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 Resistance and the Quarter-cycle Phase Shift

Let us consider, once again, a 'current element' (or an 'elementary doublet' if you like, or a 'Hertzian dipole'). It is a piece of wire of zero thickness and infinitesimal length $\delta\ell$, carrying a sinusoidal current $I(t) = I_0 \: \cos \; \omega t$, such that the current at each point on the wire is the same.

Using the position of the current element as the origin of a spherical co-ordinate system, such that the direction of $I_0$ is the $\theta=0$ direction, at any point $(r,\theta,\phi)$ in space, the electric and magnetic fields caused by the current element are as follows:

$$\vec{E}(\vec{r}, t) = \frac {2} {\epsilon_0} \left[ \frac {\sin\;\omega t'} {\omega r^3} +\frac {\cos\;\omega t'} {c r^2} \right] \frac {I_0\;\delta\ell\;\cos\:\theta} {4\pi} \hat{r} +\frac {1} {\epsilon_0} \left[ \frac {\sin\;\omega t'} {\omega r^3} +\frac {\cos\;\omega t'} {c r^2} -\frac {\omega\;\sin\;\omega t'} {c^2 r} \right] \frac {I_0\;\delta\ell\;\sin\:\theta} {4\pi} \hat{\theta} $$ $$\vec{B}(\vec{r}, t) = \mu_0 \left[ \frac {\cos\;\omega t'} {r^2} -\frac {\omega\;\sin\;\omega t'} {c r} \right] \frac {I_0\;\delta\ell\;\sin\;\theta} {4\pi} \hat{\phi} $$ Where: $$ t' = t - \frac {r} {c} $$

We know that, at any instant, the power flow through any point is given by the Poynting vector at that point:

$$\vec{P} = \frac {1} {\mu_0} \vec{E} \times \vec{B} $$

So let's work it out for our current element. We get:

$$\begin{align} \vec{P}(\vec{r}, t) &= \left( \frac {I_0\;\delta\ell} {4 \pi} \right)^2 \left[ \frac {\omega^2} {2c^3r^2} + \left( \frac {1} {cr^4} - \frac {\omega^2} {2c^3r^2} \right) \cos\;2\omega t' + \frac {1} {2} \left( \frac {1} {\omega r^5} - \frac {2\omega} {c^2r^3} \right) \sin\;2\omega t' \right] \frac {\sin^2\theta} {\epsilon_0} \hat{r} \\ & \qquad -\; \left( \frac {I_0\;\delta\ell} {4 \pi} \right)^2 \left[ \frac {1} {cr^4} \cos\;2\omega t' + \frac {1} {2} \left( \frac {1} {\omega r^5} - \frac {\omega} {c^2r^3} \right) \sin\;2\omega t' \right] \frac {\sin\;2\theta} {\epsilon_0} \hat{\theta} \\ \end{align}$$

At this point, we shall introduce the well known quantities $\lambda$, the wavelength, and $\eta_0$, the 'impedance of free space'.

$$\lambda = \frac {2\pi c} {\omega}$$ $$\eta_0 = \sqrt{ \frac {\mu_0} {\epsilon_0} } $$

Now, we consider a sphere of radius $r$ centered at the current element. Let us now compute the total power passing through the surface of this sphere at any instant.

