Radiation in the Far Field without invoking the sinusoid (via Jefimenko & McDonald)

Let us say that $\mathbb{V}_s$ is the smallest spherical volume such that at any point outside $\mathbb{V}_s$ there is no electric charge or electric current at any time: $$ \forall \vec{r}\not\in \mathbb{V}_s,\quad\forall t: \qquad \rho\left(\vec{r},t\right) = \vec{J}\left(\vec{r},t\right) = 0 $$ Since $\mathbb{V}_s$ is spherical, let us say that its diameter is $\mathcal{D}$. Further, since no coordinate system has been imposed on us so far, let us choose one such that the center of $\mathbb{V}_s$ is at the origin $\vec{0}$. In other words, $$ \mathbb{V}_s = \mathbb{B}\left(\vec{0}, \frac{\mathcal{D}} 2 \right) $$ Let us also define the following: $$ \vec{r}_s \in \mathbb{V}_s,\qquad \vec{R} = \vec{r} - \vec{r}_s,\qquad R=\left|\vec{R}\right|, \qquad \hat{R}=\frac {\vec{R}} {R}, \qquad t_R = t - \frac {R} {c} $$ If we have time-invariant charge and current densities, the expressions for $\vec{E}$ and $\vec{B}$ also do not vary with time, and are: $$ \vec{E}\left(\vec{r}\right) = \frac {1} {4 \pi \epsilon_0} \iiint_{\mathbb{V}_s} {\left[ \frac {\rho (\vec{r}_s)} {R^2} \hat{R} \right]} \space dV\left(\vec{r}_s\right) $$ $$ \vec{B}\left(\vec{r}\right) = \frac {\mu_0} {4 \pi} \iiint_{\mathbb{V}_s} {\left[ \frac {\vec{J} (\vec{r}_s)} {R^2} \times \hat{R} \right]} \space dV\left(\vec{r}_s\right) $$ If we allow the charge and current densities to change with time, the the electric field $\vec{E}$ and magnetic field $\vec{B}$ are given by Jefimenko's Equations, which are: $$ \vec{E}\left(\vec{r},t\right) = \frac {1} {4 \pi \epsilon_0} \iiint_{\mathbb{V}_s} {\left[ \frac {\rho (\vec{r}_s, t_R)} {R^2} \hat{R} + \frac {1} {R c} \frac {\partial \rho (\vec{r}_s, t_R) } {\partial t} \hat{R} - \frac {1} {R c^2} \frac {\partial \vec{J} (\vec{r}_s, t_R) } {\partial t} \right]} \space dV\left(\vec{r}_s\right) $$ $$ \vec{B}\left(\vec{r},t\right) = \frac {\mu_0} {4 \pi} \iiint_{\mathbb{V}_s} {\left[ \frac {\vec{J} (\vec{r}_s, t_R)} {R^2} \times \hat{R} + \frac {1} {R c} \frac {\partial \vec{J} (\vec{r}_s, t_R) } {\partial t} \times \hat{R} \right]} \space dV\left(\vec{r}_s\right) $$ In addition to $\rho$ and $\vec{J}$, the expression for time varying fields contain $\dfrac {\partial \rho} {\partial t}$ and $\dfrac {\partial \vec{J}}{\partial t}$. But we can immediately suspect that $\dfrac {\partial \rho} {\partial t}$ can be eliminated by the application of the Continuity Equation for electric charge: $$-\dfrac {\partial \rho (\vec{r}, t)} {\partial t} = \nabla \cdot \vec{J} (\vec{r}, t) $$ We have to be careful here because we have $\rho (\vec{r}_s, t_R)$ and $\vec{J} (\vec{r}_s, t_R)$ rather than $\rho (\vec{r}, t)$ and $\vec{J} (\vec{r}, t)$, but this paper entitled "The Relation Between Expressions for Time-Dependent Electromagnetic Fields Given by Jefimenko and by Panofsky and Phillips" by Kirk T. McDonald helps us transform the expression for $\vec{E}$ to: $$ \small{ \vec{E}\left(\vec{r},t\right) = \frac {1} {4 \pi \epsilon_0} \iiint_{\mathbb{V}_s} {\left[ \frac {\rho (\vec{r}_s, t_R)} {R^2} \hat{R} + \frac { \left(\vec{J} (\vec{r}_s, t_R) \cdot \hat{R}\right)\hat{R} + \left(\vec{J} (\vec{r}_s, t_R) \times\hat{R}\right) \times \hat{R} } {R^2 c} + \frac {1} {R c^2} \left( \frac {\partial \vec{J} (\vec{r}_s, t_R) } {\partial t} \times \hat{R} \right) \times \hat{R} \right]} \space dV\left(\vec{r}_s\right) } $$ What we would want to do now is to calculate the power flow, expressed as the Poynting vector: $$ \vec{S}\left(\vec{r},t\right) = \dfrac {\vec{E}\left(\vec{r},t\right) \times \vec{B}\left(\vec{r},t\right)} {\mu_0} $$ But it is far too complicated in the general case, so we'll look for some approximations. When discussing radiation, we are usually interested in the 'far field' region, where $R \to \infty$. We note that as $R \to \infty$, the quantities $\dfrac 1 R$ and $\hat{R}$ converge towards values that are independent of $\vec{r}_s$.

