In short

An electromagnetic wave is a self-sustaining oscillation of electric field \vec{E} and magnetic field \vec{B}, travelling through vacuum at the universal speed

\boxed{\;c \;=\; \frac{1}{\sqrt{\mu_0\varepsilon_0}} \;=\; 2.998\times 10^8\text{ m/s}\;}

Five properties characterise every such wave:

  1. Transverse: \vec{E} and \vec{B} both oscillate perpendicular to the direction of propagation \hat{k}. They never point along \hat{k}.
  2. Mutually perpendicular: \vec{E} \perp \vec{B}, and the triad (\vec{E},\,\vec{B},\,\hat{k}) obeys the right-hand rule. \vec{E}\times\vec{B} points along \hat{k}.
  3. Locked amplitudes: the field magnitudes are tied by
\boxed{\;\frac{E_0}{B_0} \;=\; c\;}

At any instant and any point along the wave, E = cB.

  1. Energy flux (intensity): averaged over a cycle, the energy crossing unit area per unit time is
\boxed{\;I \;=\; \tfrac{1}{2}\varepsilon_0\,c\,E_0^2 \;=\; \frac{c\,B_0^2}{2\mu_0}\;}
  1. Momentum flux (radiation pressure): the wave carries momentum along \hat{k}. On a perfectly absorbing surface the pressure is
\boxed{\;P_\text{rad} \;=\; \frac{I}{c}\;}

On a perfect reflector it doubles to 2I/c.

No medium required. Unlike sound or ripples, electromagnetic waves propagate through empty space — from distant quasars to your phone — because the medium is the two fields themselves, each regenerating the other as described by Maxwell's equations.

Hold a modern smartphone at arm's length on a Marine Drive evening. A 5G tower on top of a Nariman Point office building, a few kilometres away, is pouring microwaves into the air around you at a carrier frequency near 3.5 GHz. Those waves reach your phone's antenna in roughly ten microseconds. Across the same evening, All India Radio's Nagpur transmitter is sending news on the medium-wave band at 1 MHz — a radio wave about 300 metres long — that your grandmother's transistor on the next balcony decodes into Hindi speech. The sun, 150 million kilometres away, is depositing visible light on your shirt at about 1 kW per square metre. And cosmic gamma rays, born in supernovae in the Andromeda galaxy, are streaming through your body at this instant at around 10^{20} Hz.

Four wildly different frequencies, four completely different technologies to produce and detect them — and one identical physics. Each of them is a transverse wave in the electric and magnetic fields, travelling at the same speed c, obeying the same five rules. This chapter writes those rules down, derives them from Maxwell's equations, and builds the interactive picture you can actually see.

The setup: a plane wave solution to Maxwell's equations

Start from Maxwell's four equations in vacuum — no charges, no currents, just the fields regenerating themselves in empty space:

\nabla\cdot\vec{E} = 0,\qquad \nabla\cdot\vec{B} = 0
\nabla\times\vec{E} = -\frac{\partial\vec{B}}{\partial t},\qquad \nabla\times\vec{B} = \mu_0\varepsilon_0\,\frac{\partial\vec{E}}{\partial t}

Why: in empty space, charge density \rho and current density \vec{J} are both zero. The two Gauss laws kill the sources of flux; Faraday and Ampère–Maxwell remain in their source-free form — each curl is driven by the other field's time derivative.

You already met the derivation of the wave equation in the displacement-current article. The result: each component of \vec{E} and \vec{B} obeys

\frac{\partial^2 E}{\partial t^2} \;=\; \frac{1}{\mu_0\varepsilon_0}\,\frac{\partial^2 E}{\partial x^2}

with wave speed c = 1/\sqrt{\mu_0\varepsilon_0} = 2.998\times 10^8 m/s. The next five sections pick this wave apart and extract the five properties that appear on every JEE physics formula sheet.

Property 1 — EM waves are transverse

Try to build a plane wave travelling along +\hat{x} whose electric field points along \hat{x} itself. Write

\vec{E}(x,t) \;=\; E_x(x,t)\,\hat{x}

Why: "transverse" means the oscillation is perpendicular to travel. "Longitudinal" means the oscillation is along travel. You're testing the longitudinal option to see whether Maxwell's equations allow it.

