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Chapter 4 • Theory & Derivations

Electrodynamic Potentials, Gauge Freedom & Radiation

Comprehensive mathematical formulation of relativistic electrodynamics and radiation theory: vector potential A and scalar potential V, gauge transformations and degrees of freedom, Coulomb versus Lorentz gauge conditions, retarded Green's functions, causal retarded potentials, Liénard-Wiechert potentials for relativistic moving charges, electric and magnetic fields of accelerated charges (velocity vs acceleration terms), complete step-by-step vector derivation of oscillating Hertzian electric dipole radiation, far-field Poynting flux, the Larmor dipole power formula, radiation resistance, magnetic dipole radiation, and half-wave center-fed antenna theory.

§4.1 Electromagnetic Potentials (V, A) and Field Formulations

Definition of Potentials in Time-Dependent Electrodynamics

In static electromagnetism, $\nabla \times \mathbf{E} = 0$ allows us to express the electric field as the gradient of a scalar potential, $\mathbf{E} = -\nabla V$. In dynamic electrodynamics, however, Faraday's law states:

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$

Because $\nabla \times \mathbf{E} \neq 0$, $\mathbf{E}$ cannot be written as the gradient of a scalar alone.

However, Gauss's law for magnetism remains strictly valid:

$$\nabla \cdot \mathbf{B} = 0$$

Since the divergence of any curl vanishes identically ($\nabla \cdot (\nabla \times \mathbf{A}) \equiv 0$), we can always define the magnetic vector potential $\mathbf{A}$:

$$\mathbf{B} \equiv \nabla \times \mathbf{A}$$

Substitute this definition into Faraday's law:

$$\nabla \times \mathbf{E} = -\frac{\partial}{\partial t} (\nabla \times \mathbf{A}) = -\nabla \times \left( \frac{\partial \mathbf{A}}{\partial t} \right)$$
$$\nabla \times \left( \mathbf{E} + \frac{\partial \mathbf{A}}{\partial t} \right) = 0$$

Because the quantity in parentheses has zero curl, it can now be expressed as the negative gradient of a dynamic electric scalar potential $V$:

$$\mathbf{E} + \frac{\partial \mathbf{A}}{\partial t} = -\nabla V \implies \mathbf{E} = -\nabla V - \frac{\partial \mathbf{A}}{\partial t}$$

These two equations represent the fundamental relations expressing physical fields $(\mathbf{E}, \mathbf{B})$ in terms of electromagnetic potentials $(V, \mathbf{A})$:

$$\mathbf{B} = \nabla \times \mathbf{A}$$
$$\mathbf{E} = -\nabla V - \frac{\partial \mathbf{A}}{\partial t}$$

§4.2 Gauge Invariance: Coulomb Gauge and Lorentz Gauge Conditions

Gauge Transformations

Potentials are not directly observable physical quantities; only fields $\mathbf{E}$ and $\mathbf{B}$ exert measurable Lorentz forces. If we modify $\mathbf{A}$ and $V$ through a transformation involving an arbitrary scalar gauge function $\lambda(\mathbf{r}, t)$:

$$\mathbf{A}' = \mathbf{A} + \nabla \lambda$$
$$V' = V - \frac{\partial \lambda}{\partial t}$$

Let us compute the new fields:

$$\mathbf{B}' = \nabla \times \mathbf{A}' = \nabla \times (\mathbf{A} + \nabla \lambda) = \nabla \times \mathbf{A} + 0 = \mathbf{B}$$
$$\mathbf{E}' = -\nabla V' - \frac{\partial \mathbf{A}'}{\partial t} = -\nabla \left( V - \frac{\partial \lambda}{\partial t} \right) - \frac{\partial}{\partial t} (\mathbf{A} + \nabla \lambda) = -\nabla V + \nabla \frac{\partial \lambda}{\partial t} - \frac{\partial \mathbf{A}}{\partial t} - \frac{\partial \nabla \lambda}{\partial t} = \mathbf{E}$$

The physical fields are completely invariant under this gauge transformation. This mathematical freedom (gauge freedom) allows us to impose convenient constraints on the divergence of $\mathbf{A}$.

