Advanced Electrodynamics

11.1 Multipole Expansion

The scalar potential of a localised charge distribution at large distances ( $r \gg d$ Where $d$ is the size of the distribution):

$\phi(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\left[\frac{Q}{r} + \frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^2} + \frac{1}{2}\sum_{ij}Q_{ij}\frac{\hat{r}_i\hat{r}_j}{r^3} + \cdots\right]$

Monopole term: $Q = \int \rho\, dV$ (total charge).

Dipole term: $\mathbf{p} = \int \mathbf{r}"\,\rho(\mathbf{r}')\,dV'$ (electric dipole moment).

Quadrupole term: $Q_{ij} = \int (3r'_ir'_j - r'^2\delta_{ij})\,\rho(\mathbf{r}')\,dV'$ (traceless quadrupole tensor).

The quadrupole term is important for nuclei with spin $I \geq 1$ and for non-spherical charge distributions. The quadrupole moment $Q = Q_{zz}$ (in the principal axis frame) characterises the deviation from spherical symmetry.

11.2 Gauge Transformations and Potentials

The scalar and vector potentials are not unique. The gauge transformation:

$\mathbf{A}' = \mathbf{A} + \nabla\chi, \quad \phi' = \phi - \frac{\partial\chi}{\partial t}$

Leaves $\mathbf{E}$ and $\mathbf{B}$ unchanged for any scalar function $\chi(\mathbf{r}, t)$ .

Common gauges:

Gauge	Condition	Use
Coulomb	$\nabla \cdot \mathbf{A} = 0$	Static problems, quantum mechanics
Lorenz	$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial\phi}{\partial t} = 0$	Relativistic problems, radiation
Temporal	$\phi = 0$	Some scattering problems

In the Lorenz gauge, both $\mathbf{A}$ and $\phi$ satisfy wave equations with sources:

$\Box^2\mathbf{A} = -\mu_0\mathbf{J}, \quad \Box^2\phi = -\frac{\rho}{\varepsilon_0}$

Where $\Box^2 = \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}$ is the d’Alembertian.

11.3 Electromagnetic Stress-Energy Tensor

The electromagnetic stress-energy tensor $T^{\mu\nu}$ encodes the energy density, momentum density, and stress:

$T^{00} = \frac{1}{2}\left(\varepsilon_0 E^2 + \frac{B^2}{\mu_0}\right) \quad \text{(energy density)}$

$T^{0i} = \frac{1}{c}(\mathbf{E} \times \mathbf{B})_i = \frac{S_i}{c} \quad \text{(momentum density)}$

$T^{ij} = -\varepsilon_0 E_i E_j - \frac{1}{\mu_0}B_i B_j + \frac{1}{2}\delta_{ij}\left(\varepsilon_0 E^2 + \frac{B^2}{\mu_0}\right) \quad \text{(Maxwell stress tensor)}$

Conservation law: $\partial_\mu T^{\mu\nu} = -f^\nu$ where $f^\nu$ is the Lorentz force density on charges.

Radiation pressure: For a normally incident plane wave with intensity $I$ :

$P_{\text{rad} = \frac{I}{c} = \frac{1}{2}\varepsilon_0 E_0^2}$

For a perfect reflector, the radiation pressure is $2I/c$ (momentum transfer is doubled).

Worked Example 11.1: Radiation Pressure from Sunlight

Solar constant at Earth: $I = 1361$ W/m $^2$ .

Radiation pressure on a perfectly absorbing surface:

$P = \frac{I}{c} = \frac{1361}{3 \times 10^8} = 4.54 \times 10^{-6}\ \text{N}/m^2 = 4.54\ \mu\text{Pa}$

For a perfect reflector: $P = 9.07\,\mu\text{Pa}$ .

This is tiny compared to atmospheric pressure ( $10^5$ Pa), but is significant for:

Solar sails: A 100 m $\times$ 100 m sail with 90% reflectivity experiences $F \approx 0.12$ N, producing acceleration $a \approx 0.6$ mm/s $^2$ for a 100 kg sail.
Asteroid deflection: Sustained radiation pressure can perturb asteroid orbits over years.
Atom optics: Laser cooling uses radiation pressure to slow atoms to microkelvin temperatures.

Common Pitfalls (Additional)

$v_p > c$ does not violate relativity: The phase velocity in a waveguide exceeds $c$ But no information or energy travels faster than $c$ . The group velocity (signal velocity) is always $v_g < c$ . Similarly, the refracted phase front in a prism can appear to move faster than $c$ But the actual signal does not.
Gauge choice matters for potentials, not fields: Different gauges give different $\mathbf{A}$ and $\phi$ for the same $\mathbf{E}$ and $\mathbf{B}$ . In quantum mechanics, the Hamiltonian depends on the gauge, but all physical observables are gauge-invariant. The Aharonov—Bohm effect shows that even in regions where $\mathbf{E} = \mathbf{B} = 0$ The vector potential $\mathbf{A}$ has measurable physical effects.
Multipole expansion convergence: The multipole expansion converges only outside a sphere that encloses all charges. Inside the charge distribution, the expansion diverges and must not be used. The expansion parameter is $d/r$ where $d$ is the source size and $r$ is the observation distance.
Radiation fields vs. Near fields: At distances $r \ll \lambda$ (near field), the fields are dominated by $1/r^2$ (induction) and $1/r^3$ (electrostatic/magnetostatic) terms. The radiation fields ( $\propto 1/r$ ) dominate only in the far field ( $r \gg \lambda$ ). Do not apply the Larmor formula or radiation resistance in the near field.
Poynting vector is not unique: The Poynting vector $\mathbf{S} = \mathbf{E} \times \mathbf{H}$ is gauge-dependent and can be nonzero even in static situations (e.g., a charged capacitor in a constant magnetic field). Only the surface integral $\oint \mathbf{S} \cdot d\mathbf{a}$ (total power flow) is physically meaningful.

