Electromagnetic Waves

Beginner

Where does a radio wave come from? Picture a metal rod joined to an alternating voltage. Electrons are pushed up the rod, then down, then up again, so its two ends turn positive and negative in turn. Around the rod there is an electric field, and every time the charges swap, the field flips. The old field does not vanish: it moves away from the rod while a new one forms behind it. A ripple of electric field spreads outwards. And while the electrons rush along the rod they form a current, which makes a magnetic field, so a magnetic ripple travels out too. Together they are an electromagnetic wave.

In the animation the wiggling charge on the left plays the antenna. The blue wave and arrows show the electric field, the red wave the magnetic field. They rise and fall together, at right angles to each other and to the direction of travel.

Far away, the electric field shakes the electrons in a car's aerial, and the radio turns that tiny jiggle back into sound. To carry music, a station varies the strength of its wave (AM) or its frequency (FM). These waves need no air: they cross empty space, all at the speed of light, about 300,000 km every second.

Radio is one member of a big family. From long waves to short: radio, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. The colour bar shows this spectrum; our eyes see only its narrow rainbow band.

Check yourself: which crosses space faster, radio waves or X-rays? Answer: neither; both travel at the speed of light.

Intermediate

What is waving

A car aerial picks up a station far away, with no wire between them. An electromagnetic (EM) wave is a travelling pattern of an electric field $\vec{E}$, in volts per metre (V/m), and a magnetic field $\vec{B}$, in teslas (T). Nothing material moves: at each point the fields grow, shrink and reverse, and the pattern travels. The push of $\vec{E}$ on the electrons of an aerial or of your eye is how the wave is detected.

Two laws keep it going. As Electromagnetic Induction showed, a changing magnetic field produces an electric field; Maxwell saw that a changing electric field likewise produces a magnetic field. Each change creates the other field a little further on, so the disturbance runs off on its own, with no wires and no medium.

Accelerating charges start it: electrons driven up and down an aerial by an alternating voltage, or the gold charge on the left of the animation. A charge at rest or a steady current makes only steady fields.

Transverse, in phase, at one speed

In vacuum every EM wave has three properties, all visible in the animation:

• It is transverse: $\vec{E}$ (blue, vertical) and $\vec{B}$ (red, horizontal, drawn in perspective) are perpendicular to each other and to the direction of travel, the arrow $c$.
• The two fields oscillate in phase: they peak and pass through zero at the same places and the same instants.
• It travels at $c=3.00\times 10^{8}m/s$, whatever its frequency.

The wave travels along $\vec{E}\times \vec{B}$: turn the fingers of your right hand from $\vec{E}$ towards $\vec{B}$ and your thumb gives the direction. With $\vec{E}$ up the page and $\vec{B}$ out of it, the wave moves right; half a cycle later both have reversed, and it still does. The direction of $\vec{E}$ is the polarisation, vertical in the animation.

The speed follows from the two laws: $c=1/\sqrt{{\mu }_0{\epsilon }_0}$, where ${\epsilon }_0=8.85\times 10^{-12}F/m$ and ${\mu }_0=1.26\times 10^{-6}H/m$ are the constants of electric forces between charges and of magnetic forces between currents, measured with batteries and coils. They give $3.00\times 10^{8}m/s$, the speed of light; in the 1860s this convinced Maxwell that light is an electromagnetic wave.

The working equations

Three relations do most of the work at this level:

• $c=f\lambda$
• $E=cB$
• $I=\frac{1}{2}c{\epsilon }_0{E}_0^2$

The first is the rule of Wave Basics: in one period the wave advances by one wavelength. Since $c$ is fixed, doubling the frequency halves the wavelength: along the spectrum bar, to the right, $\lambda$ falls and $f$ rises.

The second holds at every point and instant. In teslas $B$ looks tiny, 10 nT beside 3.0 V/m, but quantities in different units cannot be compared. Energies can: the densities $\frac{1}{2}{\epsilon }_0{E}^2$ and $B^2/(2{\mu }_0)$ are exactly equal, so each field carries half of the energy.