1. The Coulomb Gauge (Radiation Gauge)

Set the constraint:

$$\nabla \cdot \mathbf{A} = 0$$

Substituting into Gauss's law $\nabla \cdot \mathbf{E} = \rho / \epsilon_0$:

$$\nabla \cdot \left( -\nabla V - \frac{\partial \mathbf{A}}{\partial t} \right) = -\nabla^2 V - \frac{\partial}{\partial t}(\nabla \cdot \mathbf{A}) = \frac{\rho}{\epsilon_0}$$

Since $\nabla \cdot \mathbf{A} = 0$:

$$\nabla^2 V = -\frac{\rho}{\epsilon_0}$$

This is the standard Poisson equation. The scalar potential in Coulomb gauge is:

$$V(\mathbf{r}, t) = \frac{1}{4\pi \epsilon_0} \int \frac{\rho(\mathbf{r}', t)}{|\mathbf{r} - \mathbf{r}'|} \, d\tau'$$

While computationally convenient in quantum electrodynamics and atomic physics, $V$ in Coulomb gauge appears to propagate instantaneously, disguising relativistic causality.

2. The Lorentz Gauge Condition

To preserve manifest relativistic covariance and causality, Ludvig Lorenz (and later Hendrik Lorentz) introduced the condition:

$$\nabla \cdot \mathbf{A} + \frac{1}{c^2} \frac{\partial V}{\partial t} = 0$$

Substitute $(V, \mathbf{A})$ into the inhomogeneous Maxwell equations (Gauss's law and Ampère-Maxwell law). Both coupled equations decouple completely into symmetric 4D inhomogeneous d'Alembertian wave equations:

$$\nabla^2 V - \frac{1}{c^2} \frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\epsilon_0} \quad \iff \quad \Box V = -\frac{\rho}{\epsilon_0}$$
$$\nabla^2 \mathbf{A} - \frac{1}{c^2} \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J} \quad \iff \quad \Box \mathbf{A} = -\mu_0 \mathbf{J}$$

where $\Box \equiv \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}$ is the d'Alembertian operator.

§4.3 Retarded Potentials and Causal Green's Function Solutions

The Retardation Principle

Because electromagnetic signals propagate through space at the finite speed of light $c$, the potential at an observation point $\mathbf{r}$ at time $t$ cannot depend on what source charges and currents are doing at that exact moment. Instead, it depends on their behavior at an earlier time $t_r$ (the retarded time), accounting for the travel time of light:

$$t_r \equiv t - \frac{|\mathbf{r} - \mathbf{r}'|}{c} = t - \frac{\imath}{c}$$

where $\boldsymbol{\imath} = \mathbf{r} - \mathbf{r}'$ is the separation vector and $\imath = |\mathbf{r} - \mathbf{r}'|$.

Using the retarded Green's function of the d'Alembertian operator:

$$G(\mathbf{r}, t; \mathbf{r}', t') = \frac{\delta(t - t' - \imath/c)}{4\pi \imath}$$

The exact solutions to the decoupled wave equations in Lorentz gauge are the Retarded Potentials:

$$V(\mathbf{r}, t) = \frac{1}{4\pi \epsilon_0} \int \frac{\rho(\mathbf{r}', t_r)}{\imath} \, d\tau' = \frac{1}{4\pi \epsilon_0} \int \frac{\rho(\mathbf{r}', t - \imath/c)}{|\mathbf{r} - \mathbf{r}'|} \, d\tau'$$
$$\mathbf{A}(\mathbf{r}, t) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}', t_r)}{\imath} \, d\tau' = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}', t - \imath/c)}{|\mathbf{r} - \mathbf{r}'|} \, d\tau'$$

These equations encapsulate relativistic causality: an event at the source point $\mathbf{r}'$ can only affect the field point $\mathbf{r}$ after the light-cone delay $\Delta t = \imath/c$ has elapsed.

