# Tag Info

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You're asking a classical electromagnetism question. You don't benefit from thinking about photons. If the antenna's electromagnetic field (calculated from Maxwell's equations) is right-circularly-polarized at a certain point, then you can say that every photon is right-circularly-polarized. If the field is linearly polarized, then you can say that every ...

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Photons as described by QED (quantum electro dynamics) are based on wave solutions of maxwells equations in free space (no charges, no currents). In antenna theory these may be called far field solutions. However, mathematically they constitute a complete set of solutions to maxwells equations. This implies that by making linear combinations of the wave ...

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Short answer: yes. Regarding quantum fields, one can think of the "particles" associated with them as sort of a minimum currency for exchange, i.e. to carry an interaction, it can only be done in discrete units. But like currency, it is still a currency of some type, so a photon represents the unit of interaction of the field it comes from. Different ...

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Ion drag is associated with low frequency (RF) EM waveforms, and is what causes resistive heating, not dielectric heating. Dielectric heating (e.g. microwaves) relies on dipole rotation.

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Electromagnetic field of light has two kind of angular momentum first spin angular momentum (SAM) and secondly orbital angular momentum (OAM). former one represent the dynamical rotation of electric (or magnetic) field of around propagation direction and indicate the polarization of beam. Later one represent the rotation of light around beam axes. The ...

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How a radio wave leave his source you can see her https://en.wikipedia.org/wiki/Camille_Papin_Tissot

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There are two possibilities to accelerate particles; often this are electrons, so I will talk only about them. First you can accelerate electron straight or in a curve in an electric or magnetic field. Second you can accelerate electron in a curve and with constant angular velocity. In both cases you interact with the electron with help of electromagnetic ...

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Yes, EM wave can be formed by electrons in constant velocity. As Maxwell said $$\vec{\nabla}\times \vec{B}=\mu_{0}J+\frac{1}{c^{2}}\frac{\partial \vec{E}}{\partial t}$$ Electron moving in constant velocity means a current. This current gives rise to a magnetic field. This is not instantaneous. This information travels with the speed of light. So at every ...

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Question: Why do accelerating charges radiate, and can moving charges with zero acceleration also produce electromagnetic radiation? Quick Answer: Only accelerating charges (and changing currents) produce electromagnetic radiation. The power of fields generated by charges with zero acceleration dies out at far distances. Full Answer: The full answer to ...

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This is a good question that I struggled with myself for some time. I believe that the correct answer is the following. The imaginary part of the index of refraction, i.e. $\kappa$, quantifies the dissipation of light through a medium. However, if one wants to quantify the dissipation due to nonretarded electric fields alone, the quantity that quantifies ...

