Radiation & Antennas

Step 1: Point charge density

For a charge $q$ at position $\mathbf{w}(t)$: $\rho(\mathbf{r}',t') = q\delta^3(\mathbf{r}'-\mathbf{w}(t'))$.

Step 2: Substitute into the retarded potential

$$\Phi = \frac{q}{4\pi\epsilon_0}\int\frac{\delta^3(\mathbf{r}'-\mathbf{w}(t_r))}{|\mathbf{r}-\mathbf{r}'|}\,d^3r'$$

The retarded time condition $t_r = t - |\mathbf{r}-\mathbf{r}'|/c$ makes the delta function implicit.

Step 3: Evaluate the integral carefully

Converting to a time integral using $\delta^3(\mathbf{r}'-\mathbf{w}(t'))$ and accounting for the Jacobian from the retarded time constraint introduces a factor $|1-\hat{\mathscr{r}}\cdot\boldsymbol{\beta}|^{-1}$ where $\boldsymbol{\beta} = \mathbf{v}/c$.

Step 4: Lienard-Wiechert potentials

$$\boxed{\Phi = \frac{q}{4\pi\epsilon_0}\frac{1}{\mathscr{r}(1-\hat{\mathscr{r}}\cdot\boldsymbol{\beta})}\bigg|_{t_r}, \quad \mathbf{A} = \frac{\mu_0 q c}{4\pi}\frac{\boldsymbol{\beta}}{\mathscr{r}(1-\hat{\mathscr{r}}\cdot\boldsymbol{\beta})}\bigg|_{t_r}}$$

Step 5: Physical interpretation

The factor $(1-\hat{\mathscr{r}}\cdot\boldsymbol{\beta})^{-1}$ is the "headlight effect": potentials are enhanced in the forward direction of motion. For a charge at rest, these reduce to the Coulomb potential. The fields derived from these potentials contain both velocity (near) and acceleration (radiation) terms.

Step 1: Radiation field

The radiation (far-field) electric field from the Lienard-Wiechert fields, for $v \ll c$:

$$\mathbf{E}_{\text{rad}} = \frac{q}{4\pi\epsilon_0 c^2}\frac{\hat{\mathscr{r}}\times(\hat{\mathscr{r}}\times\mathbf{a})}{\mathscr{r}}\bigg|_{t_r}$$

Step 2: Poynting vector

With $\mathbf{B}_{\text{rad}} = \hat{\mathscr{r}}\times\mathbf{E}_{\text{rad}}/c$:

$$\mathbf{S} = \frac{1}{\mu_0}|\mathbf{E}_{\text{rad}}|^2\hat{\mathscr{r}} = \frac{q^2 a^2\sin^2\theta}{16\pi^2\epsilon_0 c^3 \mathscr{r}^2}\hat{\mathscr{r}}$$

Step 3: Angular distribution

The power per solid angle is $dP/d\Omega = \mathscr{r}^2|\mathbf{S}| = q^2 a^2\sin^2\theta/(16\pi^2\epsilon_0 c^3)$. This is the characteristic $\sin^2\theta$ dipole pattern: maximum radiation perpendicular to the acceleration, zero along it.

Step 4: Integrate over all angles

$$P = \int\frac{dP}{d\Omega}\,d\Omega = \frac{q^2 a^2}{16\pi^2\epsilon_0 c^3}\int_0^{2\pi}d\phi\int_0^\pi\sin^3\theta\,d\theta$$

The angular integral gives $2\pi \times 4/3 = 8\pi/3$.

Step 5: Larmor formula

$$\boxed{P = \frac{q^2 a^2}{6\pi\epsilon_0 c^3}}$$

For an electron in a magnetic field (cyclotron radiation), $a = v_\perp\omega_c$, and the radiated power limits the electron energy in synchrotrons.

