# Tag Info

## New answers tagged electromagnetic-radiation

2

Is there anything wrong with this understanding? A few minor points: Black-body radiation does not refer to a specific microscopic process, and in particular is not necessarily "the bonds between atoms giving off energy". Black-body radiation is an explicitly macroscopic phenomenon, whose usefulness comes from the fact that it doesn't particularly care ...

-2

This is a fundamental question requiring fundamental thinking. I shall keep away from theories and concentrate on simple facts. From the day we knew of the Brownian motion and realizing that particles of matter are on a continuous motion and not at rest, we should have realized that motion and not rest is the true influential variable of nature. Velocity ...

1

The electric and magnetic field vectors of a transverse electromagnetic wave are always at right angles to each other and in phase (for waves in a vacuum or non-conducting medium), whatever the state of polarisation. The polarisation state refers to the direction of the electric field vector, from which the magnetic field vector can always be inferred. One ...

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As pointed out by Hepper and Jon Custer, at the frequencies of operation of the system, significant bending of the radio waves will occur along with reflection off the ionosphere- making over-the-horizon transmission easy.

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Unfortunately, a one-way window in the sense your talking about would contradict the second law of thermodynamics: Say you have two chambers with a one-way window between them, the direction selective transmission would then cause spontaneous energy transfer in the direction of allowed transmission. If the two chambers were initially at the same temperature, ...

0

The refractive index of air depends slightly on temperature, barometric pressure, and humidity. A simplified formula given by NIST is $$n = 1 + 7.86\cdot10^{-4}p/(273+t) - 1.5\cdot10^{-11}{\rm RH}(t^2+160)$$ where $p$ is pressure in kPa, $t$ is temperature in Celsius, and RH is relative humidity as a percentage. This formula is valid specifically for 633 ...

0

I don't see what this has to do with the title. Anyway, if you oscillate one end of a long rope (the equivalent of the electromagnetic field in this case) with your hand (the equivalent of the oscillating body in this case) at frequency $\omega$ then the propagating wave in the rope will oscillate at the same frequency. This is an intuitive analogy to ...

6

Here is an argument based on spacetime symmetry and the fact that U(1) theory is abelian (therefore photons don't interact with each other) to show that the probability is zero. Suppose we start from a uniform magnetic field in the $z$ direction. Since there is no preferred direction in the $xy$-plane, the final state photon must be emitted along the $z$ ...

0

It would be reflected back (almost) completely if the earth was metal but it is not. In fact the earth and sea present a dissipative load to the current pulse and there is also a fair amount sideways radiation that propagates very well in the cavity resonator formed by the ionosphere and the earth. It can and does propagate all around the earth and these ...

5

$k = \dfrac {2\pi}{\lambda}$ where $\lambda$ is the wavelength of the wave. $\cos(kz - \omega t)$ is the equation of a (plane wave) travelling in the positive $z$ direction. You can think of it as $\cos(\vec k \cdot \vec r - \omega t)$ where $\vec k= k \,\hat z$ and $\vec r= z \,\hat z$. The direction of the k-vector gives the direction of travel of ...

1

Sentinel 2 has three sensor ranges -- visual, near IR, and Short Wave IR. Yes, the near IR is affected by smoke, and this will degrade its image quality near forest fires. But the SW IR penetrates smoke and will give ground data even near forest fires. While the LW IR would provide slightly better smoke penetration than even SW IR, LW IR gives lower ...

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All the material in the outer region of the Sun has thermal motion, which includes vibration and continuous excitation/de-excitation of atoms and ions. This goes on at all frequencies from zero to high, with a peak around the visible spectrum and tailing off above that. This motion produces light through various mechanisms, including electrons changing ...

3

The main source of opacity that defines the visible photospheric continuum of the Sun is that due to H$^{-}$ ions. This has been understood for 80 years (Wildt 1939). These ions are in an equilibrium with neutral hydrogen atoms, where the additional electron can be ejected due to the absorption of photons (bound-free absorption) with a continuum of energies ...

1

Correction to the derivation in the question, as per the comment under it: Your error is here: "If we approximate the wave as a plane wave, so that along a given $\hat{x}−\hat{y}$ plane the E field is constant at a fixed time....", instead at a fixed time in any direction you have still a sinusoidal spatial distribution of the field components. Only ...

