Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

In the static case the potentials of given sources are the integrals of Chapters 3 and 4 (Figure 5.5.9),

A (r) V (r) R = \frac{μ}{4 π} \int_{V^{'}} \frac{J ( r ^{'} )}{R} d v^{'}, = \frac{1}{4 π ε} \int_{V^{'}} \frac{ρ _{v} ( r ^{'} )}{R} d v^{'}, = ∣ r - r^{'} ∣, (5.5.48)

where $r$ is the field point $P$ , $r^{'} = (x^{'}, y^{'}, z^{'})$ runs over the source points in the region $V^{'}$ , and $d v^{'}$ is the volume element there. These are the solutions of the two Poisson equations $\nabla^{2} A = - μ J$ and $\nabla^{2} V = - ρ_{v} / ε$ (Section 4.1). In the dynamic case the solutions of the wave equations Eq. (5.5.44) are called the retarded potentials.

Retarded potentials

A (r, t) V (r, t) = \frac{μ}{4 π} \int_{V^{'}} \frac{J ( r ^{'} , t - R / v )}{R} d v^{'}, = \frac{1}{4 π ε} \int_{V^{'}} \frac{ρ _{v} ( r ^{'} , t - R / v )}{R} d v^{'}, (5.5.49)

with $v = 1/ μ ε$ ( $= c$ in free space). They solve the inhomogeneous wave equations Eq. (5.5.44) and, for sources that conserve charge, satisfy the Lorenz gauge.

In plain words

The only change from the static integrals is when the sources are taken. The static integrals use the sources now. The retarded integrals use each source element at the retarded time $t^{'} = t - R / v$ , earlier by the time a signal needs to travel from that element to $P$ . Every source point has its own retarded time.

The geometry of the retarded potentials. The source element dv' at \vec r' contributes to the potentials at the field point P (at \vec r) the value it had at t-R/v. The dashed arcs are the spherical fronts that carry its influence outward at the speed v. — **Figure 5.5.9.** The geometry of the retarded potentials. The source element $d v^{'}$ at $r^{'}$ contributes to the potentials at the field point $P$ (at $r$ ) the value it had at $t - R / v$ . The dashed arcs are the spherical fronts that carry its influence outward at the speed $v$ .

What the retardation means

Figure 5.5.10 follows a short pulse of current at the source to two points at the distances $R_{1}$ and $R_{2} = 2.17 R_{1}$ . Three facts can be read from it.

What a source element does at $t^{'}$ is felt at the distance $R$ only at $t = t^{'} + R / v$ : the potentials travel at the speed of the wave. Nothing at all happens at $R$ before $R / v$ has elapsed.
The factor $1/ R$ is the static one: the amplitude of the potentials falls as $1/ R$ , not faster. That slow decay is what makes radio possible over large distances.
For a sinusoidal source of angular frequency $ω$ the delay $R / v$ is a phase lag $ω R / v = 2 π R / λ$ , with the wavelength $λ = 2 π v / ω$ .

What the retardation means. A short pulse of current at t_0 is felt at the distance R_1 after the delay R_1/v, and at R_2=2.17R_1 after R_2/v, smaller by the factor R_1/R_2. The shape of the pulse is kept. — **Figure 5.5.10.** What the retardation means. A short pulse of current at $t_{0}$ is felt at the distance $R_{1}$ after the delay $R_{1} / v$ , and at $R_{2} = 2.17 R_{1}$ after $R_{2} / v$ , smaller by the factor $R_{1} / R_{2}$ . The shape of the pulse is kept.

When can we forget it? If the whole region of interest is small compared with $λ /2 π$ , the phase lag $ω R / v$ is small everywhere in it, $t - R / v \approx t$ , and the static formulas Eq. (5.5.48) hold at every instant with the present values of the sources. This quasi-static regime is the domain of circuit theory: at $50 Hz$ , $λ = c / f = 6000 km$ and $λ /2 π = 955 km$ , so a laboratory, a building and even most power networks are small. At $1 MHz$ , $λ /2 π$ is only $47.7 m$ (Problem 5.5.6), and at the frequencies of mobile radio a few centimetres.

Causality

The wave equations are unchanged if $t$ is replaced by $- t$ , and they also admit the advanced potentials, with the sources taken at $t + R / v$ : a field that arrives before its cause. They are discarded on physical grounds, because effects follow their causes. The retarded solution is the one that is zero everywhere until the sources are switched on.

Quick check

A mobile phone transmits at $900 MHz$ to a base station $3 km$ away. What is the delay, and by how many periods does the field at the base station lag behind the current in the phone's antenna? Is the link quasi-static?

$R / c = 3000/ (3 \times 1 0^{8}) = 10 μ s$ , which is $900 \times 1 0^{6} \times 1 0^{- 5} = 9000$ periods. Far from quasi-static: $λ /2 π = 5.3 cm$ , and the distance is about $57000$ times that.

