Amplitude, Frequency, Wave Number, Phase Shift

Dispersion of light waves through a prism, a consequence of the wavenumber slightly altering the index of refraction of each color in the prism.

Amplitude, frequency, wavenumber, and phase shift are properties of waves that govern their physical behavior.

Each describes a separate parameter in the most general solution of the wave equation. Together, these properties account for a wide range of phenomena such as loudness, color, pitch, diffraction, and interference.

Waves propagating in some physical quantity $y$ obey the wave equation:

$$\frac{1}{v^2} \frac{\partial^2 y}{\partial t^2} = \frac{\partial^2 y}{\partial x^2},$$

where $v$ is the velocity of the wave. Solutions to this equation are written as a linear superposition of right-traveling and left-traveling waves. These can be any arbitrary functions of the form $f(x-vt)$ and $g(x+vt)$.

One set of simple examples are the so-called harmonic waves, which are sinusoidal:

$$y(x,t) = A \sin (x-vt) + B \sin (x+vt) ,$$

where $y_0$ is the amplitude of the wave and $A$ and $B$ are some constants.

A linear superposition of left-traveling (blue) and right-traveling (green) sinusoidal waves solving the wave equation [1].

Using the sum rules for trigonometric functions, this solution can also be written in the form:

$$y(x,t) = Y_0 \sin (k(x-vt)-\phi) ,$$

for $Y_0$ some new constant called the amplitude, $k$ a constant called the wavenumber, and $\phi$ a constant called the phase shift. From the equation above, $Y_0$ gives the maximum height of the wave, $k$ describes how tightly spaced the oscillations of the wave are, and $\phi$ describes how the sine function has been shifted to the left or right at time $t=0$. The wavenumber $k$ is related to the wavelength describing the distance between adjacent peaks of a wave by:

$$k = \frac{2\pi}{\lambda}.$$

To see that this is true, note that when the argument of the sine function changes by $2\pi$, one full oscillation has been completed. Therefore, comparing $x = 0$ to $x=\lambda$ should increase the argument by $2\pi$, which corresponds to the above definition of the wavenumber.

Note that some sources (particularly in chemistry) define the wavenumber differently and use a different symbol as in $\bar{\nu} = \frac{1}{\lambda}$. In terms of physical significance, both definitions are essentially equivalent.

For some types of waves, such as light, the (angular) frequency $\omega$ of the wave, which describes how rapidly the wave oscillates in time, satisfies the equation:

$$\omega = vk.$$

The angular frequency is related to a quantity often labeled $f$ and also called the frequency by $\omega = 2\pi f$. With these new definitions, solutions to the wave equations can be written in a number of different forms, for example:

$$\begin{aligned} y(x,t) &= Y_0 \sin (kx-\omega t - \phi)\\ & = Y_0 \sin (2\pi (\frac{x}{\lambda} - f t) - \phi) \\ &= Y_0 \cos \phi \sin (kx - \omega t) - Y_0 \sin \phi \cos (kx - \omega t) . \end{aligned}$$

The amplitude $Y_0$ is related to the energy per unit time per unit area (the intensity or power flux) carried by that wave. Specifically, the intensity is proportional to the square of the amplitude. This fact is usually proved on a case-by-case basis for different physical scenarios, rather than for general solutions of the wave equation.

Show that the power flux carried by a classical electromagnetic wave goes as the square of the electric or magnetic field amplitude.

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Solution:

Recall that the power flux $\vec{S}$ carried by the Poynting vector for electromagnetic waves is:

$$\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B},$$

where $\mu_0$ is the permeability of free space and $E$ and $B$ are the electric and magnetic fields respectively. For light in vacuum, the amplitude of the magnetic field goes as $B = \frac{E}{c}$ where $c$ is the speed of light, and the magnitude of the Poynting vector instantaneously goes as:

$$S = \frac{E^2}{\mu_0 c},$$

which is proportional to the square of the electric field (or magnetic field) amplitude as claimed.

Show that the average power carried by a wave on a string goes as the square of the amplitude of the wave.

