Zero reactance means voltage and current are in phase.
For example, consider a voltage source connected to a nonreactive load. This means the peaks of the applied voltage should coincide with the peaks of the resulting current.
It's equally valid to consider a current source connected to a nonreactive load. In this case the peaks of the applied current should coincide with the peaks of the resulting voltage.
Either way, reactance is part of impedance, and impedance is the relationship between voltage and current.
A dipole is just a bit of balanced transmission line that's been pulled apart. What happens when a DC voltage step is applied to the end of a section of transmission line that's open on the end opposite the voltage source?
simulate this circuit – Schematic created using CircuitLab
If we want to know the impedance of this "load" (the transmission line) we need to know how much current flows. We know that eventually the current must be zero, because the circuit is open at the end. But how can the voltage step know that, not yet having seen the open end?
So what happens is initially some current flows, in an amount defined by the surge impedance (also known as characteristic impedance) of the transmission line. But the current is constrained to zero at the open end, so a reflected wave is superimposed on the initial wave, propagating from the open end and back to the voltage source. It may help to play with a time domain transmission line simulator to get an intuition for this process.
What happens when the reflected wave gets back to the source is key. In the case of a DC step, the source will see too much voltage, and so it will reduce current. And this sets off another round of wave propagation, with each iteration getting closer to what we know the DC solution must be: zero current, that is infinite impedance.
But in the case of AC, the voltage source is not a step but rather a sinusoid. We must consider both the phase of the reflected wave, and the additional phase delay introduced by the propagation of the forward wave and then the reflected wave.
Exactly what is the cause for the zero reactance seen in the impedance at the center feed point of a resonant half wave dipole?
When the transmission line is open, the current of the reflected wave will always be equal but opposite the forward wave, because the open end always wants to cancel the current to make it zero. In other words, the reflection adds 180 degrees of phase delay.
When the length of the transmission line is 90 degrees, it is resonant. This is due to the 90 degrees of delay for the forward wave, plus 90 degrees for the reflected wave, plus 180 degrees for the phase of the reflection equals 360 or 0 degrees. Current is in phase with voltage, which means zero reactance, which means resonance.
I don't understand yet how the radiation resistance fits into all of this.
In the case of an ideal 1/4 wave transmission line, the impedance seen by the voltage source is exactly 0+0j ohms. This is because the current from each reflected wave reinforces each forward wave, and there's no loss in the system, so the current builds to infinity. But in an ideal resonant dipole some energy is lost to radiation (represented by a resistance), and so the current builds to a high but finite quantity, resulting in the low but non-zero impedance of about 70+0j ohms.
Now, what about this graphic:
At a glance, it looks like the red and blue curves, labeled "voltage" and "current" respectively, are not in phase, but quadrature. How is this reconciled with the above explanation, where voltage and current are in phase?
More confusing, but perhaps more helpful is the old version of the image which shows only the standing wave, but does not include the influence of the voltage source (perhaps it would be better if the illustration did not include a voltage source, since its effects are not illustrated):
Here, the red and blue curves are exactly in quadrature. And this is no mistake, since the standing wave is purely reactive.
I think the confusing thing about this image is it just says "V" (for voltage) without really explaining what that means. Anything measured in volts could be called voltage. That's not very specific or helpful.
If we are concerned about the feedpoint impedance, the voltage we are concerned about is more specifically the electric potential difference between the two feedpoint terminals.
If we are concerned about the electromagnetic fields around the dipole, we are probably more concerned about the electric field intensity, which is a vector quantity for some point in space around the antenna, measured in volts per meter.
The "voltage" in the graphic shows the electric potential for each point along the length of the antenna. Electric potential is the electric potential difference between the measured point, and a theoretical point infinitely far away, which is 0 volts by definition. In the case of a dipole, the electric potential right at the center is also 0 volts.
Now the question is: how can the electric potential difference between the feedpoint terminals be in phase with current when on the graphic the blue curve is clearly not in phase with the red curve?
The answer is quite simple: theoretically the feedpoint terminals are separated by only an infinitesimal distance. It doesn't actually matter what the blue curve is doing, because the electric potential difference between two points approaches zero as the two points approach zero separation.
Put another way, electric potential difference between two points in a uniform electric field is the electric field intensity (volts/meter) multiplied by the distance between the points (meter). If the distance is small, the electric potential difference can be neglected.
The full picture of what happens on the dipole is the superposition of:
- the standing waves, shown in the image above, where the electric and magnetic fields are in quadrature, and
- the influence of the voltage (or current) source driving the antenna, where voltage and current are in phase.
My understanding which is becoming more and more confused as time goes on is that at resonance the voltage of the standing wave which is 90 deg out of phase with the applied current is always zero at the feed point at resonance.
Although it is true the standing waves are associated with a high electric field intensity around the feedpoint terminals, as long as the terminals are not far apart this has negligible significance to the electric potential difference between the terminals.