Step 1. Apply \nabla\cdot\vec{E} = 0 (Gauss's law in vacuum).

\nabla\cdot\vec{E} \;=\; \frac{\partial E_x}{\partial x} \;=\; 0

Why: the only non-zero component is E_x, and it only depends on x (plane wave along \hat{x}). So the divergence reduces to \partial E_x/\partial x.

Step 2. If \partial E_x/\partial x = 0, then E_x does not vary with x. It is the same at every point along the wave's direction of travel. A wave has to vary along x — that is what makes it a wave. A uniform field is not a travelling wave.

Conclusion: Maxwell's equations forbid a longitudinal electric-field wave in vacuum. The same argument with \nabla\cdot\vec{B} = 0 forbids a longitudinal magnetic wave. Both \vec{E} and \vec{B} must oscillate in directions perpendicular to \hat{k}. Electromagnetic waves in vacuum are always transverse.

Contrast this with sound, a longitudinal wave: air molecules oscillate along the direction of travel, compressing and rarefying the medium. That works for sound because sound is a pressure wave and pressure is a scalar — there is no "direction" restricting it. For \vec{E}, Gauss's law does the restricting.

Property 2 — \vec{E}, \vec{B}, and \hat{k} form a right-handed triad

Pick the simplest transverse wave: \vec{E} along \hat{y}, travelling along +\hat{x}. Write

\vec{E}(x,t) \;=\; E_0\sin(kx - \omega t)\,\hat{y}

Why: a sinusoidal plane wave of amplitude E_0, angular frequency \omega, wavenumber k = \omega/c. The phase kx - \omega t travels at speed \omega/k = c along +\hat{x} — standard travelling-wave form.

Now ask Maxwell's equations where \vec{B} points.

Step 1. Compute \nabla\times\vec{E}. With \vec{E} having only a y-component that depends only on x,

\nabla\times\vec{E} \;=\; \begin{vmatrix}\hat{x}&\hat{y}&\hat{z}\\ \partial_x & \partial_y & \partial_z \\ 0 & E_y & 0\end{vmatrix} \;=\; \frac{\partial E_y}{\partial x}\,\hat{z}

Why: only the E_y term is non-zero in the curl determinant. The \hat{z} component picks up \partial E_y/\partial x; the other components vanish because E_y has no y or z dependence and E_x = E_z = 0.

Step 2. Plug in the sinusoidal form.

\frac{\partial E_y}{\partial x} \;=\; k E_0 \cos(kx - \omega t)

So

\nabla\times\vec{E} \;=\; k E_0 \cos(kx - \omega t)\,\hat{z}

Step 3. Use Faraday's law \nabla\times\vec{E} = -\partial\vec{B}/\partial t.

\frac{\partial\vec{B}}{\partial t} \;=\; -k E_0 \cos(kx - \omega t)\,\hat{z}

Why: whatever \vec{B} is, its time derivative must point along -\hat{z}\cdot\cos(\cdots). Integrating with respect to t gives \vec{B} itself along \hat{z}, matching the sine form.

Step 4. Integrate in time.

\vec{B}(x,t) \;=\; \frac{k E_0}{\omega}\sin(kx - \omega t)\,\hat{z} \;=\; \frac{E_0}{c}\sin(kx - \omega t)\,\hat{z}

Why: the antiderivative of -\cos(kx-\omega t) with respect to t is \sin(kx-\omega t)/\omega. The constant of integration is zero for a pure travelling wave (no static background field). And k/\omega = 1/c.

Result: \vec{E} is along \hat{y}, \vec{B} is along \hat{z}, and the wave travels along \hat{x}. Check with the right-hand rule:

\hat{y}\times\hat{z} \;=\; \hat{x}

so \vec{E}\times\vec{B} points along \hat{k}. That is the rule: the triad (\vec{E},\,\vec{B},\,\hat{k}) is always right-handed, with \vec{E}\times\vec{B} pointing in the direction of propagation.