0

This question is related to an area of study known as plasmonics, which is the coupling of light to a solid-state version of a plasma. This phenomenon is responsible for the differences in colors between silver, gold, and copper, which is related to differences in the naturally resonant frequency of the electrons in these metals' free-electron gas plasmas. ...

2

Generally, the size of the system that produces an EM wave is comparable to the wavelength of the wave. Radio waves come from antennas measured in meters, microwaves from resonant cavities measured in centimeters, light from electron orbitals, and gamma waves from nuclei. A circuit and antenna to produce light would have to be the size of a molecule. Its ...

1

Visible light cannot be generated using switched or oscillating currents with today's technology. I believe that THz radiation, with wavelengths on the order of 0.1 to 1 mm, is the closest that has been generated using high frequency currents. This band is also called submillimeter radiation, and barely reaches the far infrared range. Various high ...

3

Plugging a plane wave solution $\vec{E}=\vec{E_0}e^{i\vec{k}\cdot\vec{x}-i\omega t}$ into $\nabla\cdot\vec{E}=0$ yields $$0=\vec{E_0}\cdot\vec{k}\;e^{i\vec{k}\cdot\vec{x}-i\omega t}$$ and thus the field is necessarily orthogonal to the direction of propagation. You may also start from an unknown field $\vec{E}$ and Fourier transform $$\nabla\cdot\vec{E}... 1 I defer to: https://webhome.phy.duke.edu/~rgb/Class/safety.html They do list X-ray generation, however the primary risks are ozone (which destroys organic compounds; your body is primarily composed of organic compounds...) with insufficient ventilation and electric shock. X-rays are typically generated using 20kV to 60kV. A Tesla coil can definitely reach ... 0 An alternative to measuring the time between peaks, would be to count the number of peaks which occur each second. Either method would require an accurate internal clock. 1 inside the multimeter there is a peak-detection circuit which finds the topmost point in a sine wave that is applied to its input leads. Once it detects a peak, it starts a timer running which is shut off when the next peak in the sine wave train is detected. the meter then knows the time-between-crests of the sine wave and from that it calculates the ... 0 If it was a real physical experiment, the repulsion of the charges on the sphere would have to be compensated by some tension of the surface, eg. a rubber based balloon. Then the pulsation of the sphere would be nothing but a harmonic oscillation around the equilibrium, where the tension is equal to the repulsion. For larger radii, the tension exceeds the ... 1 The energy required to compress the sphere of charges goes into the electric field which has an associated energy density of \frac{1}{2}\epsilon_0 \mathbf E^2. As the radius of the sphere decreases, the integral of this quantity increases. Per Poynting's theorem the amount of this increase is equal to the electrical work done compressing the sphere. Note, ... 1 Your guess as to how they 'hear' the 50Hz frequency is correct. If you have fiddled with audio electronics, or played the electric guitar you will know that 'mains hum' – 50Hz (sometimes 100hz) in countries with 50Hz mains and 60Hz (sometimes 120hz) in countries with 60Hz mains – is a pervasive problem which leaks everywhere. Occasionally hum leaks into ... 0 If you have a collection of blackbodies, these will after some time be in thermal equilibrium with each other, after which they can be thought of as emitting as one single blackbody. 0 Blackbody emmitters are of macroscopic dimensions, there is no black body model for individual atoms or molecules ( chromophores a a chemistry definition ). Black body radiation is one of the main reasons quantum mechanics was needed to describe the microcosm. Have a look here. The black body radiation is built up by all possible electromagnetic emissions ... 0 it has an AC power supply connected to an unbroken (not sure what you mean by "broken") circuit called a microwave oscillator. At the frequencies generated by that circuit, the dipole antenna it feeds looks like a piece of hollow pipe with a flared-out end on it, pointing into the oven chamber. The necessary amplification is generated by the vacuum tube (... 1 Is the purpose of this screen to construct a Farraday cage so none of the radiation can escape? The purpose of the screen is to allow persons to view the contents of the microwave oven. It allows visible light to pass through but not microwaves, You are correct that the metal screen is a form of a Faraday cage. The the diameter of the openings, on the ... 0 An electron is an elementary particle of fixed mass. It can scatter off a photon, (which is also an elementary particle); if accelerated it can emit a photon, but it does not absorb it, because the electron's mass is fixed, and if it were able to absorb a photon - at the electron's center of mass - the mass would have to change, which contradicts ... 