Jefimenko's Equations and the Current Element (a.k.a. Hertzian Dipole) - Part 2

I want to re-visit the Hertzian dipole, armed with the derivations from my previous post. In the diagram below, $O$ is the origin of a spherical coordinate system $\langle {r},\;\theta,\;\phi\rangle$ where: $$ 0\le {r} \lt \infty, \qquad 0 \le \theta \le \pi, \qquad 0 \le \phi \le 2\pi, \qquad \hat{r} \times \hat{\theta} = \hat{\phi} $$ The $\theta = 0$ axis will be called the $\ell$-axis. The choice $\phi = 0$ direction, as long as it is perpendicular to the $\ell$-axis, does not affect what we are going to work out.
Our dipole is the segment $\overline{BA}$ which lies along this $\ell$-axis. At any instant $t$, the current at any point on the dipole is $I_0\cos {\omega t}\;\hat{\ell}$. The points $A$ and $B$ are assumed to have infinite capacitance.
We are interested in the fields at point $P$, so $\vec{OP} = \vec{r}$.

I'll recap the part of my last post that I use in this one.

Given the following definitions: $$ r = \left|\vec{r}\right|, \qquad \hat{r} = \dfrac {\vec{r}} {r}, \qquad t_r = t - \frac r c $$ The electric and magnetic fields in the far-field region are given by: $$ \vec{E}\left(\vec{r},t\right) \approx \frac {1} {4 \pi \epsilon_0} \left[ \frac {\mathcal{U}\left(\vec{r},t\right)} {r^2} \hat{r} + \frac {\left( \vec{\mathcal{X}}\left(\vec{r},t\right) \times \hat{r}\right) \times \hat{r}} {rc} \right] $$ $$ \vec{B}\left(\vec{r},t\right) \approx \frac {\mu_0} {4 \pi} \left[ \frac {\vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{r}} {r} \right] $$ When: $$ \mathcal{U}\left(\vec{r},t\right) = \mathcal{Y} \left( \hat{r}, t_r \right) + \frac {\vec{\mathcal{Z}} \left( \hat{r}, t_r \right) \cdot \hat{r}} {c} $$ $$ \vec{\mathcal{X}} \left(\vec{r},t\right) = \frac {\vec{\mathcal{Z}} \left( \hat{r}, t_r \right)} {r} + \frac {\vec{\mathcal{Q}} \left( \hat{r}, t_r \right)} {c} $$ And: $$ \mathcal{Y} \left(\hat{r},t\right) = \iiint_{\mathbb{V}_s} \left[ \rho \left( \vec{r}_s, t + \dfrac {\vec{r}_s \cdot \hat{r}} {c} \right) \right] \space dV\left(\vec{r}_s\right) $$ $$ \vec{\mathcal{Z}} \left(\hat{r},t\right) = \iiint_{\mathbb{V}_s} \left[ \vec{J} \left( \vec{r}_s, t + \dfrac {\vec{r}_s \cdot \hat{r}} {c} \right) \right] \space dV\left(\vec{r}_s\right) $$ $$ \vec{\mathcal{Q}} \left(\hat{r},t\right) = \iiint_{\mathbb{V}_s} \left[ \frac {\partial} {\partial t} \vec{J} \left( \vec{r}_s, t + \dfrac {\vec{r}_s \cdot \hat{r}} {c} \right) \right] \space dV\left(\vec{r}_s\right) $$

Let us now describe the source. As a consequence of the continuity equation, the points $A$ and $B$ will have a time varying electric charge. Thus, our source actually consists of three distinct parts:

1. The segment $\overline{BA}$, carrying the time-varying current $I_0\:\cos\;\omega t\;\hat{\ell}$ at every point.
2. The point $A$, having a time-varying charge of $I_0 \int_0^t \cos\;\omega t \;dt = (I_0 / \omega)\sin\;\omega t$
3. The point $B$, having a time-varying charge of $-I_0 \int_0^t \cos\;\omega t \;dt = (-I_0 / \omega)\sin\;\omega t$