The right-handed triad of an electromagnetic wave A 3D axis diagram showing the direction of propagation k along the x-axis, the electric field E oscillating along y, and the magnetic field B oscillating along z. E cross B points along k. $\hat{k}$ (direction of travel) $\vec{E}$ along $\hat{y}$ $\vec{B}$ along $\hat{z}$ origin right-hand rule: $\vec{E}\times\vec{B}$ points along $\hat{k}$
The direction of propagation $\hat{k}$, the electric field $\vec{E}$, and the magnetic field $\vec{B}$ form a right-handed triad. Point your right hand's fingers from $\vec{E}$ toward $\vec{B}$: your thumb points along $\hat{k}$.

Property 3 — E_0/B_0 = c

Look again at step 4 above:

\vec{B} \;=\; \frac{E_0}{c}\sin(kx - \omega t)\,\hat{z}

So the amplitude of the magnetic field is

\boxed{\;B_0 \;=\; \frac{E_0}{c}\,,\qquad E_0 \;=\; cB_0\;}

This is not a free parameter — Maxwell's equations forced this ratio. The same ratio holds at every instant and every point along the wave: E(x,t) = c\,B(x,t).

Numerical check. Sunlight at the surface of the Earth has an electric-field amplitude of roughly E_0 \approx 1000 V/m. Then

B_0 \;=\; \frac{1000}{3\times 10^8} \;=\; 3.3\times 10^{-6}\text{ T} \;=\; 3.3\ \mu\text{T}

The magnetic component of sunlight is tiny compared to the Earth's own static magnetic field (\approx 50\ \muT). That is why, historically, people noticed the electric effects of light long before anyone suspected a magnetic side. In SI units, E and B look wildly asymmetric because c is big — but in the natural units of the wave, they carry exactly the same information.

Watch it move

Watching a static figure of an electromagnetic wave teaches you the shape. Watching one move is what teaches you the phase lock: E and B oscillating in step, perpendicular to each other, and the whole pattern sliding along \hat{k} at speed c.

Animated: E and B oscillating in phase, perpendicular to each other and to the direction of travel Two sinusoidal curves, E along vertical axis and B along horizontal-depth axis, phase-locked and moving along the +x direction. The y-ordinate shows E amplitude; the trailing grey curve shows B on the same axis at reduced amplitude for visual compactness. $x$ $\vec{E}$ $\vec{B}$
At a fixed point $x = 2$ m in space, watch $E$ (red, above the axis) and $B$ (dark, below) oscillate in phase — both peak at the same instant, both pass through zero at the same instant. The wave travels along $+\hat{x}$ at speed $c$; you are watching one point on it change over time. Click replay to watch again.

The phase lock matters: at the moments E is maximum, B is also maximum. They peak together and vanish together. If they were out of phase, the wave would carry no net energy — so nature (and Maxwell's equations) arrange the phases so that E and B rise and fall in perfect lockstep.

Property 4 — energy flux (intensity)

An electromagnetic wave carries energy. Your skin absorbs about 1 kW per square metre of sunlight on a clear Mumbai afternoon. A solar panel on an ISRO satellite bus converts that flux into electrical power. A 5G antenna at your phone absorbs a tiny fraction of the wave passing it. All of these depend on one quantity — the intensity I, the time-averaged energy crossing unit area per unit time.

Step 1. Write the energy density stored in the fields. From the articles on electric-field energy density and magnetic-field energy density,

u_E \;=\; \tfrac{1}{2}\varepsilon_0 E^2,\qquad u_B \;=\; \frac{B^2}{2\mu_0}

Why: these two expressions are standard results for the energy stored per unit volume in an electric field and a magnetic field respectively. They add up to the total electromagnetic energy density u = u_E + u_B.

