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.