The third gives the intensity $I$, the average power per square metre (W/m²), from the amplitude $E_0$; the $\frac{1}{2}$ is the average of ${\sin }^2$ over a cycle. Double $E_0$ and the power quadruples.

Worked example: the green light in the animation

The animation models green light with $\lambda =550\mathrm{nm}$, the crest-to-crest distance marked by the ruler. In metres, $550\mathrm{nm}=5.50\times 10^{-7}\mathrm{m}$, so the frequency and the period are
$f=\frac{c}{\lambda }=\frac{3.00\times 10^8}{5.50\times 10^{-7}}\approx 5.45\times 10^{14}\mathrm{Hz}$
$T=\frac{1}{f}\approx 1.83\times 10^{-15}\mathrm{s}$

This is the frequency printed in the panel. A real cycle lasts under two femtoseconds and a drawn one 2 s, so the picture runs about $10^{15}$ times slower than the light.

The field amplitude is $E_0=3.0V/m$, so the magnetic amplitude and the intensity are
$B_0=\frac{E_0}{c}=\frac{3.0}{3.00\times 10^8}=1.0\times 10^{-8}\mathrm{T}=10\mathrm{nT}$
$I=\frac{1}{2}\cdot 3.00\times 10^8\cdot 8.85\times 10^{-12}\cdot 3.0^2\approx 1.2\times 10^{-2}W/{\mathrm{m}}^2$

At point P the readouts swing between $+3.0$ and $-3.0V/m$ and between $+10$ and $-10\mathrm{nT}$: when $E$ reads 1.5 V/m, $B$ reads 5.0 nT, and both cross zero together. The arrows, one every eighth of a wavelength (69 nm in the real wave), grow and shrink in step. The 12 mW/m² of the advanced view means dim light, some 80 000 times weaker than sunlight.

Worked example: FM radio and a microwave oven

An FM station broadcasts at 100 MHz. Convert first, $100\mathrm{MHz}=1.00\times 10^8\mathrm{Hz}$; then
$\lambda =\frac{c}{f}=\frac{3.00\times 10^8}{1.00\times 10^8}=3.0\mathrm{m}$

Reading 100 MHz as $10^7\mathrm{Hz}$ gives 30 m, ten times too long. A quarter-wave car aerial is therefore 75 cm long. About 30 km from a 100 kW transmitter the amplitude is roughly 0.10 V/m, so $B_0=0.10/(3.00\times 10^8)\approx 3.3\times 10^{-10}\mathrm{T}$ and $I=\frac{1}{2}\cdot 3.00\times 10^8\cdot 8.85\times 10^{-12}\cdot 0.10^2\approx 1.3\times 10^{-5}W/{\mathrm{m}}^2$. The signal took $3.0\times 10^4/(3.00\times 10^8)=1.0\times 10^{-4}\mathrm{s}$ to arrive.

A microwave oven at 2.45 GHz uses $\lambda =3.00\times 10^8/(2.45\times 10^9)\approx 0.122\mathrm{m}$. Reflected inside the metal box, the waves form a standing pattern with hot spots half a wavelength, about 6.1 cm, apart. Without the turntable, briefly heat a layer of chocolate and measure the gap between soft spots: $2\cdot 0.061\cdot 2.45\times 10^9\approx 3.0\times 10^{8}m/s$, the speed of light from a kitchen.

Reading the spectrum bar

The bar at the bottom orders the family by wavelength, long on the left; each band's wiggle has its own wavelength, but all move at one speed. The limits are conventions:

• Radio, longer than about 1 m: AM waves near 300 m, FM waves of 3 m.
• Microwaves, 1 mm to 1 m: radar, mobile phones, Wi-Fi and ovens.
• Infrared, 700 nm to 1 mm: heat radiation, strongest near 10 µm for your skin.
• Visible light, only 400 to 700 nm: the narrow outlined rainbow band.
• Ultraviolet, below 400 nm, then X-rays and gamma rays, the shortest far smaller than an atom.

Moving right, the energy of each photon, $hf$, rises too ($h=6.63\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$ is Planck's constant). That energy, not the intensity, decides what a wave does to a molecule: visible photons of about 2 eV drive vision, and only those above roughly 10 eV (ultraviolet below about 124 nm, X-rays, gamma rays) ionise most atoms.

