# Tag Info

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I am guessing that what you call disturbances in the fields, are the oscillations of electric and magnetic fields that you mention just before. (Well, these disturbances propagate. The solution of the Maxwell's equation are traveling fields. Let me for top simplicity the case without charges (the case with charges are covered in my answer to this question, ...

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There are several mechanisms to create/cause electromagnetic waves: Macroscopically: accelerating charge (just moving at constant velocity is not enough; this is why we drive antennae with an alternating current pushing electrons back and forth.) Microscopically: spontaneous (i.e. without cause); think of exicted atoms emitting a photon. This is what ...

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Classically speaking, electromagnetic radiation is an oscillation of electric and magnetic fields which propagates. The movement of charged particles causes those oscillations - the motion of charges and the electric and magnetic fields are coupled - see Maxwell's equations. For example, an antenna broadcasts electromagnetic radiation when an alternating ...

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Q1. Does the electromagnetic field contain any energy? Yes,it does. Example:a capacitor stores its energy in its electric field. An inductor stores its energy in its magnetic field. here is the derivation http://hyperphysics.phy-astr.gsu.edu/hbase/electric/engfie.html Q2. Electromagnetic fields are the same as waves, so does that mean that when current is ...

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Q1: yes, it does. Q2: "Electromagnetic fields are the same as waves" Not always, e.m. fields may be static - static electric field around charges and static magnetic fields around magnets or (DC) currents, or waves - e.g as emitted by an antenna. "so does that mean that when current is induced, they get the energy from electromagnetic waves being ...

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Does the electromagnetic field contain any energy? The electromagnetic field can store as well as transport energy and momentum. Electromagnetic fields are the same as waves, Not so. There can be waves in the electromagnetic field but the electromagnetic field is not a wave. then doesn't that mean that you can induce current from waves? ...

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With proper containment fields it should be possible to reconstruct and deconstruct molecular/atomic energy and matter. Universal or infinite varience of cotinual recalibration may be required because containment fields might require these additional steps to hold fields.

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The efficiency of the pumping source is $x$ means that $x$ amount of electrical power is converted to energy which is useful for pumping the laser medium. The absorption of the pump is $y$ means that $y$ amount of the energy from the pump source is actually pumped into the medium to generate the population inversion necessary for lasing. The total amount ...

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Searching that book for "lineshape function" will return this page, which explains what that means. Essentially, the atoms in the gain medium are usually able to respond to frequencies $\omega$ which are close to, but not necessarily exactly equal to, the central frequency $\omega_0$. The response is strongest at $\omega_0$ and then it tapers off over some ...

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To give an answer we have to remember the historical facts. While making a light source point like and for better visible results nearly monochromatic one can see behind an edge an intensity pattern (fringes) right and left from the geometric line of the shadow. Hence the intensity pattern are something equal to water waves light has to have wave ...

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The issue here is how much the refractive index $n$ tells you about dissipation. As you rightly said, the imaginary part of $n$, which depends on both real and imaginary parts of $\epsilon$, leads to an imaginary part in k which describes an exponentially decaying electric field. However, this doesn't necessarily correspond to dissipation (i.e. a drop in ...

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The rest mass of an electron is 0.511 MeV. When an electron and a positron annihilate their mass turns to energy (two 0.511 MeV photons) so for each annihilation an energy of 1.022 MeV is released. One electron volt is $1.602 \times 10^{-19}$ joules, so in joules the energy released is $1.637 \times 10^{-13}$ J. You ask what happens if $2.3 \times 10^{28}$ ...

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Would the human start emitting photons and die? EDIT: This answer is an answer to the original question regarding "a billion" positions. The question was subsequently edited to now read "2.3*10^28" positrons. That is not cool. The human would start emitting photons. This is exactly what happens during a PET (Positron Emission Tomography) scan at your ...

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Electromagnetic waves are called waves because there are waves (propagating disturbances), waves in the electromagnetic field. These electromagnetic waves, like material waves, transport energy. According to the Wikipedia article "Wave" In physics, a wave is disturbance or oscillation (of a physical quantity), that travels through matter or space, ...

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In addition to the other answers, back in the olden days they were thought of as oscillations in the ether. As a result of the Michelson-Morley experiment back in 1887, physicists began to think that there was no ether. But the term didn't change.

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Since no one else has mentioned it ... If you want to have a better conceptual understanding of the apparent slowing of light (and other electromagnetic waves) in materials, I strongly suggest reading Richard Feynman's lectures, especially Chapter 31 of volume I. That will give you much more explanation than is possible in this forum. All the Feynman ...

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The electromagnetic waves satisfy the Maxwell equations for waves. They don't need a medium for propagating, because these waves are their own energy-carriers, the photons. By that, they differ from water waves whose energy is propagated by the intermediation of the water molecules, or sound waves whose energy is propagated through the molecules of the ...

