A half-wave dipole isn't resonant, or is it?

"A half wave dipole is resonant."
"No, it isn't: it actually has an impedance of 73.1 + j42.5 ohms."
"To make it resonant, reduce its length by a little bit."

I decided to figure it out for myself.

Let us consider a center-fed dipole antenna of length $\ell$, where the radius of the conductor is $a$. Let us further assume that the resistivity of the antenna conductor is zero, and that it has an infinitesimal feed gap.

To express the results ahead, we will need the following trigonometric integrals: $$ \renewcommand{\Cin}{\:\mathrm{Cin}\;} \renewcommand{\Si}{\:\mathrm{Si}\;} \renewcommand{\Ci}{\:\mathrm{Ci}\;} \Si {x} = \int_0^x\; \frac{\sin t} {t} dt \qquad\qquad \Cin {x} = \int_0^x\; \frac{1-\cos t} {t} dt \qquad\qquad \Ci x = -\int_x^\infty \frac{\cos t}{t} dt = \gamma + \ln x - \Cin x $$ ($\gamma \approx 0.5772...$ is the 'Euler-Mascheroni constant'.)

To express the dimensions in terms of the wavelength $\lambda$, we define:

$$ \chi = \frac{2\pi\ell}{\lambda} \qquad\qquad \xi = \frac{4\pi a}{\lambda} $$

For the conditions $a \ll \lambda$, $a \ll \ell$ and $\frac{\lambda}{2}-\varepsilon \lt \ell \lt \frac{\lambda}{2}+\varepsilon$ (for small $\varepsilon$), the current distribution on the dipole is very close to sinusoidal. Using this assumption, the radiation resistance $R_{rad}$ and self-reactance $X_{self}$ of this antenna can be found to be as follows:

$$ R_{rad}\left(\chi\right) \approx \cfrac {\eta_0} {2\pi\;\sin^2\cfrac{\chi}{2}} \left[ \Cin\chi + \left( \Cin\chi- \frac{\Cin 2\chi}{2} \right)\cos\chi - \left( \Si\chi- \frac{\Si 2\chi}{2} \right)\sin\chi \right] $$ $$ \begin{align} X_{self}\left(\chi,\xi\right) &\approx \cfrac {\eta_0} {2\pi\;\sin^2\cfrac{\chi}{2}} \left[ \Si\chi + \left( \Si\chi- \frac{\Si 2\chi}{2} \right)\cos\chi + \left( \Cin\chi-\frac{\Cin 2\chi}{2} - \ln\chi + \ln\xi \right)\sin\chi \right] \\ &= \cfrac {\eta_0} {2\pi\;\sin^2\cfrac{\chi}{2}} \left[ \Si\chi + \left( \Si\chi-\frac{\Si 2\chi}{2} \right)\cos\chi - \left( \Ci\chi - \frac{\Ci 2\chi}{2} + \ln\sqrt {\chi} - \ln {\xi} \sqrt {\frac {e^\gamma}{2}} \right) \;\sin\chi \right] \\ \end{align} $$

(The derivation for this can be found in section 22.3 in this book.)

For the half wave dipole, $\ell=\lambda/2$, so $\chi=\pi$. Hence, for all $\xi \gt 0$, we have:

$$ R_{rad} = \cfrac {\Cin 2\pi} {4\pi} \eta_0 \approx 73.1\;\Omega $$ $$ X_{self} = \cfrac {\Si 2\pi} {4\pi} \eta_0 \approx 42.5\;\Omega $$

This tells us a couple of things:

• As $\sin\pi=0$, the impedance of a half-wave dipole is independent of its thickness.
• Since an antenna is resonant when it has zero reactance, the half-wave dipole is not resonant!

Okay, so a dipole is resonant when it is slightly less than $\lambda / 2$ in length. Can we figure out what that length is?

For all other values of $\ell$ close to $\cfrac{\lambda}{2}$, we can use the identity $\csc x + \cot x = \cot\cfrac{x}{2}$ to obtain:

$$ R_{rad}\left(\chi\right) \approx \left( \cot \cfrac {\chi} {2} \;\Cin\chi - \cot\chi\cfrac{\Cin 2\chi}{2} - \Si\chi + \cfrac{\Si 2\chi}{2} \right) \cfrac {\eta_0}{\pi} \cot \cfrac{\chi}{2} $$ $$ X_{self}\left(\chi,\xi\right) \approx \left( \cot \cfrac {\chi} {2} \;\Si\chi - \cot\chi\cfrac{\Si 2\chi}{2} - \Ci\chi + \cfrac{\Ci 2\chi}{2} - \ln\sqrt {\chi} + \ln {\xi} + \ln \sqrt {\frac {e^\gamma}{2}} \right) \cfrac {\eta_0}{\pi} \cot \cfrac{\chi}{2} $$