Common mistakes

• EM waves need no medium: sunlight crosses 150 million km of near-vacuum in 8.3 minutes.
• Higher frequency is not faster: a radio echo and a laser pulse bounced off the Moon both return after about 2.6 s.
• The magnetic field is not negligible: 10 nT beside 3.0 V/m carries equal energy. Its force on an electron at speed $v$ is, however, at most $v/c$ of the electric force, so $\vec{E}$ defines the polarisation.
• Glass changes the speed and the wavelength, not the frequency: with $n=1.5$, green light slows to $2.0\times 10^{8}m/s$ and 550 nm shrinks to about 367 nm, while $f$ stays $5.45\times 10^{14}\mathrm{Hz}$.
• “Radiation” is not “radioactive”, and power does not make a wave ionising: a 1000 W oven's photons carry $10^{-5}\mathrm{eV}$, a single X-ray photon tens of keV.

Where you meet it

Radio, television, mobile phones, Wi-Fi, satellite navigation (19 cm), radar and microwave ovens use the long-wavelength end. Remote controls, fibre-optic links (1.55 µm) and thermal cameras use infrared; ultraviolet lamps disinfect water, X-rays image bones and crystals, and gamma rays treat tumours.

This lesson bridges electricity and optics. Polarization of Light follows the direction of $\vec{E}$; Interference and Superposition and Diffraction and the Double Slit add the fields of several waves; Reflection and Refraction follows a wave into matter. Blackbody Radiation and the Quantum and Photoelectric Effect show that light is emitted and absorbed in packets of energy $hf$.

Advanced

Maxwell's equations in empty space

Light, radio waves and X-rays are all waves of electric and magnetic field. To see why such waves exist, and why they all travel at one fixed speed, start from the laws of electromagnetism in a region with no charges and no currents. There they take a strikingly symmetric form:
$\nabla \cdot \vec{E}=0,\nabla \cdot \vec{B}=0$
$\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t},\nabla \times \vec{B}={\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}$

Here $\vec{E}$ is the electric field, $\vec{B}$ the magnetic field, and ${\mu }_0$ and ${\epsilon }_0$ are the magnetic and electric constants of empty space. Read the laws one at a time. The first two are Gauss's laws: with no charges present, no field lines begin or end in the region. The third is Faraday's law: a magnetic field that changes in time produces an electric field that circulates around it. The fourth is Ampère's law with Maxwell's addition, the displacement current ${\epsilon }_0\partial \vec{E}/\partial t$: an electric field that changes in time produces a magnetic field that circulates around it. The last two together let each field keep the other going with no charge nearby, and that is exactly what a wave needs.

Maxwell's term is not a free choice: charge conservation demands it. Where charges are present the fourth law reads $\nabla \times \vec{B}={\mu }_0\vec{J}+{\mu }_0{\epsilon }_0\partial \vec{E}/\partial t$, with $\vec{J}$ the current density. Take the divergence of both sides. The left side is zero, because a curl never has a divergence. On the right, Gauss's law $\nabla \cdot \vec{E}=\rho /{\epsilon }_0$, with $\rho$ the charge density, turns the second term into ${\mu }_0\partial \rho /\partial t$. What remains is $\nabla \cdot \vec{J}+\partial \rho /\partial t=0$: charge flowing out of any small volume is exactly the charge lost inside it. Without Maxwell's term the same step would demand $\nabla \cdot \vec{J}=0$, which is false at the plate of a charging capacitor, where current arrives and none leaves.

Why the front moves at the speed of light

Legend

$\sigma$	Charge per unit area of the moving sheet.	$C/{\mathrm{m}}^2$
$v_s$	Speed of the sheet, down the page in the upper view.	$m/s$
$v$	Speed of the front, the edge of the region that has field.	$m/s$
$\vec{E}$	Electric field behind the front: up in the upper view, out of the page in the lower one.	$V/m$
$\vec{B}$	Magnetic field behind the front: out of the page in the upper view, down in the lower one.	$\mathrm{T}$
$L$	Height of the loop, measured along the front.	$\mathrm{m}$
$vdt$	How far the front advances in a short time $dt$: the width of the shaded strip.	$\mathrm{m}$

Before solving any equation, a thought experiment shows how a change in one field creates the other and how the pair moves off together. It uses only the loop forms of Faraday's law and the Ampère–Maxwell law, and it gives the speed of the front directly.

