Electromagnetic wave

An electromagnetic wave is a wave produced by the periodic variation and mutual induction of an electric and a magnetic field. It is a transverse wave, which means that the fields are always orthogonal to the direction of propagation.

Electromagnetic waves are described through Maxwell's equations. In the vacuum, with no charges whatsoever, the equations can be decoupled to read

Both of these are examples of the wave equation and thus $\mathbf{E}$ and $\mathbf{B}$ accept wave solutions. Their phase velocity must be

which is the speed of light.

In the vacuum#

Let's start by analyzing the behavior of electromagnetic sine waves. These are, as always, of paramount importance since all waves can be decomposed into a Fourier series of sine waves. Thus, understanding sine waves in theory means that you understand all other waves. In complex notation, they read

assuming propagation on the $x$ axis. The real fields can be found by taking the real part of the complex field. Now, this equation is entirely determined from the wave equation. However, as it is, it cannot be correct. This is because Maxwell's equation impose specific constraints on the possible functions that $\mathbf{E}$ and $\mathbf{B}$ can be. Here we bypassed all of that by arbitrarily picking a known solution to the wave equation. While it is true that all vacuum solutions of Maxwell's equation obey the wave equation, the converse is not necessarily true: not all solutions to the wave equation obey Maxwell's equations. In light of this, we should analyze Maxwell's equation to figure out what these additional constraints are.

Maxwell's equation determine the fields, so the conditions must be applied to $\tilde{\mathbf{E}}_{0}$ and $\tilde{\mathbf{B}}_{0}$. These conditions are the values of the Divergence and Curl. The divergence of these fields in a vacuum is zero, so it must be that

Intuitively, this is because the divergence in a way represents propagation. As such, if it must be zero, then the field cannot possibly extend in the direction of propagation. But if the oscillation cannot be in the direction of propagation, then all electromagnetic waves must be transverse!

That's it for divergence. The curls on the other hand tell us that the fields induce each other, so clearly there has to be some connection between electric waves and magnetic waves. The curl of $\mathbf{E}$ (Faraday's law) gives one connection (by looking at the components of the Vector product directly):

and also

(since the $x$ component is guaranteed zero). So, comparing the two:

which is equivalent to the vector product

The curl of $\mathbf{B}$ provides the same result. This should be enough to convince you that the electric and magnetic parts of the wave not only are symmetric, but also strongly coupled, for they must be both in phase and perpendicular (due to the vector product). Their real amplitudes must be related by

where $c=\omega/k$ is the phase velocity of the wave, coinciding with the speed of light.

For example, suppose you have a wave moving on the $x$ axis. Suppose also that the electric field moves on the $y$ axis. Then, the magnetic one must occupy the last remaining axis, $z$. Their equations in general are of the sort

Apply the previous conditions to get

Take the real part and finally you have a electromagnetic sine wave on the $x$ axis:

This is an electromagnetic wave propagating over $\hat{\mathbf{x}}$ that is polarized on $\hat{\mathbf{y}}$ (by convention, it is the direction of $\mathbf{E}$ that specifies the polarization). The magnetic field needs to be orthogonal to both propagation and $\mathbf{E}$, so it ends up on $\hat{\mathbf{z}}$.

Of course, electromagnetic waves can propagate in any direction, so we can generalize further by introducing the wavevector $\mathbf{k}$ and the polarization vector $\hat{\mathbf{n}}$:

whose real part is

Transported quantities#

Electromagnetic waves are really just traveling electromagnetic fields. As such, they carry all quantities that the fields themselves do, chiefly energy and momentum.