The resonant lengths $\ell$ of a dipole for any given conductor radius $a$ correspond to the roots $\chi$ of the following equation for a given $\xi$: $$ X_{self}\left(\chi,\xi\right) = 0 $$

Since we know that $\chi=\pi$ is not a root, the equation can be re-written as:

$$ \cot \cfrac {\chi} {2} \;\Si\chi - \cot\chi\cfrac{\Si 2\chi}{2} - \Ci\chi + \cfrac{\Ci 2\chi}{2} - \ln\sqrt {\chi} + \ln {\xi} + \ln \sqrt {\frac {e^\gamma}{2}} = 0 $$

This does not help us analytically determine the resonant value of $\chi$ given $\xi$. It can be done numerically, and like most computations for antennas, this is what is normally done. However, it works well the other way around: to analytically determine the value of $\xi$ for which we have resonance for a given $\chi$.

$$ \ln \xi = \cot\chi\cfrac{\Si 2\chi}{2} -\cot \cfrac {\chi} {2} \;\Si\chi + \Ci\chi - \cfrac{\Ci 2\chi}{2} + \ln\sqrt {\chi} - \ln \sqrt {\frac {e^\gamma}{2}} $$

A picture is worth a thousand words. Let us plot $\ln\xi$ against $\chi$ in the range $0 \lt \chi \lt 2\pi$.

The graph tells us (and we can verify analytically) that in the vicinity of $\chi=\pi$, we have:

• As $\chi \to \pi^+$, $\ln\xi \to +\infty$, or equivalently $\xi \to \infty$, and
• As $\chi \to \pi^-$, $\ln\xi \to -\infty$, or equivalently $\xi \to 0^+$

We started with the assumption that $a \ll \lambda$, which implies $\xi \ll 1$, so we need not bother with the $\xi \to \infty$ case. But the $\xi \to 0^+$ case nicely fits in with the $\xi \ll 1$ assumption, and tells us that for small values of $\xi$, resonant dipole antennas have $\chi$ that are somewhat less than $\pi$.

In other words, this confirms that a dipole is resonant when it is slightly less than $\lambda / 2$ in length!

But hang on, can we use this to investigate an infinitesimally thin dipole? Can we say that as $\chi \to \pi^-$ for $\xi \to 0^+$, the half-wave dipole is resonant in the limiting case of the infinitesimal thickness?

Unfortunately not: the approximation breaks down somewhere as the wire gets thinner. It says here (in section 16.3) that an infinitesimally thin, infinitely conductive dipole would actually be resonant for $\ell \approx 0.4857\lambda$, with a corresponding $R_{rad}=67.2\Omega$. Which seems to imply that the expressions for $R_{rad}$ and $X_{self}$ mentioned above hold for $a \ll \lambda$, but not $a \lll \lambda$. How interesting!

I'm still trying to understand exactly why the approximation does not hold for $a \lll \lambda$. The most obvious explanation is that we started out assuming a sinusoidal current distribution which is approximately, but not exactly, true.

ϵ0, μ0 and the Special Theory of Relativity

Consider two point charges $q_1$ and $q_2$, at a distance $r$ from each other. We know that there is an electrostatic force between them, along the line joining the charges, as follows:

$$ F_E \propto \frac {q_1 q_2} {r^2} $$

Now consider infinitesimally thin wire segments, each of length $\delta\ell$, parallel to each other at a distance of $r$, each oriented perpendicular to the line joining them. Let us say they are carrying currents $I_1$ and $I_2$ respectively. We know that there is a magnetic force between them, along the line joining the two wire segments, as follows:

$$ F_M \propto -\frac {I_1\delta\ell\;I_2\delta\ell} {r^2} $$

The force is considered positive if it is a repulsion, and negative if it is an attraction.

Let us express the expressions in terms of constants of proportionality $\alpha/4\pi$ and $\beta/4\pi$, as follows:

$$ F_E = \left( \frac {\alpha} {4\pi} \right) \frac {q_1 q_2} {r^2} $$ $$ F_M = -\left( \frac {\beta} {4\pi} \right)\frac {I_1\delta\ell\;I_2\delta\ell} {r^2} $$

The values of $\alpha$ and $\beta$ depend on the choice of units. For instance, in SI units, we say $\alpha = 1/\epsilon_0$ and $\beta=\mu_0$. But are $\alpha$ and $\beta$ related in any way?

The answer is yes, and to uncover the relation, we need to dip into ... the Special Theory of Relativity!

Consider two wires, each of infinitesimal and cross section $\delta a$, parallel to each other, separated by a distance $r$. Let us assume that both wires are uncharged, and that both are carrying a current $I$ in the same direction $\hat{\ell}$.

Now consider a segment in each wire, each of infinitesimal length $\delta\ell$, such that the line joining the center of the two segments is perpendicular to either wire.

The current density in each segment is $\vec{J}=(I/{\delta a})\;\hat{\ell}$ through any surface perpendicular to their respective lengths.

