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I once learned in school, and as far as I know this is also a common thing in many introductory physics books, that a parallel LC circuit can be "bent open" into a dipole antenna, like this:

A parallel LC circuit shown with magnetic and electric field lines, with the capacitor bending up into a dipole antenna, also shown with field lines

I am under the impression that this explanation is somewhat dumbed down and not the whole truth. My question is, how much truth is there to it?

To be more precise:

  1. Can (parts of) the dipole antenna be interpreted as some degenerated case of a cylindrical coil and a parallel plate capacitor? I very naively tried to relate this to the formulas $L = \frac{N^2 \mu A}{l}$ and $C= \frac{\varepsilon A}{d}$. For the inductance I can vaguely imagine that a straight wire might be considered as a cylindrical coil with $A=0$ and $N$ infinite, so if you take the correct limits this might get somewhere, but I'm really unsure if this makes any sense.

  2. A dipole antenna does have some inductance and capacity, however related or not to a cylindrical coil and capacitor. Can one compute the resonant frequency from those, similarly to how one computes $f = \frac{1}{2\pi \sqrt{LC}}$ for the LC circuit? Is there an intuitive way to think about this? For LC circuits I have a quite visual conception of how one first charges the capacitor, then this imbalance of charge creates voltage and thus causes a current from one plate to the other, but this current is in some sense delayed by the inductance and thus over-corrects the imbalance in charges, leading to the capacitor being charged the other way around, and so on. Is there a similar explanation for the oscillation of a dipole? (If this is also dumbed down and not quite correct, please tell me)

  3. An LC circuit has precisely one resonant frequency, while a dipole is also resonant at odd multiples of its base frequency. How does this fit into the picture?

Andrea Lachmann
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    Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Dec 22 '22 at 14:42

3 Answers3

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First, let me urge you to do some experiments. Physics is physical: it relates to real phenomena in the real world. Textbook physics is a poor substitute for real world physics. These days, this physics is much more accessible than it used to be: you can get a nanoVNA for about $60 and measure these things.

  1. Current in the antenna produces a magnetic field. You may model the stored energy in that magnetic field as a consequence of inductance. The current also separates charge between the antenna legs, producing an electric field. That may be modeled as capacitance.

  2. The inductance and capacitance are not "lumped" in an inductor and capacitor, but distributed around the structure.

  3. The lumped model, with a single inductor and capacitor, has a single resonant mode. The distributed model effectively has an infinite collection of inductors and capacitors, and thus an infinite collection of resonant modes. Reality is somewhere between.

Modeling distributed L/C circuits is the domain of transmission line theory. Thus, it is common to treat the legs of a dipole as segments of transmission line. That's where we get our intuition. You may relate it back to inductance and capacitance via transmission line theory.

John Doty
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You have a lot of questions here, and I'm not sure I can answer all of them. I can respond to one of the points here, though: a straight wire is not the same as a solenoid (coil) with $0$ area and infinite turn density. The difference is that in a solenoid, the electrons follow a very specific spiraling path around the central axis which gives them angular momentum about that axis (and thereby produces a magnetic field in the direction of the axis). In a straight wire, the electron travel straight along the wire, and therefore there is no magnetic field about the axis.

slithy_tove
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    There's still a magnetic field, and thus inductance. And forget electrons here: they add no value to your model. – John Doty Dec 22 '22 at 14:54
  • There is a magnetic field, of course, I'm just pointing out that it's a different kind of field than if the current was moving through a solenoid (even one with area approaching $0$). – slithy_tove Dec 22 '22 at 16:28
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It is a common misunderstanding that an antenna is a resonator. The drawing that supposedly explains how an "LC" circuit radiates assumes that you get better and better radiation by opening up the capacitor plates presumably because when they are parallel it radiates badly or none at all. This is not true. It is true though that if you properly match the radiation impedance of a pair of plates then it will happily radiate as any microstrip antenna of nearly rectangular shape proves it. Actually, the purpose of the inductor is to supply a back-and-forth current to the plates not to "resonate" them. This inductor, therefore, can be replaced with a current source and it will work just as well as the inductor. See also 1 and 2

hyportnex
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    The common "half-wave" dipole antenna is a resonator. A non-resonant antenna is also possible, and even a resonant antenna can be made to work off-resonance, but that doesn't appear to be the issue here. – John Doty Dec 22 '22 at 15:18
  • @JohnDoty when an antenna is referred to as a "resonator" it is meant that in order to radiate an inductor and a capacitor are needed to be driven at frequency $1/\sqrt{LC}$, ie., behave as a filter. My point is that the reactance is spurious to be "matched away" because the purpose of an antenna is to be an impedance transformer between the source and $377\Omega$; everything else is to be eliminated and not taken advantage of as one would have it in a filter. – hyportnex Dec 22 '22 at 15:33
  • To my reading, the question appears to refer to the self-resonant properties of the half-wave dipole. Radiation isn't mentioned. So here, the reactance isn't "spurious": it's the subject of the question. – John Doty Dec 22 '22 at 15:40
  • @JohnDoty an "LC" cannot refer to a half-wave dipole; if it models anything it would model a very short dipole. To drive a capacitor into radiation with a voltage source first you would have to get rid of its capacitance with a n inductor and then match the usually low source impedance to the radiation resistance with an impedance transformer, both can be done with some inductive circuit without which next to nothing will be radiated by the cap. You can also get the inductor radiate by matching it capacitively to the source. – hyportnex Dec 22 '22 at 16:12
  • Again, the question doesn't mention radiation: it's apparently about self-resonance. And, of course "LC" can refer to the half-wave dipole, with the complication that the inductance and capacitance are distributed. Read my answer. – John Doty Dec 22 '22 at 16:31
  • You may use a very short dipole for reception without an inductor. That doesn't fit your model. – John Doty Dec 22 '22 at 16:32
  • @JohnDoty of course it does, it just needs high impedance input lna – hyportnex Dec 22 '22 at 23:40
  • Let us continue this discussion in chat. – John Doty Dec 23 '22 at 00:59