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Solution:

Recall the formula for power $P$ exerted by a force $\vec{F}$ on an object traveling velocity $\vec{v}$:

$$P = \vec{F} \cdot \vec{v}.$$

For small-amplitude waves on a string, only the forces and displacements in the vertical direction matter. If the string is at angle $\theta$ with respect to the horizontal, Newton's second law gives:

$$P = -T\sin \theta \frac{\partial y}{\partial t} = -T \frac{\partial y}{\partial x} \frac{\partial y}{\partial t}.$$

Substituting in for the derivatives from the solution to the wave equation $y(x,t) = Y_0 \sin (k(x-vt) - \phi)$, one finds the power:

$$P = -T k (-vk) Y_0^2 \cos^2 (k(x-vt) - \phi) = Y_0^2 vTk^2 \cos^2 (k(x-vt) - \phi),$$

which indeed goes as the square of the amplitude $Y_0$.

A third example is from electric current: it is well-known from Ohm's law that the power dissipated in a resistor goes as the square of the amplitude of the current through the resistor:

$$P = I^2 R.$$

In the above examples, the power instantaneously went as the square of the peak amplitude. However, the peak amplitude rarely has physical meaning by itself. Typically, the power supplied by a wave is averaged over many periods; since the wave is not at peak amplitude for most of oscillation, it does not make sense for the peak amplitude to be the only important factor. Furthermore, for waves that are not harmonic (not sinusoidal), there may not be a single well-defined peak amplitude.

In these cases, it is more correct to use the root-mean-square amplitude derived by taking the square root of the average of $y^2 (x,t)$ over a period. When the waves are harmonic, averaging the square of the sine or cosine function over a period typically contributes a factor of $\frac12$.

If the solution to the wave equation describes sound waves, the intensity directly corresponds to the loudness of the wave, as typically measured in decibels. Since the intensity goes as the square of the amplitude $A$, the loudness $L$ can be defined:

$$L = 20 \log_{10} \frac{A}{A_0},$$

where $A_0$ is the amplitude of a sound wave at the threshold of human hearing, corresponding to a power of $2 \times 10^{-12} \text{ W}$ [2]. The decibel scale therefore measures the loudness of sounds only as relative to the threshold of human hearing.

The frequency of a wave describes how rapidly the wave oscillates in time. As an example, for light waves in vacuum, the frequency and wavelength are interchangeable definitions. This is because light in vacuum obeys the relation:

$$\lambda f = c$$

and the speed of light $c$ is a constant. Therefore, the frequency and wavelength of light are equivalent definitions, which are used to define the spectrum of types of electromagnetic radiation:

Spectrum of electromagnetic radiation, ranging from radio waves at long wavelength / low frequency to gamma rays at short wavelength / high frequency [3].

Light also obeys the relation from quantum mechanics:

$$E = h f$$

That is, quantum mechanics predicts that the energy $E$ of a photon is proportional to the frequency of light that the photon represents, with the proportionality constant $h$ called Planck's constant. Therefore, the wavelength, frequency, and energy of light described by a small number of photons (quantum mechanically, not classicaly) are all interchangeable definitions.

Human perception of color is directly related to the quantum-mechanical behavior of electronic transitions in atomic elements in the inner structure of the eye. Essentially, color corresponds to the energy of the incoming of photon and therefore to its frequency. Therefore, the frequency of light waves essentially represents color, which is consistent with the usual drawing of the spectrum of electromagnetic radiation.

Direct visualization of the correspondence between color and frequency for light waves. Higher frequency corresponds to the blue/purple end of the spectrum and vice versa.

Because of the relation $\lambda f = c$, colors of light are often labeled by their wavelength: about $400 \text{ nm}$ for purple light and $700 \text{ nm}$ for red light. However, it is more precise to label colors by their frequency. This is an important clarification because the effective speed of light is not always $c$. In materials with an index of refraction greater than one (anything other than vacuum), the constant absorption, emission, and scattering of light from atoms in the material causes the effective speed of light to appear lower than $c$ (in reality, the speed of light is always $c$, but heuristically one can think of extra time for light to propagate being added by the time between absorption and reemission by an atom). Specifically, the speed of light is scaled down by the index of refraction $n$: $c \to \frac{c}{n}$. When the speed of light is lower than $c$, the wavelength of the light changes in the material, but the frequency does not. Since the color does not change when entering the material (consider viewing objects underwater: the color is not altered), the frequency more fundamentally describes the color.

