Electromagnetic waves

The previous lesson ended with two equations that describe empty space and refuse to be trivial, each field driving the other with no charge anywhere in sight.

This lesson solves them. The mathematics needs one new piece of notation, the partial derivative, introduced below in a sentence, and the result is that optics stops being a separate subject. The lesson also names the places where the classical theory runs out, because it does, and knowing where is what makes the rest of it trustworthy.

Guessing the shape of the answer

Solving a pair of field equations in full generality is a large task. Solving them for one particular shape of field is not, and if that shape works it is a solution, which is all that is needed to establish that such things exist.

So assume the simplest possible arrangement. Let the electric field point along $y$ and the magnetic field along $z$ , and let both depend only on the coordinate $x$ and on time. That describes a plane wave travelling along $x$ : at any instant the fields are uniform across every plane perpendicular to $x$ and vary only as you move along it.

The notation $\partial E / \partial x$ means the rate of change of $E$ with respect to $x$ at a fixed instant, and $\partial E / \partial t$ the rate of change with time at a fixed place. That is all a partial derivative is: an ordinary derivative with the other variable held still.

Two thin rectangles, and a wave equation

The strategy is to apply each of the two surviving equations to a rectangle so thin that the result becomes a statement about the fields at a point.

Draw the first rectangle in the $x y$ plane with one pair of sides of length $a$ running along $y$ , at positions $x$ and $x + Δ x$ , and the other pair, of length $Δ x$ , running along $x$ .

Take the circulation of $E$ round it. The electric field points along $y$ , so the two sides running along $x$ contribute nothing, since the field is perpendicular to them. The two sides of length $a$ contribute $E$ at their own position times $a$ , with opposite signs because the circuit traverses them in opposite directions, so the total is

a [E (x + Δ x) - E (x)] = a \frac{\partial E}{\partial x} Δ x

Now the flux of $B$ through the rectangle. The magnetic field points along $z$ , perpendicular to the rectangle, so the flux is $B$ times the area $a Δ x$ , and its rate of change is $(\partial B / \partial t) a Δ x$ .

Faraday's law equates the first to minus the second, and $a Δ x$ cancels from both sides:

\frac{\partial E}{\partial x} = - \frac{\partial B}{\partial t}

The area of the rectangle dropping out is what makes this a statement about the fields at a point rather than about a particular loop.

Now repeat with a rectangle in the $x z$ plane, sides of length $b$ along $z$ at $x$ and $x + Δ x$ . The magnetic field points along $z$ , so the same argument gives its circulation as $b (\partial B / \partial x) Δ x$ . The electric flux through this rectangle is $E$ times $b Δ x$ , since $E$ is perpendicular to it.

There are no charges and no currents, so the only term on the right is the displacement one, and

\frac{\partial B}{\partial x} = - ε_{0} μ_{0} \frac{\partial E}{\partial t}

The two minus signs are what the right hand rules give when the orientations are tracked consistently, and the reader who works them through and gets a different pair of signs has probably chosen the opposite direction of travel, which is equally valid.

That leaves two coupled equations, each relating one field to the other. Eliminate $B$ in the standard way: differentiate the first with respect to $x$ ,

\frac{\partial^{2} E}{\partial x^{2}} = - \frac{\partial}{\partial x} \frac{\partial B}{\partial t} = - \frac{\partial}{\partial t} \frac{\partial B}{\partial x}

where the order of the two derivatives can be swapped for any well behaved function. Substituting the second equation for $\partial B / \partial x$ , the two minus signs cancel:

\frac{\partial^{2} E}{\partial x^{2}} = ε_{0} μ_{0} \frac{\partial^{2} E}{\partial t^{2}}

That is the wave equation. Any function of the form $f (x - v t)$ satisfies it, provided $v^{2} = 1 / (ε_{0} μ_{0})$ : such a function is a fixed shape sliding along the $x$ axis at speed $v$ without changing, which is what a wave is. The identical elimination the other way gives the same equation for $B$ , so both fields travel together.

So electromagnetic waves exist, they are predicted by the four equations with nothing else assumed, and they travel at

v = \frac{1}{\sqrt{ε_{0} μ_{0}}}

The number

Put in the constants. $ε_{0} = 8.8542 \times 10^{- 12}$ F/m, measured with capacitors, and $μ_{0} = 1.2566 \times 10^{- 6}$ T m/A, measured with the force between current-carrying wires. Their product is $1.1127 \times 10^{- 17}$ , and

v = \frac{1}{\sqrt{1.1127 \times 10^{- 17}}} = 2.9979 \times 10^{8} m/s

That is the speed of light, obtained from two electrical measurements involving no light whatsoever.

Maxwell had the comparison in 1862. The electrical constants came from Weber and Kohlrausch's 1856 experiment, in which they measured a capacitor's charge twice, once electrostatically with a torsion balance and once magnetically by discharging it through a galvanometer, and took the ratio: 310,740,000 m/s, high by 3.7 per cent. The speed of light came from Fizeau's 1849 toothed wheel, 314,858,000 m/s, high by 5.0 per cent. Maxwell wrote that the velocity of transverse undulations in his hypothetical medium agreed so exactly with the velocity of light that we can scarcely avoid the inference that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena.