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The definition of a wave is not that it is the oscillation of a medium. Waves are called waves because they are solutions to a wave equation, which is, for a generic "excitation" $A(t,x)$ depending on the time $t$ and some spatial coordinate $x\in\mathbb{R}^n$, of the general form $$\frac{\partial^2 A}{\partial t^2} = c^2\Delta A$$ where $\Delta$ is the ...

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Of course something is wrong: if you want to preserve the total intensity you feed each antenna with a current $$I(t) = \frac {A}{\sqrt {2}} cos(2\pi f_c t). \tag{i}$$ About the electric field beware, it is as you say, $\vec E = \vec E_1 + \vec E_2$, but if you follow my $\text {(i)}$, it becomes $$\vec E = \frac {\vec E_1 + \vec E_2}{\sqrt {2}}. ... 2 So, this is an old post that I came across when I had a similar question. Here's a paper where they dissolve different amounts of ions in the water and found that the ability for the microwave oven to heat the water actually reduces as more ions are introduced. 0 Speed of light is constant in vacuum but different electromagnetic waves travel at different speeds in different media due to different refractive index. 3 An oscillating magnetic field is always accompanied by an electric field. This is because Maxwell's equations tells us that (amongst other things):$$ \nabla \times {\bf E} = - \frac{\partial{\bf B}}{\partial t} On the right hand side of this equation the symbol \partial{\bf B}/\partial t means the rate of change of the magnetic field with time, and ... 0 In a bar magnet, if its a metal magnet, there is no electric field. Its the alignement of the iron that creates the magnetic field. If you talk about an electro-magnet, the magnet is obtained by circulating current in circular motion in a coil. the following image represents this : You cannot shield the electrical field, because its inside the wire that ... 1 The Cosmic Microwave Background includes photons that will not be absorbed before the universe inflates to the point where there is nothing to hit ever again. If parts of the last scattering were somehow barred from releasing that energy, I think we would notice. The photon carries energy. Particles do that. How is it any different from the electron, which ... 0 I recall a tech presentation on YouTube that I cannot find to link to after an hour of searching. It was about a laser being built by the military, and he said that it's only an order of magnitude (or something like that) to scale up the technology to a something that would be an interplanetary searchlight, and could also signal over interstellar distances ... 4 This paper contains an important analysis of the different trade-off between bandwidth and energy efficiency. The interesting conclusion from that paper is that the most energy-efficient way to send and receive interstellar messages (over flat spacetime) that maximise the bit-rate requires making the bandwidth of transmission very large. In particular, this ... 14 Some numbers come from a review paper by Cullers (2000), who discusses the SETI Phoenix project. There, it is claimed that the Arecibo dish is capable of detecting a narrow band, coherent signal of f=10^{-27} W/m^2 given a 1000 second observation. Assuming that this is an isotropic signal, then the implied power at distance d is p=4\pi d^2 f, which ... 0 1 - The main reason is that your car RESEMBLES (not IS) a Faraday cage (even though, hey, we are talking about 10 meters - the smallest - wavelength here! The wave doesn't exactly "see" the car - it sees a material that is a mix of air, metal and silica). In the car there are also electronics that COULD also produce noise... But it wouldn't be the main ... 1 The pot may have been steel (not that good a conductor) and thin. The lid may not have fit well, leaving gaps. Cell phone signals are short wavelength, meaning a small gap will not completely block them. Cell phones are good at picking up weak signals. Next time, just turn it off? Or answer it and tell the poor guy where his phone is? 0 An ideal blackbody has a unique emission spectrum for a given temperature: With a uniquely located peak, estimated by Wien's law: \lambda_\text{max} = \frac{2.898\times 10^{-3}\text{m K}}{T}  A cheap estimate might interpolate the spectral maximum from several singular spectral measurements and infer the temperature from Wien's law, assuming the ...

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Emissivity of the material certainly does play a role. Also, if the material is particularly reflective, you might get stray energy bouncing in from the surrounding (i've seen this happen with aluminum). If your sensor absorbed all radiation, you'd use the Stephan-Boltzman law, as you've mentioned. The trick that you need is to find the fraction of the ...

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"Looking at diagrams of Electromagnetic Waves, it b would appear to me that at certain times the waves have zero amplitude, and consequently zero energy. Indeed, substituting in the sinusoidal terms into the Poynting Vector equation, It would seem that at certain times the energy disappears. Why is this not the case?" Anthony. You are right Anthony, the ...

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An electromagnetic wave doesn't violate the conservation of energy. Maxwell's electromagnetism does. When you apply the Maxwell's equations to two beams of light of the same frequency, collinear, out of phase 180 degree, occurs the phenomenon of destructive interference. In this special case the electric field of one wave is canceled with the electric field ...