In addition to representing colors, frequencies of sound/pressure waves are also well-known to correspond directly to pitch. Again, this is a consequence of the physics of the human body: the human ear is configured in such a way so that high frequencies are received in a different part of the ear than low frequencies, and the location of reception corresponds to pitch. High frequencies correspond to higher pitches and vice versa:

The harmonics of a guitar string. The high-frequency pitches correspond to higher harmonics of the string. Higher harmonics are colored towards the blue/purple end of the spectrum to make the analogy with the frequencies of light [4].

As described above, when waves enter media, their effective velocity can change and so the effective wavelength changes as well. Since $k = \frac{2\pi}{\lambda}$, these effects are equivalently captured by the change in wavenumber.

The equation that relates the wavenumber to the angular frequency, $\omega(k)$ is called a dispersion relation, and describe how the speed of waves varies at different wavelengths. The dispersion relation of waves in a particular medium is of paramount importance for describing both a) how information is transferred in waves through that medium and b) in describing the allowed energies for waves traveling through that medium (e.g. if the waves are the matter waves of electrons in a periodic crystal where the allowed values of momentum are restricted).

When the dispersion relation of waves in a medium is nontrivial, the velocity of the wave is no longer uniquely defined and so it is nontrivial to describe how information is transferred by the wave. This is especially important in ultrarelativistic physics where depending on how the velocity is described, some ways of describing the wave velocity may exceed the speed of light without violating causality. The two types of velocity are defined as follows:

Group velocity: $$v_g = \frac{d\omega }{dk}.$$

Phase velocity: $$v_p = \frac{\omega}{k}.$$

Roughly, the phase velocity describes the speed of an individual "piece" of the wave while the group velocity describes the speed at which the overall shape of the wave propagates. Depending on context, either may correspond to the signal velocity which determines how the wave transmits information.

Find the dispersion relation for light waves and compute the phase and group velocities.

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Solution:

Light obeys the relation between frequency and wavelength:

$$\lambda f = c.$$

Substituting in for the definitions of angular frequency and wavenumber and rearranging, one arrives at the dispersion relation:

$$\omega = ck.$$

Computing the group and phase velocities, one finds:

$$v_g = v_p = c.$$

Since the group and phase velocities are the same, light waves are non-dispersive. An animation visualizing what it means for a wave to be non-dispersive is displayed below:

The red dots travel at the phase velocity while the green dots travel at group velocity. Notably, the two are equal and the positions of the red dots with respect to the green dots do not change [5].

Given the dispersion relation for deep water waves:

$$\omega = \sqrt{gk},$$

with $g$ the acceleration due to gravity on Earth's surface, compute the phase and group velocities.

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Solution:

It is straightforward to compute the velocities from their definitions, but the result is instructive:

$$v_g = \frac12 \sqrt{\frac{g}{k}},$$

$$v_p = \sqrt{\frac{g}{k}}.$$

The phase velocity is twice the group velocity, and the waves are dispersive:

Red dots again travel at phase velocity while green dots travel at group velocity. Notably, the red dots move with respect to the green dots, corresponding to the dispersion [5].

When waves of multiple wavelengths are superimposed, the wave shape more obviously disperses: The different velocities resulting from the dispersion relation of deep water waves (first three, light blue) causes dispersion of the wave shape (bottom, dark blue) [5].

Dispersion is also literally responsible for the dispersion of light waves through a prism, as reproduced below:

Dispersion of light waves through a prism, a consequence of the wavenumber slightly altering the index of refraction of each color in the prism.

In a material of index of refraction $n$, light obeys the dispersion relation:

$$\omega = \frac{ck}{n}.$$

In most media, the index of refraction $n(k)$is a very weakly increasing function of wavenumber. The group velocity can thus be written in terms of the phase velocity as:

$$v_g = \frac{c}{n} - \frac{ck}{n^2} \frac{dn}{dk} = v_p \left(1- \frac{k}{n} \frac{dn}{dk}\right).$$

Since $n(k)$ is weakly increasing, the group velocity is slightly less than the phase velocity. This fact is responsible for the increased deflection of higher frequencies in a prism.