Two independently measured constants of electricity, each known to a few per cent, combining to give a third quantity from an entirely different branch of physics: that is the strongest kind of evidence a physical theory can offer, because there was no adjustable parameter anywhere to make it come out right.

The confirmation took twenty-five years. Heinrich Hertz, between 1887 and 1888, built an oscillator from a spark gap between two metal spheres and a detector from a loop with a smaller gap, and saw sparks in the detector when the transmitter fired across the room. He measured the wavelength by finding the nodes of a standing wave reflected off a metal sheet, computed the speed, and got light's. He then showed the waves reflect, refract through a pitch prism, and can be polarised by a grid of wires. Asked what use it was, he said none whatsoever, and died in 1894 at thirty-six, seven years before Marconi sent a signal across the Atlantic.

What the wave looks like

The solution carries more information than its speed.

$E$ and $B$ are perpendicular to each other and both perpendicular to the direction of travel: the wave is transverse, which is why light can be polarised, and polarisation is direct evidence for the transverse character since a longitudinal wave has no orientation to filter.

They are in phase, peaking and vanishing together, and their magnitudes are locked. Substituting a travelling sine wave into either coupled equation gives

E = c B

so the fields are far from equal in size in SI units, which is a fact about the units rather than about the wave: the energy in the two is identical, as the next section shows.

Nothing in the derivation fixed the frequency. Any $f (x - c t)$ works, so the equations permit waves of every wavelength, all travelling at the same speed in vacuum, and the electromagnetic spectrum is one continuum. Radio at a hundred megahertz has a wavelength of three metres; a microwave oven at 2.45 GHz gives 12.2 cm, and the half-wavelength spacing of 6.1 cm between the nodes of the standing wave inside is why an oven has a turntable. Green light at 550 nm is $5.45 \times 10^{14}$ Hz. X-rays and gamma rays continue the same list. Radio and gamma rays differ by nineteen orders of magnitude in frequency and obey the identical four equations.

Example. A microwave oven operates at 2.45 GHz. What is the wavelength, and how far apart are the hot spots of the standing wave inside?

$λ = c / f = (2.998 \times 10^{8}) / (2.45 \times 10^{9}) = 0.122$ m, so 12.2 cm. A standing wave has antinodes every half wavelength, so the hot spots sit 6.1 cm apart, which is a substantial fraction of a dinner plate.

Now you. A mobile phone transmits at 1800 MHz. What is the wavelength?

Answer

$λ = c / f = (2.998 \times 10^{8}) / (1.80 \times 10^{9}) = 0.167$ m, about 16.7 cm. A quarter of that, 4.2 cm, is a typical internal antenna length.

Energy, and how much

The two energy densities from earlier lessons apply at every point of the wave: $\frac{1}{2} ε_{0} E^{2}$ from the fifth and $B^{2} / 2 μ_{0}$ from the tenth. Substituting $B = E / c$ and $c^{2} = 1 / (ε_{0} μ_{0})$ into the second gives $ε_{0} E^{2} / 2$ , identical to the first. A light wave carries exactly half its energy in the electric field and half in the magnetic, at every point and every instant.

The energy passing through unit area per second is the density times the speed, $u_{total} c = ε_{0} c E^{2}$ , usually written as the Poynting vector $S = E B / μ_{0}$ , which points in the direction of travel. For a sine wave the average of $E^{2}$ over a cycle is half the peak squared, so the intensity is $\frac{1}{2} ε_{0} c E_{0}^{2}$ , or equivalently $ε_{0} c E_{rms}^{2}$ .

This is the promissory note of the fifth lesson being redeemed. The energy density $\frac{1}{2} ε_{0} E^{2}$ was derived there from a capacitor and could have been dismissed as an algebraic rearrangement. Here the same expression describes energy in a region with no charge in it, travelling, arriving somewhere else, and warming a thermometer. The energy is genuinely in the field.

Example. Sunlight above the atmosphere delivers 1361 W/m². What are the electric and magnetic field strengths in it?

From $S = ε_{0} c E_{rms}^{2}$ , $E_{rms}^{2} = 1361 / [(8.854 \times 10^{- 12}) (2.998 \times 10^{8})] = 1361 / (2.654 \times 10^{- 3}) = 5.13 \times 10^{5}$ , so $E_{rms} = 716$ V/m and the peak is $\sqrt{2}$ times that, 1013 V/m. The magnetic field follows from $B = E / c$ : $B_{rms} = 716 / (2.998 \times 10^{8}) = 2.39 \times 10^{- 6}$ T, about five per cent of the Earth's field. Sunlight is a 700 V/m electric field oscillating $5 \times 10^{14}$ times a second, which is not how it feels.