The energy density of an electromagnetic field is

In case of a sine wave, we can use $(1)$ to state

and so

Evidently, only the electric field matters in the energy transfer, as the magnetic field's contribution is identical to that of the electric field¹. We can say more about this by looking at the Poynting vector, which represent the energy flux density (energy per unit area per unit time). Plugging in the previous equation for $B$ gives

Perhaps unsurprisingly, the energy moves in the exact direction of the wave. Finally, the momentum density carried by the field is

Generally speaking though, electromagnetic wave frequencies are fast. So fast in fact that any measurement done at a macroscopic level will surely encompass a large number of cycles of oscillation. As such, a reading of any of these quantities is going to be smeared across so many cycles it essentially just becomes the time average of the quantity. As such, it's in our interest to take a look at the mathematical predictions of these averages. Since the only time-variable part in these quantities is the square cosine, we just need its average over a full cycle of period $T$:

Thus, the average energy density, Poynting vector and momentum density are

Notably, energy density and momentum density (and by extension energy and momentum themselves) are related by

The energy of a wave is given by the momentum it carries times the speed of light. "Neat," you might say, but keep this in mind when you look at quantum physics, as it might just reappear somewhere else (and if you've already studied quantum physics, you might recognize a sneaky first appearance of the Planck formula and the de Broglie formula).

In the context of electromagnetic waves, the irradiance of the wave is the time average of the Poynting vector on an incident surface. If the surface is perpendicular to the wave, its normal unit vector is $\hat{\mathbf{k}}$, so

(which coincides with the norm of the Poynting vector, but it's a useful reminder that irradiance only makes sense when projected over a surface, so orientation matters.) Since momentum is conserved, whenever light collides with something, a part of its momentum will be transferred to it. If light lands perpendicularly (normal incidence) and the receiver is a so-called "perfect absorber" (meaning, it absorbs 100% of the momentum that hits it), we can give a mathematical expression of this transfer. In a time $\Delta t$, the transfer is determined by the momentum density as $\Delta \mathbf{p}=\langle \mathbf{g} \rangle Ac\Delta t$. But remember that by Newton's second law, a variation in momentum is a force, so when spread over the area $A$ that's being lit up, we get a pressure:

This pressure is known as radiation pressure and, in general, it is extremely weak. Weak enough that it's fair if you assumed that electromagnetic radiation was simply incapable of applying a force to something. But it is there, all because momentum conservation must be preserved. By the way, if the receiver was a "perfect reflector" (meaning, it reflects 100% of the momentum), the pressure would be doubled since the effective $\Delta \lvert \mathbf{p} \rvert$ would itself be doubled (since we'd be sending $\mathbf{p}\mapsto-\mathbf{p}$ instead of $\mathbf{p}\mapsto0$). Fun fact: radiation pressure is actually a viable method of spacecraft propulsion. Basically, you make a very large reflective surface, point it a powerful light source (a star, realistically) and wait for the light to exert pressure to accelerate the ship. This kind of propulsion device is called a lightsail. It's weak propulsion of course, but the benefit is that it is "free", in that it requires no fuel, just like a normal wind sail.

In dielectrics#

To develop a theory of electromagnetic waves in matter, we turn to the Maxwell equations in matter. We start off in an analogous way as the vacuum case by setting the free charge and free current to zero; this corresponds to claiming with are working with dielectrics, where free charges are scarce. Conductors require more refined analysis, which we'll defer to later. Still, even in this no-free-charge setting we quickly hit an impasse. The behavior of waves in matter is, unsurprisingly, dependent on the nature of the material itself, namely how the electric displacement and auxiliary field behave. We can however restrict ourselves to the usual and generally quite useful case of linear, homogeneous and isotropic materials, in which the shapes of $\mathbf{D}$ and $\mathbf{H}$ are well-known. In such a setting, we get

This form is near-identical to the vacuum one, with the difference of using general permittivity and permeability instead of their vacuum versions. This leads to a different shape for the phase velocity:

The quantity $n=\sqrt{ \varepsilon_{r}\mu_{r} }\geq1$ is called the refractive index of the material and, among other things, determines a slowdown of the wave when in matter. Most materials are not really magnetic, so in many cases $\mu_{r}\simeq1$ and $n\simeq \sqrt{ \varepsilon_{r} }$.

The transported quantities of these waves are largely the same and can be found by substituting $\varepsilon_{0}\to \varepsilon$, $\mu_{0}\to \mu$ and $c\to c/n$, so

In case of a sine wave, we have the same condition as $(1)$, so

and then averaged

You might notice the suspicious lack of momentum density here. That's because the momentum of a wave in matter is, as of today, still a matter of debate, so no formula here.

