**Table of Contents**:

The following section attempts to give a bit more of an explanation of some of the elementary-level electromagnetics concepts important for optics, but that perhaps would not be discussed in an introductory graduate class.

One of the most beautiful truths of optics is that all of the equations we use in optics can be derived from four equations, postulated by Maxwell in 1865. If Maxwell could understand the world around him using the simple instruments accessible in the day, surely we can also!

We can start by considering a charged particle in a dielectric medium. Believe it or not, that charged particle will feel a force exerted on it directly proportional to the amount of charge the particle contains, as well as the strength of the electric field present in the dielectric medium. This is the Lorentz law, given by:

\begin{equation} F=q(E + v\times{B}) \end{equation}

where “F” is the force, “q” is the charge of the particle, “E” is the electric field, “v” is the velocity of light, and “B” is the magnetic field. Most times though, we can ignore the cross term, as it is negligible compared to the magnitude of the electric field (more on this later, though). But how there an electric field present in the medium in the first place, one may ask? It turns out there are two methods, which we describe below here:

**Electric Field:**

**Gauss’s Law:** Following the formulation of Griffith’s “Introduction to Electrodynamics”, we will begin with the first way to create an electric field in a medium, which starts with a single charge at the origin. Every charged particle produces an electric field: for positively charged particles, such as protons, the electric field lines point away from the particle; for negatively charged particles, such as electrons, the electric field lines point towards the particle. The electric field in the case of a single charged particle is then given in \eqref{static_electric}.

\begin{equation} E(r) = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} \label{static_electric} \end{equation}

If we then imagine that the charged particle is bound by a closed surface, then we can write the flux of the electric field through that surface as:

\begin{align} \Phi_e & = \oint E \cdot da \\ & = \int \frac{1}{4\pi\epsilon_0}(\frac{q}{r^2}\hat{r}) (r^2sin(\theta)~d\theta~{d\phi}~\hat{r}) \\& = \frac{q}{\epsilon_0} \label{q_micro}\end{align}

where all we have done is found the amount of “E” passing through a spherical closed surface. We can see, though, that the $r^2$ dependence cancels, which makes sense because as the circle grows larger, the electric field will also drop off as $1/r^2$. This then means, though, that it actually doesn’t matter what shape the enclosing surface is; the field lines don’t care and will continue to propagate in the same way! This works out well for us.

Now, in the case of multiple charged particles enclosed in a surface, the principle of superposition of electric fields tells us that we can have a summation of the electric field contribution from the individual particles. This means that Equation \eqref{q_micro} can be rewritten as:

\begin{equation} \sum_{i=1}^{n}\oint E_i \cdot da = \sum_{i=1}^{n}(\frac{1}{\epsilon_0}q_i) \end{equation}

which Griffith’s then writes for any closed surface as:

\begin{equation} \oint_S E_i \cdot da = \frac{1}{\epsilon_0}Q_{enc} \end{equation}

which is Gauss’s Law in integral form. We can also easily rewrite this in the normal differential form by applying Gauss’s divergence theorem:

\begin{equation} \oint_S E \cdot da = \int_V(\nabla \cdot E)d\tau \label{div_theo}\end{equation}

which states that the electric field flux through a surface is equivalent to the total sum of the divergence of the field within a volume. We can also rewrite the total enclosed charge in terms of the sum of the charge density within a volume:

\begin{equation} Q_{enc} = \int_V\rho d\tau \label{q_enc}\end{equation}

Now, using Equations \eqref{div_theo} and \eqref{q_enc}, and keeping in mind that this theorem has to hold for any volume so that the integral volume dependence should vanish, Gauss’s Law is given as:

\begin{equation} \nabla \cdot E = \frac{1}{\epsilon_0}\rho \label{diff_gauss} \end{equation}

The interpretation of Gauss’s law is that the electric field flux through the *surface* of the medium is equal to the total number of charges within the volume contained within the surface.