Step 2. Use the phase lock B = E/c to rewrite u_B.

u_B \;=\; \frac{E^2/c^2}{2\mu_0} \;=\; \frac{E^2}{2\mu_0 c^2}

Since c^2 = 1/(\mu_0\varepsilon_0),

u_B \;=\; \frac{E^2 \cdot \mu_0\varepsilon_0}{2\mu_0} \;=\; \tfrac{1}{2}\varepsilon_0 E^2

Why: substituting c^2 = 1/(\mu_0\varepsilon_0) cancels \mu_0 in the denominator and produces exactly the same form as u_E. The magnetic energy density equals the electric energy density at every point along the wave.

Step 3. Add them.

u \;=\; u_E + u_B \;=\; \varepsilon_0 E^2

This is the total energy density at one instant. To get intensity (energy per area per time), multiply by the wave speed c — because in one second a slab of the wave c metres thick crosses unit area:

I_\text{inst} \;=\; u\,c \;=\; \varepsilon_0 c\,E^2

Step 4. Take the time average. With E = E_0\sin(kx-\omega t),

\langle E^2\rangle \;=\; E_0^2\,\langle\sin^2(kx-\omega t)\rangle \;=\; \tfrac{1}{2}E_0^2

Why: the time average of \sin^2 over one full period is 1/2. This is a standard integral — \int_0^T \sin^2(\omega t)\,dt = T/2.

Therefore

\boxed{\;I \;=\; \langle I_\text{inst}\rangle \;=\; \tfrac{1}{2}\varepsilon_0 c\,E_0^2\;}

Or, using E_0 = cB_0,

I \;=\; \tfrac{1}{2}\varepsilon_0 c \cdot c^2 B_0^2 \;=\; \tfrac{1}{2}\varepsilon_0 c^3 B_0^2 \;=\; \frac{c B_0^2}{2\mu_0}

Why: in the last step, \varepsilon_0 c^2 = 1/\mu_0, which folds into c B_0^2/(2\mu_0).

Either form gives you the intensity. Numerically, for sunlight at the Earth's surface, E_0\approx 1000 V/m, \varepsilon_0 = 8.854\times 10^{-12} C²/(N·m²), c = 3\times 10^8 m/s:

I \;=\; \tfrac{1}{2}(8.854\times 10^{-12})(3\times 10^8)(1000)^2 \;\approx\; 1328\text{ W/m}^2

which matches the measured solar constant (the flux at the top of the atmosphere, about 1361 W/m²) to within the rounding.

Property 5 — momentum and radiation pressure

Electromagnetic waves carry momentum as well as energy. This is the subtle one, and it is the property that makes solar sails possible and predicts the tail of Halley's comet pointing away from the sun.

Step 1. Where does the momentum come from? Consider a free charge q sitting in the path of an EM wave travelling along +\hat{x}, with \vec{E} along \hat{y}. The electric field pushes the charge along \hat{y}:

\vec{F}_E \;=\; q\vec{E}

The charge picks up velocity \vec{v} along \hat{y}. Now the magnetic part of the wave, \vec{B} along \hat{z}, exerts a Lorentz force:

\vec{F}_B \;=\; q\vec{v}\times\vec{B}

Step 2. Compute \hat{y}\times\hat{z} = \hat{x}. With \vec{v} along \hat{y} and \vec{B} along \hat{z}, \vec{v}\times\vec{B} points along +\hat{x} — the direction of wave propagation. The magnetic part of the wave pushes the charge along the direction the wave is travelling.

Why: the electric field sets the charge into transverse motion; the magnetic field converts a fraction of that motion into longitudinal force. This is the mechanism by which a wave transfers momentum to anything that absorbs it.

Step 3. Quantitatively, if the wave delivers energy U to an absorber, it also delivers momentum

p \;=\; \frac{U}{c}

Derivation of this relation. The rate at which the electric field does work on the charge is

\frac{dU}{dt} \;=\; qE\,v

Why: power equals force times velocity. The electric force does the work; the magnetic force (perpendicular to \vec{v}) does no work but creates the longitudinal push.

