each other, which is hardly surprising. It also appears as though there is no phase difference between them. If there had been, we would have seen a factor of $e^{i\delta}\;$ appear.

If you are observant, you will note that you can rewrite those two equations in one:

$\tilde{\vec{B_0}} = {1 \over c} (\hat{i} \times \tilde{\vec{E_0}})$

Example 3

Let's have E wave in the y direction. That means B waves in the z direction. Let's draw what this looks like.

(Drawing of E and B fields)

note that these are plane waves: the values for E and B are the same anywhere along the yz plane. It isn't focused just on the axis.

By convention, we say that this wave is polarized in the y direction, because we use the E field to determine polarization. Note that whether it's in the positive or negative y direction is irrelevant: that only changes the phase constant by 180 degrees. Polarization vectors are weird this way: they give you both the direction and anti-direction. Or rather, two polarization vectors pointing in opposite directions are pretty much the same thing.

Let's suppose our wave is not traveling in the x direction, but along some vector $\vec{\kappa}$. This vector's magnitude will represent kappa, the wave number. Now we have the wave number and direction all wrapped in one. For a plane wave, what is the E and B field at any given point described by the vector r? We can use the scalar product kappa dot r to describe how quickly the wave moves along the r direction. If r is perpendicular to kappa, then there is no waving at all, but if r aligns with the propagation, then it waves as quickly as r moves. Let me draw that out.

(draw a stack of parallel planes, with the kappa vector perpendicular to it.) When the r points along the kappa direction, then you're supposed to get the same equations we got earlier. However, when you point at different directions, you can see how the wave changes more slowly, until you hit the perpendicular where it changes not at all depending on the magnitude of r.

We also have to introduce n-hat, the polarization vector. This points in the direction of the E field.

The equations for E and B look like this with our modifications:

...

And we can just take the real parts, with our cosines:

...

Though you'll not that working with complex numbers is much easier, so stick with the former when you can.

Next, we'll have a brief description of the energy and momentum of an EM wave, the Poynting Vector and why it is so aptly named, and introduce intensity.