8c. Light, spectroscopy

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Getting back now to physics, a natural question would be that of going back to the equations that we started with, namely the wave and heat equations in 2D, and see if our accumulated knowledge about harmonic functions, which is after all not that bad, can help there. Unfortunately, this is not exactly the case, and we are still a long way to go, from solving that equations. So, obviously, time to ask the cat. And cat says: \begin{cat} If you don't understand physics, do more physics. \end{cat} Thanks cat, and this seems wise indeed, let's have some fun with more physics, still related of course to the Laplace operator [math]\Delta[/math], and for difficult mathematics, which is still to be solved, we will see later, towards the end of the present book. Getting started now, we will need for what we want to talk about something quite heavy, namely:

Theorem (Maxwell theory)

In regions of space where there is no charge or current present the Maxwell equations for electrodynamics read

[[math]] \lt \nabla,E \gt = \lt \nabla,B \gt =0 [[/math]]

[[math]] \nabla\times E=-\dot{B}\quad,\quad \nabla\times B=\dot{E}/c^2 [[/math]]
and both the electric field [math]E[/math] and magnetic field [math]B[/math] are subject to the wave equation

[[math]] \ddot{\varphi}=c^2\Delta\varphi [[/math]]
where [math]\Delta=\sum_id^2/dx_i^2[/math] is the Laplace operator, and [math]c=299,792,458[/math].


Show Proof

This is something fundamental, appearing as a tricky mixture of physics facts and mathematical results, the idea being as follows:


(1) To start with, electrodynamics is the science of moving electrical charges. And this is something quite complicated, because unlike in classical mechanics, where the Newton law is good for both the static and the dynamic setting, the Coulomb law, which is actually very similar to the Newton law, does the job when the charges are static, but no longer describes well the situation when the charges are moving.


(2) The problem comes from the fact that moving charges produce magnetism, and with this being visible when putting together two electric wires, which will attract or repel, depending on orientation. Thus, in contrast with classical mechanics, where static and dynamic problems are described by a unique field, the gravitational one, in electrodynamics we have two fields, namely the electric field [math]E[/math], and the magnetic field [math]B[/math].


(3) Fortunately, there is a full set of equations relating the electric field [math]E[/math] and the magnetic field [math]B[/math]. These are the Maxwell equations, which look as follows:

[[math]] \lt \nabla,E \gt =\frac{\rho}{\varepsilon_0}\quad,\quad \lt \nabla,B \gt =0 [[/math]]

[[math]] \nabla\times E=-\dot{B}\quad,\quad\nabla\times B=\mu_0J+\mu_0\varepsilon_0\dot{E} [[/math]]


(4) To be more precise, regarding first the math, the dots denote derivatives with respect to time, and [math]\nabla[/math] is the gradient operator, or space derivative, given by:

[[math]] \nabla=\begin{pmatrix} \frac{d}{dx}\\ \frac{d}{dy}\\ \frac{d}{dz} \end{pmatrix} [[/math]]

(5) Regarding the physics, the first formula is the Gauss law, [math]\rho[/math] being the charge, and [math]\varepsilon_0[/math] being a constant, and with this Gauss law more or less replacing the Coulomb law from electrostatics. The second formula is something basic, and anonymous. The third formula is the Faraday law. As for the fourth formula, this is the Ampère law, as modified by Maxwell, with [math]J[/math] being the volume current density, and [math]\mu_0[/math] being a constant.


(6) Without bothering too much about the precise meaning of all this, we can see right away that under the circumstances in the statement, namely in the regions of space where there is no charge or current present, the Maxwell equations take a simple and comprehensible form, readable even without much physics background, namely:

[[math]] \lt \nabla,E \gt = \lt \nabla,B \gt =0 [[/math]]

[[math]] \nabla\times E=-\dot{B}\quad,\quad \nabla\times B=\mu_0\varepsilon_0\dot{E} [[/math]]


(7) Thus, we have reached to the equations in the statement, modulo a discussion about the constant [math]\mu_0\varepsilon_0[/math]. And the point here is that, according to a remarkable discovery of Biot and Savart, the main electrodynamics constants [math]\mu_0,\varepsilon_0[/math] are magically related to the observed speed of light in vacuum [math]c=299,792,458[/math] by the following formula:

[[math]] \mu_0\varepsilon_0=\frac{1}{c^2} [[/math]]


(8) Summarizing, we have our equations. Leaving aside the first two equations, by applying the curl operator to the last two equations, we obtain:

[[math]] \nabla\times(\nabla\times E)=-\nabla\times\dot{B}=-(\nabla\times B)'=-\ddot{E}/c^2 [[/math]]

[[math]] \nabla\times(\nabla\times B)=\nabla\times\dot{E}/c^2=(\nabla\times E)'/c^2=-\ddot{B}/c^2 [[/math]]


But the double curl operator is subject to the following formula:

[[math]] \nabla\times(\nabla\times\varphi)=\nabla \lt \nabla,\varphi \gt -\Delta\varphi [[/math]]


Now by using the first two equations, we are led to the conclusion in the statement.

