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The parabolic equation for a beam propagating in the $z$ direction comes from inserting $$\varphi(x,y,z,t) = \psi(x,y,z)e^{i(kz-\omega t)}$$ into the wave equation $$\frac{\partial^2 \varphi}{\partial x^2}+\frac{\partial^2 \varphi}{\partial y^2}+\frac{\partial^2 \varphi}{\partial z^2}- \frac 1{c^2}\frac{\partial^2 \varphi}{\partial t^2}=0$$ to get $$... 3 Your question is somewhat meandering, but I'll try to answer what I can. First, I advise you to stop thinking about antennas in terms of photons, or quantum mechanical phenomena in general. The first antennas were built using only a classical understanding of electromagnetism, and such an understanding is all that's needed to understand most of their ... 3 Let me first comment a sentence from SuperCiocia’s answer. The photodetector clicks ... are caused by the photoelectric effect, that is bound electrons in the photodetector are in quantised orbits and are only capable of discrete energy jumps. (1) In addition to this statement, please recapitulate that any observation of the wave behaviour of light during ... 3 I will try to summarize the classic results from 1, 2, 3. In short, a receiving antenna does not scatter as much power as it absorbs except in a few very special cases such as a very short dipole or a small loop. For these small dipoles, when the incident wave induces the same current distribution as would be seen when the antenna is used and driven as a ... 2 I had come to the agreement that to understand how antennas work, it is best to assume light/electromagnetic waves act like photons/particles in the presence of recieving antennas. You're way off base here. Thinking about quantum effects won't help you understand RF antennas at all. antennas are nothing but metal wires which work on photoelectric effect ... 2 The simple answer is no. For a photon to be observed, all its energy must be collected. You cannot observe half a photon, either you observe it or you do not. The observation or detection can only happen in one place. This is often referred to as "the collapse of the wave function". As an electromagnetics engineer I sometimes monitored very faint ... 2 Light does not behave like a wave some times and like a particle some other times. Light behaves as light. Trying to categorise some behaviour as "wave-like" or "particle-like" is just an attempt to build an intuitive understanding for quantum phenomena by relating them to simpler everyday things like water waves or marbles hitting a wall.... 1 EM waves are generated in a circuit having a physical inductor and capacitor but in an ideal "LC" circuit charges move and oscillate without EM waves being generated. The reason for no waves is the assumption that the circuit consists of infinitesimally small elements and connections, in practice, having linear extent negligible compared to \... 1 It will originate from everywhere: the wires, the capacitor and all the elements carrying the current or the displacement current. Indeed, thee emission of the EM waves is described by the Maxwell equations where the sources are the time-dependent currents and fields. What is confusing in LC circuit is that it is not suitable for the description of the ... 1 The maximum operating frequency of a waveguide is typically limited by the higher order modes. If you try to use the waveguide at a frequency where multiple modes can propagate, you will get some coupling between these modes, which will cause some signal power to travel faster than the main signal. This generally causes problems depending on how the signal ... 1 The explanation is on page 80, where the source, transmission line, etc. of Figure 3.1 are shown to be enclosed by a hemi-sphere over which the contribution of the fields can be neglected; this figure is referenced on page 87. 1 The reflection efficiency is given by$$1 - |\Gamma|^2 = 1 - \left|\frac{Z_L - Z_0}{Z_L + Z_0}\right|^2 where $\Gamma$ is the reflection coefficient, $Z_L$ is the load (antenna) impedance and $Z_0$ is the characteristic impedance of the transmission line. Your third expression gives the correct efficiency.

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It comes from the freespace impedance, convert it to same scale (20*log10(377Ω)) = 51.5 In dB multiplication becomes adition, realizing this, you have something that looks like ohms law.

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In short: One diagram is a side view, the other is a top view. Longer answer: In the figures showing the intensity due to two dipoles [a) and b)], the direction of dipole moment oscillation (and thus acceleration of the charged particles involved) is not in the plane of the paper(screen), but perpendicular to it. So the far fields from both fields add up ...

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As your expression for $E_\theta$ suggest, this is in fact the $\hat \theta$ component of a plane wave (in the far field) expressed in spherical coordinates. In the far field the curvature of the wavefronts is negligible and all wavefronts can be considered as planes. The (nice) figure you give is really in the approximation of a very very long slit. ...

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If your circuit was oscillating since the beginning of time: Many photons are generated simultaneously, and the collection of these packets of multiple photons being produced at a constant rate produce the E-field. (As a technical point, it's actually the coherence between these possibilities that produce a nonzero E-field, but you can think of it like their ...

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This is rather a comment on the "VERY long" wavelength. About the "inside the radiantsphere" you think about a liquid and the inner tension at which a drop can leave the liquid. Let us not start with your LC-circuit, but only with a resistor. The resistor is connected to a DC power source. After switching on the source, the temperature of the resistor ...

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You can think of two antennas in each other's near field as two halves of an air-core transformer. As such, they will load each other in ways that don't happen in the far field. This principle can be used to couple RF power to an antenna in a manner that prevents induced currents from other nearby antennas from propagating in the antenna leads. Instead of ...

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This question was asked a long time ago. I hope the subject is still active. The issue touches the very basics of what capacitance means. Let us not talk, at this stage, of a dipole antenna. Consider two lengths of wires of any length, say 10 meters each, connected to a a plain old 1.5 V AA galvanic cell. Before the connection, the two wires are charge ...

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There’s no question that power is being reflected from the mismatch: if you look at the fields there, Maxwell’s equations say that’s happening. That reflected wave does give rise to a standing wave. But it’s a mistake to say that’s all that happens a result. The reflected power has to go somewhere. During the startup it goes into energizing the standing ...

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