Voltage gradient across an antenna (and how it drives an AM radio)
New Question:

To answer your newer question, I think you'd be better off studying a little bit about the distributed element model and transmission-line theory first. From the linked Wikipedia article (emphasis mine):


  The distributed element model is more accurate but more complex than the lumped element model. The use of infinitesimals will often require the application of calculus whereas circuits analysed by the lumped element model can be solved with linear algebra. The distributed model is consequently only usually applied when accuracy calls for its use. Where this point is dependent on the accuracy required in a specific application, but essentially, it needs to be used in circuits where the wavelengths of the signals have become comparable to the physical dimensions of the components. An often quoted engineering rule of thumb (not to be taken too literally because there are many exceptions) is that parts larger than one tenth of a wavelength will usually need to be analysed as distributed elements.


If you were to excite your antenna with a low frequency source (or a DC source, let's say), then your intuition is correct: current would flow for a very short period of time until the voltage across the antenna was stabilized. But when exciting it with an AC source of a suitable frequency (i.e. a frequency where the wavelength is, let's say four times as large as the antenna, making the antenna a quarter-wave in length), the voltage will never be the same across the whole antenna, since the source is changing as fast as it takes the wave to reach the top of the antenna. In other words, there is no way for the bottom of the antenna to be fixed at a total of 0V, but it can have a DC potential fixed at 0V (with an AC potential swinging above and below 0V). This is what The Photon was mentioning in his comment talking about driving the antenna relative to the actual earth ground:







simulate this circuit &ndash; Schematic created using CircuitLab

Here, the DC source is 0V (i.e. it could be omitted), meaning your antenna's bottom end is fixed at earth potential on average. The AC source will excite a sinusoidal voltage at the bottom of the antenna, swinging from +1V to -1V at 1kHz. We'll assume the antenna is a quarter-wavelength (at 1GHz, that's about 3 inches long). I've also included the typical distributed model for a transmission line. That is how regular wires are modeled once the frequency of excitation becomes fast enough.

In your second schematic, with diode-detector circuit, you say the circuit is connected at two points to the antenna, but that's not true. The antenna is only connected to the node at the top-left of the circuit. The resistors are not part of the antenna.



Old question:

A quick answer: I think a \$\frac{\lambda}{4}\$ (quarter-wave) monopole antenna precisely describes what you're asking about. In the case of a monopole antenna, the end of the monopole facing the ground plane is actually a node of the voltage standing wave. Call that point GND, and that point will "maintain \$0\text{V}\$"; it is the node of a standing wave (i.e. no change in magnitude), and by calling it GND you've defined it to be \$0\text{V}\$, full-stop.

I'm still not sure I entirely understand your question, and schematics don't necessarily convey what you're trying to get across. By definition, ground is \$0\text{V}\$. Any part of your schematic that touches a GND is fixed at \$0\text{V}\$, since a wire in a schematic is defined to be a perfect conductor. For example, in your first picture, exactly zero current flows because there is no loop. GND is just a symbol we use to define where \$0\text{V}\$ is, and nothing more. You can't have a GND in a schematic that has any potential other than \$0\text{V}\$.

I think your confusion stems from the following:


  But it looks like that in most real cases ground is merely a theoretical construct...


In all cases, ground is merely a theoretical construct.