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.