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.

Maxwell, Jefimenko & Heras?

Recently, I came across a paper by José A. Heras that was published in the American Journal of Physics. In this paper, there was the the statement and proof of the following theorem:

Theorem

If, confined to a finite volume of space, there exists a scalar field $\rho(\vec{r}, t)$ and a vector field $\vec{J}(\vec{r}, t)$ related to each other as follows: $$ \nabla \cdot \vec{J} (\vec{r}, t) = - \frac {\partial \rho (\vec{r}, t)} {\partial t} $$ And if, for any chosen speed $v$, and for any chosen values of constants $\alpha$ and $\beta$, a pair of fields $\vec{E}(\vec{r},t)$ and $\vec{B}(\vec{r},t)$ are defined for all points in space as follows: $$\vec{E}(\vec{r},t) = \frac {\alpha} {4 \pi} \iiint {\left( \frac {\rho (\vec{r}_s, t_r)} {R^3} \vec{R} + \frac {1} {R^2 v} \frac {\partial \rho (\vec{r}_s, t_r) } {\partial t} \vec{R} - \frac {1} {R v^2} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \right)} \;d^3 \vec{r}_s $$ $$\vec{B}(\vec{r},t) = \frac {\beta} {4 \pi} \iiint {\left( \frac {\vec{J} (\vec{r}_s, t_r)} {R^3} \times \vec{R} + \frac {1} {R^2 v} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \times \vec{R} \right)} \;d^3 \vec{r}_s $$ Where: $$\vec{R} = \vec{r} - \vec{r}_s, \qquad R = \left | \vec{R} \right |, \qquad t_r = t - \frac {R} {v}$$ Then, at any point in space, the following relations are satisfied: $$ \nabla \cdot \vec{E} (\vec{r}, t) = \alpha \rho (\vec{r}, t) $$ $$ \nabla \cdot \vec{B} (\vec{r}, t) = 0 $$ $$ \nabla \times \vec{E} (\vec{r}, t) + \frac {1} {v^2} \left( \frac {\alpha} {\beta} \right) \frac {\partial \vec{B} ( \vec{r}, t)} {\partial t} = 0 $$ $$ \nabla \times \vec{B} (\vec{r}, t) - \left( \frac {\beta} {\alpha} \right) \frac {\partial \vec{E} ( \vec{r}, t)} {\partial t} = \beta \vec{J} (\vec{r}, t) $$

The theorem makes a mathematical statement, not a physical one. The only physical quantities in it are the position $\vec{r}$ (in Euclidean space $\mathbb{R}^3$) and time $t$. Except for $v$, which can be considered the speed at which the effects (on $\vec{E}$ and $\vec{B}$) due to any change in the sources ($\rho$ and $\vec{J}$) travel to the point of observation, we cannot deduce the physical interpretation of any of the other terms. We don't know the value of $v$, or what $\rho$ and $\vec{J}$ actually are, or what $\vec{E}$, $\vec{B}$, $\alpha$ and $\beta$ physically mean. What we have here is an 'existence theorem'.

But there is a reason why we have chosen use the names $\rho$, $\vec{J}$, $\vec{E}$ and $\vec{B}$ in the statement of the theorem. And there is a reason why a physicist or mathematician would take the trouble to twist one simple equation into six complicated ones.

Before we step out into the physical world, let us consider what happens to the theorem when $\rho$ and $\vec{J}$ are, in fact, time-invariant. In that case, for any interval $\delta$ we have:

$$ \rho(\vec{r},t-\delta) = \rho(\vec{r}, t)$$ $$ \vec{J}(\vec{r},t-\delta) = \vec{J}(\vec{r}, t)$$

This immediately implies that, for all values of $\vec{r}$ and $t$,

$$ \frac {\partial\rho(\vec{r}, t)} {\partial t} = 0$$ $$ \frac {\partial\vec{J}(\vec{r}, t)} {\partial t} = 0$$

In such a case, the time $t$ is literally 'out of the equation', decoupling $\rho$ and $\vec{J}$ from each other. As they are no longer functions of time, we can call them $\rho(\vec{r})$ and $\vec{J}(\vec{r})$ instead. Our theorem now breaks up into two parts, each of which can be easily proved even without using this theorem:

Corollary 1

If, confined to a finite volume of space, there exists a time-invariant scalar field $\rho(\vec{r})$, and if, for any chosen value of the constant $\alpha$, a time-invariant field $\vec{E}(\vec{r})$ is defined for all points in space as follows: $$ \vec{E}(\vec{r}) = \frac {\alpha} {4 \pi} \iiint { \left( \frac {\rho (\vec{r}_s)} {R^3} \vec{R} \right) } \;d^3 \vec{r}_s $$ Then, at any point in space, the following relations are satisfied: $$ \nabla \cdot \vec{E} (\vec{r}) = \alpha \rho (\vec{r}) $$ $$ \nabla \times \vec{E} (\vec{r}) = 0 $$

and

Corollary 2

If, confined to a finite volume of space, there exists a time-invariant vector field $\vec{J}(\vec{r})$, and if, for any chosen value of the constant $\beta$, a time-invariant field $\vec{B}(\vec{r})$ is defined for all points in space as follows: $$ \vec{B}(\vec{r}) = \frac {\beta} {4 \pi} \iiint {\left( \frac {\vec{J} (\vec{r}_s)} {R^3} \times \vec{R} \right)} \;d^3 \vec{r}_s $$ Then, at any point in space, the following relations are satisfied: $$ \nabla \cdot \vec{B} (\vec{r}) = 0 $$ $$ \nabla \times \vec{B} (\vec{r}) = \beta \vec{J} (\vec{r}) $$

Let us now indulge in some purely imaginary history. None of this actually happened, and possibly couldn't have happened. We're just having fun here.

Let us pretend we are a physicist in the early part of 19th century, and let us imagine that a mysterious piece of paper with the theorem and the corollaries above written on it has landed on our desk. Never mind just how—a crack in space-time, maybe?

In the beginning, we don't pay much attention to it. We are, at the moment, engrossed in our study of electricity.

The first thing we are investigating is electric charge, as may be produced by an electrophorus. We call it $q$. We know that two electrically charged bodies attract or repel each other, and we are trying to analyze it. We have figured out that the force between the two is:

$$ \vec{F}_q \propto \frac {q_1 q_2} {\left| \vec{r}_2 - \vec{r}_1 \right| ^3} \left( \vec{r}_2 - \vec{r}_1 \right) $$

These forces can be modeled by the interaction between the charge with a 'field' that exists in space, where the 'field' is produced by the other charge.

For the force between electric charges, we say that an 'electric field' $\vec{E}$ is involved, and say that the constant of proportionality is $\alpha/4\pi$: $$\vec{E}(\vec{r}) = \frac {\alpha} {4 \pi} \left( \frac {q} { \left| \vec{r} \right|^3} \right) \vec{r}$$ $$\vec{F}_q(\vec{r}) = q\;\vec{E}(\vec{r})$$

The other thing we are looking at is electric current, as may be produced by a Voltaic cell. We call it $I$. We have found that two wires carrying current will also attract or repel one another, and we are looking into that too.

$$ \vec{F}_I \propto \int_{\ell_1} \int_{\ell_2} \frac {I_1 \; I_2} {\left| \vec{r}_2 - \vec{r}_1 \right| ^3} d\vec{\ell}_1 \times \left[ d\vec{\ell}_2 \times \left( \vec{r}_2 - \vec{r}_1 \right) \right] $$

Here, too, we can model the phenomenon as an interaction between the current in one wire and a 'field' that is produced by the current of the other wire. We call the field a 'magnetic field' $\vec{B}$, and say that the constant of proportionality is $\beta/4\pi$: $$\vec{B}(\vec{r}) = \frac {\beta} {4 \pi} \int_{\ell}\; \frac {I} { \left| \vec{r} - \vec{r}_\ell \right|^3} d\vec{\ell} \times \left( \vec{r} - \vec{r}_\ell \right)$$ $$\vec{F}_I(\vec{r}) = \int_{\ell}\; I\;d\vec{\ell} \times \vec{B}(\vec{r}) $$

We can use $q$ and $I$ for working with idealized 'point charges' and currents that move in wires of zero thickness, but to describe real-life structures, we need to generalize these two quantities to the charge density $\rho$ over a volume and current density $\vec{J}$ over a surface. $$ q = \iiint \rho(\vec{r}) \; d\mathbb{v}$$ $$ I = \iint \vec{J}(\vec{r}) \cdot d\vec{s}$$