Now consider the fact that a current is a flow of charge. A flow is a motion, and motion has a velocity. A current density $\vec{J}$ can be represented by a charge density $\rho$ moving at a velocity $v$, where $\vec{J} = \rho\vec{v}$.

So each wire can be considered to have a charge density $+\rho$ moving with a velocity $\vec{v} = v\;\hat{\ell}$. But since each wire has a zero net charge, it also simultaneously has a charge density of $-\rho$ that is not moving. Since $(I/{\delta a})\;\hat{\ell} = \rho\;v\;\hat{\ell}$, we have $(I/v) = \rho\;\delta a$

Let us call the positive charge density the 'red cloud', and the negative charge density the 'blue cloud'.

Inertial Reference Frame A

'Frame A' is a frame of reference that is stationary with respect to the blue clouds.

The electrostatic force between the two segments is zero as both wires have zero net charge each.

$$ F_E = 0 $$

From classical electromagnetism, we know that the magnetic force between the two wire segments is:

$$ F_M = -\left( \frac {\beta} {4\pi} \right) \frac {(I\;\delta\ell)^2} {r^2} $$

The '$-$' sign indicates that the force is attractive. The direction of the force is along the line joining the two segments.

For the next step, we will need the Special Theory of Relativity, in the form of the Lorentz-Fitzgerald contraction. The 'Lorentz Factor' corresponding to a speed $v$ is:

$$ \gamma = \cfrac {1} {\sqrt {1 - \cfrac {v^2} {c^2} } } $$

... where $c$ is the relativistic speed limit, a.k.a. speed of light in vacuum. A moving charge density appears to have increased, as compared to its value when static, by the Lorentz factor (because lengths along the direction of motion appear to have shrunk by that factor).

Inertial Reference Frame B

'Frame B' is moving at a velocity $v\;\hat{\ell}$ relative to the first frame. In this frame, the red clouds are stationary, while the blue clouds are moving with a velocity $-v\;\hat{\ell}$.

In Frame A, the red cloud was moving, while in Frame B it isn't. So in Frame B, its charge density will appear to be $+\rho/\gamma(v)$. On the other hand, in Frame A the blue cloud was static while in Frame B it is moving. So its charge density will appear to be $-\rho\;\gamma(v)$. Therefore, in Frame B, each segment has a net non-zero charge density which is:

$$ \rho_B = \rho \left( \frac {1} {\gamma} - \gamma \right) $$

This will cause an electrostatic force between the two wire segments, which is:

$$ F_E = \left( \frac {\alpha} {4\pi} \right) \frac {(\rho\;\delta a\;\delta\ell)^2} {r^2} \left( \frac {1} {\gamma} - \gamma \right)^2 = \left( \frac {\alpha} {4\pi} \right) \frac {(I\;\delta\ell)^2} {r^2} \left( \frac {1} {\gamma} - \gamma \right)^2\frac {1} {v^2} $$

This force is positive because it is repulsive in nature.

Let us now look at the current. It is now a result of the blue clouds moving. As we've seen before, in Frame B, the charge density of the blue cloud is $-\rho\;\gamma$. As the velocity of this cloud is $-v \hat{\ell}$, the currents are now $I\;\gamma$. The magnetic force between the two segments is now:

$$ F_M = -\left( \frac {\beta} {4\pi} \right) \frac {(I\;\delta\ell)^2} {r^2}\;\gamma^2 $$

Now, as per the Special Theory of Relativity, the sum of the forces $F_E$ and $F_M$ should be the same irrespective of the frame. So we have:

$$ -\left( \frac {\beta} {4\pi} \right) \frac {(I\;\delta\ell)^2} {r^2} + 0 = -\left( \frac {\beta} {4\pi} \right) \frac {(I\;\delta\ell)^2} {r^2}\;\gamma^2 + \left( \frac {\alpha} {4\pi} \right) \frac {(I\;\delta\ell)^2} {r^2} \left( \frac {1} {\gamma} - \gamma \right)^2\frac {1} {v^2} $$

This means:

$$ \beta\left(\gamma^2-1\right) = \frac {\alpha} {v^2} \left( \frac {1} {\gamma} - \gamma \right)^2 $$

For all $0 \lt v \lt c$, this implies:

$$ \frac {\beta} {\alpha} = \frac {\gamma^2 - 1} {\gamma^2 v^2} = \frac {1} {v^2} \left( 1 - \frac {1} {\gamma^2} \right) = \frac {1} {v^2} \left( \frac {v^2} {c^2} \right) = \frac {1} {c^2} $$

Which means:

$$ \beta = \frac {\alpha} {c^2} $$

For SI units, this translates to:

$$ \mu_0\epsilon_0 = \frac {1} {c^2} $$

... and we have deduced this without measuring $\epsilon_0$, $\mu_0$ or $c$!