We can see that the maximum possible difference between values of $\dfrac 1 R$ for any two points $\vec{r}_s \in \mathbb{V}_s$ is: $$ \frac{1}{R} - \frac{1}{R+\mathcal{D}} = \frac{\mathcal{D}}{R\left(R+\mathcal{D}\right)} = \frac{1}{R}\cfrac{ \color{red}{\cfrac{\mathcal{D}}{R}}}{1+\color{red}{\cfrac{\mathcal{D}}{R}}} $$ Similarly, the maximum possible angle between values of $\hat{R}$ for any two points $\vec{r}_s \in \mathbb{V}_s$ is: $$ 2\;{\tan}^{-1}\left(\frac12 \color{red} {\frac{\mathcal{D}}{R}}\right) $$ Now $\dfrac {\mathcal{D}} R \to 0$ as $R \to \infty$, so for large values of $R$ each of the two terms above become negligible.

We can,therefore, bring all $\dfrac 1 R$ and $\hat{R}$ factors outside the integral sign whenever it is convenient to do so. Now we can define:

$$ \mathcal{U}\left(\vec{r},t\right) = \iiint_{\mathbb{V}_s} \left[ \rho \left( \vec{r}_s, t_R \right) + \frac { \vec{J} \left( \vec{r}_s, t_R \right) \cdot \hat{R} } {c} \right] \space dV\left(\vec{r}_s\right) $$ $$ \vec{\mathcal{X}} \left(\vec{r},t\right) = \iiint_{\mathbb{V}_s} \left[ \frac { \vec{J} \left( \vec{r}_s, t_R \right) } {R} + \frac 1 c \frac {\partial \vec{J} \left( \vec{r}_s, t_R \right) } {\partial t} \right] \space dV\left(\vec{r}_s\right) $$ $\mathcal{U}$ and $\vec{\mathcal{X}}$ depend on $\vec{r}$ and $t$ because $t_R$ does: $$ t_R = t - \dfrac R c = \bbox[yellow]{t} - \dfrac {\left|\bbox[yellow]{\vec{r}}-\vec{r}_s\right|} c $$ $\rho$ and $\vec{J}$ also depend on $\vec{r}_s$, but since we have integrated over $\vec{r}_s$, it doesn't appear in $\mathcal{U}$ and $\vec{\mathcal{X}}$.

Substituting, we get:

$$ \vec{E}\left(\vec{r},t\right) \approx \frac {1} {4 \pi \epsilon_0} \left[ \frac {\mathcal{U}\left(\vec{r},t\right)} {R^2} \hat{R} + \frac {\left( \vec{\mathcal{X}}\left(\vec{r},t\right) \times \hat{R}\right) \times \hat{R}} {Rc} \right] $$ $$ \vec{B}\left(\vec{r},t\right) \approx \frac {\mu_0} {4 \pi} \left[ \frac {\vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R}} {R} \right] $$ $$ \vec{S}\left(\vec{r},t\right) \approx \mathcal{U} \left(\vec{r},t\right) \frac {\hat{R} \times \left( \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right) } {16 \pi^2 \epsilon_0 R^3} + \frac {\left| \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right|^2} {16 \pi^2 \epsilon_0 R^2 c} \hat{R} $$

We can see immediately that the power flow along the direction $\hat{R}$ is:

$$ \vec{S}\left(\vec{r},t\right)\cdot \hat{R} \approx \frac {\left| \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right|^2} {16 \pi^2 \epsilon_0 R^2 c} $$

Let us now consider another sphere $\mathbb{V}$ concentric with $\mathbb{V}_s$, and let us say that the radius of $\mathbb{V}$ is much larger than the radius $\dfrac {\mathcal{D}} {2}$ of $\mathbb{V}_s$.

We define: $$ r = \left|\vec{r}\right|, \qquad \hat{r} = \dfrac {\vec{r}} {r}, \qquad t_r = t - \frac r c $$ Reasoning like we did earlier, we can see that given any point $\vec{r} \in \partial\mathbb{V}$ (i.e. on the surface of $\mathbb{V}$), the quantities $\dfrac {1} {R}$ and $\hat{R}$ can be considered independent of $\vec{r}_s \in \mathbb{V}_s$. Since $\vec{0} \in \mathbb{V}_s$, we can use the following approximations: $$\dfrac{1} {R} \approx \dfrac {1} {\left|\vec{r}\right|} = \dfrac {1} {r}$$ $$\hat{R} \approx \dfrac {\vec{r}} {\left|\vec{r}\right|} = \hat{r} $$ This simplification does not extend directly to $R$, but a little vector arithmetic tells us that, when $r \gg \dfrac {\mathcal{D}} {2}$, we get: $$ R = \dfrac {r - \left( \vec{r}_s \cdot \hat{r} \right)} {\hat{R} \cdot \hat{r}} \approx r - \left( \vec{r}_s \cdot \hat{r} \right) $$ ... and consequently: $$ t_R \approx t_r + \dfrac {\vec{r}_s \cdot \hat{r}} {c} $$

Now we can apply the approximations we have just deduced.

