Index | Math in Matter

This entry is dedicated to my cousin Joe and his wife Michelle. Congratulations on tying the knot you two!

Every summer, my family would travel up to Sybil Lake in Minnesota to go fishing. When we weren’t out on the boat with our parents and Grandpa Ervin, we kids would be on the dock with our lines in the water. We’d be waiting for a bite from the same rock bass we had caught the previous day with reasonable success.
I would sometimes dip my fishing rod into the water and exclaim something like “oh no, my fishing rod is bent!” It wasn’t bent though; it was the water refracting the light from the part of the rod that was submerged, making the rod appear bent. A classic prank.
You can reproduce this effect with a cup of water and a straw, as seen in the picture below.

Light goes from one medium (air) to another medium (water) and its behavior changes as a result. The most noticeable change here is the angle. Light comes in at an angle $\theta$ and is redirected to travel along a different angle $\theta^\prime$ by the nature of the substance it is passing through. How does this happen?

On the atomic scale, the light vibrates the charges in the substance that it is passing through. These vibrating charges themselves, emit light which then interferes with the initial light, causing it to slow down and change its trajectory. The degree to which light is affected by a substance depends on how influenced it is by electric and magnetic fields. Keep in mind that light is composed of oscillating electric and magnetic fields. This phenomenon is why we have lenses and rainbows.

Fermat’s Principle
It can be shown that this bent path is actually the shortest path from one medium to another. In a previous entry, I talked about the principle of least action, and how the distance between two points is not always a straight line. In this case, it can be shown to be two straight lines. In general, one medium has a refractive index of $n$ , and the other has a refractive index of $n^\prime$ . These could be water, glass, air, a vacuum, etc. The refractive index is a metric of how influenced a medium is by electric and magnetic fields.

Apply some geometry and trig to this problem.

To show how this bent path between media minimizes the action, we break down the problem using geometry to parameterize the path taken by the light by a distance variable $x$ . The light will go from point A to point B, but we want to minimize the amount of time it takes to do so.
The time it takes to go from A to B is the distance traveled above the water $D$ , divided by the speed above the water $v$ , added to the distance traveled below the water $D^\prime$ , divided by the speed below the water $v^\prime$ .
$t = \frac{D}{v} + \frac{D^\prime}{v^\prime}$
We can then parameterize this equation in terms of distance $x$ using geometry shown in the image above. Also the fact that the speed of light in a medium is dependent on the nature of that medium $n$ , such that $v = c/n$ .
$t = \frac{1}{c} ( n \sqrt{x^2 + h^2} + n^\prime \sqrt{(q-x)^2 + y^2})$
To minimize the time, we take the derivative with respect to $x$ and set it equal to zero.
$\frac{dt}{dx} = 0$
We then get the equation:
$\frac{dt}{dx} = \frac{1}{c} \left( n\frac{x}{\sqrt{x^2 + h^2}} + n^\prime \frac{-(q-x)}{\sqrt{(q-x)^2 + y^2}} \right) = 0$
Note that these terms are equivalent to the sines of the angles in the problem:
$\sin\theta = \frac{x}{\sqrt{x^2 + h^2}}$
$\sin\theta^\prime = \frac{(q-x)}{\sqrt{(q-x)^2 + y^2}}$
This leaves us with the following equation:
$n \sin\theta = n^\prime \sin\theta^\prime$
which is Snell’s law! This law shows how the angle of a light beam changes from one medium to another. There are actually a number of ways to derive this law.

Wavelength and Trig
It can be shown that the wavelength of the light passing through a medium changes. This can also be shown using geometry. Copy the original image and shift it over by an arbitrary distance $d$ , so there are two light beams entering the medium in the same way. You can then imagine the wavelengths along the light beams meeting at the boundary between the two media. Where one wavelength $\lambda$ ends at the surface, another one $\lambda^\prime$ begins. The speed of light in a medium is related to wavelength and period time by
$v = \lambda/t$ .

Finding the relationship between the wavelength change, the refraction angle, the incident angle then becomes a trigonometry problem. Note that in the image above, the wavelengths and distance $d$ form two triangles sharing a side, with angles $\theta, \theta^\prime$ :
$d \sin\theta = \lambda = ct$
$d \sin\theta^\prime = \lambda^\prime = vt$
Take the ratio of these two equations (the arbitrary distance $d$ cancels out):
$\frac{\sin\theta}{\sin\theta^\prime} = \frac{\lambda}{\lambda^\prime} = \frac{c}{v}$
The ratio between the speed of light from one medium to another can be thought of as an refractive index $n$ :
$n = c/v$
$\frac{\sin\theta}{\sin\theta^\prime} = n$
Or in general, with two media, one with refractive index $n$ and the other with $n^\prime$ .
$\frac{\sin\theta}{\sin\theta^\prime} = \frac{v}{v^\prime} = \frac{n^\prime}{n} = \frac{\lambda}{\lambda^\prime}$
This also gives us Snell’s law!
$n\sin\theta = n^\prime\sin\theta^\prime$

Note, there is no solution if:
$1 < \frac{n}{n^\prime} \sin\theta$
What does this mean? There is a critical angle where refraction stops happening.
There’s no loss of generality by saying $n^\prime > n$
$\sin^{-1} \frac{n^\prime}{n} = \theta_c < \theta$
Beyond this critical angle, there is no refraction, only reflection. In general, a portion of the incoming light is refracted and a portion is reflected.