Now let's look what the electric and magnetic fields look like for time invariant $\rho$ and $\vec{J}$. Integrating over the volume of the source, we get: $$\vec{E}(\vec{r}) = \frac {\alpha} {4 \pi} \iiint \frac {\rho(\vec{r}_s)} { \left| \vec{r} - \vec{r}_s \right|^3} (\vec{r} - \vec{r}_s) \; d^3\vec{r}_s$$ $$\vec{B}(\vec{r}) = \frac {\beta} {4 \pi} \iiint \frac {\vec{J}(\vec{r}_s)} { \left| \vec{r} - \vec{r}_s \right|^3} \times (\vec{r} - \vec{r}_s) \; d^3\vec{r}_s$$

We suddenly remember that mysterious scrap of paper on our desk, and realize that we have a pair $(\rho,\;\vec{E})$ that matches corollary 1, and a pair $(\vec{J},\;\vec{B})$ that matches corollary 2 of the theorem. We can directly conclude, therefore, that:

$$ \nabla \cdot \vec{E} (\vec{r}) = \alpha \rho (\vec{r}) $$ $$ \nabla \times \vec{E} (\vec{r}) = 0 $$ $$ \nabla \cdot \vec{B} (\vec{r}) = 0 $$ $$ \nabla \times \vec{B} (\vec{r}) = \beta \vec{J} (\vec{r}) $$

So far we have been looking at charge and current independently, without suspecting any relation between the two. But one day, we discover that they are not independent at all: an electric current is a flow of electric charge. And what's more, we also discover that for any volume in space, at any instant, the current flowing through the surface bounding that volume equals the rate of change of charge contained within that volume:

$$ \oint \vec{J} (\vec{r}, t) \cdot d\vec{s} = - \frac {\partial} {\partial t} \iiint \rho(\vec{r}, t) \; d\mathbb{v}$$

This can be expressed in an exactly equivalent 'differential form' as follows: $$ \nabla \cdot \vec{J} (\vec{r}, t) = - \frac {\partial \rho (\vec{r}, t)} {\partial t}$$

That scrap of paper again! Are we on to something? Now, time $t$ has been introduced into the mix, and we start thinking about electric and magnetic fields produced time-varying charges and currents.

But we don't get too excited. This is, after all, the early 19th century, and the world-view of physics is primarily Galilean. There is no reason whatsoever to expect $\vec{E}(\vec{r},t)$ and $\vec{B}(\vec{r},t)$ to depend on time-varying $\rho(\vec{r},t)$ and $\vec{J}(\vec{r},t)$ in a manner any different from the way $\vec{E}(\vec{r})$ and $\vec{B}(\vec{r})$ depend on time-invariant $\rho(\vec{r})$ and $\vec{J}(\vec{r})$.

For all that our human senses and instruments can tell us, this does appear to be the case. When we perturb $\rho$ and measure $\vec{E}$ across the laboratory, the effect seems to appear immediately. We get the same result if we perturb $\vec{J}$ and measure $\vec{B}$. Apparently, the theorem on the scrap of paper is indeed satisfied, but with the trivial value of $v = \infty$.

But there is another possibility: maybe $v$ is actually finite, but very large: so large that the time interval between the cause and the effect is too small for our instruments to measure. How can we tell?

Fortunately, the theorem on the scrap of paper helps us test whether $v = \infty$ without having to measure very small time intervals.

Consider the expression for $\vec{E}(\vec{r},t)$. We can see that if $v = \infty$, it depends only on $\rho$. If $v \ne \infty$, however, $\vec{E}(\vec{r},t)$ depends not only on $\rho$, but also on ${\partial \vec{J}}/{\partial t}$.

So let us devise an experiment.

Experiment

Let's have a thin conductor of length $\ell$. Let's take a very small and light pith ball, charge it to $+q$, and suspend it using a silk thread just above the midpoint of the wire, at a distance of $d$ from it. Let us now enclose the entire setup in a glass bubble, with only the two ends of the conductor sticking out.

Now, we connect the two ends of the conductor, through a switch (which is initially open), to a battery bank. The steady state current $I$ will depend on the battery voltage, the battery internal resistance, and the resistance of the wire. We ensure that the two wires connecting the ends of our conductor to the battery bank and switch are:

• Straight
• Parallel to each other
• Perpendicular to the conductor
• Of a length $D \gg \ell$

When it is all set up, we evacuate the glass bubble, and wait for everything to settle down.