Interface reflection and transmission#

As all waves, electromagnetic waves too exhibit interesting behavior when they reach the "end of the line", that is, when they collide with an interface between two different materials. Say we've got a wave traveling over the $x$ direction and there's an interface on the $yz$ plane at $x=0$ so that $x<0$ is one material with refractive index $n_{1}$ and $x>0$ is another with refractive index $n_{2}$. Say also for simplicity that the wave is linearly polarized on the $y$ axis and it has an angular frequency of $\omega$ and a wavenumber $k_{1}$. The wave equations are

We know from the general theory of waves (see Wave equation > Boundary conditions) that an incident wave to a surface gives rise to a reflected and a transmitted wave, whose nature depends on the boundary conditions. These waves are

for the reflected one and

The boundary conditions are given by Maxwell's equations

Now, what these tell us depends on how the wave strikes the surface. If, as we chose above, have a wave that hits the surface at an exact 90° angle, we say that the wave is at normal incidence. This is the easiest case and in general not too unrealistic either in some artificial conditions, so that's the one we start with.

Normal incidence#

When incidence is normal, the two perpendicular boundary conditions don't matter since there is no perpendicular component. The parallel electric field condition implies

and the parallel magnetic field implies

If we define

this simplifies to

We can solve these to express reflected and transmitted amplitude in terms of the incident one:

These are the amplitudes we were looking for. These are true in general for electromagnetic sine waves at normal incidence. We can however squeeze more physical significance out of these. This is because in most materials, the relative permeability is essentially one, and so $\mu \simeq \mu_{0}$. Thus, $\beta$ simplifies down to $\beta \simeq n_{2}/n_{1}$ and our amplitudes become

Don't let the approximate equals sign fool you: these are perfectly valid equations in most materials. These of course fail when $\mu_{r}\gg1$, as is the case in many metals and generally all strong magnetic amplifiers, but most other materials have quite negligible relative permeabilities, so these are safe to use. If $n_{1}<n_{2}$, the reflected wave is flipped upside down compared to the incident wave (a minus sign on complex amplitude corresponds to a $\pi$ phase shift). If $n_{1}>n_{2}$, then the phase is the same. The real parts are

The amplitudes themselves are theoretically important, but in practice one generally cares about the irradiance of the wave. The amplitude uniquely determines the irradiance according to

More so than the irradiances themselves though, it's interesting to see how they change after reflection or transmission. We do this by finding the ratios between incident and reflected/transmitted irradiance:

These are known as the reflection and transmission coefficients (or reflectance and transmittance) and represent the fraction of the irradiance that is transferred to the reflected and transmitted waves on interface interaction. The ratios of the electric fields are instead called the Fresnel coefficients, and the reflectance and transmittance are their squares. Note that, since energy must be conserved, these coefficients abide by

as can be shown by doing the math directly.

Now, to be a little pedantic, calling $I_{R}$ and $I_{T}$ "irradiances" is technically wrong here because both reflected and transmitted waves are outgoing, not ingoing. Irradiance is specifically the energy flux incident on a surface. Flux that is exiting a surface is called radiant exitance². For correctness, we can call $I_{R}$ reflected exitance and $I_{T}$ transmitted exitance. If you are confused at the nomenclature, don't worry, radiometry (the study of electromagnetic radiation) is confusing. I have written a Radiometric nomenclature survival guide to help.

Oblique incidence#

Now that we have developed a theory in normal incidence, let's twist the surface a bit so that it is no longer perpendicular to the radiation. We call $\theta_{I}$ the angle of incidence, the angle between the wavevector and the surface normal. Normal incidence is when $\theta_{I}=0$. The reflected and transmitted waves then emanate at their own angles of reflection $\theta_{R}$ and transmission $\theta_{T}$ (also called angle of refraction).

Suppose we have a generic incident sine wave

which induces a reflected wave

and a transmitted wave

all at the same frequency $\omega$.