First the source. An infinite flat sheet lies in the $yz$-plane; Figure 1 shows it edge on, as the orange line at the left. Take $x$ to the right, away from the sheet, $y$ up the page and $z$ out of it. The sheet carries a charge $\sigma$ per square metre and lies on a fixed sheet with the opposite charge, so at rest there is no field outside.

At $t=0$ the charged sheet is jerked into motion down the page, along $-y$, at a constant speed $v_s$. A moving sheet of charge is a current, so a current $\sigma v_s$ per metre of width switches on at once. Right next to the sheet, Ampère's law gives the field of a current sheet, $B={\mu }_0\sigma v_s/2$, pointing out of the page, along $z$.

Farther out the field cannot appear everywhere at the same instant. So suppose it fills a region behind a flat front that moves away from the sheet at some unknown speed $v$ (the dashed line): uniform behind the front, zero ahead of it. The two loop laws will fix both $v$ and the electric field that must come with $\vec{B}$.

Figure 1 draws the same front from two directions, each time with a rectangular loop of height $L$ lying across it. In a short time $dt$ the front advances by $vdt$, and the loop gains the shaded strip of field, of area $Lvdt$.

The upper view looks along $z$. The loop lies in the $xy$-plane, so $\vec{B}$ (the circled dots, out of the page) passes through it, and the magnetic flux through the loop grows by $BLvdt$, that is, at the rate $BLv$. Faraday's law says that a changing flux needs an electric field circulating around the loop. The sides along $x$ contribute nothing, because $\vec{E}$ along $y$ is perpendicular to them, and the side ahead of the front has no field. Only the side of length $L$ behind the front counts, so $EL=BLv$.

The lower view looks down along $y$, from above. Now the loop lies in the $xz$-plane, $\vec{E}$ passes through it (out of the page here) and $\vec{B}$ runs along the loop's side behind the front. The same strip adds electric flux at the rate $ELv$. No current crosses this loop, so the Ampère–Maxwell law says the circulation of $\vec{B}$ is ${\mu }_0{\epsilon }_0$ times that rate; again only the side behind the front counts, so $BL={\mu }_0{\epsilon }_{0}ELv$. The directions drawn are the ones that give both laws the right signs; only the sizes matter here.

Cancel $L$, and the two conditions read
$E=vB,B={\mu }_0{\epsilon }_{0}vE$
Put the first into the second: $B={\mu }_0{\epsilon }_0{v}^{2}B$. Both can hold only if $v^2=1/({\mu }_0{\epsilon }_0)$:
$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}\approx 3.00\cdot 10^{8}m/s$

Nothing in this depends on $\sigma$, on $v_s$ or on how suddenly the sheet started: however hard the jerk, the front has this one speed, built from two constants that are measured with charges and currents. Maxwell saw that it equals the measured speed of light and concluded that light is an electromagnetic wave.

Behind the front $E=cB=\sigma v_s/(2{\epsilon }_{0}c)$, since ${\mu }_{0}c=1/({\epsilon }_{0}c)$. $\vec{E}$ points against the sheet's velocity, up the page in Figure 1, and $\vec{E}\times \vec{B}$ points away from the sheet, the way the front moves.

Notice what the field records: how fast the sheet moves, not where it is. So if the sheet stops after a time $\tau$, a second front of the opposite sign follows the first, and between them a slab of field $c\tau$ thick travels on with no charge anywhere near it. Shake the sheet up and down and the outgoing field copies its velocity history, delayed by $x/c$ at a distance $x$: a plane wave.