The rate at which the magnetic force pushes the charge along \hat{x} is

\frac{dp}{dt} \;=\; qvB \;=\; qv\cdot\frac{E}{c} \;=\; \frac{1}{c}\cdot qEv \;=\; \frac{1}{c}\cdot\frac{dU}{dt}

Why: substitute B = E/c. The magnetic push per unit time is exactly 1/c times the electric work per unit time. Integrate both sides over the duration of interaction, and p = U/c for any energy transfer.

Step 4. Convert to pressure. The momentum delivered to one square metre per second is the force per square metre — the pressure. Energy delivered per unit area per unit time is the intensity I. So for a perfectly absorbing surface,

\boxed{\;P_\text{rad} \;=\; \frac{I}{c}\,\text{ (absorber)}\;}

For a perfect reflector, the wave bounces back carrying momentum -U/c; the change in momentum of the wave is 2U/c, so the surface receives a push of

P_\text{rad} \;=\; \frac{2I}{c}\,\text{ (reflector)}

Numerical size. Sunlight at Earth has I\approx 1360 W/m². Radiation pressure on a perfect absorber:

P_\text{rad} \;=\; \frac{1360}{3\times 10^8} \;\approx\; 4.5\times 10^{-6}\text{ Pa}

Atmospheric pressure is 10^5 Pa — ten billion times larger. That is why you do not feel sunlight pushing you. But on a solar sail of area 1 km² in interplanetary space, the total force is 4.5\times 10^{-6}\times 10^6 = 4.5 N — enough, over months, to accelerate a spacecraft to tens of kilometres per second without any fuel.

Property 6 — no medium required

The most philosophically strange property. Water waves need water. Sound waves need air. Every wave in everyday life needs a medium whose atoms carry the disturbance. Light does not.

What makes an EM wave self-supporting? Look at the two source-free Maxwell equations:

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

A changing \vec{B} creates a circulating \vec{E}.

\nabla\times\vec{B} \;=\; \mu_0\varepsilon_0\,\frac{\partial\vec{E}}{\partial t}

A changing \vec{E} creates a circulating \vec{B}.

Each field's time change is the source of the other field. Nothing else is required — no atoms, no ether, no "something" to vibrate. The two fields are the medium; they sustain each other as they propagate.

This is the conceptual leap Maxwell was forced to take in 1865. His predecessors (including Michael Faraday) imagined an all-pervading "luminiferous ether" whose mechanical vibrations were light. The Michelson–Morley experiment in 1887 searched for that ether and found nothing. Einstein's special relativity (1905) confirmed there is no such medium. Light propagates in itself.

Operationally, this is why you can receive signals from Voyager 1 — now 24 billion kilometres from Earth, beyond the solar system, in almost perfect vacuum. Between Voyager's antenna and the Goldstone Deep Space Network dish in California there is essentially nothing. And yet, every 17 hours, faint microwave signals arrive. No medium. Just Maxwell.

Worked examples

Example 1: Electric and magnetic amplitudes of a Mumbai FM radio wave

A Mumbai FM transmitter broadcasts at 98.3 MHz with an effective transmitted power that produces an electric-field amplitude E_0 = 0.012 V/m at a receiver 5 km away. Find (a) the magnetic-field amplitude B_0, (b) the intensity at the receiver, (c) the power captured by a small dipole antenna of effective area 0.1\text{ m}^2.

FM transmitter to receiver, 5 km apart Tower on the left transmits a 98.3 MHz wave; a dipole antenna on the right (5 km distance) picks up the wave with E amplitude 0.012 V/m. FM tower 98.3 MHz dipole $A = 0.1\text{ m}^2$ $E_0 = 0.012\text{ V/m}$ 5 km
An FM tower radiates a plane wave; at a 5 km distance a dipole antenna samples it. The amplitudes and intensity follow from Maxwell's equations alone.

Step 1. Find B_0 from E_0 = cB_0.

B_0 \;=\; \frac{E_0}{c} \;=\; \frac{0.012}{3\times 10^8} \;=\; 4.0\times 10^{-11}\text{ T}

Why: the amplitudes are locked by Maxwell's equations. One number, the speed of light, converts between them.

Step 2. Compute the intensity.

