A good gift for your buddies….

Maxwell's equations, named after James Clerk Maxwell who first expressed them together are a set of four equations from which all electromagnetic theory can be derived.

The integral form of Maxwell's equations in free space (ie., in the absence of dielectric or magnetic materials) are

$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{0}}$ (Gauss's Law)

$\oint \vec{B}\cdot d\vec{A}=0$ (Magnetic equivalent of Gauss's Law)

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}$ (Faraday's Law)

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I+\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$ (Modified form of Ampere's Law).

In the absence of moving charges and changing currents (stationary charges and steady currents), the electric and magnetic fields do not change with time at any point.

$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{0}}$

$\oint\vec{E}\cdot d\vec{l}= 0$

$\oint \vec{B}\cdot d\vec{A}=0$

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I$

Notice that electric and magnetic fields are decoupled, each with two equations.

When we want to find the gradient of a vector field we use an operator $\nabla$ which is read as del

$\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}$

When applied to a scalar field we can use the word grad, so we would say “$E$ is equal to minus *grad* $V$”

$\vec{E}=-\nabla V $

$\vec{E}=-\hat{i}\frac{\partial V}{\partial x}-\hat{j}\frac{\partial V}{\partial y}-\hat{k}\frac{\partial V}{\partial z}$

The $\nabla$ operator can also be applied to a vector, but as it is a vector itself there are two operations, that can be performed, the scalar product “$\cdot$” and the vector product “$\times$”.

The scalar product of the $\nabla$ operator with a vector field is called the Divergence of the field, for example the divergence of the electric field is

$\nabla\cdot \vec{E}=\frac{\partial E_{x}}{\partial x}+\frac{\partial E_{y}}{\partial y}+\frac{\partial E_{z}}{\partial z}$

The divergence of a field gives a measurement of the magnitude of the source or sink of the field at a point, and we will discuss the relationship between this and charge in a moment.

The vector product of the $\nabla$ operator with a vector field,(e.g. $\nabla\times\vec{E}$) is called the Curl of the field, and gives a measurement of the rotation of the field.

This page has a nice description of the idea behind divergence and curl.

Using the divergence theorem

$\oint \vec{F}\cdot d\vec{A}=\int\nabla\cdot \vec{F}\, dV$

and Stokes' theorem

$\oint \vec{F}\cdot d\vec{l}=\int\nabla\times\vec{F}\cdot dA$

(Some links with a nice discussion of the idea behind the divergence theorem and Stokes' theorem).

Each of Maxwell's equations can also be written in differential form (the form that is most useful will depend on the problem to be solved!)

$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{0}}$ → $\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_{0}}$

$\oint \vec{B}\cdot d\vec{A}=0$ → $\nabla\cdot\vec{B}=0$

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}$ → $\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I+\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$ → $\nabla\times\vec{B}=\mu_{0}\vec{J}+\mu_{0}\varepsilon_{0}\frac{\partial \vec{E}}{\partial t}$

$\oint\vec{E}\cdot d\vec{A}=0$ → $\nabla\cdot\vec{E}=0$

$\oint \vec{B}\cdot d\vec{A}=0$ → $\nabla\cdot\vec{B}=0$

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}$ → $\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$ → $\nabla\times\vec{B}=\mu_{0}\varepsilon_{0}\frac{\partial \vec{E}}{\partial t}$

One of the reason that Maxwell's equations are considered to be so important is that they can explain how electromagnetic waves can be produced from changing electromagnetic fields.

This only works because of Maxwell's correction to Ampere's Law.

We should note that electromagnetic waves were predicted by Maxwell before they were observed experimentally by Heinrich Hertz (unfortunately after Maxwell had died).

A key point to take into account is that electromagnetic fields take time to reach distant points, there is a speed of propagation, which we will later show is the speed of light $c$.

Here is a nice applet to show the propagation of electric and magnetic fields from antennas.

Here is a nice demo to demonstrate that energy is propagated via electromagnetic waves and the waves have a definite alignment with respect to orientation of the antenna.

$\nabla\times\vec{E}=(\frac{\partial}{\partial y}E_{z}-\frac{\partial}{\partial z}E_{y})\hat{i}+(\frac{\partial}{\partial z}E_{x}-\frac{\partial}{\partial x}E_{z})\hat{j}+(\frac{\partial}{\partial x}E_{y}-\frac{\partial}{\partial y}E_{x})\hat{k} $

If we consider a electric wave oscillating in the $y$ direction and moving in the $x$ direction then this reduces to $\frac{\partial E_{y}}{\partial x}\hat{k}$

The 3rd Maxwell equation

$\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$

tells us that there must then be a time varying magnetic field $B_{z}$ as

$\frac{\partial E_{y}}{\partial x}\hat{k}=-\frac{\partial B_{z}}{\partial t}\hat{k}$

A similar exercise with the 4th equation

$\nabla\times\vec{B}=\mu_{0}\vec{J}+\mu_{0}\varepsilon_{0}\frac{\partial \vec{E}}{\partial t}$

gives us

$-\frac{\partial B_{z}}{\partial x}\hat{j}=\mu_{0}\varepsilon_{0}\frac{\partial E_{y}}{\partial t}\hat{j}$

We now have two equations

$\frac{\partial E_{y}}{\partial x}=-\frac{\partial B_{z}}{\partial t}$ and $-\frac{\partial B_{z}}{\partial x}=\mu_{0}\varepsilon_{0}\frac{\partial E_{y}}{\partial t}$

which can be used to obtain the wave equation for the electric and magnetic fields. To find the wave equation for $E$ we take $\frac{\partial}{\partial x}$ of the first equation and $\frac{\partial}{\partial t}$ of the second, giving.