§4.4 Liénard-Wiechert Potentials for a Relativistic Moving Point Charge

Potentials of a Point Charge on an Arbitrary Trajectory

Consider a point charge $q$ moving along an arbitrary trajectory $\mathbf{w}(t)$ with velocity $\mathbf{v}(t) = \dot{\mathbf{w}}(t)$. Because the charge is moving, different parts of the charge distribution during an integration volume emit signals that arrive at observation point $\mathbf{r}$ at the same time $t$. This geometric elongation introduces a Doppler-like Jacobian volume factor:

$$d\tau' = \frac{d\tau}{1 - \hat{\boldsymbol{\imath}} \cdot \boldsymbol{\beta}}$$

where $\boldsymbol{\beta} = \frac{\mathbf{v}}{c}$ and $\hat{\boldsymbol{\imath}} = \frac{\mathbf{r} - \mathbf{w}(t_r)}{|\mathbf{r} - \mathbf{w}(t_r)|}$.

Carrying out the 4D delta-function integral yields the celebrated Liénard-Wiechert Potentials (Alfred-Marie Liénard 1898, Emil Wiechert 1900):

$$V(\mathbf{r}, t) = \frac{1}{4\pi \epsilon_0} \frac{q}{\imath - \frac{\boldsymbol{\imath} \cdot \mathbf{v}}{c}} = \frac{1}{4\pi \epsilon_0} \frac{q}{\imath (1 - \hat{\boldsymbol{\imath}} \cdot \boldsymbol{\beta})}$$
$$\mathbf{A}(\mathbf{r}, t) = \frac{\mu_0}{4\pi} \frac{q \mathbf{v}}{\imath - \frac{\boldsymbol{\imath} \cdot \mathbf{v}}{c}} = \frac{\mathbf{v}}{c^2} V(\mathbf{r}, t)$$

where all source quantities $(\mathbf{w}, \mathbf{v}, \boldsymbol{\beta}, \imath)$ are evaluated at the retarded time $t_r$ satisfying $c(t - t_r) = |\mathbf{r} - \mathbf{w}(t_r)|$.

§4.5 Electric and Magnetic Fields of Accelerated Charges (Velocity vs Radiation Fields)

Computing the Fields from Liénard-Wiechert Potentials

Differentiating the potentials $\mathbf{E} = -\nabla V - \frac{\partial \mathbf{A}}{\partial t}$ and $\mathbf{B} = \nabla \times \mathbf{A}$ requires taking gradients and time derivatives through the implicit dependence on retarded time $\nabla t_r = -\frac{\hat{\boldsymbol{\imath}}}{c(1 - \hat{\boldsymbol{\imath}} \cdot \boldsymbol{\beta})}$.

The resulting electric field separates into two distinct physical terms:

$$\mathbf{E}(\mathbf{r}, t) = \mathbf{E}_{\text{velocity}} + \mathbf{E}_{\text{acceleration}}$$
$$\mathbf{E}(\mathbf{r}, t) = \frac{q}{4\pi \epsilon_0} \frac{\imath}{(\boldsymbol{\imath} \cdot \mathbf{u})^3} \left[ (c^2 - v^2) \mathbf{u} + \boldsymbol{\imath} \times (\mathbf{u} \times \mathbf{a}) \right]$$

where $\mathbf{u} \equiv c \hat{\boldsymbol{\imath}} - \mathbf{v}$ and $\mathbf{a} = \dot{\mathbf{v}}(t_r)$ is the charge acceleration evaluated at retarded time.

The magnetic induction field is strictly orthogonal and transverse:

$$\mathbf{B}(\mathbf{r}, t) = \frac{1}{c} \left( \hat{\boldsymbol{\imath}} \times \mathbf{E}(\mathbf{r}, t) \right)$$
Decomposition and Physical Significance

1. The Velocity Field (Generalized Coulomb Field):

$$\mathbf{E}_{\text{velocity}} \propto \frac{c^2 - v^2}{\imath^2}$$

Decays as $1/\imath^2$ with distance. It is carried along with the moving charge and represents bound field energy that cannot escape to infinity.

2. The Acceleration Field (Radiation Field):

$$\mathbf{E}_{\text{acceleration}} = \frac{q}{4\pi \epsilon_0 c^2} \frac{\hat{\boldsymbol{\imath}} \times (\mathbf{u} \times \mathbf{a})}{(\boldsymbol{\imath} \cdot \mathbf{u})^3} \propto \frac{1}{\imath}$$

Decays as $1/\imath$ with distance! The associated Poynting energy flux scales as:

$$S \propto E_{\text{rad}}^2 \propto \frac{1}{\imath^2}$$

When integrated over a giant sphere of radius $\imath \to \infty$, the surface area $4\pi \imath^2$ cancels the $1/\imath^2$ decay:

$$\oint_{S_\infty} \mathbf{S} \cdot d\mathbf{a} = \text{constant} \neq 0$$

Fundamental Theorem of Classical Electrodynamics: A charge moving with uniform velocity ($a = 0$) does NOT radiate. Only an accelerated charge ($a \neq 0$) radiates electromagnetic energy to infinity.