$$ \begin{align} P_{total}(r, t) &= \oint \vec{P} (\vec{r}, t) \cdot d\vec{s} \\ &= \frac {2\pi} {3} \eta_0 \left( \frac {I_0\;\delta\ell} {\lambda} \right)^2 \left[ \left( \frac {\lambda} {2\pi r} \right)^3 \frac {\sin\;2\omega t'} {2} + \left( \frac {\lambda} {2\pi r} \right)^2 \cos\;2\omega t' - \left( \frac {\lambda} {2\pi r} \right) \sin\;2\omega t' \right] \\ & \qquad +\; \frac {2\pi} {3} \eta_0 \left( \frac {I_0\;\delta\ell} {\lambda} \right)^2 \sin^2\omega t' \\ &= \frac {2\pi} {3} \eta_0 \left( \frac {I_0\;\delta\ell} {\lambda} \right)^2 \sin^2\omega t' \\ & \qquad +\; \frac {2\pi} {3} \eta_0 \left( \frac {I_0\;\delta\ell} {\lambda} \right)^2 \left[ \frac {\kappa^3-2\kappa} {2} \sin\;2\omega t' + \kappa^2 \cos\;2\omega t'\right] \\ &= \frac {2\pi} {3} \eta_0 \left( \frac {I_0\;\delta\ell} {\lambda} \right)^2 \sin^2\omega t' \\ & \qquad +\; \frac {2\pi} {3} \eta_0 \left( \frac {I_0\;\delta\ell} {\lambda} \right)^2 \frac {\kappa\sqrt{\kappa^4+4}} {2} \sin\left(2\omega t' + \tan^{-1} \frac {2\kappa} {\kappa^2 - 2} \right) \\ \end{align}$$ Where: $$ \kappa=\frac {\lambda} {2\pi r} $$

We now have two expressions that can be sliced and diced in a variety of ways. Consider the following limit:

$$ \lim\limits_{r \to \infty} P_{total} (r,t) = \frac {2\pi} {3} \eta_0 \left( \frac {I_0\;\delta\ell} {\lambda} \right)^2 \sin^2\omega t' $$

This is the total power that, as far away from the current element as we like, can be found to be flowing away from it at any instant. This power is never negative, so it never returns to the current element. In other words, it is the radiation from our current element.

Now, consider the time-averaged power flowing through any point $(r, \theta, \phi)$.

$$ \left< \vec{P}( \vec{r}, t ) \right> = \left( \frac {I_0\;\delta\ell} {4 \pi} \right)^2 \frac {\omega ^2} {2\epsilon_0 c^3 r^2} \sin^2 \theta \; \hat{r} $$

This gives us the radiation pattern of the current element.

Next, consider the time-averaged total power at any distance $r$.

$$ \left< P_{total} (r,t) \right> = \frac {\pi} {3} \eta_0 \left( \frac {I_0\;\delta\ell} {\lambda} \right)^2 $$

We can use this value to determine an equivalent 'radiation resistance' $R_{rad}$ that would have dissipated the same amount of power ohmically.

$$ R_{rad} = \frac {2\pi} {3} \eta_0 \left( \frac {\delta\ell} {\lambda} \right)^2 $$

Now, we'll compute one final limit:

$$ \begin{align} \lim\limits_{r \to 0} P_{total} (r,t) &= \frac {2\pi} {3} \eta_0 \left( \frac {I_0\;\delta\ell} {\lambda} \right)^2 \sin^2\omega t + \frac {2\pi} {3} \eta_0 \left( \frac {I_0\;\delta\ell} {\lambda} \right)^2 \left( \lim\limits_{\kappa \to \infty} \frac {\kappa\sqrt{\kappa^4+4}} {2} \right) \sin \; 2\omega t \\ &= {I_0}^2 R_{rad}\;\cos^2 \left( \omega t - \frac {\pi} {2} \right) + {I_0}^2 \frac {R_{rad}\;\lim\limits_{\kappa \to \infty} \left( \kappa\sqrt{\kappa^4+4} \right) } {2} \sin \; 2\omega t \\ \end{align} $$

What does this limit mean? It is the total power passing through a sphere of zero radius centered at the current element. In other words, it is the instantaneous power passing through a surface just outside the current element.