Lastly, note (as explored in the below problem) that matter waves representing particles in quantum mechanics have a non-trivial dispersion relation. Since $p = \hbar k$ in quantum mechanics according to the de Broglie relation, we see that wavenumber corresponds directly to momentum in this context and the dispersion corresponds to the behavior of energy as a function of momentum in quantum mechanics.

Dispersion of the matter waves of electrons at different group velocities, with highest group velocity on top and lowest on the bottom [5].

The phase shift $\phi$ in solutions to the wave equation at first glance seems unimportant, since coordinates may always be shifted to set $\phi = 0$ for one particular solution. However, what is important is the relative phase shift $\Delta \phi$ between two different solutions to the wave equation, which is responsible for interference and diffraction patterns.

To see why relative phase shift is important, consider the superposition of two identical waves that have a relative phase shift of $\pi$:

Destructive interference of waves (solid red and dashed red) at a relative phase shift of $\pi$, giving the net result of zero (blue) everywhere.

These waves are called out of phase to denote the fact that the phase shift puts the peaks of one wave exactly opposite the peaks of the other. The result of the superposition is that the positive and negative peaks cancel, obtaining zero, which is called destructive interference.

If two waves are in phase, however, the peaks line up. This always occurs when the relative phase shift is zero, but also effectively occurs for small phase shifts. The result is constructive interference, where the peaks of the result are at a height given by the sum of the two original peaks:

Constructive interference of two waves (solid red and dashed red) that are perfectly in phase, giving a result of larger amplitude (blue).

Below, some examples of how superposition of waves at different phase shifts cause important interference and diffraction effects in physics are explored.

Photons corresponding to light of wavelength $\lambda$ are fired at a barrier with two thin slits separated by a distance $d$ as shown in the diagram below. After passing through the slits, they hit a screen at a distance $D$ away, $D \gg d$ and the point of impact is measured. Remarkably, both the experiment and theory of quantum mechanics predict that the number of photons measured at each point along the screen follows a complicated series of peaks and troughs called an interference pattern as below. The photons must exhibit the wave behavior of a relative phase shift somehow to be responsible for this phenomenon. Find the condition for which maxima of the interference pattern occur on the screen.

Left: actual experimental two-slit interference pattern of photons, exhibiting many small peaks and troughs. Right: schematic diagram of the experiment as described above [6].

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Solution:

Since $D \gg d$, the angle from each of the slits is approximately the same and equal to $\theta$. If $y$ is the vertical displacement to an interference peak from the midpoint between the slits, it is therefore true that:

$$D\tan \theta \approx D\sin \theta \approx D\theta = y.$$

Furthermore, there is a path difference $\Delta L$ between the two slits and the interference peak. Light from the lower slit must travel $\Delta L$ further to reach any particular spot on the screen, as in the diagram below:

Light from the lower slit must travel further to reach the screen at any given point above the midpoint, causing the interference pattern.

The condition for constructive interference is that the path difference $\Delta L$ is exactly equal to an integer number of wavelengths. The phase shift of light traveling over an integer $n$ number of wavelengths is exactly $2\pi n$, which is the same as no phase shift and therefore constructive interference. From the above diagram and basic trigonometry, one can write:

$$\Delta L = d\sin \theta \approx d\theta = n\lambda.$$

The first equality is always true; the second is the condition for constructive interference.

Now using $\theta = \frac{y}{D}$, one can see that the condition for maxima of the interference pattern, corresponding to constructive interference, is:

$$n\lambda = \frac{dy}{D},$$

i.e. the maxima occur at the vertical displacements of:

$$y = \frac{n\lambda D}{d}$$

When light shines on a thin film like a soap bubble, an interference pattern results. This is because the light that reflects of the top surface of the thin film has a small phase shift from the light that reflects back out off the bottom surface of the thin film, which has traveled an extra distance related to the thickness of the film (see below diagram).