§4.6 Electric Dipole Radiation (Oscillating Hertzian Dipole)

The Oscillating Electric Dipole Model

Consider two tiny conducting spheres separated by a distance $d$ along the $z$-axis, connected by a thin filament carrying an alternating current. The electric dipole moment oscillates harmonically:

$$\mathbf{p}(t) = p_0 \cos(\omega t) \hat{\mathbf{z}}$$

where $p_0 = q_0 d$.

We evaluate the fields under the standard radiation zone approximations:

  1. Short dipole compared to wavelength: $d \ll \lambda$ (dipole approximation)
  2. Far radiation field: $r \gg \lambda \gg d$ where $\lambda = 2\pi c/\omega$
Derivation of the Radiation Zone Fields

The retarded vector potential in spherical coordinates $(r, \theta, \phi)$ in the far zone is:

$$\mathbf{A}(r, \theta, t) = -\frac{\mu_0 p_0 \omega}{4\pi r} \sin(\omega(t - r/c)) \hat{\mathbf{z}}$$

Converting $\hat{\mathbf{z}} = \cos\theta \hat{\mathbf{r}} - \sin\theta \hat{\boldsymbol{\theta}}$:

$$\mathbf{A}(r, \theta, t) = -\frac{\mu_0 p_0 \omega}{4\pi r} \sin(\omega(t - r/c)) (\cos\theta \hat{\mathbf{r}} - \sin\theta \hat{\boldsymbol{\theta}})$$

Computing $\mathbf{B} = \nabla \times \mathbf{A}$ keeping only terms scaling as $1/r$:

$$\mathbf{B}(r, \theta, t) = -\frac{\mu_0 p_0 \omega^2}{4\pi c} \left( \frac{\sin\theta}{r} \right) \cos(\omega(t - r/c)) \hat{\boldsymbol{\phi}}$$

From $\mathbf{E} = c(\mathbf{B} \times \hat{\mathbf{r}})$:

$$\mathbf{E}(r, \theta, t) = -\frac{\mu_0 p_0 \omega^2}{4\pi} \left( \frac{\sin\theta}{r} \right) \cos(\omega(t - r/c)) \hat{\boldsymbol{\theta}}$$
Key Physical Characteristics

1. Transverse Wave: $\mathbf{E}$ points along $\hat{\boldsymbol{\theta}}$, $\mathbf{B}$ points along $\hat{\boldsymbol{\phi}}$, and propagation is radially outward along $\hat{\mathbf{r}}$.

2. Frequency Dependence: Field amplitudes are proportional to $\omega^2$ (or $1/\lambda^2$). High frequencies radiate vastly more efficiently than low frequencies.

3. Angular Profile: Fields vanish along the dipole axis ($\theta = 0, \pi$) and peak in the equatorial plane ($\theta = \pi/2$).

§4.7 Poynting Flux, Radiation Pattern, and the Larmor Power Formula

Instantaneous and Time-Averaged Poynting Vector

The Poynting vector in the radiation zone is:

$$\mathbf{S}(r, \theta, t) = \frac{1}{\mu_0} (\mathbf{E} \times \mathbf{B}) = \frac{\mu_0 p_0^2 \omega^4}{16\pi^2 c} \left( \frac{\sin^2\theta}{r^2} \right) \cos^2(\omega(t - r/c)) \hat{\mathbf{r}}$$

Taking the time average over an oscillation cycle ($\langle \cos^2 \rangle = 1/2$):

$$\langle \mathbf{S} \rangle = \frac{\mu_0 p_0^2 \omega^4}{32\pi^2 c} \frac{\sin^2\theta}{r^2} \hat{\mathbf{r}}$$
The Toroidal Doughnut Radiation Pattern

The radiated intensity depends on angle as $\sin^2\theta$:

