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:
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:
Since the divergence of any curl vanishes identically ($\nabla \cdot (\nabla \times \mathbf{A}) \equiv 0$), we can always define the magnetic vector potential $\mathbf{A}$:
Substitute this definition into Faraday's law:
Because the quantity in parentheses has zero curl, it can now be expressed as the negative gradient of a dynamic electric scalar potential $V$:
These two equations represent the fundamental relations expressing physical fields $(\mathbf{E}, \mathbf{B})$ in terms of electromagnetic potentials $(V, \mathbf{A})$:
§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 $\lambda(\mathbf{r}, t)$:
Let us compute the new fields:
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:
Substituting into Gauss's law $\nabla \cdot \mathbf{E} = \rho / \epsilon_0$:
Since $\nabla \cdot \mathbf{A} = 0$:
This is the standard Poisson equation. The scalar potential in Coulomb gauge is:
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:
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:
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:
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:
The exact solutions to the decoupled wave equations in Lorentz gauge are the Retarded Potentials:
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:
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):
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:
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:
Decomposition and Physical Significance
1. The Velocity Field (Generalized Coulomb Field):
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):
Decays as $1/\imath$ with distance! The associated Poynting energy flux scales as:
When integrated over a giant sphere of radius $\imath \to \infty$, the surface area $4\pi \imath^2$ cancels the $1/\imath^2$ decay:
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:
where $p_0 = q_0 d$.
We evaluate the fields under the standard radiation zone approximations:
- Short dipole compared to wavelength: $d \ll \lambda$ (dipole approximation)
- 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:
Converting $\hat{\mathbf{z}} = \cos\theta \hat{\mathbf{r}} - \sin\theta \hat{\boldsymbol{\theta}}$:
Computing $\mathbf{B} = \nabla \times \mathbf{A}$ keeping only terms scaling as $1/r$:
From $\mathbf{E} = c(\mathbf{B} \times \hat{\mathbf{r}})$:
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:
Taking the time average over an oscillation cycle ($\langle \cos^2 \rangle = 1/2$):
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 $r$:
The standard integral is $\int_0^\pi \sin^3\theta \, d\theta = \frac{4}{3}$:
In terms of the speed of light in vacuum ($c = 1/\sqrt{\epsilon_0 \mu_0}$):
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:
Equating to the equivalent dissipated circuit power $P = \frac{1}{2} I_0^2 R_{\text{rad}}$:
For a short dipole where $d \ll \lambda$ (e.g., $d = 0.05\lambda$):
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:
The total radiated power is:
Comparison Between Electric and Magnetic Dipoles
For comparable geometric dimensions ($p_0 \sim q d, m_0 \sim I d^2 \sim q \omega d^2$):
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 $z = \pm L/2$:
where $k = \frac{2\pi}{\lambda} = \frac{\pi}{L}$.
Integrating across the antenna length yields the far-field radiation pattern:
Integrating over a sphere gives the total power radiated:
Solving for the radiation resistance of a resonant half-wave antenna:
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 & proofsA 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.
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.
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).
The ratio $|E|/|B| = \omega / k = c$, correctly reproducing a propagating plane electromagnetic wave in vacuum.
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 $R_{\text{loss}} = 2.50\,\Omega$.
The antenna height $h = 10\text{ m}$ is $h/\lambda = 10/300 = 1/30 \ll 1$, so the short antenna approximation applies.
The radiation resistance is only 0.44 Ohms.
The antenna successfully broadcasts 49.3 watts of RF power into the air.
Over 85% of transmitter power is wasted as heat in ground resistance because the antenna is physically much shorter than $\lambda/4$ ($75\text{ m}$).
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.
The atomic collision produces a staggering deceleration of over $8.8 \times 10^{21}\text{ m/s}^2$.
During the deceleration instant, a single microscopic electron emits 0.44 nanowatts of continuous X-ray photon radiation.
The sudden deceleration creates an ultrashort femtosecond burst of continuous X-ray radiation (Bremsstrahlung, 'braking radiation').
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 ($\theta = 90^\circ$).
The half-wave dipole is exactly 1.5 meters from tip to tip.
The peak current driving the center feedpoint is approximately 37 Amperes.
A field strength of 88.7 millivolts per meter provides crystal-clear reception for automobile and mobile FM receivers.