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If you have a thin circuit with a total resistance $R$, and place it in an external (changing) $\vec{B}$ field, then there is flux through the ring. First, there is flux $\Phi_1$ from the external, changing $\vec{B}$ field. Since that $\vec{B}$ field is changing, there is an emf due to that. Second, the current from the ring itself produces its own ...

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This is simply a rotating magnetic dipole, much like a pulsar. Yes, it will radiate electromagnetic energy, as per standard M1 radiation formulas. "But this does explain the way the torque is applied to the magnet." Why and how the magnet was set up to rotate in the first place has nothing to do with what happens next. "One way of looking at it would be ...

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In the book "Physics of the Plasma Universe" Dr. Anthony Peratt puts candle flames near the bottom of "energy in electronvolts" portion of the 'plasma spectrum'. If you look at the chart below, you'll see candles flames about midway (ok, cosmologically) between the ends: solar bodies and laser radiation terrestrial flames interstellar charged gases ...

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The red, orange, yellow, and white parts of a candle flame results from glowing soot. The color in this part of the flame is indicative of the temperature. The spectrum in this part of the flame is fairly close to that of a black body. The blue part of the candle flame at the bottom of the flame results from chemiluminescence. Chemiluminescence is not black ...

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The optical isolator component is active. It consumes energy and so is no different (thermodynamically speaking) from the heat-pump in a refrigerator. If you are talking about a passive component that draws no power then, a surface that allowed light pass in one direction only would violate the second law of thermodynamics. To see why, imagine two rooms, ...

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It's like saying sound produces sound: actually, particles move other particles, and this entire process is sound. Huygens' principle talks about propagation, not "emission" in the conventional sense. The light is a wave in the electromagnetic field and the changing field at a point in space induces field in the surrounding space when time moves forward. ...

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Huygens assumed the existence of "ether particles". And every point in wave front, which your statement of Huygens principle is saying to be the source of secondary wavelets, are nothing but the ether particles, which Huygens assumed to give rise to its own individual wave. Huygens said that light waves were longitudinal as they passed through a ...

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The key phrase is "may be considered". That is, Huygens principle, as a mathematical procedure, gives (approximately) the right answer. The principle does not say that light is actually being generated at each point on the wavefront. Another way to look at it: the light at a phase front already exists there. There is no need to generate it all over ...

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I think that the thermal light is a good example. It is emitted by any object of temperature above absolute zero. Thermal light is a chaotic state, not described by a quantum state but by a diagonal density matrix whose elements describe the probability of different wavelengths, different polarizations, and in general no polarization is preferred, i.e. any ...

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The device on the armature is a compact, commercial, bremsstrahlung-xray source. These have been used in dental officess for decades with both film and solid-state detector system Presumably the object you placed inside your mouth was a battery powered, pixelated solid-state detector of some kind. They can be as simple as CCDs like those in a digital ...

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By definition, an electromagnetic wave is a solution to Maxwell's equations in vacuum. The electric field of a EM wave solution is always perpendicular to the direction of propagation. Let me denote this electric field by $\vec{E}_{EM}$. If $\vec{v}$ is the velocity of the wave, then we must have $\vec{E}_{EM} \cdot \vec{v} = 0$. However, Maxwell's ...

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An electromagnetic wave satisfies $\mathbf E = \mathbf v \times \mathbf B$ and therefore the electric and magnetic fields are always perpendicular to the direction of motion in vacuum. Any electric and magnetic field must satisfy Maxwell's equations, for they won't be physically allowable otherwise.

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Something similar does exist. It's called a "directional light filter": https://www.sciencenews.org/article/light-filter-lets-rays-through-only-one-direction Note: I said "similar" as you asked in your question. This isn't exactly what you're looking for but it's step in the right direction. This might be more on track. A "wave diode": ...

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An interesting type of camera has become more popular in recent years because of simplified technology, both for still and video, in industrial and military settings. It's hyper-spectral imaging. These cameras have the ability to discriminate thousands of wavelengths from near IR to UV and selectively display them, usually with false color. It is useful ...

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What you wrote is a solution normally used far away from its source and in a limited volume. These solutions make sense as an external force in the equation of motion of another system with charges. This wave may well be absorbed and turned into heat. In other words, we always deal with emitters and absorbers so the wave equations are not "free" and are ...

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I am asking whether the field preexists? Imagine a particle which is at rest atfirst.The particle has electric field around it to infinity.(As we know electric field has infinit range and the particle has no magnetic field). NOW lets suppose somehow the particle begin to oscillate. And also imagine a region far away from the charge. An observer in that ...

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The solution you have written down is the solution for an infinite plane wave. The wave described by this solution has existed for an infinite time and will continue to exist for an infinite time, and it has an infinite extent in space. This is an idealised solution that doesn't exist in the real world because in the real world EM waves are created, ...

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