Let us now consider a current $I_0\;\cos\;\omega t$ flowing through a circuit which has a resistance $R$ and reactance $X$. The instantaneous power drawn by this circuit will be:

$$ \begin{align} P(t) &= V(t)I(t) \\ &= \left[ \sqrt{R^2 + X^2} I_0\;\cos\left( \omega t + \tan^{-1} \frac {X} {R} \right) \right]I_0\;\cos\;\omega t \\ &= {I_0}^2 \sqrt{R^2 + X^2} \cos\;\omega t \left[ \cos\;\omega t\;\cos\left( \tan^{-1} \frac {X} {R} \right) - \sin\;\omega t\;\sin\left( \tan^{-1} \frac {X} {R} \right) \right] \\ &= {I_0}^2 \sqrt{R^2 + X^2} \cos\;\omega t \left[ \cos\;\omega t\frac {R} {\sqrt{R^2 + X^2}} - \sin\;\omega t\frac {X} {\sqrt{R^2 + X^2}} \right] \\ &= {I_0}^2R\;\cos^2\omega t - {I_0}^2X\;\cos\;\omega t\;\sin\;\omega t \\ &= {I_0}^2R\;\cos^2\omega t - {I_0}^2 \frac {X} {2} \sin\;2\omega t \\ \end{align} $$

If we were to model the current element as a circuit component, in addition to a resistive component $R_{rad}$, it would also have a reactive component $X_{self}$ to account for the sinusoidal 'near field' terms (those containing $\lambda / 2\pi r$) in $P_{total}$. This means, we would expect the instantaneous power drawn by the circuit element to be

$$ P(t) = {I_0}^2R_{rad}\;\cos^2\omega t - {I_0}^2 \frac {X_{self}} {2} \sin\;2\omega t $$

Let us compare this with $\lim\limits_{r \to 0} P_{total} (r,t)$. The reactive term maps nicely between the two expressions, telling us $X_{self}=-\infty$ (or, in any practical approximation, a very high capacitive reactance.)

The resistive term, however, turns out to be ${I_0}^2R_{rad}\;\cos^2 \left(\omega t - \pi/2 \right)$ instead of ${I_0}^2R_{rad}\;\cos^2 \omega t$. So we find that, in a current element, the radiated power lags the driving current by a quarter cycle!

Another way of looking at it is, instead of $I(t)$, the power corresponds to $I(t - \pi/2\omega)$. So we have a frequency dependent delay in the radiation. Of course, the delay is not the only thing which is frequency dependent. Let us take another look at the fields, but this time let us look at only the radiation components (i.e. the 'far-field', or the $r \gg \lambda$ approximation)

$$\vec{E}(\vec{r}, t) \approx -\omega\; I\left(t'-\frac {\pi} {2\omega}\right) \;F_{\vec{r}}\left( r, \theta \right) \; \hat{\theta} $$ $$ \vec{B}(\vec{r}, t) \approx -\omega\; I\left(t'-\frac {\pi} {2\omega}\right) \frac {F_{\vec{r}}\left( r, \theta \right)} {c} \hat{\phi} $$ Where: $$ F_{\vec{r}}\left( r, \theta \right) = \frac {\mu_0\;\delta\ell\;\sin\theta} {4\pi r} $$

So we have a delay that is inversely proportional to the frequency, and an amplitude that is directly proportional to the frequency. If the driving current has more than one frequency component, this results in distortion which becomes more pronounced as the bandwidth of the signal increases. This is linear distortion, and can be compensated for.

The following figure shows an example of the distortion. We show $\cos \omega t + \cos 2\omega t$ in red, and $\cos \left(\omega t - \pi/2\right) + 2\cos \left(2\omega t - \pi/2 \right)$ in blue. They are obviously two different waveforms!

Now here is some speculation. Given the law of conservation of energy, the following expression describes the energy that should be somewhere within the current element. It has been drawn from the source driving the current, and is notionally 'in flight' before it bursts out of the surface!

$$ \begin{align} \mathscr{E}(t) &= {I_0}^2\;R_{rad}\;\int_{t-\frac{\pi}{2\omega}}^t \cos^2 \omega t \;dt \\ &= \frac {{I_0}^2\;R_{rad}} {2\omega} \left( \frac {\pi} {2} + \sin 2\omega t \right) \\ &= \left( \frac {\pi} {2} + \sin 2\omega t \right) \frac {\eta_0} {12\pi c^2} \left( {{I_0}\delta\ell} \right)^2\omega \\ \end{align} $$

So is this real? I haven't found it mentioned anywhere. I don't know of any practical implication of this energy, if it does exist at all.