The wave equation

The sheet produced a flat front. The differential form of the laws shows that a field pattern of any shape travels at the same speed. Take the curl of Faraday's law. On the left use the identity $\nabla \times (\nabla \times \vec{E})=\nabla (\nabla \cdot \vec{E})-{\nabla }^2\vec{E}$, whose first term vanishes because $\nabla \cdot \vec{E}=0$. On the right, swap the curl with the time derivative and insert the fourth law for $\nabla \times \vec{B}$. The minus signs cancel, leaving
${\nabla }^2\vec{E}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}$

This is the wave equation of Wave Basics with $1/({\mu }_0{\epsilon }_0)$ in place of $v^2$, so every solution travels at $c$. The same steps, starting from the fourth law, give the same equation for $\vec{B}$. The minus sign in Faraday's law is essential: it says that induced fields oppose the change that causes them. With a plus sign the solutions would grow or die away exponentially in time instead of travelling, and there would be no waves.

A plane wave: two fields in phase, at right angles

Legend

$x$	Axis along the direction of travel.	$\mathrm{m}$
$y$	Axis along which $\vec{E}$ oscillates.
$z$	Axis along which $\vec{B}$ oscillates; it points towards you, drawn down-left.
$\vec{E}$	Electric field of the wave.	$V/m$
$\vec{B}$	Magnetic field of the wave, in phase with $\vec{E}$.	$\mathrm{T}$
$E_0$	Amplitude of the electric field.	$V/m$
$B_0$	Amplitude of the magnetic field, equal to $E_0/c$.	$\mathrm{T}$
$\lambda$	Wavelength: the distance from one crest to the next.	$\mathrm{m}$
$c$	Speed of the wave, along the direction of travel.	$m/s$

The simplest solution, a sinusoidal plane wave, shows how $\vec{E}$ and $\vec{B}$ are arranged in all light. Figure 2 freezes one travelling along $x$ at the moment $t=0$:
$E=E_0\sin (kx-\omega t),\omega =ck$
Here $k=2\pi /\lambda$ is the wave number and $\omega$ the angular frequency; $\omega =ck$ is what the wave equation requires of this shape.

In the figure the axes are right-handed, with $z$ pointing towards you (drawn down-left on the page). The electric field $\vec{E}$ (blue) oscillates along $y$, the magnetic field $\vec{B}$ (red) along $z$. Both are sine curves with the same wavelength $\lambda$, the distance from one crest to the next, and they cross zero and peak at the same places. Their amplitudes $E_0$ and $B_0$ have different units, so the lengths drawn for them cannot be compared.

Why is $\vec{E}$ across the direction of travel? In a plane wave the fields depend only on $x$ and $t$. Gauss's law then reduces to $\nabla \cdot \vec{E}=\partial E_x/\partial x=0$, and the $x$-component of the Ampère–Maxwell law gives $\partial E_x/\partial t=0$, because the curl of such a field has no $x$-component. So $E_x$ is at most a constant, which is no part of the wave: the wave is transverse, and we may take $\vec{E}$ along $y$.

Why is $\vec{B}$ along $z$, and in step with $\vec{E}$? The $z$-component of Faraday's law gives $\partial B_z/\partial t=-\partial E_y/\partial x=-k{E}_0\cos (kx-\omega t)$. Integrating in time, the oscillating solution is $B_z=(k/\omega )E_0\sin (kx-\omega t)$; the $y$-component gives $\partial B_y/\partial t=0$, so there is no oscillating $B_y$. Since $\omega /k=c$,
$E=cB$

So $\vec{B}$ is perpendicular to $\vec{E}$ and in phase with it, with $B_0=E_0/c$, and $\vec{E}\times \vec{B}$ points along $+x$, the direction of travel marked $c$ in the figure. For a wave moving towards $-x$ the same steps give $B_z=-E_y/c$: reversing the direction of travel reverses $\vec{B}$ relative to $\vec{E}$.

The expert graph of the animation shows the same wave in time rather than in space: $E$ at the point P, a sine of amplitude 3.0 V/m whose two cycles each last 1.83 fs, one period of 550 nm green light. P lies two wavelengths along the wave, so $kx=4\pi$ there and the trace is $E=E_0\sin (4\pi -\omega t)=-E_0\sin \omega t$: it starts at zero and dips first. This is the snapshot read backwards. The wave moves towards $+x$, so what reaches P next is the part of the wave just to its left, which at $t=0$ lies below the axis. The $B$ readout follows the trace, with $E/B=c$ at every instant.