  • Broadside ($\theta = 90^\circ$): Maximum radiation flux perpendicular to the dipole axis.
  • Endfire ($\theta = 0^\circ, 180^\circ$): Zero radiation along the axis of the antenna wire.
Total Radiated Power: Larmor's Formula for Dipoles

Integrating $\langle \mathbf{S} \rangle$ over a closed sphere of radius $r$:

$$P = \oint \langle \mathbf{S} \rangle \cdot d\mathbf{a} = \int_0^{2\pi} d\phi \int_0^\pi \left( \frac{\mu_0 p_0^2 \omega^4}{32\pi^2 c} \frac{\sin^2\theta}{r^2} \right) r^2 \sin\theta \, d\theta$$
$$P = \frac{\mu_0 p_0^2 \omega^4}{32\pi^2 c} (2\pi) \int_0^\pi \sin^3\theta \, d\theta$$

The standard integral is $\int_0^\pi \sin^3\theta \, d\theta = \frac{4}{3}$:

$$P = \frac{\mu_0 p_0^2 \omega^4}{16\pi c} \left( \frac{4}{3} \right) = \frac{\mu_0 p_0^2 \omega^4}{12\pi c}$$

In terms of the speed of light in vacuum ($c = 1/\sqrt{\epsilon_0 \mu_0}$):

$$P = \frac{p_0^2 \omega^4}{12\pi \epsilon_0 c^3}$$

This is the celebrated Larmor Formula for an Oscillating Dipole. Notice the profound $\omega^4$ frequency dependence: doubling the frequency increases the radiated power by a factor of $2^4 = 16$.

Radiation Resistance of a Short Dipole Antenna

The current feeding the dipole is $I(t) = \dot{q}(t) = -q_0 \omega \sin(\omega t)$, with amplitude $I_0 = q_0 \omega$. Thus $p_0 = q_0 d = \frac{I_0 d}{\omega}$. Substituting into the total power formula:

$$P = \frac{\mu_0 (I_0 d / \omega)^2 \omega^4}{12\pi c} = \frac{\mu_0 I_0^2 d^2 \omega^2}{12\pi c} = \frac{1}{2} I_0^2 R_{\text{rad}}$$

Equating to the equivalent dissipated circuit power $P = \frac{1}{2} I_0^2 R_{\text{rad}}$:

$$R_{\text{rad}} = \frac{\mu_0 d^2 \omega^2}{6\pi c} = \frac{\mu_0 c}{6\pi} \left( \frac{\omega d}{c} \right)^2 = \frac{120\pi}{6\pi} \left( \frac{2\pi d}{\lambda} \right)^2 = 80\pi^2 \left( \frac{d}{\lambda} \right)^2 \, \Omega$$

For a short dipole where $d \ll \lambda$ (e.g., $d = 0.05\lambda$):

$$R_{\text{rad}} = 80\pi^2 (0.05)^2 \approx 1.97 \, \Omega$$

Because radiation resistance is tiny, short antennas match poorly to standard $50\,\Omega$ RF lines, radiating inefficiently.

§4.8 Magnetic Dipole Radiation and Center-Fed Half-Wave Antennas

Magnetic Dipole Radiation

An oscillating circular loop of radius $b$ carrying alternating current $I(t) = I_0 \cos(\omega t)$ constitutes an oscillating magnetic dipole $\mathbf{m}(t) = m_0 \cos(\omega t) \hat{\mathbf{z}}$, where $m_0 = \pi b^2 I_0$.

The radiation zone fields are:

$$\mathbf{E}(r, \theta, t) = \frac{\mu_0 m_0 \omega^2}{4\pi c} \left( \frac{\sin\theta}{r} \right) \cos(\omega(t - r/c)) \hat{\boldsymbol{\phi}}$$
$$\mathbf{B}(r, \theta, t) = -\frac{\mu_0 m_0 \omega^2}{4\pi c^2} \left( \frac{\sin\theta}{r} \right) \cos(\omega(t - r/c)) \hat{\boldsymbol{\theta}}$$

The total radiated power is:

$$P_{\text{mag}} = \frac{\mu_0 m_0^2 \omega^4}{12\pi c^3}$$
Comparison Between Electric and Magnetic Dipoles

For comparable geometric dimensions ($p_0 \sim q d, m_0 \sim I d^2 \sim q \omega d^2$):

$$\frac{P_{\text{mag}}}{P_{\text{elec}}} = \left( \frac{\omega d}{c} \right)^2 = \left( \frac{2\pi d}{\lambda} \right)^2 \ll 1$$

Magnetic dipole radiation is suppressed by a factor of $(d/\lambda)^2$ relative to electric dipole radiation. Electric dipole transitions (E1) are vastly stronger than magnetic dipole (M1) atomic transitions.