Now, you may be saying, “This is great and all, but how exactly does this relate to creating an electric field in a medium? When are you getting to the point, Optics Girl?!” Fair enough, I’ll stop rambling and get to the bottom of this. There are two main ways that one can describe how an electric field appears in a medium without having to go into any magnetic field descriptions as well. One of these ways is when you have a conducting medium, in which case you have some free charges floating around, denoted by their free charge density as $\rho_{free}$, that contribute to the overall charge density. Another contribution to the total charge density that appears in Gauss’s Law is when you have the presence of light on a dielectric medium. When an electromagnetic wave is incident on a dielectric medium, the atoms within the material become polarized i.e. the electromagnetic wave induces a dipole moment on the particle by displacing the electrons relative to the nucleus (which houses the protons). The total dipole moment per unit volume is defined as the polarization **P**, which can be written in terms of the charge density as:

\begin{equation} \rho_p = -\nabla \cdot P(r) \end{equation}

The total charge density that appears in Gauss’s Law is then the addition of these two contributions, i.e.

\begin{align} \rho_{total} & = -\nabla \cdot P + \rho_{free}\\ & =\rho_p + \rho_{free} \end{align}

There is some more math involved with showing this, but I think this explanation should suffice! For the interested reader, though, this online resource walks through a lot of the math. We’re almost done with Gauss’s Law, I promise, but one last useful equation is Gauss’s Law written in terms of the electric displacement **D**. The electric displacement is given as:

\begin{equation} D = \epsilon_0{E} + P \label{disp} \end{equation}

If we solve Equation \eqref{disp} for “**E**” and substitute this into Equation \eqref{diff_gauss} where “$\rho=\rho_{total}$”, then we can rewrite Gauss’s Law as:

\begin{equation} \nabla \cdot D = \rho \end{equation}

which says now that the total electric displacement is proportional to the charge density.

**Faraday’s Law:** Remember I said that there were two ways to have an electric field inside of a medium? We have finally reached the second method to do this, which involves a changing magnetic field. For those of you with some background in electromagnetism, you may have seen a demonstration of Faraday’s Law of Electromagnetic Induction using a magnet moving through a loop of wires. For those of you without a background in electromagnetism, check out this video below, as we will be using the form of Faraday’s Law of Electromagnetic Induction that can be derived from this video to obtain a more useful form for us later on.

The above video goes through a derivation of Faraday’s Law in terms of magnetic flux and voltage, but we will go through a different derivation here to have a better form to derive Faraday’s Law. Using the above video to derive our relation, we have a bar magnet *moving* through a coil of copper wires (a well known conductor), and as Dr. Melloch notes, the faster the magnet moves through the coil of wires, the more current is produced. This implies that we will have a $\frac{\partial{B}}{\partial{t}}$ term in our answer somewhere, as the amount of current produced depends on the rate of change of the magnetic field. It is also important to note that with the larger number of coils, not only did we have more length of wire for the current to pass through, but we also had a larger surface area parallel to the magnet. This then implies that we need to find the rate of change of the magnetic field per each unit area on the coil.

Now, holding on to this thought, within the wire we have a current produced as a result of the changing magnetic field, which we will denote as “i”. As Dr. Melloch states in the video, if we were to also monitor the resistance in the wire, we would find that the current goes down as the resistance goes up and vice versa. This would given in an equation as $i \propto \frac{1}{R}$, so that if we multiply both sides by “R”, we would have the left hand side equals “iR”, which we know from Ohm’s law is equal to V, or the voltage. The electric field is given in units of volts/m (in mks units), so we can see that if we are finding the total amount of voltage created over a length of wire, we would need an integral of the electric field over a length.

Finally, we can put these two thoughts together to find Faraday’s Law of Electromagnetic Induction, given in Equation \eqref{fara_em_ind} as:

\begin{equation} \oint E\cdot dl = – \iint_s \frac{dB}{dt}\cdot{dS} \label{fara_em_ind} \end{equation}

where we have put together all of our observations from the video, just like Faraday would have done!

Now using Equation \eqref{fara_em_ind}, we can then find the form of Faraday’s Law that will be much more useful for us when deriving the wave equation shortly. Stoke’s theorem provides a relationship between the line integral and the surface integral of a field vector, so we can use Stoke’s theorem to rewrite the left hand side of Equation \eqref{fara_em_ind} as:

\begin{align} \oint E\cdot{dl} = \iint (\nabla \cdot E)\cdot {dS} \end{align}

Since the relationship between **E** and **B** must hold at all points in space, we can do away with the surface integrals to obtain the following form of Faraday’s Law:

\begin{equation} \nabla \times E = -\frac{\partial B}{\partial{t}} \end{equation}

This is the form we want, and we shall see why in a bit! First, we have to find some other relations to find the rest of Maxwell’s Laws.