Step 1: Vector potential in the radiation zone

For a localized oscillating source with $r \gg d \gg \lambda$ not required (just $r \gg \lambda \gg d$), the vector potential reduces to:

$$\mathbf{A} = \frac{\mu_0}{4\pi r}\dot{\mathbf{p}}(t_r)$$

Step 2: Compute the fields

In the far field, only terms falling as $1/r$ contribute to radiation. Taking the curl:

$$\mathbf{B} = -\frac{\mu_0}{4\pi c r}\hat{r}\times\ddot{\mathbf{p}}(t_r), \quad \mathbf{E} = c\mathbf{B}\times\hat{r}$$

Step 3: For harmonic oscillation

With $\mathbf{p} = p_0\hat{z}e^{-i\omega t}$: $\ddot{\mathbf{p}} = -\omega^2 p_0\hat{z}e^{-i\omega t}$. The radiation field magnitude is:

$$|\mathbf{E}_{\text{rad}}| = \frac{\mu_0 p_0\omega^2\sin\theta}{4\pi r}$$

Step 4: Power per solid angle

$$\frac{dP}{d\Omega} = \frac{\mu_0 p_0^2\omega^4}{32\pi^2 c}\sin^2\theta$$

Step 5: Total radiated power

Integrating: $$\boxed{\langle P\rangle = \frac{\mu_0 p_0^2\omega^4}{12\pi c}}$$

The $\omega^4$ dependence explains why the sky is blue (Rayleigh scattering) and why RF antennas must be comparable to a wavelength for efficient radiation.

Step 1: Expand the retarded potential

For a source of size $d \ll \lambda$, expand $|\mathbf{r}-\mathbf{r}'| \approx r - \hat{r}\cdot\mathbf{r}' + \ldots$ in the exponent:

$$\mathbf{A} = \frac{\mu_0}{4\pi r}\int\mathbf{J}(\mathbf{r}')e^{-ik\hat{r}\cdot\mathbf{r}'}\,d^3r' \times e^{i(kr-\omega t)}$$

Step 2: Zero-th order (electric dipole)

Setting $e^{-ik\hat{r}\cdot\mathbf{r}'} \approx 1$: $\int\mathbf{J}\,d^3r' = \dot{\mathbf{p}}$. This gives the electric dipole term.

Step 3: First-order term

Expanding to first order in $k\hat{r}\cdot\mathbf{r}'$: $-ik\int(\hat{r}\cdot\mathbf{r}')\mathbf{J}\,d^3r'$. This separates into antisymmetric (magnetic dipole) and symmetric (electric quadrupole) parts.

Step 4: Magnetic dipole contribution

The antisymmetric part gives $\mathbf{A}_m = -i\mu_0 k/(4\pi r)\,(\mathbf{m}\times\hat{r})e^{i(kr-\omega t)}$, producing the same angular pattern as the electric dipole but with E and B roles swapped.

Step 5: Hierarchy of multipoles

Each successive term is suppressed by a factor of $(d/\lambda)$. For atoms, $d/\lambda \sim \alpha \sim 1/137$, so electric dipole transitions dominate. Magnetic dipole and electric quadrupole transitions are "forbidden" (suppressed by $\alpha^2$).

Step 1: Current distribution

A half-wave dipole ($L = \lambda/2$) has a sinusoidal current: $I(z) = I_0\cos(kz)$ for $|z| \leq \lambda/4$.

Step 2: Far-field pattern

Computing the vector potential integral and taking the far-field limit:

$$E_\theta = \frac{i\mu_0 c I_0}{2\pi r}\frac{\cos\!\left(\frac{\pi}{2}\cos\theta\right)}{\sin\theta}e^{i(kr-\omega t)}$$

Step 3: Total radiated power

$$P = \frac{\mu_0 c I_0^2}{4\pi^2}\int_0^\pi\frac{\cos^2\!\left(\frac{\pi}{2}\cos\theta\right)}{\sin\theta}\,d\theta$$

Step 4: Evaluate the integral

The integral equals approximately 1.2188 (the "Cin" integral). Therefore $P \approx 1.2188 \times \mu_0 c I_0^2/(4\pi^2)$.

Step 5: Radiation resistance

Defining $P = \frac{1}{2}R_{\text{rad}}I_0^2$: $$\boxed{R_{\text{rad}} = \frac{\mu_0 c}{2\pi^2}\times 1.2188 \approx 73.1\ \Omega}$$

This is conveniently close to 75 $\Omega$, enabling direct connection to standard coaxial cable. The directivity is 1.64 (2.15 dBi).