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.