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

Somehow, in the web, I've not come across a derivation of the fields of the infinitesimal current element using Jefimenko's equations. So here is my own attempt.

Consider a piece of wire of zero thickness and infinitesimal length $\delta\ell$, carrying a sinusoidal current $I(t) = I_0 \: \cos\omega t$, such that at any instant, the current flowing through each point on the wire is the same. This is the 'current element', also known as the 'elementary doublet'. It is also sometimes called the 'Hertzian Dipole', after the original antenna used in Heinrich Hertz's classic experiment.

The current element is the simplest radiating system. It is traditionally analyzed using scalar and vector potentials. We'll do it somewhat differently, and determine the electric and magnetic fields caused by the current element using Jefimenko's equations. (This paper covers the same ground, but uses complex arithmetic for the sinusoid, and also expresses the fields in terms of dipole moments instead of electric current.)

Jefimenko's Equations

$$\vec{E}(\vec{r},t) = \frac {1} {4 \pi \epsilon_0} \iiint {\left( \frac {\rho (\vec{r}_s, t_r)} {R^3} \vec{R} + \frac {1} {R^2 c} \frac {\partial \rho (\vec{r}_s, t_r) } {\partial t} \vec{R} - \frac {1} {R c^2} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \right)} d^3 \vec{r}_s $$ $$\vec{B}(\vec{r},t) = \frac {\mu_0} {4 \pi} \iiint {\left( \frac {\vec{J} (\vec{r}_s, t_r)} {R^3} \times \vec{R} + \frac {1} {R^2 c} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \times \vec{R} \right)} d^3 \vec{r}_s $$ Where: $$\vec{R} = \vec{r} - \vec{r}_s, \qquad R = \left | \vec{R} \right |, \qquad t_r = t - \frac {R} {c}$$

Let us begin with some geometry. In the following diagram, the segment $BA$ is our current element. We define a spherical co-ordinate system, and place the current element such that the current direction during the positive half cycle is aligned with the $\theta=0$ direction, and the mid-point of $BA$ is at the origin $O$. We define a unit vector $\hat{\ell}$ in the $\theta = 0$ direction. We want to determine the electric and magnetic fields at point $P$ which is at a distance $r$ from the origin.

Applying the 'cosine rule' to $\triangle APO$ and $\triangle BPO$, and noting that $\cos\;(\pi-\theta)=-\cos\;\theta$, we get:
$${r_A}^2 = r^2 + \left( \frac {\delta\ell} {2} \right)^2 - r \; \delta \ell \; \cos\:\theta $$ $${r_B}^2 = r^2 + \left( \frac {\delta\ell} {2} \right)^2 + r \; \delta \ell \; \cos\:\theta $$

Applying the 'sine rule', and noting that $\sin\;(\pi-\theta)=\sin\;\theta$, we get:

$$\sin\;\psi_A = \left( \frac {\delta\ell} {2} \right) \frac {\sin\;\theta} {r_A} $$ $$\sin\;\psi_B = \left( \frac {\delta\ell} {2} \right) \frac {\sin\;\theta} {r_B} $$

Applying the identity $\cos^2 x + \sin^2 x = 1$, we get:

$$\cos\;\psi_A = \frac {r} {r_A} - \left( \frac {\delta\ell} {2} \right) \frac {\cos\;\theta} {r_A} $$ $$\cos\;\psi_B = \frac {r} {r_B} + \left( \frac {\delta\ell} {2} \right) \frac {\cos\;\theta} {r_B} $$

