Carrying Power: Poynting Vector & Wave Impedance

Where does the energy actually flow?

Ask an engineer how a battery powers a light bulb and most will say, without hesitation, that energy travels down the wire like water down a pipe. It is one of the most natural pictures in all of physics — and it is almost entirely wrong. The electrons in a copper wire drift astonishingly slowly, a few metres per *hour*, far too sluggish to deliver the watts a bulb burns in real time. So if the energy is not riding the electrons through the metal, where is it?

The answer, first worked out by John Henry Poynting in 1884, is that the energy flows in the fields that surround the wires — through the air, the vacuum, the insulation — and merely *guided* by the copper. The wire is not a pipe carrying energy; it is a rail that steers a river of energy flowing alongside it. The electric field points across the gap between conductors, the magnetic field wraps around each wire in circles, and where those two fields coexist, power streams in the direction perpendicular to both.

The Poynting vector: a compass for power

Poynting captured this in one elegant cross product. The Poynting vector S = E × H does two jobs at once. Its *direction* is the direction power flows; its *magnitude* is the power crossing one square metre, measured in watts per square metre (W/m²). Because it is a cross product, S is perpendicular to both the electric field E and the magnetic field H — exactly the geometry of a plane wave, where E, H and the direction of travel form a right-handed set like your thumb, index, and middle fingers.

        E (electric field)
        ^
        |
        |          S = E x H   (power flow, into the page)
        |
        +---------> H (magnetic field)
       /
      / direction of travel = E x H
     v

  Right-hand rule:  point fingers along E, curl toward H,
                    thumb points the way the wave (and power) goes.

For a sinusoidal wave the fields swing up and down, so S pulses too. What we usually care about is the time-averaged power density. For a wave with peak field E₀, the average Poynting magnitude is S_avg = E₀² / (2·Z₀), where Z₀ is the impedance we meet next. (If you quote E as an RMS value instead of a peak, drop the factor of 2: S = E_rms² / Z₀.) The squaring is the same one you see in P = V²/R — power always goes as field squared, just as it goes as voltage squared.

377 ohms: the resistance of empty space

In an electromagnetic wave the electric and magnetic fields are not independent — they are locked together in a fixed ratio. Maxwell's equations demand that at every instant E / H equals the same number, no matter how strong the wave or what its frequency. That ratio is the wave impedance of free space, written Z₀ (or η₀, eta), and for a vacuum it comes out to a famous value:

  Z0 = E / H = sqrt( mu_0 / eps_0 )

     = sqrt( 4*pi*1e-7  /  8.854e-12 )  ohms

     = 376.730...  ohms   ~=  377 ohms

  (exactly 119.9169832 * pi ohms in the pre-2019 SI definition)

Look at how this parallels Ohm's law. There, voltage divided by current gives resistance: R = V / I. Here, electric field divided by magnetic field gives an impedance: Z₀ = E / H. The electric field plays the role of 'voltage per metre', the magnetic field plays 'current per metre', and 377 Ω is the resistance the vacuum itself presents to a passing wave. It is not a resistor that dissipates heat — empty space loses nothing — but the *bookkeeping* is identical to Ohm's law, which is why engineers reach for it instinctively.

Worked example 1: how much power is in sunlight?

Sunlight is just an electromagnetic wave that travelled 150 million kilometres to reach you, so the Poynting vector applies to it exactly as it does to a radio signal. Above the atmosphere, the Sun's power density — the 'solar constant' — is measured at about 1360 W/m². At ground level, after clouds and air, a bright noon delivers roughly 1000 W/m². Let us work backwards and find the electric field that carries that power.

  Given:  S_avg = 1000 W/m^2  (bright noon at the surface)
  Use:    S_avg = E0^2 / (2 * Z0)   ->   E0 = sqrt(2 * Z0 * S_avg)

  E0 = sqrt( 2 * 377 * 1000 )
     = sqrt( 754000 )
     ~= 868 V/m        (peak electric field)

  Magnetic field:  H0 = E0 / Z0 = 868 / 377 ~= 2.3 A/m
  (or B0 = mu_0 * H0 ~= 2.9 microtesla)

  Check the area scaling:
  A 2 m^2 solar panel intercepts  1000 * 2 = 2000 W of raw sunlight.
  At 20% efficiency it yields  ~400 W of electricity.

Notice what just happened: from a single measured power density we recovered the actual electric and magnetic field strengths of sunlight, then scaled up by area to size a solar panel. The Poynting vector and Z₀ are the entire toolkit. That 868 V/m sounds enormous, but it is spread thinly and oscillates 500 trillion times a second; you feel its time-average as warmth, not as a shock.

Worked example 2: a radio transmitter's far field

Now point the same tools at a transmitter. A radio station radiates power outward in all directions (or focused into a beam). Far enough away — the 'far field' — the wavefronts look flat and local, just like sunlight, so S = E²/(2Z₀) holds again. If a source radiates P watts equally in every direction, that power spreads over the surface of an expanding sphere of area 4πr², so the power density falls off as 1/r². Double the distance, quarter the power.

1. Start with the radiated power and distance. Take a P = 1000 W (1 kW) FM transmitter and stand r = 10 km away.
2. Spread it over the sphere. S = P / (4πr²) = 1000 / (4π · (10 000)²) ≈ 8.0 × 10⁻⁷ W/m² — under a microwatt per square metre.
3. Back out the field with Z₀. E₀ = √(2·Z₀·S) = √(2 · 377 · 8.0×10⁻⁷) ≈ 0.025 V/m, i.e. about 25 millivolts per metre.
4. Sanity-check against the receiver. A 1 m antenna sees ~25 mV — small, but a thousand times the microvolts a good FM tuner needs. The link closes with margin to spare.

Why impedance is the bridge to what comes next

Hold on to the idea that a wave has an impedance — a fixed E-to-H ratio that the medium dictates. Free space offers 377 Ω. A coaxial cable, a circuit-board trace, even a pair of parallel wires each present their own wave-like impedance, typically 50 or 75 Ω, set by their geometry rather than by the vacuum. The next rung builds on exactly this: a guided wave travelling down a line, with its own impedance, and the trouble that erupts when two impedances don't match.