I \;=\; \tfrac{1}{2}\varepsilon_0 c\, E_0^2 \;=\; \tfrac{1}{2}(8.854\times 10^{-12})(3\times 10^8)(0.012)^2
I \;=\; \tfrac{1}{2}\times 2.656\times 10^{-3}\times 1.44\times 10^{-4} \;\approx\; 1.9\times 10^{-7}\text{ W/m}^2

Why: plug the given E_0 into the intensity formula. Note how tiny the number is — FM radio delivers about 200 nanowatts per square metre to a receiver 5 km from the tower, yet your car radio turns that into an audible signal.

Step 3. Power captured by the antenna.

P \;=\; I\cdot A \;=\; 1.9\times 10^{-7}\times 0.1 \;=\; 1.9\times 10^{-8}\text{ W} \;=\; 19\text{ nW}

Why: intensity times capture area gives the total power intercepted. 19 nanowatts is enough, after amplification by the radio's low-noise amplifier (gain \sim 10^6), to drive a speaker.

Result: B_0 \approx 4\times 10^{-11} T, I\approx 1.9\times 10^{-7} W/m², captured power \approx 19 nW.

What this shows: even a vanishingly small electric-field amplitude (12 mV/m) carries a measurable energy flux when integrated over an antenna. The FM radio in your Maruti Swift is detecting fields that are a billion times weaker than the Earth's own atmospheric electric field.

Example 2: Solar-cell Poynting flux on an ISRO satellite

A solar panel on ISRO's Chandrayaan-3 orbiter, operating near Earth (solar constant I_\odot = 1361 W/m²), has area A = 15\text{ m}^2 and absorbs sunlight perfectly. Find (a) the amplitude E_0 of the incident sunlight, (b) the total power delivered to the panel, (c) the radiation pressure on the panel and the total force it produces.

Solar panel on a satellite receiving sunlight The sun on the left radiates outward; a rectangular solar panel on the right intercepts the flux perpendicular to the direction of propagation, absorbing the full 1361 W per square metre. Sun panel, $A = 15\text{ m}^2$ $I_\odot = 1361\text{ W/m}^2$
Sunlight (plane wave, $I_\odot = 1361$ W/m²) strikes the Chandrayaan-3 panel perpendicular to its surface. The panel absorbs the flux and feels radiation pressure.

Step 1. Electric-field amplitude. Invert I = \tfrac{1}{2}\varepsilon_0 c\,E_0^2.

E_0 \;=\; \sqrt{\frac{2I_\odot}{\varepsilon_0 c}} \;=\; \sqrt{\frac{2\times 1361}{(8.854\times 10^{-12})(3\times 10^8)}}
E_0 \;=\; \sqrt{\frac{2722}{2.656\times 10^{-3}}} \;=\; \sqrt{1.025\times 10^6} \;\approx\; 1013\text{ V/m}

Why: the intensity formula inverts cleanly. A kilovolt per metre is the "voltage drop per metre" inside a sunbeam — surprisingly large for something you can hold your hand in without feeling an electric shock. You do not feel the shock because the field oscillates at 10^{14} Hz; your nerves cannot respond.

Step 2. Total power delivered.

P \;=\; I_\odot\cdot A \;=\; 1361\times 15 \;=\; 20{,}415\text{ W} \;\approx\; 20.4\text{ kW}

Why: intensity times area. A 15 m² panel at 100% efficiency would drain 20 kW from the sunbeam. Real Chandrayaan panels run at about 25% efficiency, so the usable electrical output is around 5 kW — plenty for radio, camera, and instruments.

Step 3. Radiation pressure (absorbing panel).

P_\text{rad} \;=\; \frac{I_\odot}{c} \;=\; \frac{1361}{3\times 10^8} \;\approx\; 4.5\times 10^{-6}\text{ Pa}

Step 4. Total force.