Energy, intensity and momentum

Legend

$c\Delta t$	Length of the tube: how far the wave advances in the time $\Delta t$.	$\mathrm{m}$
$A$	Cross-section of the tube: the patch the energy crosses.	${\mathrm{m}}^2$
$u$	Energy per unit volume stored in the field in the tube.	$J/{\mathrm{m}}^3$
$\vec{E}$	Electric field of the wave inside the tube.	$V/m$
$\vec{S}$	Power per unit area through the face, along the direction of travel.	$W/{\mathrm{m}}^2$
$c$	Speed of the wave and of the energy it carries.	$m/s$

A wave carries energy away from its source; this is how sunlight warms the Earth. To find how much power crosses each square metre, Figure 3 follows the energy through a patch of area $A$ held across the wave's path.

The fields store energy with density $u=\frac{1}{2}{\epsilon }_0{E}^2+B^2/(2{\mu }_0)$, joules per cubic metre: the same expressions as for the field in a capacitor and in a coil. The whole pattern moves at $c$ without changing shape, so in a time $\Delta t$ the face $A$ is crossed by all the energy in the tube of cross-section $A$ and length $c\Delta t$ behind it, which is $uAc\Delta t$. Dividing by $A$ and by $\Delta t$, the power per square metre is $cu$.

Now use the plane-wave link $B=E/c$ together with $c^2=1/({\mu }_0{\epsilon }_0)$. The magnetic term becomes $B^2/(2{\mu }_0)=E^2/(2{\mu }_0{c}^2)=\frac{1}{2}{\epsilon }_0{E}^2$, so the two halves of $u$ are equal, $u={\epsilon }_0{E}^2$, and the power per square metre is $c{\epsilon }_0{E}^2$. Poynting's theorem, energy conservation for fields, writes this in general as the vector $\vec{S}$ of the figure:
$\vec{S}=\frac{\vec{E}\times \vec{B}}{{\mu }_0}$

For the plane wave its size is $EB/{\mu }_0=E^2/({\mu }_{0}c)=c{\epsilon }_0{E}^2$, as found above, and it points along the direction of travel.

The field of visible light reverses about $10^{15}$ times a second, far too fast for any detector to follow, so what is measured is the average over many cycles. The average of ${\sin }^2$ over a cycle is one half, so the intensity is
$I=\frac{1}{2}c{\epsilon }_0{E}_0^2$

For the animation's 3.0 V/m this is the 12 mW/m² of the panel, with an average energy density of $4.0\times 10^{-11}J/{\mathrm{m}}^3$. Sunlight at the ground, about 1.0 kW/m², corresponds to $E_0\approx 870V/m$ and a magnetic amplitude of 2.9 µT.

The fields also carry momentum, $S/c^2$ per cubic metre. Moving at $c$, that momentum arrives at a surface at the rate $S/c$ per square metre. A black surface absorbs it and so feels a pressure $I/c$; a mirror sends it back and feels twice that. Sunlight on a black surface at the ground pushes with 3.3 µPa; on a mirror above the atmosphere, where the intensity is 1361 W/m², the force is about 9 N per square kilometre. This tiny, unceasing push drives solar sails and blows comets' dust tails away from the Sun.

Radiation from accelerating charges

Legend

$q$	The charge that is pushed.	$\mathrm{C}$
$O$	Where the charge rested before the push.
$a$	Acceleration of the charge during the push.	$m/{\mathrm{s}}^2$
$vt$	Distance the charge has moved from $O$ by the time $t$.	$\mathrm{m}$
$ct$	Radius of the outer sphere: how far the news of the push has travelled in the time $t$.	$\mathrm{m}$
$c\tau$	Thickness of the shell: how far the news travels while the push lasts, a time $\tau$.	$\mathrm{m}$
$\theta$	Angle between a field line and the line of the push.	$\mathrm{rad}$
$\vec{E}$	Sideways field in the shell, the radiation field: opposite to the sideways part of the acceleration.	$V/m$
$c$	Speed at which the shell, and the kink in it, spreads outward.	$m/s$

So far the waves were simply given. Where do they come from? From charges whose velocity changes. A charge at rest has its Coulomb field. A charge moving at constant velocity carries that field along, since in its own frame it is at rest. Only an acceleration sends out a wave, and Figure 4 shows why.