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Electromagnetic radiation in a medium propagates according to the law $$\mathbf E,\mathbf B \propto e^{\imath(\pm k_xx-\omega t)}$$ where $$k_x^2 = \frac{n^2\omega^2}{c^2}\;.$$ The refractive index $n$ can also be complex, in which case its imaginary part describes the absorption of the EM wave in the medium. But the oscillating part is in any case $$... -1 This is basically about dispersion: EM waves with different frequencies travel at different speed in a medium because of interaction. Usually, as in a standard textbook experiment wherein a light beam is bent by a prism, higher frequency waves bends more. The more it bends, the more slowly it travels. So, lower frequency waves usually go faster. However, ... 2 Yes, it an extremely small effect but it exists in Einsteins general relativity. There is one case of a double star where there rotation around each other seems to lose energy at rate that this phenomena should give according to general relativity 1 what does orbits with definite energy mean? It means that each orbit has an amount of energy associated with it. If you move between 2 orbits, it requires a certain amount of energy. It will be the same amount of energy for the same transition in the same atoms. Different transitions have different amounts of energy and different atoms have different ... 0 My question is what does orbits with definite energy mean? It means that electron circumscribes an ellipse and the total energy of the system nucleus + electron is constant; when the electron is closer to the nucleus, its potential energy is lower and the kinetic energy is higher, but their sum remains the same. why do the electron in their ground ... 0 Here is an excellent tool for "visualising" the electric and magnetic fields of waves being produced by a variety of sources (including antennae). This applet allows you to slow down time or even freeze it at a particular instant. You can show the fields, but also the rate of change of the fields and the currents. What is doesn't do is show this in full 3-D, ... 1 Part of the genius of Maxwell was to realise that it did not require the presence of currents or charges to generate a magnetic field. In analogy to the way that changes in magnetic field generating an electric field (or macroscopically we say that changes in the magnetic flux linked with an electric circuit can produce an EMF and hence a current), it turns ... 0 There exists a very good formulation in this blog entry of Lubios Motl on how classical fields emerge from quantum theory, . The link between the quantized photons and the classical field are the Maxwell equations , Classical solutions of the equations give rise to the classical fields. Used with the operator formalism the wavefunction describes a photon. ... 0 For a classical discussion on waves I would refer you to review the Wikipedia site on Maxwell's equations first and then also to discussion on the Continuity equation for the purpose of understanding conservation laws. Now for the quantum part. First we need to get you up to speed on what a field is. Simply put it is an abstraction used in physics to ... 2 Perhaps the best way to understand this is to start simply: Consider a function f(x). Now, let's try to take the Fourier transform of its derivative f'(x). Just use the definition of the Fourier transform:$$\mathscr{F}(f'(x))(k)=\frac{1}{\sqrt{2\pi}}\int dx\ e^{-ikx}f'(x) $$and now use integration by parts (assuming f(x) \to 0 as |x|\to \infty, a ... 4 Every star or galaxy contains some elements, and each element emits a particular frequency. Here are the lines of the Sun (https://en.wikipedia.org/wiki/Fraunhofer_lines) In particular, Hydrogen is present almost everywhere and Hydrogen lines are visible in most galaxy spectra. The Hydrogen-alpha line is particularly strong in many galaxies. This ... -1 The visible and radio waves can cause heating effect when absorbed by the living cells. Human mind is effected by the electromagnetic waves and can change moods of a person. Higher radiations like Ultraviolet, X rays and Gama radiations can produce damage to life at powers that produce little heating. 1 No. When matter and antimatter annihilate, they do so particle by particle. Each electron annihilates with an anti-electron, protons with anti-protons, etc. And in each annihilation the total rest mass of the two particles is converted to energy in photons. Even the lightest annihilation, that of an electron with an anti-electron, must put over 1\ ... 0 The lowest frequency limit is provided by the size of the universe. If we could make an antenna that size, the frequency would be "very close" to zero. The highest EM frequency limit is provided by the smallest antenna we could make. I believe that would be the size of a hydrogen atom, giving us,$$F_u = \frac{2.997x10^8}{6.28x5.29x$10^{-11}} = ... 1 There is a good explanation of this in Matter and Interactions vol II by Sherwood and Chabay. I no longer have the text; I will try to summarize its explanation as I remember it. The electrons in a substance are analogous to charged masses on springs. The electrons in insulators are relatively tightly bound; those in conductors are loosely bound or unbound. ... 3 An overview in layman's terms: First, it is important to note that not any electric field will induce current in a conductor, because other than the fact the intensity of the field defines the speed of each charge (bigger difference of potential), the oscillation frequency of the$\mathbf{E}$also plays a very important role, if the frequency is too high, ... 8 The reasoning has to be the other way around: Light acts on the metal and makes the electrons move. This, however, results in an energy loss, as the electrons feel a resistance and thus the radiation loses energy. This can be formulated more precisely with counteracting electric fields. That's why all good conductors are opaque. In insulators this can not ... 0 To get to the heart of the matter, you need to understand two things: Maxwell's equations Fourier analysis Maxwell's equations tell us the relationship between electricity and magnetism. In particular, they tell us that a change in an electric field causes a magnetic field. Similarly, they tell us that a change in a magnetic field causes a change in an ... 3 It's because most materials have (many) natural resonances. I assume you are alright with the phase velocity being different in different media, that is I assume you are alright with something of the form $$\frac{ \omega }{ k } = \frac{c}{n} = \frac{ c}{\sqrt{ \mu \epsilon }} \sim \frac{c}{\sqrt \epsilon}$$ where$k$is the wavenumber of a wave, ... 