Problems (Additional)

Problem 19: TE$_{10}$ Mode Field Patterns

For a rectangular waveguide ( $a \times b$ ) operating in TE $_{10}$ mode at frequency $f$ :

(a) Write the complete expressions for all six field components ( $E_x, E_y, E_z, B_x, B_y, B_z$ ).

(b) Sketch the field pattern: show the direction and relative magnitude of $\mathbf{E}$ and $\mathbf{B}$ in the $xy$ -plane at $z = 0$ .

(c) Find the positions of maximum surface current density on the walls and explain why the waveguide loss is minimised by making the broad wall dimension $a$ as large as possible (for a given $f$ ).

Solution:

(a) For TE $_{10}$ : $B_z = B_0\cos(\pi x/a)\,e^{i(\beta z - \omega t)}$ .

$E_x = 0, \quad E_y = \frac{i\omega\mu_0 a}{\pi}B_0\sin\!\left(\frac{\pi x}{a}\right)e^{i(\beta z - \omega t)}$

$E_z = 0$

$B_x = -\frac{i\beta a}{\pi}B_0\sin\!\left(\frac{\pi x}{a}\right)e^{i(\beta z - \omega t)}, \quad B_y = 0$

$B_z = B_0\cos\!\left(\frac{\pi x}{a}\right)e^{i(\beta z - \omega t)}$

(b) The electric field $E_y$ is purely vertical, with a $\sin(\pi x/a)$ profile: zero at the side walls ( $x = 0, a$ ) and maximum at the centre ( $x = a/2$ ). The magnetic field forms closed loops in the $xz$ -plane.

(c) Surface current $\mathbf{K} = \hat{\mathbf{n}} \times \mathbf{H}$ . On the broad walls ( $y = 0, b$ ): $\mathbf{K}$ has components from $B_x$ and $B_z$ With maximum at $x = a/2$ (where $\sin(\pi x/a) = 1$ ). The power loss per unit length is:

$P_{\text{loss} = \frac{R_s}{2}\oint |\mathbf{K}|^2\, dl}$

Where $R_s = \sqrt{\omega\mu_0/(2\sigma)}$ is the surface resistance. For fixed $f$ Increasing $a$ reduces the current density on the broad walls and increases the power-handling capacity.

Problem 20: Antenna Radiation Pattern

A half-wave dipole antenna of length $\ell = \lambda/2$ carries a sinusoidal current distribution:

$I(z) = I_0\cos(kz), \quad -\lambda/4 \leq z \leq \lambda/4$

(a) Calculate the radiation fields $\mathbf{E}$ and $\mathbf{B}$ in the far field.

(b) Find the angular distribution of radiated power $dP/d\Omega$ .

(c) Calculate the total radiated power and the radiation resistance. Compare with the short-dipole result $R_{\text{rad} = 197(\ell/\lambda)^2\,\Omega}$ .

Solution:

(a) The vector potential in the far field:

$A_z = \frac{\mu_0}{4\pi}\frac{e^{ikr}}{r}\int_{-\lambda/4}^{\lambda/4}I_0\cos(kz')\,e^{-ikz'\cos\theta}\,dz'$

The integral evaluates to:

$A_z = \frac{\mu_0 I_0}{4\pi}\frac{e^{ikr}}{r}\frac{2\cos\!\left(\frac{\pi}{2}\cos\theta\right)}{k\sin^2\theta}$

The radiation fields:

$E_\theta = ikA_z\sin\theta = \frac{i\mu_0 c I_0}{4\pi}\frac{e^{ikr}}{r}\frac{\cos\!\left(\frac{\pi}{2}\cos\theta\right)}{\sin\theta}$

$B_\phi = E_\theta/c$

(b) The angular distribution:

$\frac{dP}{d\Omega} = \frac{r^2}{2Z_0}|E_\theta|^2 = \frac{Z_0 I_0^2}{32\pi^2}\frac{\cos^2\!\left(\frac{\pi}{2}\cos\theta\right)}{\sin^2\theta}$

$P = \frac{Z_0 I_0^2}{32\pi^2}\int_0^{2\pi}\!\!\int_0^\pi \frac{\cos^2\!\left(\frac{\pi}{2}\cos\theta\right)}{\sin^2\theta}\sin\theta\,d\theta\,d\phi$

$= \frac{Z_0 I_0^2}{16\pi}\int_0^\pi \frac{\cos^2\!\left(\frac{\pi}{2}\cos\theta\right)}{\sin\theta}\,d\theta$

With the substitution $u = \cos\theta$ : $\int_{-1}^{1}\frac{\cos^2(\pi u/2)}{1-u^2}\,du = 1.2188$ (the Siegel integral).

$P = \frac{377 \times 1.2188}{16\pi}I_0^2 = 9.16\,I_0^2$

Radiation resistance: $R_{\text{rad} = 2P/I_0^2 = 18.3\,\Omega}$ .

For comparison, a short dipole ( $\ell \ll \lambda$ ) of length $\lambda/2$ would give $R_{\text{rad} = 197 \times 0.25 = 49.3\,\Omega}$ . The half-wave dipole has lower radiation resistance because the current distribution (cosine) has less total effective acceleration than a uniform current.

The directivity of the half-wave dipole is $D = 1.64$ (2.15 dBi), slightly higher than the short dipole ( $D = 1.5$ ).

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