A charge $q$ rests at $O$ until $t=0$. It is then pushed along $x$ with acceleration $a$ for a short time $\tau$, and afterwards drifts on at the small speed $v=a\tau$, so a time $t$ later it is a distance $vt$ from $O$. The figure exaggerates $v/c$; the argument needs $v\ll c$.

News of the push spreads outward at $c$, so at the time $t$ the field lines (drawn in cross-section) fall into three zones. Outside the sphere of radius $ct$ about $O$ the news has not arrived yet: the lines still point away from $O$, where the charge would be had it never moved. Inside the sphere of radius $c(t-\tau )$ the charge has been moving steadily all along, as far as the field knows, and its lines point away from where it is now. In the thin shell of thickness $c\tau$ between the two spheres each line must join its outer part to its inner part, and since the two parts are offset, the line runs sideways across the shell. This sideways stretch is the kink.

Read its size off the figure. A line leaving at the angle $\theta$ to the line of the push is displaced sideways by $vt\sin \theta$, the part of the offset $vt$ across the line, and it makes this shift within a shell only $c\tau$ thick. Within the shell the field follows the line, so the ratio of its sideways part $E_{\perp }$ to its radial part $E_r$ is the ratio of these two lengths. Using $v=a\tau$, and $r=ct$ for the radius of the shell,
$\frac{E_{\perp }}{E_r}=\frac{vt\sin \theta }{c\tau }=\frac{ar\sin \theta }{c^2}$
The radial part is still the Coulomb field, $E_r=q/(4\pi {\epsilon }_0{r}^2)$, because by Gauss's law the shell lets through as many field lines as before. Multiplying the two gives the sideways field, the arrow $\vec{E}$ in the figure.

Look at the powers of $r$. The radial part falls as $1/r^2$, but the sideways part falls only as $1/r$, because the offset $vt$ grows as the shell spreads while its thickness stays $c\tau$. Far from the charge the kink therefore dominates. It is the radiation field, and it moves outward at $c$ with the shell.

For a push that varies smoothly in time, and for speeds well below $c$, the same argument gives the field at a distance $r$ and an angle $\theta$ from the line of the acceleration:
$E(r,t)=\frac{qa(t-r/c)\sin \theta }{4\pi {\epsilon }_0{c}^{2}r}$
Three things stand out. The field is transverse, opposite to the sideways part of the acceleration for a positive charge. It is set by the acceleration at the earlier time $t-r/c$, when the news set out. And its intensity, proportional to $E^2$, goes as ${\sin }^2\theta /r^2$: zero along the line of the acceleration and greatest at right angles to it. That is why a vertical radio mast, whose charges surge up and down, beams its signal out along the ground.

The total power comes from adding the intensity $c{\epsilon }_0{E}^2$ over a sphere of radius $r$. The sphere's area, proportional to $r^2$, cancels the $1/r^2$, and ${\sin }^2\theta$ averages to two thirds over all directions, which gives the Larmor formula:
$P=\frac{q^2{a}^2}{6\pi {\epsilon }_0{c}^3}$

The $1/r$ law is what lets energy escape: the power through a sphere, proportional to $E^2{r}^2$, is the same at every radius, whereas a field falling as $1/r^2$ would carry nothing to great distances. That is why starlight still arrives after many light-years.

In the animation the gold charge oscillates, so its acceleration $a=-{\omega }^{2}y$ is always opposite to its displacement $y$. The radiated field is opposite to the acceleration, so it follows the displacement: the blue wave beside the charge rises and falls with it, and the point P, two wavelengths away, receives each movement two periods (3.7 fs) later. Synchrotron light and X-ray tubes obey the same rule.