Let us now express the unit vectors $\hat{\ell}$, $\hat{r}_A$ and $\hat{r}_B$ in terms of $\hat{r}$ and $\hat{\theta}$ at $P$.

$$\hat{\ell} = \cos\;\theta\;\hat{r} - \sin\;\theta\;\hat{\theta}$$ $$\hat{r}_A = \cos\;\psi_A\;\hat{r} + \sin\;\psi_A\;\hat{\theta} = \frac {r} {r_A} \hat{r} - \frac {\delta\ell} {2} \: \frac {\cos\;\theta} {r_A} \hat{r} + \frac {\delta\ell} {2} \: \frac {\sin\;\theta} {r_A} \hat{\theta} $$ $$\hat{r}_B = \cos\;\psi_B\;\hat{r} - \sin\;\psi_B\;\hat{\theta} = \frac {r} {r_A} \hat{r} + \frac {\delta\ell} {2} \: \frac {\cos\;\theta} {r_B} \hat{r} - \frac {\delta\ell} {2} \: \frac {\sin\;\theta} {r_B} \hat{\theta}$$

So far, all the expressions have been exact: no approximations have been used. From this point onwards, however, we will use approximations which, because they involve terms containing the infinitesimal $\delta\ell$, are asymptotically exact. The basic approximation is, for any $\varepsilon$ such that $| \varepsilon | \ll 1$, as long as at least one of $a_0$ and $a_1$ is non-zero,

$$\sum_{i=0}^{n} a_i \varepsilon^i \approx a_0 + a_1 \varepsilon$$

Using this as the basis, and using the Taylor Series or the Binomial Theorem as appropriate, we get the following approximations:

$$\sqrt{1 + \varepsilon} \approx 1 + \frac {\varepsilon} {2}$$ $$\sin\;\varepsilon \approx \varepsilon$$ $$\cos\;\varepsilon \approx 1$$ $$(1 - \varepsilon)^n + (1 + \varepsilon)^n \approx 2$$ $$(1 - \varepsilon)^n (1 + \varepsilon)^n \approx 1$$

Let us begin by obtaining approximate values for $r_A$ and $r_B$. We ignore the $\delta\ell^2$ terms, and use $\sqrt{1 + \varepsilon} \approx 1 + \frac {\varepsilon} {2}$.

$$r_A \approx r \left( 1 - \frac {\delta\ell} {2r} \cos\;\theta\right)$$ $$r_B \approx r \left( 1 + \frac {\delta\ell} {2r} \cos\;\theta\right)$$

Using $(1 - \varepsilon)^n + (1 + \varepsilon)^n \approx 2$, we get:

$$\frac {1} {{r_A}^2} + \frac {1} {{r_B}^2} \approx \frac {2} {r^2}$$ $$\frac {1} {{r_A}^3} + \frac {1} {{r_B}^3} \approx \frac {2} {r^3}$$

Using $(1 - \varepsilon)^n (1 + \varepsilon)^n \approx 1$, and by ignoring $\delta\ell^3$ terms when they appear, we get:

$$\frac {1} {{r_A}^2} - \frac {1} {{r_B}^2} \approx \frac {2\;\delta\ell\;\cos\:\theta} {r^3}$$ $$\frac {1} {{r_A}^3} - \frac {1} {{r_B}^3} \approx \frac {3\;\delta\ell\;\cos\:\theta} {r^4}$$

The groundwork is ready.