The wavevectors are all connected by

These waves must obey the boundary conditions, so $\tilde{\mathbf{E}}_{I}+\tilde{\mathbf{E}}_{R}$ and $\tilde{\mathbf{B}}_{I}+\tilde{\mathbf{B}}_{R}$ must be joined with $\tilde{\mathbf{E}}_{T}$ and $\tilde{\mathbf{B}}_{T}$ in a way that satisfies the conditions. Altogether we get

where $A,B$ and $C$ are some values that we'll determine later. Right now we care about the exponentials. For this equality to be true, we need

Notice how the right hand side is the same as the starting equation, just multiplied by $c$. So, we can say

$$$$

\mathbf{k}{I}\cdot \mathbf{r}=\mathbf{k}{R}\cdot \mathbf{r}=\mathbf{k}_{T}\cdot \mathbf{r}\quad\text{on the surface}\tag{2}

yk_{I,y}+zk_{I,z}=yk_{R,y}+zk_{R,z}=yk_{T,y}+zk_{T,z}

zk_{I,z}=zk_{R,z}=zk_{T,z}

yk_{I,y}=yk_{R,y}=yk_{T,y}\tag{3}

k_{I}\sin \theta_{I}=k_{R}\sin \theta_{R}=k_{T}\sin \theta_{T}

This is also known as Snell's law. These are the three laws of classical geometric optics. These have been known for a very long time, even before the rise of modern electrodynamics, and it is interesting to note how they arise spontaneously only though the wave form of Maxwell's equations. In fact, none of this derivation even really cares about electric or magnetic fields. The only prerequisite here is the wave equation and the quite generic boundary conditions we used, which can represent a whole host of waves, like sea waves or sound waves or what have you. In fact, these laws are essentially universal across all kinds of waves, provided you can express them as sinusoidal plane waves.

According to $(2)$, the exponential factors cancel out and we are left with the leading coefficients. These obey the boundary conditions specific to electrodynamics

which using our values and our frame of reference become

Suppose that the incident wave is polarized parallel to the plane of incidence. Then, the reflected and transmitted waves must themselves be polarized in the same plane³. In this case, we get

Using the law of reflection and Snell's law, top left and bottom right both become

where $\beta=\mu_{1}n_{2}/\mu_{2}n_{1}$. Meanwhile, top right becomes

where $\alpha=\cos \theta_{T}/\cos \theta_{I}$. Solving these to express the amplitudes in terms of $\tilde{E}_{0,I}$ yields

These are known as Fresnel's equations for polarization in the plane of incidence. There are two more, for polarization perpendicular to the plane of incidence. From the sign of the coefficients, we can see that the transmitted wave must always be in phase with the incident one (since $2/(\alpha+\beta)>0$), but the reflected may or may not be depending on the sign of $\alpha-\beta$. If $\alpha>\beta$, then it is in phase, otherwise it is out of phase by precisely half a cycle (it's "upside down") since changing the sign of the amplitude in a sine wave is like shift by half a cycle ($\pi$ or 180°). Of course, if set $\theta_{I}=0$, then $\alpha=1$ and we get the equations we found in the normal incidence case.

Note how since $\cos \theta_{I}$ is at the denominator in $\alpha$, if $\theta_{I}$ approaches $\pi/2$ (or 90°), the $\cos \theta_{I}$ approaches zero and $\alpha$ diverges. In that case, the coefficient on the transmitted wave collapses to zero and only the reflected wave remains: this is called total reflection and you can pretty easily experience it yourself by taking any partially reflective object and rotating it while under a light. Eventually, you'll see nothing but reflected light.

Curiously, there is another angle worth discussing. When $\alpha=\beta$, we get

In most materials, $\mu_{1}\simeq \mu_{2}$, so $\beta \simeq n_{2}/n_{1}$ and this simplifies down to

This angle is known as Brewster's angle and it is special because here, the reflected wave is fully extinguished. This is the other half of the medal: at 90°, the transmitted wave vanishes, and at Brewster's angle, the reflected wave vanishes.