Worked problem: a 1 W green lamp

A small lamp emits $P=1.0\mathrm{W}$ of 550 nm green light equally in all directions. At what distance is its field amplitude the animation's 3.0 V/m? What changes at twice that distance? What force acts on a black card of 1.0 cm² held face-on there, and how many photons does it absorb each second?

At a distance $r$ the watt is spread over a sphere of area $4\pi r^2$, so $I=P/(4\pi r^2)$. Set this equal to $\frac{1}{2}c{\epsilon }_0{E}_0^2$ and solve for $r$:
$r=\sqrt{\frac{P}{2\pi c{\epsilon }_0{E}_0^2}}=\sqrt{\frac{1.0}{2\pi \cdot 3.00\times 10^8\cdot 8.85\times 10^{-12}\cdot 3.0^2}}\approx 2.6\mathrm{m}$

At 5.2 m the same watt crosses a sphere four times larger, so the intensity drops fourfold, to 3.0 mW/m². The amplitude goes as the square root of the intensity, so it only halves, to 1.5 V/m, as the $1/r$ law requires.

Back at 2.6 m the intensity is 12 mW/m², so the card absorbs $1.2\times 10^{-2}\cdot 1.0\times 10^{-4}=1.2\times 10^{-6}\mathrm{W}$. An absorber feels the pressure $I/c$, so the force is the absorbed power divided by $c$: $F=1.2\times 10^{-6}/(3.00\times 10^8)=4.0\times 10^{-15}\mathrm{N}$. The pressure, $4.0\times 10^{-11}\mathrm{Pa}$, equals the beam's average energy density, as it does for any absorber, since both are $I/c$.

The light's frequency is $f=c/\lambda =5.45\times 10^{14}\mathrm{Hz}$, and each photon carries the energy $hf$, with $h$ Planck's constant: $hf=6.63\times 10^{-34}\cdot 5.45\times 10^{14}\approx 3.6\times 10^{-19}\mathrm{J}$, about 2.3 eV. The card therefore absorbs $1.2\times 10^{-6}/(3.6\times 10^{-19})\approx 3.3\times 10^{12}$ photons a second. That is far too many for their graininess to show, so the smooth classical wave is an excellent description here.

Limits of validity

The classical theory handles propagation, interference, diffraction and polarisation at every frequency. It fails where matter exchanges energy with the field, because that happens in quanta $hf$: 2.3 eV for green light, $4\times 10^{-7}\mathrm{eV}$ for FM radio, tens of keV for medical X-rays. Most atoms need roughly 10 eV to lose an electron, so only wavelengths below about 124 nm ionise, however weak the beam. When photons are few, the wave gives only the probability of detecting one.

It fails inside atoms too. An electron circling as in a classical hydrogen atom, with $a\approx 9.0\times 10^{22}m/{\mathrm{s}}^2$, would radiate $4.7\times 10^{-8}\mathrm{W}$ by Larmor's formula and spiral into the nucleus in about 16 ps. Atoms are stable; explaining why begins in Atomic Models: the Bohr Atom.

In matter the wave slows to $c/n$, where the refractive index $n$ depends on frequency. Conductors absorb it: seawater cuts a 10 kHz wave by a factor $e$ every 2.5 m or so, which is why submarines use very low frequencies. In vacuum the equations are linear, so beams cross without disturbing each other. Finally, $c$ is exact: since 1983 the metre has been defined so that $c=299792458m/s$.

Where this leads

Maxwell's equations fix a speed $c$ without saying relative to what; Special Relativity makes it the same for every inertial observer and mixes $\vec{E}$ and $\vec{B}$ when the frame changes. Polarization of Light follows the direction of $\vec{E}$; Interference and Superposition and Diffraction and the Double Slit add fields; Prisms, Dispersion and Colour traces the slower speed in glass to electrons re-radiating like the sheet above. Blackbody Radiation and the Quantum, Photoelectric Effect and Wave–Particle Duality take up the quanta $hf$; Particles and Forces: the Standard Model makes the photon the carrier of the electromagnetic force. Gravitational waves (Gravity as Curved Spacetime) also travel at $c$: in 2017 those of a neutron-star merger arrived within 2 s of its gamma rays.