So, question now, what is light? Light is the wave predicted by Theorem 8.15, travelling at speed [math]c[/math], and with the important extra property that it depends on a real positive parameter, that can be called, upon taste, frequency, wavelength, or color: \begin{fact} An accelerating or decelerating charge produces electromagnetic waves, travelling in vacuum at speed [math]c=299,792,458[/math],

[[math]] \ddot{\varphi}=c^2\Delta\varphi [[/math]]

and in non-vacuum at a lower speed [math]v \lt c[/math]. These waves are called light, whose frequency and wavelength can be explicitly computed. \end{fact} This phenomenon can be observed is a variety of situations, such as the usual light bulbs, where electrons get decelerated by the filament, acting as a resistor, or in usual fire, which is a chemical reaction, with the electrons moving around, as they do in any chemical reaction, or in more complicated machinery like nuclear plants, particle accelerators, and so on, leading there to all sorts of eerie glows, of various colors.


Let us try now to understand the simplest solutions of the wave equation [math]\ddot{\varphi}=c^2\Delta\varphi[/math], or more generally [math]\ddot{\varphi}=v^2\Delta\varphi[/math], from Fact 8.16. In 1D, the situation is as follows:

Theorem

The 1D wave equation, with speed [math]v[/math], namely

[[math]] \ddot{\varphi}=v^2\,\frac{d^2\varphi}{dx^2} [[/math]]
has as basic solutions the following functions,

[[math]] \varphi(x)=A\cos(kx-wt+\delta) [[/math]]
with [math]A[/math] being called amplitude, [math]kx-wt+\delta[/math] being called the phase, [math]k[/math] being the wave number, [math]w[/math] being the angular frequency, and [math]\delta[/math] being the phase constant. We have

[[math]] \lambda=\frac{2\pi}{k}\quad,\quad T=\frac{2\pi}{kv}\quad,\quad \nu=\frac{1}{T}\quad,\quad w=2\pi\nu [[/math]]
relating the wavelength [math]\lambda[/math], period [math]T[/math], frequency [math]\nu[/math], and angular frequency [math]w[/math]. Moreover, any solution of the wave equation appears as a linear combination of such basic solutions.


Show Proof

There are several things going on here, the idea being as follows:


(1) Our first claim is that the function [math]\varphi[/math] in the statement satisfies indeed the wave equation, with speed [math]v=w/k[/math]. For this purpose, observe that we have:

[[math]] \ddot{\varphi}=-w^2\varphi\quad,\quad \frac{d^2\varphi}{dx^2}=-k^2\varphi [[/math]]


Thus, the wave equation is indeed satisfied, with speed [math]v=w/k[/math]:

[[math]] \ddot{\varphi}=\left(\frac{w}{k}\right)^2\frac{d^2\varphi}{dx^2}=v^2\,\frac{d^2\varphi}{dx^2} [[/math]]


(2) Regarding now the other things in the statement, all this is basically terminology, which is very natural, when thinking how [math]\varphi(x)=A\cos(kx-wt+\delta)[/math] propagates.


(3) Finally, the last assertion is standard, coming from Fourier analysis, and we will leave this as an instructive exercise, based on the material from chapter 7.

As a first observation, the above result invites the use of complex numbers. Indeed, we can write the solutions that we found in a more convenient way, as follows:

[[math]] \varphi(x)=Re\left[A\,e^{i(kx-wt+\delta)}\right] [[/math]]


And we can in fact do even better, by absorbing the quantity [math]e^{i\delta}[/math] into the amplitude [math]A[/math], which becomes now a complex number, and writing our formula as:

[[math]] \varphi=Re(\widetilde{\varphi})\quad,\quad \widetilde{\varphi}=\widetilde{A}e^{i(kx-wt)} [[/math]]


Moving ahead now towards electromagnetism and 3D, let us formulate:

Definition

A monochromatic plane wave is a solution of the 3D wave equation which moves in only 1 direction, making it in practice a solution of the 1D wave equation, and which is of the special from found in Theorem 8.17, with no frequencies mixed.