Actually, there is yet another interesting angle to talk about, although it is not universal. When a wave moves from an "optically dense" material (say, water) to an "optically thin" one (say, air), which is to say when $n_{1}>n_{2}$, the transmitted wavevector tends to move away from the normal and tends to become parallel to the interface. This isn't asymptotic: the wavevector becomes exactly parallel with the surface ($\theta_{T}=\pi/2$ or 90°) at the critical angle

as per Snell's law. On incidence at any angle greater than the critical one ($\theta_{I}>\theta_{C}$), there is no transmission, just like normally when reaching 90°. Since air is quite optically thin, this typically happens when a ray of light attempts to leave a material (almost certainly optically denser) and move into the air. As such, in some conditions, it's possible that a ray of light might remain "trapped" inside the material if it only ever bounces at angles greater than the critical one. This is the idea behind total internal reflection, a technique used to transfer light beams from one place to another without energy loss by making sure all internal incidence is at an angle greater than critical (if you're familiar with fiber optics, this is how it works).

This said, while the wave is fully reflected, the fields are actually not zero in material 2. They form a so-called evanescent wave, a wave that attenuates extremely rapidly and transports no energy into material 2. It can be constructed by setting $k_{T}=\omega n_{2}/c$ and $\mathbf{k}_{T}=k_{T}(\sin \theta_{T}\ \hat{\mathbf{y}}+\cos \theta_{T}\ \hat{\mathbf{z}})$. This causes Snell's law

to report an imaginary transmission angle when $\theta_{I}>\theta_{C}$. Of course, it no longer makes sense as an angle (at least a real one).

Finally, we close this off with the irradiance of the incident wave:

The reflected and transmitted exitances:

And the reflectance and transmittance:

In conductors#

Now that we're done with dielectrics, we move on to conductors. Maxwell's wave equations for conductors are

They differ by having an additional first derivative in time on top of the second derivative. These are no longer the standard wave equation we've gotten used to and are in fact a generalization of it. Nevertheless, they still admit plane wave solutions, so that's what we'll follow through with:

You'll notice an unfamiliar tilde over the $k$: this is because in order for a plane wave to satisfy one of these equations, it must have a complex wavenumber. Before moving on, we should probably investigate this odd quantity further.

Substituting the plane wave, our wave equations become (we'll use the electric field, but the magnetic field is identical):

Be careful with $\tilde{k}^{2}$! It's a complex value, not a real one, so the square is a little more complicated. When plugging this Laplacian in the wave equation, we get

We'll split the real and imaginary parts into $\tilde{k}=k+i\kappa$ so that the square is

We now have two results for $\tilde{k}^{2}$ that we can compare to get:

Solving for real and imaginary parts separately gives us

The real and imaginary wavenumbers end up being

(The only difference between the two is the plus or minus at the end). These are significant results for a couple of reasons, especially the imaginary part. In fact, if we go back to the oscillating fields

we have the usual plane wave with a real wavenumber $k$, but now it's weighed by $e^{-\kappa x}$. Specifically, this term says that as $x$ grows larger (i.e. the wave propagates), it is suppressed, with a larger $\kappa$ incurring a more significant suppression; this phenomenon is known as the skin effect:

$1/e$ is qualitatively around one third, so the wave's amplitude cuts down by a third every skin depth. This leads to a very fast suppression of the wave, doubly so since skin depths are typically very small, generally millimeters or fractions of millimeters. Lower frequencies have higher skin depths and viceversa⁴.

Moreover, we see that the wavevector is dependent on $\omega$ multiplied by a factor made up of $\varepsilon$, $\mu$ and $\sigma$. These three parameters determine the materials behavior against incoming electromagnetic waves. You could consider them constants, but you'd be slightly wrong. These are actually measured to be mildly frequency-dependent: $\varepsilon\equiv \varepsilon(\omega)$, $\mu\equiv \mu(\omega)$ and $\sigma\equiv \sigma(\omega)$. As such, $\tilde{k}$ is nonlinearly dependent on $\omega$. This leads to a phenomenon known as dispersion, which occurs when the phase velocity of the wave in the material is dependent on the frequency, $v_{p}\equiv v_{p}(\omega)$. The relation $k\equiv k(\omega)$ is called the dispersion relation. In our case, since $v_{p}=c/n$, this is like saying that dispersion occurs when the refractive index depends on frequency, $n\equiv n(\omega)$. (And in fact, remember that $n=\sqrt{ \varepsilon_{r}\mu_{r} }\simeq \sqrt{ \varepsilon_{r} }$, so if $\varepsilon_{r}$ depends on $\omega$, as we've claimed, so does $n$.)