Application 5.5.2 · Light-speed delay in satellite links

A radio signal is the field of the retarded potentials of the current in the transmitting antenna: it arrives at the distance $R$ only $R / c$ later. On the ground this is one microsecond per $300 m$ and nobody notices. In space it is the dominant delay of the link (Figure 5.5.11). For a hop from one ground station up to a satellite and down to another station, and for a range measured from a delay,

t_{hop} = \frac{R _{up} + R _{down}}{c}, R = c t . (5.5.50)

GEO, the geostationary orbit at $35786 km$ above the equator. The satellite seems fixed in the sky and can be served by fixed antennas, but one hop up and down takes at least $2 \times 35786 km / c = 239 ms$ . A question and its answer need two hops: about half a second of silence in a telephone call.
LEO, a low orbit near $550 km$ . The one-way delay is only $1.8 ms$ at the zenith and $3.3 ms$ at $3 0^{\circ}$ elevation. This is why internet constellations fly low, at the price of thousands of satellites, since each one is in view of a station for a few minutes only.
GPS, at $20200 km$ , $67 ms$ away at the zenith. Here the receiver measures the delay of the signals of several satellites and converts it into ranges, $R = c \times$ delay. One nanosecond of timing error is $30 cm$ of position error.

Application 5.5.2. The three orbits drawn roughly to scale (Earth radius 6371\,\mathrm{km}). The red rays are one GEO hop between two ground stations. At the speed of light the hop takes about a quarter of a second, while a LEO satellite is a few milliseconds away. — **Figure 5.5.11.** Application 5.5.2. The three orbits drawn roughly to scale (Earth radius $6371 km$ ). The red rays are one GEO hop between two ground stations. At the speed of light the hop takes about a quarter of a second, while a LEO satellite is a few milliseconds away.

Typical numbers. The slant range from a station that sees a GEO satellite at $3 0^{\circ}$ elevation is $38609 km$ , so a hop between two such stations takes $257 ms$ and a question with its answer more than half a second (Worked example 5.5.2). The atmosphere adds a little: the ionosphere, whose delay depends on frequency, adds metres of equivalent path (at times tens of metres), which dual-frequency receivers remove by comparing two carriers, and the troposphere about $2.5 m$ at the zenith, which is modelled.

Application 5.5.3 · Retardation in radio broadcasting

Digital radio (DAB) and digital terrestrial television (DVB-T) cover a country with many transmitters on one frequency, a single-frequency network. All transmitters send the same symbols at the same instant, synchronised by GPS. A receiver hears every transmitter, each one retarded by its own $R / c$ (Figure 5.5.12). The signal is cut into long OFDM symbols, each preceded by a guard interval $T_{g}$ , a copy of the end of the symbol. The receiver starts its processing window at the end of the guard interval of the first copy. As long as the later copy arrives within $T_{g}$ , it only adds to the wanted signal, and the condition is

Δ t = \frac{∣ R _{2} - R _{1} ∣}{c} \leq T_{g} . (5.5.51)

A later copy spills into the next symbol and acts as interference.

Application 5.5.3. Two synchronised transmitters on one frequency. The receiver gets the same OFDM symbols twice, the copy from the farther transmitter T_2 delayed by \Delta t=(R_2-R_1)/c. Each symbol starts with a guard interval T_g (dark). If \Delta t\le T_g the late copy still lies inside the same symbol when the processing window opens, and it adds to the signal instead of disturbing the next symbol. — **Figure 5.5.12.** Application 5.5.3. Two synchronised transmitters on one frequency. The receiver gets the same OFDM symbols twice, the copy from the farther transmitter $T_{2}$ delayed by $Δ t = (R_{2} - R_{1}) / c$ . Each symbol starts with a guard interval $T_{g}$ (dark). If $Δ t \leq T_{g}$ the late copy still lies inside the same symbol when the processing window opens, and it adds to the signal instead of disturbing the next symbol.

Table 5.5.3 gives the guard intervals of two common systems and the largest path difference $c T_{g}$ that they absorb. The spacing of the transmitters of a network is planned from such numbers, a few tens of kilometres.

Table 5.5.3. Guard intervals of single-frequency networks and the path difference they absorb.

System	$T_{g}$	$c T_{g}$ (largest $R_{2} - R_{1}$ )
DVB-T, 8k mode, guard $1/4$	$224 μ s$	$67.2 km$
DVB-T, 8k mode, guard $1/8$	$112 μ s$	$33.6 km$
DAB, mode I	$246 μ s$	$73.8 km$

The 1/R of the potentials helps. A distant transmitter is not only late but also weaker. Far from the antennas the fields, like the retarded potentials, fall as $1/ R$ , so two equal transmitters at $10 km$ and $50 km$ arrive with the amplitude ratio $50/10 = 5$ , that is $20 lo g_{10} 5 = 14 dB$ (Worked example 5.5.3). Exercise 5.5.9 shows what happens on one frequency without the digital safeguards: the two signals interfere and a moving receiver passes through deep fades.

Table 5.5.4. The applications of this section.

Application	Physics used	Worked out in
Signals on a chip, a board and a cable	$v = c / ε_{r}$ , $t_{d} = L / v$	Worked example 5.5.1, Exercise 5.5.2
Light-speed delay in satellite links	delay $R / c$ of the retarded potentials	Worked example 5.5.2
Retardation in radio broadcasting	delay $R / c$ and amplitude $1/ R$	Worked example 5.5.3, Exercise 5.5.9

5.5.10 Retarded potentials

What the retardation means