Thin-film interference on a soap bubble [7]. The color dependence goes as the thickness of the bubble; for monochromatic light the pattern would be of light and dark bands.

Schematic diagram of thin-film interference. Some light entering at angle $\theta_1$ reflects off the top surface, incurring a $\pi$ phase shift. The rest of the light enters the film at an angle dictated by Snell's law, reflects off the bottom, and exits again with a phase shift relative to the originally reflected wave.

To complicate things, when light reflects off a medium of higher index of refraction, Maxwell's equations require that the phase of the light shift by $\pi$.

If the thin film is of thickness $d$, find the condition for destructive interference, in terms of $d$, the wavelength $\lambda$ of the light, the index of refraction $n$ of the film, and the angle $\theta_1$ of incidence with respect to the normal, when light entering from air shines on the film. Note that the index of refraction of the film is greater than that of air (for which $n_{air} = 1$).

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Solution:

For destructive interference, the total extra distance traveled (scaled by the index of refraction) must be an integer number of wavelengths of the light. This is because the ray that reflects off the top surface of the film picks up a phase shift of $\pi$. If the extra distance traveled (scaled by index of refraction) is an integer number of wavelengths, this extra phase shift puts the two rays perfectly out of phase, resulting in destructive interference. The reason for the scaling by index of refraction is because the effective velocity of light is slower when $n \neq 1$, so more phase is accumulated by traveling the same distance (frequency is the same, but velocity is slower, so there is more time for the frequency to accumulate phase $\Delta \phi = \omega \Delta t$.

To find the extra distance traveled in terms of $\theta_1$, first use Snell's law to find the angle $\theta_2$ at which the light enters the film:

$$\sin \theta_1 = n\sin \theta_2.$$

From the diagram, one can see that the extra distance traveled inside the film is $\Delta L_{film} = AB+BC$:

$$\Delta L_{film} = \frac{2d}{\cos \theta_2}.$$

There is an extra path difference, from the amount the light that reflects off the top travels before the second ray exits the film parallel to it. This is segment $AD$ in the diagram. Some plane geometry (try it yourself!) gives the length of $AD$ as:

$$AD = 2d \tan \theta_2 \sin \theta_1.$$

The total extra path difference accounting for the index of refraction is therefore:

$$\frac{2nd}{\cos \theta_2}-2d \tan \theta_2 \sin \theta_1 = 2nd \cos \theta_2.$$

Using the expression for $\theta_2$ in terms of $\theta_1$ from Snell's law and the fact that the $\pi$ phase shift puts the rays perfectly out of phase, one finds the condition for destructive interference, where $m$ is any integer:

$$2nd \cos \theta_2= m\lambda \implies 2nd \cos ( \sin^{-1} (\sin(\theta_1)/n)) = m\lambda.$$

The concept of a relative phase shift is also responsible for the experimental technique of interferometry, which was for instance used at LIGO to discover gravitational waves. Interferometers send laser light down and back along two perpendicular tubes and measure the interference pattern where the light rays recombine. If the length of either arm is slightly longer or shorter than the other, the light picks up a small relative phase which is measured by the interference pattern.

[1] By Lookang (Own work) [CC BY-SA 4.0 (http://creativecommons.org/licenses/by-sa/4.0)], via Wikimedia Commons. https://en.wikipedia.org/wiki/Wave#Traveling_waves

[2] Hass, Jeffrey. An Acoustics Primer: Chapter 6. Center for Electronic and Computer Music, School of Music, Indiana University. http://www.indiana.edu/~emusic/acoustics/amplitude.htm.

[3] Image from https://en.wikipedia.org/wiki/Light under Creative Commons licensing for reuse and modification.

[4] Image from https://en.wikipedia.org/wiki/Guitar_harmonics under Creative Commons licensing for reuse and modification.

[5] Images from https://en.wikipedia.org/wiki/Dispersion_relation under GFDL licensing for reuse.

[6] Image from https://en.wikipedia.org/wiki/Double-slit_experiment under Creative Commons licensing for reuse and modification.

[7] Image from https://en.wikipedia.org/wiki/Thin-film_interference under CC-2.5.