The positive charge on the pith ball would have induced a small negative charge on the conductor, so there would be a small electric field, pointing directly downwards, where the ball is. This would lightly pull the ball down. Other than gravitational attraction, this is the only other force on the ball in a steady state.

Now, we close the switch.

This causes the current in the conductor to rapidly rise from zero to $I$, which should mean a large ${dI}/{dt}$.

If $\vec{E}$ and $\vec{B}$ are actually as stated in the theorem, this would produce a component of $\vec{E}$ parallel to the conductor:

$$ -\frac {\alpha} {4 \pi} \iiint {\left( \frac {1} {R v^2} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \right)} \;d^3 \vec{r}_s = - \frac {\alpha} {4 \pi v^2} \left[ \int_{-\ell/2}^{+\ell/2}\;\frac{1}{\sqrt{d^2+z^2}}\;\frac {d} {dt} I\left( t - \frac {\sqrt{d^2+z^2}} {v} \right) \;dz \right] \hat{z} $$

Where:

$$ k=\cfrac {\ell} {2d} $$

If our Galilean worldview is correct, and the theorem applies with $v = \infty$, this component of $\vec{E}$ will be zero. As $\vec{B}$ and ${\partial \vec{B}}/{\partial t}$ will not affect a static charged body, we would expect to see no effect on the pith ball.

If the theorem applies with $v \ne \infty$, however, this component of $\vec{E}$ will not be zero, and the pith ball would feel a 'kick' parallel to the conductor. The resultant motion can be observed. When the switch is subsequently opened, a kick in the opposite direction should also be seen.

So we do the experiment, and there is a kick when the switch is opened or closed! If the theorem does apply, this rules out $v = \infty$.

What is the value of $v$ then?

At this point, we can stop and do a little dimensional analysis. As a basis, we will use charge $Q$, force $F$, length $L$ and time $T$. We get the following for the variables:

$I:$	$ \frac {Q} {T} $
$\rho:$	$ \frac {Q} {L^3} $
$\vec{J}:$	$ \frac {Q} {TL^2} $
$\vec{E}:$	$ \frac {F} {Q} $
$\vec{B}:$	$ \frac {FT} {LQ} $

And the following for the constants:

$v:$	$ \frac {L} {T} $
$\alpha:$	$ \frac {FL^2} {Q^2} $
$\beta:$	$ \frac {FT^2} {Q^2} $

Interestingly, this tells us that $\cfrac {\alpha} {\beta}$ has the same dimensions as $v^2$. That is, $\sqrt{\cfrac{\alpha}{\beta}}$ has the dimensions of speed.

At this point, we'll indulge in a 'thought experiment'.

Thought Experiment

Consider a pair of infinitely long conductors (shown in black in the figure above) that are parallel to each other separated by a distance of $d$. Each of these conductors has a circular cross section of infinitesimal area $\delta a$. Let us assume that they have equal but opposite electric charge densities $+\rho$ and $-\rho$ respectively.

We will define a quantity $\lambda$, the 'linear charge density', which is: $$ \lambda = \rho\;\delta a $$ So the linear charge densities on the two conductors are $+\lambda$ and $-\lambda$.

Now since the conductors are infinite in length, the total charge on each is infinite, and so the attractive force between them is infinite too. However, it can be shown that the force per unit length isn't. So let us devise a contraption to measure this 'force per unit length'. We will assume that friction, wherever it may be expected, is zero.

In the diagram above, every unit length, we have a wheel (in orange) rolling over each conductor. The axle of each wheel is connected to one end of a bar (shown in blue). The bar has a vertical slot cut out of it. The other end of the bar is attached to an instrument that can measure the compressive force between its two sides. The wheel and bar are both perfect insulators.

Behind the bar, we have a rigid rod 'ladder' (in green), which has two studs sticking through each slot that is cut into the bars. This means that the bars and the ladder can move freely relative to each other in the up/down direction, but not in the left/right direction. In other words, if the ladder moves left or right, the bars (and the meters) move with it. There are two studs per bar to keep each bar perpendicular to the conductors.

So what does each meter read? Working out the integral, we find that the reading should be:

$$ \frac {\alpha} {2\pi d} \left( \lambda \right)^2 $$

Now, let us consider the situation where the conductors are moving at a speed of $v$ relative to the ladder and the measuring devices. What is the force per unit length now?