Wave variables
We know a wave’s speed is related to its wavelength and period by:
$v = \lambda/t$
Also note that frequency is the reciprocal of period.
$f = 1/t$
Another way to think about wavelength is by taking its reciprocal to get the wavenumber $k$ .
$k = 2\pi/\lambda$
(This is related to momentum in quantum mechanics by Planck’s constant $p = \hbar k$ )
Frequency can be scaled to angular frequency $\omega$
$f = 2\pi \omega$
So we can express speed as:
$v = \omega/k$

Phase and group velocity
What if there is a change in frequency $\omega$ with respect to wavenumber $k$ ? In general, frequency is a function of wavenumber $\omega(k)$ . The relationship between these two quantities depends on the nature of the media the wave is propagating through.
Initially, we have a sine wave propagating at a speed of $\omega/k$ .
$\sin(kx - \omega(k) t)$
Then the wave number starts to change from passing through a dispersive medium.
$k \rightarrow k^\prime = k+ \Delta k$
In response, the frequency changes:
$\omega \rightarrow \omega^\prime = \omega + \Delta \omega$
This change transforms trig function terms: $e^{i kx - \omega t} \rightarrow e^{i kx - \omega t}e^{i \Delta kx - \Delta\omega t}$
The sine wave becomes:
$\sin(kx - \omega(k) t) \rightarrow \sin([k + \Delta k ]x - [\omega + \Delta \omega] t)$
This can also be written as:
$= \sin(kx -\omega t) \cos(\Delta k x - \Delta \omega t) + \cos(kx-\omega t)\sin(\Delta k x - \Delta \omega t)$
which is a convolution of two waves. One with wavelength $2\pi/k$ , and the other with a wavelength of $2\pi/\Delta k$ . One with the speed of $\omega/k$ as mentioned previously, and the other wave with a speed of $\Delta \omega/\Delta k$ .
These quantities are known as the phase velocity and group velocity, respectively:
$v_p = \frac{\omega}{k}$
$v_g = \frac{d\omega}{dk}$
The ratio of angular frequency and wavenumber is phase velocity and the derivative of angular frequency with respect to wavenumber is group velocity. The “groups” are thought of as envelopes or packets, wrapping the smaller wavelength wave.

This image has an empty alt attribute; its file name is image-6.png

For more on how trig functions work, check out my entry about them.

Dispersion
If group velocity is different from phase velocity, we get dispersion. For refraction, this means the refractive index will be dependent on wavelength. This is why we have rainbows: different colors of light will refract at different angles in water.

This image has an empty alt attribute; its file name is image-8.png — Light of different wavelengths (colors) refracts at different angles.

Suppose we have light in a vacuum propagating towards a medium made of matter. While in the vacuum it has a wavelength of:
$\lambda_0 = \frac{2\pi c}{\omega}$
Once it enters the medium the wavelength becomes:
$\lambda = \frac{2\pi}{k} = 2\pi \frac{v_p}{\omega}$
From before, we know that the refractive index is the ratio of the phase velocities (or the ratio of the wavelengths).
$n = \frac{c}{v_p} = \frac{\lambda_0}{\lambda}$
We also saw that phase velocity is the ratio between angular frequency and wavenumber. This gives us an equation relating angular frequency to wavenumber, light speed in a vacuum and the refractive index.
$\omega = k v_p = k c /n$
Now suppose the refractive index is dependent on the frequency of the light $n = n(\omega)$ .
The group velocity can be found by taking the derivative.
$\frac{d\omega}{dk} = c \frac{1}{n} - kc \frac{1}{n^2} \frac{dn}{dk}$
$v_g = v_p \left(1 - \frac{k}{n} \frac{dn}{dk}\right)$
This equation can be expressed in terms of wavelength as well:
$k = \frac{2\pi}{\lambda}$
$dk = - \frac{2\pi}{\lambda^2} d\lambda$
$v_g = v_p \left(1 + \frac{\lambda}{n} \frac{dn}{d\lambda} \right)$
Note that if the refractive index is independent of frequency that group velocity will be equivalent to phase velocity.
if: $v_p \neq v_g$
$1 \geq 1 + \frac{\lambda}{n} \frac{dn}{d\lambda}$
$0 \geq \frac{dn}{d\lambda}$
We can use Snell’s law to then show angular dependence on wavelength:
$\sin \theta_0 = n(\lambda) \sin\theta$
The incoming light is at an angle $\theta_0$ , and the refracted light is at an angle $\theta$ , which is dependent on the wavelength of the light. This is because the refractive index is dependent on wavelength.
$\theta(\lambda) = \arcsin \left( \frac{\sin\theta_0}{n(\lambda)} \right)$
$\frac{d\theta}{d\lambda} = -\frac{1}{\sqrt{1- \frac{\sin^2\theta_0}{n^2}}} \frac{\sin\theta_0}{n^2} \frac{dn}{d\lambda}=\frac{-\sin\theta_0}{n \sqrt{n^2 - \sin^2\theta_0}} \frac{dn}{d\lambda}$
The way in which $n$ depends on $\lambda$ is theorized with Cauchy’s equation:
$n(\lambda) = \sum\limits_{j=0}^J \frac{C_j}{\lambda^{2j}}$
where the coefficients $C_j$ depend on the material.
This allows rainbows to appear in the sky when sunlight refracts through raindrops.

This image has an empty alt attribute; its file name is rainbow.png — Photo of a rainbow I took in Minnesota years ago.

If light is a wave of electric and magnetic fields, then how does it travel at the speed it does? How does a refractive index relate to these fields? How much light is reflected and how much is refracted? To address these questions, we will solve Maxwell’s equations to derive waves of light as a consequence.
Maxwell’s equations:
$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon}$
$\nabla \cdot \vec{B} = 0$
$\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}$
$\nabla \times \vec{B} = \mu \left(\vec{J} + \epsilon \frac{\partial \vec{E}}{\partial t} \right)$
Stay tuned!