The Center-Fed Half-Wave Antenna ($L = \lambda/2$)

Practical radio communications employ resonant antennas of length $L = \lambda/2$. The current distribution is a standing wave that vanishes at the endpoints $z = \pm L/2$:

$$I(z) = I_0 \cos(kz) e^{-i\omega t}$$

where $k = \frac{2\pi}{\lambda} = \frac{\pi}{L}$.

Integrating across the antenna length yields the far-field radiation pattern:

$$\langle \mathbf{S} \rangle = \frac{\eta_0 I_0^2}{8\pi^2 r^2} \left[ \frac{\cos\left( \frac{\pi}{2}\cos\theta \right)}{\sin\theta} \right]^2 \hat{\mathbf{r}}$$

Integrating over a sphere gives the total power radiated:

$$P = \frac{\eta_0 I_0^2}{4\pi} \int_0^\pi \frac{\cos^2\left(\frac{\pi}{2}\cos\theta\right)}{\sin\theta} \, d\theta = \frac{120\pi I_0^2}{4\pi} (1.2188) = 36.56 I_0^2 = \frac{1}{2} I_0^2 R_{\text{rad}}$$

Solving for the radiation resistance of a resonant half-wave antenna:

$$R_{\text{rad}} = 2 \times 36.56 \, \Omega \approx 73.13 \, \Omega$$

Engineering Significance: A radiation resistance of $73\,\Omega$ provides an outstanding impedance match to standard coaxial transmission cables ($50\,\Omega$ or $75\,\Omega$), making half-wave dipoles the universal cornerstone of modern telecommunications.

📝 Chapter Worked Examples & Exercises

Complete analytical solutions & proofs
Medium Example 4.1: Gauge Transformation Connecting Coulomb and Lorentz Gauges

A vector potential in a region is given by $\mathbf{A}_1 = A_0 \cos(kz - \omega t) \hat{\mathbf{x}}$ and $V_1 = 0$. (a) Determine whether this potential satisfies the Coulomb gauge condition. (b) Determine whether it satisfies the Lorentz gauge condition. (c) Construct a gauge function $\lambda(\mathbf{r}, t)$ that transforms $\mathbf{A}_1$ into a new vector potential $\mathbf{A}_2$ and scalar potential $V_2$ such that the Lorentz gauge condition is explicitly satisfied.

Step 1: Check Coulomb Gauge Condition
$$\nabla \cdot \mathbf{A}_1 = \frac{\partial A_{1x}}{\partial x} = \frac{\partial}{\partial x}[A_0 \cos(kz - \omega t)] = 0$$

Since $\mathbf{A}_1$ points along $\hat{\mathbf{x}}$ and varies only with $z$, its spatial divergence is identically zero. Thus, $\mathbf{A}_1$ satisfies the Coulomb gauge.

Step 2: Check Lorentz Gauge Condition
$$\nabla \cdot \mathbf{A}_1 + \frac{1}{c^2}\frac{\partial V_1}{\partial t} = 0 + 0 = 0$$

Remarkably, because both $\nabla \cdot \mathbf{A}_1 = 0$ and $\frac{\partial V_1}{\partial t} = 0$, this specific potential simultaneously satisfies both the Coulomb and Lorentz gauge conditions (this is the transverse radiation gauge).

Step 3: Verification of Physical Fields
$$\mathbf{B} = \nabla \times \mathbf{A}_1 = \left| \begin{matrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \partial_x & \partial_y & \partial_z \\ A_0\cos(kz-\omega t) & 0 & 0 \end{matrix} \right| = k A_0 \sin(kz - \omega t) \hat{\mathbf{y}}\n\mathbf{E} = -\nabla V_1 - \frac{\partial \mathbf{A}_1}{\partial t} = -\omega A_0 \sin(kz - \omega t) \hat{\mathbf{x}}$$

The ratio $|E|/|B| = \omega / k = c$, correctly reproducing a propagating plane electromagnetic wave in vacuum.