Now you. A laser delivers 5000 W/m² onto a surface. What is the rms electric field in the beam?

Answer

$E_{rms}^{2} = 5000 / [(8.854 \times 10^{- 12}) (2.998 \times 10^{8})] = 5000 / (2.654 \times 10^{- 3}) = 1.88 \times 10^{6}$ , so $E_{rms} = 1.37 \times 10^{3}$ V/m.

Momentum and pressure

A wave that carries energy also carries momentum, at $p = U / c$ per unit energy, a result that comes out of Maxwell's equations and that relativity later makes inevitable for anything moving at $c$ . Absorbing a wave therefore delivers momentum, which is a force, which over an area is a pressure:

P = \frac{S}{c}

for an absorbing surface, and twice that for a perfect mirror, because the reflected wave carries momentum away in the opposite direction.

The numbers are tiny. Sunlight at 1361 W/m² pushes on a black surface with $1361 / (2.998 \times 10^{8}) = 4.54 \times 10^{- 6}$ Pa, ten orders of magnitude below atmospheric pressure. It is nonetheless real, it was measured by Lebedev in 1900 and by Nichols and Hull in 1901, and it is what shapes a comet's tail, which points away from the Sun regardless of which way the comet is going.

Example. A perfectly reflecting solar sail of 200 m² is deployed at the Earth's distance from the Sun on a 10 kg spacecraft. What thrust does it produce, and what speed change over a year of continuous thrust?

The pressure on a mirror is $2 S / c = 2 (1361) / (2.998 \times 10^{8}) = 9.08 \times 10^{- 6}$ Pa. Over 200 m² that is a force of $1.82 \times 10^{- 3}$ N, under two millinewtons, roughly the weight of a grain of rice. On 10 kg it gives an acceleration of $1.82 \times 10^{- 4}$ m/s², and over a year of $3.16 \times 10^{7}$ s the speed change is $5.7 \times 10^{3}$ m/s. A thrust too small to feel, applied for a year, beats a chemical rocket stage. The Japanese probe IKAROS demonstrated this in 2010 with a sail of about 196 m², measuring a thrust near 1.12 mN, below the ideal figure because a real sail is neither perfectly reflecting nor perfectly flat.

Now you. A perfectly absorbing sheet of 50 m² is held at the same distance from the Sun. What force does the sunlight exert on it?

Answer

An absorber feels $S / c = 1361 / (2.998 \times 10^{8}) = 4.54 \times 10^{- 6}$ Pa, so over 50 m² the force is $2.27 \times 10^{- 4}$ N, about a fifth of a millinewton.

Where this stops

The theory just completed is one of the most successful in physics, and it has edges, three of which were found within twenty years of Hertz.

There is no medium. Maxwell believed his waves were undulations of a material ether, and the whole nineteenth century framework assumed one. Michelson and Morley in 1887 looked for the Earth's motion through it, with an interferometer sensitive enough to detect a hundredth of the expected effect, and found nothing. The equations turn out not to need a medium: the fields are the wave. Einstein's 1905 paper, which opened with the induction problem of the ninth lesson, disposed of the ether and made the constancy of $c$ a postulate, and Maxwell's equations then turn out to be exactly relativistic already, which is why they needed no correction when mechanics did.

Energy is not continuous. The classical wave carries as little energy as you like, arriving smoothly. Shine dim ultraviolet on a metal and electrons come off immediately with an energy fixed by the frequency, not by the intensity, which is impossible on the classical picture and which Einstein explained in the same year by quantising the wave into photons of energy $h f$ . The classical theory works when the photon count is enormous, which for sunlight and radio it always is, and fails when it is small.

Accelerating charges radiate, which classically destroys the atom. An orbiting electron is accelerating, and the equations of this course say it must radiate, losing energy and spiralling into the nucleus in about $10^{- 11}$ seconds. Atoms exist. That contradiction is one of the doors into quantum mechanics, and Atoms and Elements goes through it.

None of this makes the four equations wrong. Quantum electrodynamics, the theory that replaces them at small scales, reproduces every result in this course in the limit of many photons, and it is the most precisely tested theory ever constructed, agreeing with measurement of the electron's magnetic moment to twelve significant figures. What has changed is the domain, not the content.

What was built

The course began with two charged rods and a torsion balance, and a force law that said what two charges do and could not say how.

Replacing the force with a field made the how expressible. Symmetry and energy gave two ways of computing that field, one closed surface at a time and one scalar function at a time. Charge in motion needed a second field, defined by a sideways velocity-dependent force, and produced by currents. Changing either field made the other, which gave induction, the generator and the whole electrical industry, and which cost the electric potential its existence. One missing term, forced by a capacitor and a choice of surface, completed the set and let the two fields sustain each other with no matter present at all.

Then the four equations, containing nothing but charges, currents and two constants measured with wire and glass, turned out to describe light. That was not a goal anyone had set. It is what the equations said when they were solved, and the fact that a theory can know more than the people who built it is the best argument there is for taking a derivation seriously enough to follow it to the end.