haael said:
Let's say it emitted a very tiny amount of energy, equal to a single photon.
In this case, the emission can't really be viewed as a wave. To get observed wavelike behavior you need large numbers of photons. The single photon has a wave *function*, but that's not really the same thing.
haael said:
The radio wave propagates radially, in all directions.
Not really. The photon's wave *function* has a part that looks like a spherical wavefront expanding radially at c, yes. But it also has longitudinal and timelike components that do *not* propagate at c. In the case of an ordinary EM wave, these components cancel out when summed over large numbers of photons.
haael said:
One may get an impression that the energy is distributed equally throughout space. But when we put a receiver antenna in some neighbourhood and it happens to absorb the wave, then all the energy goes to it. Now it looks like a discrete photon flied between the emitter and the receiver. This is exactly the particle-wave duality.
Sort of, but there are complications. See above, and further comments below.
haael said:
When we see the radio signal as a wave, there is a problem with the energy smeared everywhere and not in a single point.
No, when we see the radio signal as a wave, it's because there are large numbers of photons, each of which can get absorbed at a different point in the receiving antenna. The pattern of absorptions looks like a wave.
haael said:
When we see the radio signal as a photon, there is a problem how the photon knew the place of the receiver in the first place.
If that's a problem, it's a problem for the wave case too, because, as above, the observed "wave" is composed of a large number of individual photon absorption events.
haael said:
So, waves expand in all directions, but they only carry probability of interaction, not the energy.
Then where is the energy while they are traveling? Is conservation of energy violated? If so, how? All of the quantum mechanical equations are consistent with energy conservation.
haael said:
Then, when the bodies have interacted, it looks like a discrete quantum of energy passed between them
Yes.
haael said:
knowing their position in advance.
Not really.
haael said:
This not only applies to electromagnetic waves (excitations), but also to static sources.
Quantum mechanically, static sources interact by means of virtual particles (at least, this is one common view, which also has limitations). For example, a static charged body interacts with other charged bodies by exchanging virtual photons. (The main component of these photons that contributes to the interaction is the timelike one, IIRC: the components that appear in an observed EM wave don't contribute much to the static interaction amplitude.)
haael said:
A static charged body spreads a static field around, that doesn't hold energy, but only probability.
On the virtual particle view, there is no "field"; there are just amplitudes for different possible virtual particle exchanges. The field is something that appears when you sum over a large number of virtual particle interactions. See below.
haael said:
Only when another charged body happens to interact with it, it looks like it interchanged energy with the field.
If we sum over a large number of virtual particle interactions, yes. But as you say, the body interchanges energy *with the field*. In other words, at this level of description, it no longer looks like the charged bodies are interchanging energy directly; instead, each charged body only exchanges energy with the field at its location. The field can change if its source changes, but those changes are propagated at the speed of light, so there's no causality violation.
haael said:
But again, if we wanted to say that the energy was stored in the field from the beginning, then we would have to admit that spread energy suddenly concentrated in one point.
No, we wouldn't, because the energy exchange is not between the objects; it's between each object and the field at its location. Everything is local. See above and my previous posts.