Easy Example 4.2: Radiated Power and Radiation Resistance of a Short AM Radio Antenna

A vertical short monopole antenna of height $h = 10.0\text{ m}$ operates at an AM broadcasting frequency of $f = 1000\text{ kHz}$ ($1.00\text{ MHz}$). The antenna carries a peak base current of $I_0 = 15.0\text{ A}$. Calculate: (a) the free-space wavelength $\lambda$, (b) the effective radiation resistance $R_{\text{rad}}$ of the antenna (modeled as a short monopole over a conducting ground plane, $R_{\text{rad}} = 40\pi^2 (h/\lambda)^2$), (c) the total time-averaged radiated power $P_{\text{rad}}$, and (d) the radiation efficiency if the ohmic ground loss resistance is $R_{\text{loss}} = 2.50\,\Omega$.

Step 1: Wavelength Calculation
$$\lambda = \frac{c}{f} = \frac{3.00 \times 10^8\text{ m/s}}{1.00 \times 10^6\text{ Hz}} = 300.0\text{ m}$$

The antenna height $h = 10\text{ m}$ is $h/\lambda = 10/300 = 1/30 \ll 1$, so the short antenna approximation applies.

Step 2: Radiation Resistance Calculation
$$R_{\text{rad}} = 40\pi^2 \left( \frac{h}{\lambda} \right)^2 = 40\pi^2 \left( \frac{10}{300} \right)^2 = 40\pi^2 \left( \frac{1}{900} \right) = \frac{40(9.8696)}{900} = 0.4386\,\Omega \approx 0.439\,\Omega$$

The radiation resistance is only 0.44 Ohms.

Step 3: Total Radiated Power
$$P_{\text{rad}} = \frac{1}{2} I_0^2 R_{\text{rad}} = \frac{1}{2} (15.0\text{ A})^2 (0.4386\,\Omega) = \frac{1}{2}(225)(0.4386) = 49.34\text{ W}$$

The antenna successfully broadcasts 49.3 watts of RF power into the air.

Step 4: Radiation Efficiency
$$\eta = \frac{R_{\text{rad}}}{R_{\text{rad}} + R_{\text{loss}}} = \frac{0.4386}{0.4386 + 2.50} = \frac{0.4386}{2.9386} = 0.1492 \implies 14.9\%$$

Over 85% of transmitter power is wasted as heat in ground resistance because the antenna is physically much shorter than $\lambda/4$ ($75\text{ m}$).

Hard Example 4.3: Bremsstrahlung Radiated Power from a Decelerating Relativistic Electron

An electron ($q = -e = -1.602 \times 10^{-19}\text{ C}$, $m_e = 9.109 \times 10^{-31}\text{ kg}$) in an X-ray medical tube is accelerated by a potential difference $V = 100\text{ kV}$. Upon hitting a tungsten anode target, it decelerates to a complete stop over a distance $\Delta x = 2.0\,\mu\text{m}$ under approximately uniform deceleration $a$. (a) Calculate the kinetic energy and deceleration $a$ of the electron. (b) Use Larmor's relativistic formula $P = \frac{e^2 a^2 \gamma^6}{6\pi \epsilon_0 c^3}$ (or non-relativistic $P = \frac{e^2 a^2}{6\pi \epsilon_0 c^3}$) to calculate the peak radiated power. (c) Estimate the duration of the deceleration pulse $\Delta t$ and total radiated energy.

Step 1: Electron Velocity and Deceleration
$$K = e V = (1.602 \times 10^{-19}\text{ C})(10^5\text{ V}) = 1.602 \times 10^{-14}\text{ J} = 100\text{ keV}\nv_0 = \sqrt{\frac{2K}{m_e}} = \sqrt{\frac{2(1.602 \times 10^{-14})}{9.109 \times 10^{-31}}} = 1.875 \times 10^8\text{ m/s} = 0.625 \, c\na = \frac{v_0^2}{2 \Delta x} = \frac{(1.875 \times 10^8\text{ m/s})^2}{2(2.0 \times 10^{-6}\text{ m})} = 8.79 \times 10^{21}\text{ m/s}^2$$

The atomic collision produces a staggering deceleration of over $8.8 \times 10^{21}\text{ m/s}^2$.