In other words, we are making here two assumptions on our wave. First is the 1-dimensionality assumption, which gets us into the framework of Theorem 8.17. And second is the assumption, in connection with the Fourier decomposition result from the end of Theorem 8.17, that our solution is of “pure” type, meaning a wave having a well-defined wavelenght and frequency, instead of being a “packet” of such pure waves.


Summarizing, we have now a decent intuition about what light is, and more on this later. Let us discuss now the examples. The idea is that we have various types of light, depending on frequency and wavelength. These are normally referred to as “electromagnetic waves”, but for keeping things simple and luminous, we will keep using the familiar term “light”. The classification, in a rough form, is as follows:

[[math]] \begin{matrix} {\rm Frequency}&\ \ \ \ \ \ \ \ \ \ {\rm Type}\ \ \ \ \ \ \ \ \ \ &{\rm Wavelength}\\ &-\\ 10^{18}-10^{20}&\gamma\ {\rm rays}&10^{-12}-10^{-10}\\ 10^{16}-10^{18}&{\rm X-rays}&10^{-10}-10^{-8}\\ 10^{15}-10^{16}&{\rm UV}&10^{-8}-10^{-7}\\ &-\\ 10^{14}-10^{15}&{\rm blue}&10^{-7}-10^{-6}\\ 10^{14}-10^{15}&{\rm yellow}&10^{-7}-10^{-6}\\ 10^{14}-10^{15}&{\rm red}&10^{-7}-10^{-6}\\ &-\\ 10^{11}-10^{14}&{\rm IR}&10^{-6}-10^{-3}\\ 10^9-10^{11}&{\rm microwave}&10^{-3}-10^{-1}\\ 1-10^9&{\rm radio}&10^{-1}-10^8\\ \end{matrix} [[/math]]


Observe the tiny space occupied by the visible light, all colors there, and the many more missing, being squeezed under the [math]10^{14}-10^{15}[/math] frequency banner. Here is a zoom on that part, with of course the comment that all this, colors, is something subjective:

[[math]] \begin{matrix} {\rm Frequency\ THz}=10^{12}\ {\rm Hz}&\ \ \ {\rm Color}\ \ \ &{\rm Wavelength\ nm}=10^{-9}\ {\rm m}\\ &-\\ 670-790&{\rm violet}&380-450\\ 620-670&{\rm blue}&450-485\\ 600-620&{\rm cyan}&485-500\\ 530-600&{\rm green}&500-565\\ 510-530&{\rm yellow}&565-590\\ 480-510&{\rm orange}&590-625\\ 400-480&{\rm red}&625-750 \end{matrix} [[/math]]


With this in hand, we can now do some basic optics. Light usually comes in “bundles”, with waves of several wavelenghts coming at the same time, from the same source, and the first challenge is that of separating these wavelenghts. In order to discuss this, from a practical perspective, let us start with the following fact: \begin{fact} When travelling through a material, and hitting a new material, some of the light gets reflected, at the same angle, and some of it gets refracted, at a different angle, depending both on the old and the new material, and on the wavelength. \end{fact} Again, this is something deep, and there are many things that can be said here, ranging from various computations based on the Maxwell equations, to all sorts of considerations belonging to advanced materials theory. As a basic formula, we have the famous Snell law, which relates the incidence angle [math]\theta_1[/math] to the refraction angle [math]\theta_2[/math], as follows:

[[math]] \frac{\sin\theta_2}{\sin\theta_1}=\frac{n_1(\lambda)}{n_2(\lambda)} [[/math]]


Here [math]n_i(\lambda)[/math] are the refraction indices of the two materials, adjusted for the wavelength, and with this adjustment for wavelength being the whole point, which is something quite complicated. Now as a simple consequence of the above, we have:

Theorem

Light can be decomposed, by using a prism.


Show Proof

This follows from Fact 8.19. Indeed, when hitting a piece of glass, provided that the hitting angle is not [math]90^\circ[/math], the light will decompose over the wavelenghts present, with the corresponding refraction angles depending on these wavelengths. And we can capture these split components at the exit from the piece of glass, again deviated a bit, provided that the exit surface is not parallel to the entry surface. And the simplest device doing the job, that is, having two non-parallel faces, is a prism.

As an application of this, we can study events via spectroscopy, by capturing the light the event has produced, decomposing it with a prism, carefully recording its “spectral signature”, consisting of the wavelenghts present, and their density, and then doing some reverse engineering, consisting in reconstructing the event out of its spectral signature.

General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].