Let us define: $$ \mathcal{Y} \left(\hat{r},t\right) = \iiint_{\mathbb{V}_s} \left[ \rho \left( \vec{r}_s, t + \dfrac {\vec{r}_s \cdot \hat{r}} {c} \right) \right] \space dV\left(\vec{r}_s\right) $$ $$ \vec{\mathcal{Z}} \left(\hat{r},t\right) = \iiint_{\mathbb{V}_s} \left[ \vec{J} \left( \vec{r}_s, t + \dfrac {\vec{r}_s \cdot \hat{r}} {c} \right) \right] \space dV\left(\vec{r}_s\right) $$ $$ \vec{\mathcal{Q}} \left(\hat{r},t\right) = \iiint_{\mathbb{V}_s} \left[ \frac {\partial} {\partial t} \vec{J} \left( \vec{r}_s, t + \dfrac {\vec{r}_s \cdot \hat{r}} {c} \right) \right] \space dV\left(\vec{r}_s\right) $$ We can then say: $$ \mathcal{U}\left(\vec{r},t\right) = \mathcal{Y} \left( \hat{r}, t_r \right) + \frac {\vec{\mathcal{Z}} \left( \hat{r}, t_r \right) \cdot \hat{r}} {c} $$ $$ \vec{\mathcal{X}} \left(\vec{r},t\right) = \frac {\vec{\mathcal{Z}} \left( \hat{r}, t_r \right)} {r} + \frac {\vec{\mathcal{Q}} \left( \hat{r}, t_r \right)} {c} $$ And: $$ \vec{E}\left(\vec{r},t\right) \approx \frac {1} {4 \pi \epsilon_0} \left[ \frac {\mathcal{U}\left(\vec{r},t\right)} {r^2} \hat{r} + \frac {\left( \vec{\mathcal{X}}\left(\vec{r},t\right) \times \hat{r}\right) \times \hat{r}} {rc} \right] $$ $$ \vec{B}\left(\vec{r},t\right) \approx \frac {\mu_0} {4 \pi} \left[ \frac {\vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{r}} {r} \right] $$ $$ \vec{S}\left(\vec{r},t\right) \approx \mathcal{U} \left(\vec{r},t\right) \frac {\hat{r} \times \left( \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{r} \right) } {16 \pi^2 \epsilon_0 r^3} + \frac {\left| \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{r} \right|^2} {16 \pi^2 \epsilon_0 r^2 c} \hat{r} $$

It should be obvious that $\hat{r}$ is the outward-pointing unit normal at any point on the surface of $\mathbb{V}$. Now the term $\vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{r}$ is always perpendicular to $\hat{r}$, and thus tangential to the surface $\partial\mathbb{V}$. By the Hairy Ball Theorem, this quantity must be zero for at least one value of $\hat{r}$.

We can, therefore, conclude that at any given instant there will always be at least one direction relative to $\mathbb{V}_s$ where $\vec{S}$, $\vec{B}$ and $\vec{E}$ will all be zero.

Considering that $\dfrac a r + b \to b$ as $r \to \infty$, for sufficiently large $r$ this formulation lets us simplify further. What is 'sufficiently large', however, depends on how the magnitudes $\left|\rho\right|$ and $\left|\vec{J}\right|$ compare with $\left| \dfrac {\partial\vec{J}}{\partial t} \right|$.

The expressions we have encountered before simplify to the following: $$ \vec{E}\left(\vec{r},t\right) \approx \frac {1} {4 \pi \epsilon_0 r c^2} \; \left[ \vec{\mathcal{Q}}\left(\hat{r},t_r\right) \times \hat{r} \right] \times \hat{r} $$ $$ \vec{B}\left(\vec{r},t\right) \approx \frac {\mu_0} {4 \pi r c} \; \vec{\mathcal{Q}} \left(\hat{r},t_r \right) \times \hat{r} $$ $$ \vec{S}\left(\vec{r},t\right) \approx \frac {1} {16 \pi^2 \epsilon_0 r^2 c^3} \; \left| \vec{\mathcal{Q}} \left(\hat{r},t_r \right) \times \hat{r} \right|^2 \; \hat{r} $$ If we choose a spherical coordinate system $\langle 0\le \mathrm{r} \lt \infty, \;0 \le \phi \le 2\pi,\; 0 \le \theta \le \pi\rangle$ around the origin we have already chosen, we can express the total power crossing the surface $\partial\mathbb{V}$ as follows: $$ \oint_{\partial\mathbb{V}} \vec{S}\left(\vec{r},t\right) \cdot \hat{r} \;ds\left(\vec{r}\right) = \frac 1 {16 \pi^2 \epsilon_0 c^3} \int_0^{2\pi} \int_0^{\pi} \left| \vec{\mathcal{Q}}\left(\hat{r}, t_r\right) \times \hat{r} \right|^2 \;\sin \theta \;d\theta \;d\phi $$