Step 2: Radiated Power via Larmor Formula
$$P = \frac{e^2 a^2}{6\pi \epsilon_0 c^3} = \frac{(1.602 \times 10^{-19})^2 (8.79 \times 10^{21})^2}{6\pi (8.854 \times 10^{-12}) (3.00 \times 10^8)^3}\nP = \frac{(2.566 \times 10^{-38})(7.726 \times 10^{43})}{(1.669 \times 10^{-10})(2.70 \times 10^{25})} = \frac{1.983 \times 10^6}{4.506 \times 10^{15}} = 4.40 \times 10^{-10}\text{ W}$$

During the deceleration instant, a single microscopic electron emits 0.44 nanowatts of continuous X-ray photon radiation.

Step 3: Pulse Duration and Bremsstrahlung Emission
$$\Delta t = \frac{v_0}{a} = \frac{1.875 \times 10^8\text{ m/s}}{8.79 \times 10^{21}\text{ m/s}^2} = 2.13 \times 10^{-14}\text{ s} = 21.3\text{ fs}\nE_{\text{rad}} = P \times \Delta t = (4.40 \times 10^{-10}\text{ W})(2.13 \times 10^{-14}\text{ s}) = 9.37 \times 10^{-24}\text{ J} \approx 58.5\text{ meV}$$

The sudden deceleration creates an ultrashort femtosecond burst of continuous X-ray radiation (Bremsstrahlung, 'braking radiation').

Medium Example 4.4: Input Current and Total Power Radiated by a Half-Wave Dipole Antenna

A commercial FM broadcasting station transmits at $f = 100.0\text{ MHz}$ using a resonant half-wave dipole antenna ($L = \lambda/2$). The station transmitter delivers a total radiated RF power of $P_{\text{rad}} = 50.0\text{ kW}$. (a) Calculate the physical length $L$ of the antenna. (b) Using the half-wave radiation resistance $R_{\text{rad}} = 73.13\,\Omega$, find the peak antenna feedpoint input current $I_0$. (c) Calculate the peak electric field amplitude $E_0$ detected by an FM radio receiver at a distance of $r = 25.0\text{ km}$ in the broadside direction ($\theta = 90^\circ$).

Step 1: Antenna Length Calculation
$$\lambda = \frac{c}{f} = \frac{3.00 \times 10^8\text{ m/s}}{1.00 \times 10^8\text{ Hz}} = 3.00\text{ m}\nL = \frac{\lambda}{2} = \frac{3.00\text{ m}}{2} = 1.50\text{ m}$$

The half-wave dipole is exactly 1.5 meters from tip to tip.

Step 2: Peak Input Current
$$P_{\text{rad}} = \frac{1}{2} I_0^2 R_{\text{rad}} \implies I_0 = \sqrt{\frac{2 P_{\text{rad}}}{R_{\text{rad}}}}\nI_0 = \sqrt{\frac{2(50,000\text{ W})}{73.13\,\Omega}} = \sqrt{1367.4} = 36.98\text{ A}$$

The peak current driving the center feedpoint is approximately 37 Amperes.

Step 3: Detected Electric Field at 25 km Distance
$$\text{Broadside (}\theta = 90^\circ\text{): } \langle S \rangle = \frac{\eta_0 I_0^2}{8\pi^2 r^2} \left[ \frac{\cos(0)}{1} \right]^2 = \frac{(376.73)(36.98)^2}{8\pi^2 (2.50 \times 10^4\text{ m})^2}\n\langle S \rangle = \frac{5.152 \times 10^5}{4.935 \times 10^{10}} = 1.044 \times 10^{-5}\text{ W/m}^2 = 10.44\,\mu\text{W/m}^2\nE_0 = \sqrt{2 \eta_0 \langle S \rangle} = \sqrt{2(376.73)(1.044 \times 10^{-5})} = \sqrt{7.866 \times 10^{-3}} = 0.0887\text{ V/m} = 88.7\text{ mV/m}$$

A field strength of 88.7 millivolts per meter provides crystal-clear reception for automobile and mobile FM receivers.