F \;=\; P_\text{rad}\cdot A \;=\; 4.5\times 10^{-6}\times 15 \;\approx\; 6.8\times 10^{-5}\text{ N} \;=\; 68\ \mu\text{N}

Why: pressure times area. Sixty-eight micronewtons is the weight of a seven-microgram speck of dust on Earth — unnoticeable in everyday terms, but ISRO's orbital analysts do model it when predicting long-term drift. Over a ten-year mission, even a microforce shifts the orbit by tens of kilometres if uncompensated.

Result: E_0\approx 1013 V/m; P\approx 20.4 kW; F\approx 68\ \muN.

What this shows: every property from the top of this article showed up in one calculation. The amplitude came from intensity (§4), the intensity was the Poynting flux (§4), and the radiation pressure was the momentum flux (§5). One chapter, one panel, five properties.

Example 3: A 5G beam pushing on a window

A 5G base station on top of a Bandra East office building emits a focused beam at 3.5 GHz with intensity I = 2 W/m² at a building window 200 m away. The window reflects 4% of the incident light. Find the radiation pressure on the window and compare it to atmospheric pressure.

A 5G base station beam on a glass window A 5G tower 200 m away sends a horizontal microwave beam at a building window. 96 percent is absorbed, 4 percent is reflected. 5G tower 3.5 GHz window $I = 2\text{ W/m}^2$ 4% reflected
A microwave beam from a 5G cell tower strikes a window. Most of the energy passes through (glass is transparent at 3.5 GHz in effect); only a small fraction reflects.

Step 1. Absorbed and reflected intensity.

I_\text{abs} \;=\; 0.96\,I, \qquad I_\text{refl} \;=\; 0.04\,I

Step 2. Radiation pressure. For the absorbed portion, pressure is I_\text{abs}/c. For the reflected portion, pressure is 2I_\text{refl}/c.

P_\text{rad} \;=\; \frac{I_\text{abs}}{c} + \frac{2I_\text{refl}}{c} \;=\; \frac{I}{c}\bigl(0.96 + 2\times 0.04\bigr) \;=\; \frac{I}{c}\cdot 1.04

Why: the reflected wave carries away momentum, so the window receives 2\times the momentum per reflected photon compared to an absorbed one. Weighted by reflection fraction, the net factor is 0.96 + 2\times 0.04 = 1.04.

Step 3. Plug in numbers.

P_\text{rad} \;=\; \frac{2\times 1.04}{3\times 10^8} \;=\; \frac{2.08}{3\times 10^8} \;\approx\; 6.9\times 10^{-9}\text{ Pa}

Step 4. Compare to atmospheric pressure, P_\text{atm}\approx 10^5 Pa.

\frac{P_\text{rad}}{P_\text{atm}} \;\approx\; \frac{6.9\times 10^{-9}}{10^5} \;=\; 6.9\times 10^{-14}

Why: radiation pressure from a 5G beam is fourteen orders of magnitude smaller than atmospheric pressure. The window does not move — a few nanopascals cannot push a pane of glass.

Result: P_\text{rad}\approx 7\times 10^{-9} Pa, negligible compared to atmospheric pressure.

What this shows: electromagnetic waves always carry some momentum, but for communication-grade beams the push is completely unmeasurable. Radiation pressure matters only at astrophysical scales (solar sails, photon rockets, comet tails) or in very high-intensity labs (laser cooling, optical tweezers).

Common confusions

You have the properties and the formulas. What follows is for readers who want the Poynting vector (the vector form of intensity), the full three-dimensional derivation, and the subtle link to photon momentum.

The Poynting vector

The intensity I is a scalar — a magnitude. But energy flows in a direction, and Maxwell's equations give you a vector that describes both.

Definition. The Poynting vector is

\boxed{\;\vec{S} \;=\; \frac{1}{\mu_0}\vec{E}\times\vec{B}\;}

Its magnitude is the instantaneous energy flux (in W/m²), and it points along the direction of energy flow, which is \hat{k} for a plane wave.

Why this works. Maxwell's equations imply the local conservation law

\frac{\partial u}{\partial t} + \nabla\cdot\vec{S} \;=\; -\vec{J}\cdot\vec{E}

where u = \tfrac{1}{2}\varepsilon_0 E^2 + B^2/(2\mu_0) is the electromagnetic energy density. In the absence of currents (\vec{J} = 0), energy density decreases wherever \vec{S} flows outward — \vec{S} is the flux of electromagnetic energy.