Let us now describe the source. We can see that it consists of three distinct parts:

1. The segment $BA$, carrying the time-varying current $I_0\:\cos\;\omega t$
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 $-I_0 \int_0^t \cos\;\omega t \;dt = (-I_0 / \omega)\sin\;\omega t$

The charges on $A$ and $B$ are a consequence of the continuity equation $\oint\vec{J}\cdot d\vec{s}=-\frac{\partial}{\partial t}\iiint \rho\;d\mathbb{v}$

To apply Jefimenko's equations, the charge and current density expressions must first be expressed using 'retarded time' $t_r$ as seen from $P$. Let us define a new variable:

$$t' = t - \frac {r} {c} $$

We'll need the trigonometric identities $\sin\;(x+y)= \sin\;x\;\cos\;y + \cos\;x\;\sin\;y$ and $\cos\;(x+y)= \cos\;x\;\cos\;y - \sin\;x\;\sin\;y$. We'll also need also need the following approximations that are based on $\sin\;\varepsilon \approx \varepsilon$ and $\cos\;\varepsilon \approx 1$.

$$\sin \frac {\omega\;\delta\ell\;\cos\;\theta} {2c} \approx \frac {\omega\;\delta\ell} {2c} \cos\;\theta$$ $$\cos \frac {\omega\;\delta\ell\;\cos\;\theta} {2c} \approx 1$$

For source $BA$, we have:

$$t_r = t - \frac {r} {c} = t'$$ $$\iiint\vec{J}(\vec{r}_s, t_r)\:d^3\vec{r}_s = I_0\:\delta\ell\:\cos\;\omega t'\;\hat{\ell}$$ $$\iiint \frac {\partial\vec{J}(\vec{r}_s, t_r)} {\partial t} d^3\vec{r}_s = -\omega I_0\:\delta\ell\:\sin\;\omega t'\;\hat{\ell}$$ $$ \iiint\rho(\vec{r}_s, t_r)\:d^3\vec{r}_s = \iiint \frac {\partial\rho(\vec{r}_s, t_r)} {\partial t} d^3\vec{r}_s = 0 $$

For source $A$, we have:

$$t_r = t - \frac {r_A} {c} = t' + \frac {\delta\ell} {2c} \cos\;\theta$$ $$\iiint\rho(\vec{r}_s, t_r)\:d^3\vec{r}_s \approx \frac {I_0} {\omega} \left( \sin\;\omega t' + \frac {\omega\:\delta\ell\:\cos\;\theta} {2c} \cos\;\omega t'\right)$$ $$\iiint \frac {\partial\rho(\vec{r}_s, t_r)} {\partial t} d^3\vec{r}_s \approx I_0 \left( \cos\;\omega t' - \frac {\omega\:\delta\ell\:\cos\;\theta} {2c} \sin\;\omega t'\right)$$ $$ \iiint\vec{J}(\vec{r}_s, t_r)\:d^3\vec{r}_s = \iiint \frac {\partial\vec{J}(\vec{r}_s, t_r)} {\partial t} d^3\vec{r}_s = 0 $$

For source $B$, we have:

$$t_r = t - \frac {r_A} {c} = t' - \frac {\delta\ell} {2c} \cos\;\theta$$ $$\iiint\rho(\vec{r}_s, t_r)\:d^3\vec{r}_s \approx -\frac {I_0} {\omega} \left( \sin\;\omega t' - \frac {\omega\:\delta\ell\:\cos\;\theta} {2c} \cos\;\omega t'\right)$$ $$\iiint \frac {\partial\rho(\vec{r}_s, t_r)} {\partial t} d^3\vec{r}_s \approx -I_0 \left( \cos\;\omega t' + \frac {\omega\:\delta\ell\:\cos\;\theta} {2c} \sin\;\omega t'\right)$$ $$ \iiint\vec{J}(\vec{r}_s, t_r)\:d^3\vec{r}_s = \iiint \frac {\partial\vec{J}(\vec{r}_s, t_r)} {\partial t} d^3\vec{r}_s = 0 $$

The fields for the entire current element will be the sum of the fields caused by its constituents:

Thus, $$\vec{E}(\vec{r}, t) = \vec{E}_{BA}(\vec{r}, t) + \vec{E}_A(\vec{r}, t) + \vec{E}_B(\vec{r}, t) $$ Where: $$ \vec{E}_{BA}(\vec{r}, t) = \frac {1} {4\pi\epsilon_0} \left[ \frac {\omega I_0\:\delta\ell\:\sin\;\omega t'} {r c^2} \hat{\ell} \right] $$ $$ \vec{E}_{A}(\vec{r}, t) = \frac {1} {4\pi\epsilon_0} \left[ \frac {I_0} {\omega{r_A}^2} \left( \sin\;\omega t' + \frac {\omega\:\delta\ell\:\cos\;\theta} {2c} \cos\;\omega t'\right)\hat{r}_A + \frac {I_0} {{r_A}c} \left( \cos\;\omega t' - \frac {\omega\:\delta\ell\:\cos\;\theta} {2c} \sin\;\omega t'\right)\hat{r}_A \right] $$ $$ \vec{E}_{B}(\vec{r}, t) = - \frac {1} {4\pi\epsilon_0} \left[ \frac {I_0} {\omega{r_B}^2} \left( \sin\;\omega t' - \frac {\omega\:\delta\ell\:\cos\;\theta} {2c} \cos\;\omega t'\right)\hat{r}_B + \frac {I_0} {{r_B}c} \left( \cos\;\omega t' + \frac {\omega\:\delta\ell\:\cos\;\theta} {2c} \sin\;\omega t'\right)\hat{r}_B \right] $$ And, $$\vec{B}(\vec{r}, t) = \vec{B}_{BA}(\vec{r}, t) + \vec{B}_A(\vec{r}, t) + \vec{B}_B(\vec{r}, t) $$ Where: $$ \vec{B}_{BA}(\vec{r}, t) = \frac {\mu_0} {4\pi} \left[ \frac {I_0\:\delta\ell\:\cos\;\omega t'} {r^2} \left( \hat{\ell} \times \hat{r} \right) - \frac {\omega I_0\:\delta\ell\:\sin\;\omega t'} {r c} \left( \hat{\ell}\times \hat{r} \right) \right] $$ $$ \vec{B}_{A}(\vec{r}, t) = 0 $$ $$ \vec{B}_{B}(\vec{r}, t) = 0 $$

We substitute unit vectors $\hat{\ell}$, $\hat{r}_A$ and $\hat{r}_B$ into the expression for $\vec{E}$, and expand. We face an expression with 26 terms, but never fear! Ignore all terms where $\delta\ell^2$ appears: this shrinks the expression down to 18 terms. Now reduce using the approximations for $ {1} / {{r_A}^n} \pm {1} / {{r_B}^n}$ derived earlier: this will bring it down further to 10 terms. Five of these cancel out, finally leaving us with:

$$\vec{E}(\vec{r}, t) = \frac {2} {\epsilon_0} \left[ \frac {\sin\;\omega t'} {\omega r^3} +\frac {\cos\;\omega t'} {c r^2} \right] \frac {I_0\;\delta\ell\;\cos\:\theta} {4\pi} \hat{r} +\frac {1} {\epsilon_0} \left[ \frac {\sin\;\omega t'} {\omega r^3} +\frac {\cos\;\omega t'} {c r^2} -\frac {\omega\;\sin\;\omega t'} {c^2 r} \right] \frac {I_0\;\delta\ell\;\sin\:\theta} {4\pi} \hat{\theta} $$

Then, we substitute into the expression for $\vec{B}$. This time, it is much easier: we note that $\hat{\ell} \times \hat{r} = \cos\;\theta (\hat{r}\times\hat{r})-\sin\;\theta (\hat{\theta}\times\hat{r})=\sin\;\theta\;\hat{\phi}$, giving us:

$$\vec{B}(\vec{r}, t) = \mu_0 \left[ \frac {\cos\;\omega t'} {r^2} -\frac {\omega\;\sin\;\omega t'} {c r} \right] \frac {I_0\;\delta\ell\;\sin\;\theta} {4\pi} \hat{\phi} $$

These are the same results that are obtained by the traditional approach.