Equation $(5)$ is universally correct for the conductor wave equations. Maxwell's equations impose further constraints on which waves are permitted; this is the same argument as in the beginning of > In the vacuum. Assuming the wave propagates on $x$, due to $\nabla\cdot \mathbf{E}=0$ and $\nabla\cdot \mathbf{B}=0$, there can be no $x$ component to the fields. Similarly, using the curls we get that the fields are perpendicular to each other at all times. For instance, setting the electric field as polarized over $\hat{\mathbf{y}}$ and using its curl we find

We can further analyze the complex wavenumber $\tilde{k}$ by expressing it in polar form:

The modulus is

and the angle is

Using $(6)$, we see that the complex amplitudes $\tilde{E}_{0}=E_{0}e^{i\delta_{E}}$ and $\tilde{B}_{0}=B_{0}e^{i\delta_{B}}$ are related by

Evidently, the phase is shifted by an angle $\phi$ determined by the wavevector and the fields are out of phase by precisely this angle: $\delta_{B}-\delta_{E}=\phi$. Specifically, the magnetic field is lagging behind the electric one.

The real parts of waves incident on a conductor are

The irradiance can be found to be

where $\langle E^{2} \rangle_{T}$ is the time average over a period $T$, and

$\alpha=1/2\kappa$ is the called the penetration depth of the wave. It is the characteristic length of suppression of irradiance when moving through a conductor; it's basically the skin depth of irradiance. Just like skin depth, it is usually less than or a few millimeters. For UV waves on a typical conductor, it's around $0.6\text{ mm}$ and for infrared waves, it's about $6\text{ mm}$.

Interface reflection and transmission#

Like all surfaces, an electromagnetic wave can be reflected by a conducting surface too. Unlike dielectrics however, conductors show some remarkable behavior due to their near-immediate reallocation of charge. The boundary conditions of a conductor are the general ones

since we cannot arbitrarily choose for $\sigma_{f}$ and $\mathbf{K}_{f}$ to be zero (don't confuse $\sigma_{f}$, free surface charge, with $\sigma$, conductivity, and $\hat{\mathbf{n}}$, the surface normal, with the polarization vector!).

We'll work with ohmic conductors. These work with Ohm's law $\mathbf{J}_{f}=\sigma \mathbf{E}$ and therefore implies $\mathbf{K}_{f}=0$ since if if were nonzero, we'd need an infinite electric current at the boundary. In fact, consider a slice of infinitesimal thickness $l$ of the material right on the surface. The surface current is approximately $\mathbf{K}_{f}=\mathbf{J}_{f}l=\sigma l\mathbf{E}$. For this to be exact we need $l\to 0$, which implies $\mathbf{K}_{f}\to 0$ unless $\mathbf{E}\to \infty$, which is of course unphysical, so $\mathbf{K}_{f}=0$.

Normal incidence#

We start by investigating normal incidence, $\theta_{I}=0$. As usual, we pick a wave traveling forwards on $\hat{\mathbf{x}}$ with the conductor's surface on the $yz$ plane. We polarize the wave on $\hat{\mathbf{y}}$. The coupled fields are

The reflected wave will be

and the transmitted one

The complex wavevector $\tilde{k}_{2}$ denotes the attenuation of the wave. At $x=0$, the waves must abide by the boundary conditions. $E^{\perp}=0$ on both sides since we are at normal incidence, which implies $\rho_{f}=0$ by the top left condition. Top right becomes

$B^{\perp}=0$ is also true on both sides for the same reason, so bottom left becomes trivial. Bottom right becomes

or, if we define

we get

Solving these yields

The coefficients maintain the same shape as the dielectric case, but the $\beta$ parameter is now complex to take the attenuation into account. In case of a perfect conductor, $\sigma=\infty$, so $k_{2}=\kappa_{2}=\infty$ using $(4)$, so $\tilde{\beta}=\infty$ (in the complex sense). In this case we get

The wave is totally reflected with an accompanying 180° phase flip. In other words, the conductor is a perfect mirror. If you've ever wondered why metal is so shiny, this is why!