Since we know the current, we can see that each point on segment $BA$, we have: $$ \vec{I}\left(t\right) = I_0\:\cos\;\omega t\;\hat{\ell} $$ $$ \frac {d \vec{I}\left(t\right)}{dt} = -\omega I_0\:\sin\;\omega t\;\hat{\ell} $$ The points $\vec{r}_s$ are points $\left(\ell,\; 0,\; 0\right)$ on the dipole, $A$ is $\left(\delta\ell/2,\; 0,\; 0\right)$ and $B$ is $\left(-\delta\ell/2,\; 0,\; 0\right)$. Therefore: $$ \vec{r}_s\cdot\hat{r} = \ell \cos {\theta} $$ $$ t + \frac {\vec{r}_s\cdot\hat{r}} {c} = t + \frac {\ell \cos {\theta}} {c} $$ Considering that $q = \iiint\rho\;dV$ and $\int\vec{I}\;d\ell = \iiint\vec{J}\;dV$, we now get: $$ \mathcal{Y} \left(\hat{r},t\right) = \frac {I_0} {\omega} \sin {\left( \omega t + \frac {\omega\;\delta\ell\;\cos\theta} {2c} \right)} - \frac {I_0} {\omega} \sin {\left( \omega t - \frac {\omega\;\delta\ell\;\cos\theta} {2c} \right)} $$ $$ \vec{\mathcal{Z}} \left(\hat{r},t\right) = I_0 \left[ \int_{-{\delta\ell}/{2}}^{+{\delta\ell}/{2}} \cos {\left( \omega t + \frac {\omega\ell\;\cos\theta} {c} \right)} \;d\ell \right] \hat{\ell} $$ $$ \vec{\mathcal{Q}} \left(\hat{r},t\right) = -\omega I_0 \left[ \int_{-{\delta\ell}/{2}}^{+{\delta\ell}/{2}} \sin {\left( \omega t + \frac {\omega\ell\;\cos\theta} {c} \right)} \;d\ell \right] \hat{\ell} $$ Working out the integrals, we get: $$ \mathcal{Y} \left(\hat{r},t\right) = \frac {I_0} {\omega} \left[ \sin {\left( \omega t + \frac {\omega\;\delta\ell\;\cos\theta} {2c} \right)} -\sin {\left( \omega t - \frac {\omega\;\delta\ell\;\cos\theta} {2c} \right)} \right] $$ $$ \vec{\mathcal{Z}} \left(\hat{r},t\right) = \frac {I_0\; c} {\omega\cos\theta} \left[ \sin {\left( \omega t + \frac {\omega\;\delta\ell\;\cos\theta} {2c} \right)} -\sin {\left( \omega t - \frac {\omega\;\delta\ell\;\cos\theta} {2c} \right)} \right] \hat{\ell} $$ $$ \vec{\mathcal{Q}} \left(\hat{r},t\right) = \frac {I_0\; c} {\cos\theta} \left[ \cos {\left( \omega t + \frac {\omega\;\delta\ell\;\cos\theta} {2c} \right)} -\cos {\left( \omega t - \frac {\omega\;\delta\ell\;\cos\theta} {2c} \right)} \right] \hat{\ell} $$ Since $\delta\ell$ is very small, $\left|\dfrac {\omega\;\delta\ell\;\cos\theta} {2c}\right| \ll 1$. This allows us to use the following approximations: $$ \sin {\dfrac {\omega\;\delta\ell\;\cos\theta} {2c}} \approx \dfrac {\omega\;\delta\ell\;\cos\theta} {2c} $$ $$ \cos {\dfrac {\omega\;\delta\ell\;\cos\theta} {2c}} \approx 1 $$ We can also deduce that: $$ \hat{\ell} = \cos\;\theta\;\hat{r} - \sin\;\theta\;\hat{\theta} $$ $$ \hat{\ell}\times\hat{r} = -\sin{\theta}\;\left(\hat{\theta}\times\hat{r}\right) = \sin{\theta}\;\hat{\phi} $$ $$ \left(\hat{\ell}\times\hat{r}\right)\times\hat{r} = \sin{\theta}\;\left(\hat{\phi}\times\hat{r}\right) =\sin{\theta}\;\hat{\theta} $$ So we can work out the trigonometry: $$ \mathcal{Y} \left(\hat{r},t\right) = \frac {I_0\;\delta\ell\;\cos\theta\;\cos {\omega t}} {c} $$ $$ \vec{\mathcal{Z}} \left(\hat{r},t\right) = \left( I_0\; \delta\ell \;\cos {\omega t}\right)\; \hat{\ell} = \left( I_0\; \delta\ell \;\cos {\omega t}\;\cos\theta\right)\; \hat{r} - \left( I_0\; \delta\ell \;\cos {\omega t}\;\sin\theta\right)\; \hat{\theta} $$ $$ \vec{\mathcal{Q}} \left(\hat{r},t\right) = -\left( I_0\; \omega\;\delta\ell \;\sin{\omega t}\right)\; \hat{\ell} $$ We can now substitute to get the following: $$ \mathcal{U}\left(\vec{r},t\right) = \frac {2I_0\;\delta\ell\;\cos\theta\;\cos {\omega t_r}} {c} $$ $$ \vec{\mathcal{X}} \left(\vec{r},t\right) = I_0\;\delta\ell\left[ \frac {\cos{\omega t_r}} {r} - \frac {\omega\;\sin{\omega t_r}} {c} \right]\hat{\ell} $$ Substituting this, we get: $$ \vec{E}\left( \vec{r}, t \right) \approx \frac {2} {\epsilon_0} \left( \frac {\cos {\omega t_r}} {r^2c} \right) \frac {I_0\;\delta\ell\;\cos\theta} {4\pi} \hat{r} + \frac {1} {\epsilon_0} \left( \frac {\cos{\omega t_r}} {r^2c}- \frac {\omega\;\sin{\omega t_r}} {rc^2} \right) \frac {I_0\;\delta\ell\;\sin\theta} {4\pi} \hat{\theta} $$ $$ \vec{B}\left( \vec{r}, t \right) \approx \mu_0 \left( \frac {\cos{\omega t_r}} {r^2} - \frac {\omega\;\sin{\omega t_r}} {rc} \right) \frac {I_0\;\delta\ell\;\sin\theta} {4\pi} \hat{\phi} $$ How good an approximation is this? The 'exact' expressions without the far-field approximation are: $$ \vec{E}(\vec{r}, t) = \frac {2} {\epsilon_0} \left( \color{red} {\frac {\sin\;\omega t_r} {\omega r^3}} + \frac {\cos\;\omega t_r} {r^2 c} \right) \frac {I_0\;\delta\ell\;\cos\:\theta} {4\pi} \hat{r} +\frac {1} {\epsilon_0} \left( \color{red} {\frac {\sin\;\omega t_r} {\omega r^3} } + \frac {\cos\;\omega t_r} {r^2 c} - \frac {\omega\;\sin\;\omega t_r} {r c^2} \right) \frac {I_0\;\delta\ell\;\sin\:\theta} {4\pi} \hat{\theta} $$ $$ \vec{B}(\vec{r}, t) = \mu_0 \left( \frac {\cos\;\omega t_r} {r^2} -\frac {\omega\;\sin\;\omega t_r} {r c} \right) \frac {I_0\;\delta\ell\;\sin\;\theta} {4\pi} \hat{\phi} $$ Thus we are off only by a couple of $\mathcal{O}\left(r^{-3}\right)$ terms. That's not bad, considering the tedious math we need to work out the exact expression.

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: $$ \oint_{\partial\mathbb{V}} \vec{S}\left(\vec{r},t\right) \cdot \hat{r} \;ds\left(\vec{r}\right) = \frac 1 {16 \pi^2 \epsilon_0 c^3} \int_0^{2\pi} \int_0^{\pi} \left| \vec{\mathcal{Q}}\left(\hat{r}, t_r\right) \times \hat{r} \right|^2 \;\sin \theta \;d\theta \;d\phi $$