From the point or view of the measuring device, we now not only have two equal and opposite charge densities, but also two equal and opposite currents $+v\lambda$ and $-v\lambda$. This would translate to a magnetic force of repulsion, the value of which can be found by working out another integral. The net force of attraction per unit length would therefore be:

$$ \frac {\alpha} {2\pi d} \left( \lambda \right)^2 - \frac {\beta} {2\pi d} \left( v\lambda \right)^2 $$

Now let us imagine that the conductors are entirely uniform: in appearance and otherwise. Since we also have zero friction, it is impossible for an observer stationary relative to the meters to determine the relative speed $v$. Suppose such an observer reads a force $F$ at each meter. All he knows is that the conductors are at rest relative to each other, and carry equal but opposite linear charge densities. What can he conclude about $v$ and $\lambda$?

We can see that any force $F$ can be caused by an infinite number of combinations of $v$ and $\lambda$. Let the $\lambda$ corresponding to $v = 0$ be designated $\lambda_0$, and the $\lambda$ corresponding to an arbitrary $v$ be designated $\lambda_v$. Then,

$$ \begin{align} &\frac {\alpha} {2\pi d} \left( \lambda_0 \right)^2 = \frac {\alpha} {2\pi d} \left( \lambda_v \right)^2 - \frac {\beta} {2\pi d} \left( v\lambda_v \right)^2 \\ \therefore \qquad &{\lambda_0} = {\lambda_v}\; \sqrt{1 - \cfrac {v^2}{c^2}} \\ \end{align} $$ Where: $$ c = \sqrt{ \cfrac {\alpha} {\beta} } $$

Interestingly, we find that when $\left| v \right| \to c$, $\lambda_v \to \infty$, and for $\left| v \right| \gt c$, $\lambda_v$ is imaginary! Since neither is physically possible, this kind of establishes $c$ as a universal speed limit that cannot be exceeded, and can only be reached asymptotically by any real observer.

Plugging this finding back into our equations, we have:

$$ \nabla \times \vec{E} (\vec{r}, t) = - \frac {c^2} {v^2} \frac {\partial \vec{B} ( \vec{r}, t)} {\partial t}, \qquad\qquad 0 \lt v \le c $$

If $\vec{E}$ and $\vec{B}$ are actually related to time varying $\rho$ and $\vec{J}$ as given in the statement of the theorem, $v \ne \infty$ implies $\nabla \times \vec{E} \propto -{\partial \vec{B}} / {\partial t}$.

We now conduct rigorous laboratory work, measuring $\nabla \times \vec{E}$ and ${\partial \vec{B}} / {\partial t}$ everywhere. We find that, as far as we can tell:

$$\nabla \times \vec{E} (\vec{r}, t) = - \frac {\partial \vec{B}(\vec{r}, t)} {\partial t}$$

Thus,

$$ v = c = \sqrt{ \cfrac {\alpha} {\beta} } $$

So while $v$ isn't quite infinite, it is as large as it can be in this universe.

(The author of the theorem above, José A. Heras, has another paper published in 2010 in the European Journal of Physics that could be of interest)

In today's convention, we say:

$$ \beta = \mu_0 $$ $$ \alpha= \frac {1} {\epsilon_0} $$ $$ v = c $$

In SI Units,

$$\mu_0 \equiv 4\pi \times 10^{-7}\;H/m\approx 1.2566 \times 10^{-6} \: H/m$$ $$\epsilon_0 \approx 8.8542 \times 10^{-12} \: F/m$$ $$c \approx 2.9979 \times 10^8 \: m/s$$

We have:

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: $$\vec{R} = \vec{r} - \vec{r}_s, \qquad R = \left | \vec{R} \right |, \qquad t_r = t - \frac {R} {c}$$

Such that:

Maxwell's Equations

$$ \nabla \cdot \vec{E} (\vec{r}, t) = \frac {\rho (\vec{r}, t)} {\epsilon_0} $$ $$ \nabla \cdot \vec{B} (\vec{r}, t) = 0 $$ $$ \nabla \times \vec{E} (\vec{r}, t) = - \frac {\partial \vec{B} ( \vec{r}, t)} {\partial t} $$ $$ \nabla \times \vec{B} (\vec{r}, t) = \mu_0 \vec{J} (\vec{r}, t) + \mu_0 \epsilon_0 \frac {\partial \vec{E} ( \vec{r}, t)} {\partial t}$$