1 I understand that voltage is the movement of electrons No, the movement of electrons is (one type of) electric current. There can be a voltage without the movement of charge. and that the antenna acts as a light bulb No, an antenna is a resonant system that, ideally, has zero resistance while an incandescent light bulb has a resistive element ... 1 Electrons and photons definitely are not the same. For instance electrons have rest mass and photons do not. Also, electrons have charge, while photons do not. I could go on about their very different statistical behavior in quantum mechanics (electrons have a quantum mechanical property called "spin" that is half-integer, photons have integer "spin" ) but ... 7 You should look at the form of the advanced fundamental solution of D'Alembert equation, built up in geodesically convex open sets including the source localized at the event$y$and the test point localized at the enent$x$receiving the wave generating by the source. The construction, at least for analytic manifolds with analytic metrics, is obtained by ... 3 It generally does not work in curved spacetime. There is a quite thick book almost completely devoted to study this issue by P. Günther: Huygens' Principle and Hyperbolic Equations. Some discussions can be found in Friedlander's book about the wave equation in curved spacetime. A necessary condition for the validity of the Huygens principle is that the ... 0 It is possible for spacetime curvature to scatter and reflect light. The most obvious case of this is gravitational lensing. It's probably best to just solve the wave equation for the underlying light against the correct metric than to appeal to a simplifying principle like Huygen's principle. 1 Unless you are using an oversized microwave (like industrial or scientific size) and a large protective shield for the camera, I doubt it would be practical to film the inside of a microwave oven while it is running, without destroying said camera. If you are only looking to simulate the effect, you may disconnect the magnetron circuitry from the microwave ... 15 The easiest solution is to use a fiber-optic camera, i.e. one with a fiber-optic connection between the front lens and the actual camera electronics. You can easily bend the fiber (but not too much!). You can now make a small hole in the microwave (smaller than the wavelength, insert the fiber. Bend the fiber and wrap it in aluminium foil. There will be ... 3 You correctly surmise that you could make a cage with small holes, so that the visible light could pass through but the microwaves would be blocked. However, you could have some major problems if you decide to put a metal object inside a microwave oven - see the same wikipedia page you got your frequency information from. You would have to construct your ... 1 You could put your camera in an invisibility cloak for microwaves. I don't know how that would distort the images, but it has probably the advantage of not disturbing anything else inside the microwave. If you use a metal cage, there might be interference effects, and the existence of the camera would influence the thing you want to examine. 2 This is just a complement to the previous answers which give the correct response. If you want to think about it in an intuitive way, imagine that the interaction between electrons and photons becomes weaker. In the limit when it becomes nearly zero, the light will be almost not scattered at all and will continue in a straight path. 3 It is momentum that defines the incoming direction and momentum transfer the outgoing one. The photons, quantum mechanically carry momentum equal to p=h*nu/c . Momentum is a vector and defines directions. An electromagnetic field is an emergent classical quantity built up by innumerable photons. There exists also a momentum defined for the classical field ... 2 Let me offer you a slightly modified version of your question to illustrate a way of re-formulating it your thought process. How does a pool ball know from which direction the cue ball hit it? The answer is the same in the sense that "the particle" does not know all by itself, "the system"1 has certain invariant quantities (like momentum and energy) ... 1 An analogy (possibly from Griffiths' textbook): imagine you're out in desert. Many miles away you see a truck moving on the highway. In which case is it easier to judge its speed: if it's moving radially towards or away from you, so that you have only its change in apparent size to go on, or if it's moving laterally across your line of vision? 0 Remember that when a crystal is cooled to absolute zero, the atoms don't stop moving, they still jiggle. Why? If they stopped moving, we would know where they were and that they had zero motion, and that is against the uncertainty principle. We can't know where they are and how fast they are moving, so they must be continually wiggling in there!$_1\$ The ...

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If the object has a temperature at absolute zero ( within the quantum uncertainties related to this statement) it means that all the atoms and molecules that compose it are at the lowest possible energy level. Supplying energy to heat an object above absolute zero means increasing the kinetic energy of the component parts and raising them to higher energy ...

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Theoretically, the shortest wavelengths of light would be limited by the Planck length, at some point the space 'closed' by the wavelength would be so small that gravitational effects would dominate, in the same way that black holes can bend light passing near their event horizon at very small scales the wavelength would be so small that it might be at the ...

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Do keep in mind that the frequency of light is reference frame dependent. So, for example, the cosmic background microwave radiation would appear as a concentrated gamma radiation source 'in front' to an observer with ultra-relativistic speed relative to the CMB. In other words, light emitted from a body of a particular frequency in that body's frame of ...

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The electromagnetic spectrum does range between (almost) zero and (almost) infinity. It's just that your eyes are sensitive to a very small part of it (from about 380 nm to about 800 nm). At the lowest frequencies, it becomes difficult to recognize the signal from background fluctuations. From this site: "Gamma-rays are detected by observing the effects ...

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