$\frac{\partial^{2} E_{y}}{\partial x^{2}}=-\frac{\partial^{2} B_{z}}{\partial t \partial x}$ and $-\frac{\partial^{2} B_{z}}{\partial t \partial x}=\mu_{0}\varepsilon_{0}\frac{\partial^{2} E_{y}}{\partial t^{2}}$

which can be combined to give

$$\frac{\partial^{2} E_{y}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} E_{y}}{\partial x^{2}}$$

To find the wave equation for $B$ we take $\frac{\partial}{\partial t}$ of the first equation and $\frac{\partial}{\partial x}$ of the second, giving.

$\frac{\partial^{2} E_{y}}{\partial x \partial t}=-\frac{\partial^{2} B_{z}}{\partial t^{2}}$ and $-\frac{\partial^{2} B_{z}}{\partial x^{2}}=\mu_{0}\varepsilon_{0}\frac{\partial^{2} E_{y}}{\partial x \partial t}$

which can be combined to give

$$\frac{\partial^{2} B_{z}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} B_{z}}{\partial x^{2}}$$

In order to have a wave which at time $t=0$ has a displacement function $D(x)=A\sin\frac{2\pi}{\lambda}x$ propagate with $v$ we can write

$D(x,t)=A\sin\frac{2\pi}{\lambda}(x-vt)$

which using $v=\frac{\lambda}{T}=f\lambda$ can be written

$D(x,t)=A\sin(\frac{2\pi}{\lambda}x-\frac{2\pi t}{T})=A\sin(kx-\omega t)$

We have here defined a new quantity $k$ which is the wave number, $k=\frac{2\pi}{\lambda}$ and expressed the frequency in terms of the angular frequency $\omega=2\pi f$

The velocity of the waves propagation, which we call the phase velocity can be expressed in terms of $\omega$ and $k$

$v=f\lambda=\frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$

The equation for a wave's motion we have just considered

$D(x,t)=A\sin\frac{2\pi}{\lambda}(x-vt)$

is a solution to the a differential equation that we call the wave equation.

$v^2\frac{\partial^{2} D}{\partial x^2}=\frac{\partial^{2} D}{\partial t^2}$

We can check that this is the case by substituting in our equation of motion in to the differential equation

$-v^2(\frac{2\pi}{\lambda})^{2}A\sin\frac{2\pi}{\lambda}(x-vt)=-v^{2}(\frac{2\pi}{\lambda})^{2}A\sin\frac{2\pi}{\lambda}(x-vt)$

$v^2\frac{\partial^{2} D}{\partial x^2}=\frac{\partial^{2} D}{\partial t^2}$

with a solution

$D(x,t)=A\sin\frac{2\pi}{\lambda}(x-vt)$

where $D$ is some kind of displacement.

Our equations

$\frac{\partial^{2} E_{y}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} E_{y}}{\partial x^{2}}$ and $\frac{\partial^{2} B_{z}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} B_{z}}{\partial x^{2}}$

thus give us two waves in phase with one another, but at right angles to one another

$E_{y}(x,t)=E_{0}\sin\frac{2\pi}{\lambda}(x-\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$

$B_{z}(x,t)=B_{0}\sin\frac{2\pi}{\lambda}(x-\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$

The equations

$E_{y}(x,t)=E_{0}\sin\frac{2\pi}{\lambda}(x-\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$

$B_{z}(x,t)=B_{0}\sin\frac{2\pi}{\lambda}(x-\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$

tell us that electromagnetic waves have velocity

$v=\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}=3.00\times10^{8}\,\mathrm{m/s}=c$

At the time Maxwell derived this result it was known that light had wavelike properties, but the speed of light was not known.

The speed of light was first measured by Michelson, who is also famous for one of the most influential “failed” experiments, the Michelson Morley experiment which aimed to find the ether, the reference frame in which light moves at the speed of light, but ended up showing that there was no ether and that light has the same speed in all reference frames.

Either of the equations

$\frac{\partial E_{y}}{\partial x}=-\frac{\partial B_{z}}{\partial t}$ and $-\frac{\partial B_{z}}{\partial x}=\mu_{0}\varepsilon_{0}\frac{\partial E_{y}}{\partial t}$

can be used to show that

$\frac{E_{0}}{B_{0}}=c\to\frac{E}{B}=c$

The equation we used for waves in general $v=f\lambda$ can be applied to electromagnetic waves, $c=f\lambda$, to find the frequency from the wavelength and vice versa. The full electromagnetic spectrum is much broader than the relatively narrow range we can see.