Verification for a plane wave. With \vec{E} = E_0\sin(kx-\omega t)\hat{y} and \vec{B} = (E_0/c)\sin(kx-\omega t)\hat{z},

\vec{S} \;=\; \frac{1}{\mu_0}E_0\sin\cdot\frac{E_0}{c}\sin\,\hat{y}\times\hat{z} \;=\; \frac{E_0^2}{\mu_0 c}\sin^2(kx-\omega t)\,\hat{x}

Using \mu_0 c = 1/(\varepsilon_0 c),

|\vec{S}| \;=\; \varepsilon_0 c\,E_0^2\sin^2(kx-\omega t)

Time-averaged: \langle|\vec{S}|\rangle = \tfrac{1}{2}\varepsilon_0 c E_0^2 = I. The Poynting vector's time average is exactly the intensity we derived above. Its direction is \hat{x} — along \hat{k}, as expected.

Momentum density of the wave

Define the momentum density of the electromagnetic field as

\vec{g} \;=\; \frac{\vec{S}}{c^2}

Then \vec{g} has units of kg/(m²·s) — momentum per unit volume. Integrate over a slab of the wave and you recover p = U/c.

Why \vec{S}/c^2? Because an electromagnetic wave travels at c, the energy in unit volume takes time 1/c to cross unit area, giving flux |\vec{S}| = u\cdot c. By the same geometry, momentum in unit volume is flux divided by the speed once more: |\vec{g}| = |\vec{S}|/c^2. This is consistent with the relativistic relation E^2 = (pc)^2 + (mc^2)^2 — for massless photons (m=0), E = pc, i.e. momentum equals energy over c.

Photon picture

Quantum mechanics says an EM wave is a stream of photons, each carrying energy E_\gamma = hf and momentum p_\gamma = hf/c = h/\lambda.

Check intensity. If n is the photon number density (photons per m³), the energy density is u = nhf and the flux is I = nhfc. For sunlight with hf = 2.5 eV = 4\times 10^{-19} J,

n \;=\; \frac{I}{hfc} \;=\; \frac{1361}{(4\times 10^{-19})(3\times 10^8)} \;\approx\; 1.1\times 10^{13}\text{ photons/m}^3

About 10^{13} visible photons fill every cubic metre of Mumbai sunlight — the quantised grainy structure of light is utterly invisible at this density.

Check radiation pressure. An absorber catches nc photons per m² per second, each delivering momentum hf/c. Pressure:

P_\text{rad} \;=\; n c\cdot\frac{hf}{c} \;=\; nhf \;=\; u \;=\; I/c

— exactly the classical result. The photon picture and Maxwell's picture agree numerically, which is a non-trivial confirmation that both are right.

Why transverse, at the deepest level

Gauss's law in vacuum, \nabla\cdot\vec{E} = 0, says the electric field has no sources inside the wave. For a plane wave along \hat{x}, with \vec{E} having only spatial dependence on x, the divergence reduces to \partial E_x/\partial x. Setting it to zero forces E_x to be spatially uniform — but a uniform component does not oscillate and is not part of the travelling wave. It must be zero. The same holds for \vec{B}.

At the deepest level, transversality is a statement about the absence of magnetic monopoles and free charges in vacuum. If magnetic monopoles existed, longitudinal EM waves would be possible. They have never been observed, and Maxwell's equations (as written) ban them.

Group velocity and phase velocity

In vacuum, both the phase velocity (\omega/k) and the group velocity (d\omega/dk) equal c. In a dispersive medium (glass, water, plasma), they differ. This is why light slows down in glass — but the speed that slows is the phase velocity; the group velocity, which carries information and energy, also reduces but can be computed separately from the medium's dispersion relation \omega(k). The upshot: nothing travels faster than c in vacuum, but the "speed of light in a material" is a more nuanced object. (See the refraction article for more.)

Where this leads next