**Magnetic Field:**

For the first law for magnetic fields, which I have chosen to be Gauss’s Law, I will just give the result here and then explain what it means. Gauss’s Law for Magnetism is given as:

\begin{equation} \nabla\cdot{B} = 0 \end{equation}

which says basically that there are no magnetic monopoles, or magnetic charges. What does this mean physically? Well, we can think back to earlier on this page when we found that the divergence of the electric field gives the total number of charges enclosed in a surface. Hence, there is a charge present, and it may not be counterbalanced by an opposite charge. Now, for the magnetic field, it is the case where there will not be an enclosed charge according to the results I just gave. We can take the example of a bar magnet. Every person who has played with a bar magnet knows that there is a north and a south pole to the magnet. Have you ever wondered what would happen if you cut that bar magnet in half? Would you now just have a separate north and south end? It turns out, we can find this answer from our above equation! Since there are no “magnetic charges”, then each end of the magnet that is now in two will have a new north and south pole; if you cut it again, the same thing would happen! This is because each end must perfectly balance the other side.

The next magnetic field equation we need to know is Ampere’s Law, which is given as:

\begin{equation} \oint_C B\cdot dl = \mu\iint_S J\cdot dS \end{equation}

This equation means that we can obtain a magnetic field via a steady state current. However, just like with an electric field, there are in fact two ways to obtain a magnetic field. The derivation for the second method to obtain a magnetic field comes from the constitutive relations for **B **and **H**, which is $**B** = \mu**H**$. Thus, the differential form of Ampere’s Law is given as:

\begin{equation} \nabla \times **H** = **J** \label{ampere}\end{equation}

However, this is an incomplete equation because when we go to verify another fundamental axiom of physics, conservation of charge, we will find that Equation \eqref{ampere} actually violates the conservation of charge principle. To prove this, we will first need the actual equation stating the conservation of charge, given as:

\begin{equation} \nabla \cdot J = -\frac{\partial\rho}{\partial{t}} \label{charge_cont}\end{equation}

which just means that a divergence in the current density must mean a change in the charge density. However, if we use Equation \eqref{charge_cont} in Equation \eqref{ampere}, we will see that we have $-\frac{\partial\rho}{\partial{t}} = 0$, which means that the charge continuity is violated. This is where our great hero, James Maxwell, stepped in and saved the day yet again. Maxwell pointed out that Ampere’s Law needed another term, namely the displacement current, in order to satisfy the charge continuity condition. Thus, the full Ampere-Maxwell equation is given as:

\begin{equation} \nabla \times H = J + \frac{\partial{D}}{\partial{t}} \end{equation}

Note that this equation gives us our second source for the magnetic field: a time-varying electric field (in the form of $D$, which is of course related to $E$ by $D = \epsilon{E}$)! We might have guessed that from our results before.

**All Together Now:** This is what we came here for! Using the knowledge outlined above, we can now begin to derive Maxwell’s Equations.

Maxwell’s Equations are given in Equations \eqref{Gauss}-\eqref{Ampere_Maxwell}.

\begin{align} \nabla \cdot D &= \rho \label{Gauss} \end{align}

\begin{align} \nabla \cdot B &= 0 \label{Gauss_Law_Mag} \end{align}

\begin{align} \nabla \times E & = -\frac{\partial{B}}{\partial{t}} \label{Faraday} \end{align}

\begin{align} \nabla \times H & = J + \frac{\partial{D}}{\partial{t}} \label{Ampere_Maxwell} \end{align}

Let’s go through these one by one. Equation \eqref{Gauss} is known as Gauss’s Law. Gauss’s law describes the relationship between the static electric field and the charges that cause it. Basically, it is saying that the net outflow of $\vec{E}$ through a closed surface is proportional to the enclosed charge

Equation \eqref{Gauss_Law_Mag} was also first derived by Gauss, and it is known as Gauss’s Law for Magnetism. Gauss’s Law for Magnetism is relatively simple to understand: it states that there are no “magnetic charges”, or charges equivalent to an electron in magnetism.

Equation \eqref{Faraday} is known as Faraday’s Law, and it states that a time-varying magnetic field gives rise to an electric field. We will see important consequences of this exchange in our last equation.

Equation \eqref{Ampere_Maxwell} is known as the Ampere-Maxwell Law. The Ampere-Maxwell Law gives us a couple different insights. It tells us that a magnetic field can be induced by an electric current and a changing electric field. It also echoes Faraday’s Law in that it states that a changing electric field can create a magnetic field, and vice versa. This is how electromagnetic fields propagate through space! Interestingly, the oscillation of the electric and magnetic fields as they repeatedly recreate one another is also a classical example of a harmonic oscillator.

**Extra: Gauge Equations:**