2 The concept of a blackbody is an idealization to describe a surface that absorbs any photon with which it comes in contact - no matter what the energy (i.e. wavelength) of that photon and no matter what its angle of incidence. It is a perfect absorber. An early empirical law (Kirchoff's law) states that this ideal body is then the most perfect of emitters at ... 1 The solar constant is about 1360\,\mathrm{W/m2}. This is the amount of energy which reaches the top of the atmosphere (and it varies a little depending on where the Earth is in its orbit &c &c). The amount of energy actually reaching the surface is lower, and can vary between about 500\,\mathrm{W/m2} and 1000\,\mathrm{W/m2} (obviously on a ... 0 Scattering experiment of thin milk: Mix a small amount of milk with water in cubic glass container. In the dark, we use a flashlight to illuminate the liquid from one side of the container. On the other side, we can see that the light passing through is reddish. On the top of the container, we can see that the scattered light is blue. The color of the sky ... 3 When the Sun is directly overhead, the light is going through about the equivalent of 8km of sea-level air, even though the atmosphere is roughly 50km high. When the Sun is just on the horizon, it’s 15 times as far to that 50km height, with approximately the same density profile: there about 120km of sea-level-equivalent air, or about 15 times as much. ... 3 Short answer: Yes, you pretty much summed it up. However, I wouldn't go as far as describing a wiggler as an "underperforming" undulator. Both these devices convert kinetic energy into radiation. A wiggler can produce magnitudes more radiation than an undulator, precisely because it radiates in a wider spectrum. Its overall output power is therefore much ... 0 As you have problems with the quantised picture of light, let us start from the quantised light picture and work toward the wave picture. Let us assume that we have photons (=quantised light particles). Furthermore, let us assume that each photon has the energy of 2eV, which is larger than the needed 1.8eV from your question. What happens if an electron ... 0 You are correct, if a perfect mirror is moving away from the radiation source, the reflected radiation is red-shifted, and vice versa. The frequency of the reflected waves, as observed by the detector, is$$ f = \frac{1 - v/c}{1 + v/c} f_0 $$where f_0 is the original frequency and v is the speed at which the mirror is receding from the source. There ... 0 The collision between an object and a photon can be modelled like any other collision in physics; you take a reference frame (I recommend the rest frame of the object being struck) and give the photon an energy h\nu and a momentum \frac{h\nu}{c}. If the photon is absorbed, then the object gains velocity \frac{h\nu}{M}, or if it is reflected then the ... 1 Light bodies like bacteria and viruses are pushed out of the solar system by radiation pressure. So yes, they can be transported by light but not to us as long as the sun shines. 0 The key is to work with multimode fields. Basically, the reason that field vanishes is because you are integrating field (or intensity) from an atom over a fixed area, like the size of your detector. If instead you were looking at the intensity over a fixed solid angle, you would probably find the intensity does not change with distance away from the ... 0 The expectation value of the electric field on a Fock's state is zero. It is a well known result which can be verified by taking into account that the electric field is a linear combination of a photon creation and a photon annihilation operators. But the expectation value of both operators is zero on any eigenstate of the number operator. In oder to compare ... 2 All objects in space emit a spectrum of electromagnetic radiation which closely approximates a black body spectrum that depends on their temperature. In addition, space is filled with electromagnetic radiation (also approximating a black body spectrum) corresponding to a temperature of about 2.7 degrees K, which approaches a true blackbody spectrum to a few ... 1 Yes if you look at a powerful laser you can see dust being pushed by the laser, so they can be strong enough to move bacteria or viruses. 0 The diagrams of diffraction you have shared don't show light as a longitudinal wave, they are a completely different representation of waves. The lines represent wave fronts NOT compression & rarefaction. Wave fronts can be thought of as corresponding to wave crests, but really mean points on waves that are in phase with one another. -2 You asked a very reasonable question. The EM spectrum from gamma to infrared is the emission of photons from excited electrons and other subatomic particles: Bombardment of a nucleus with high-energy particles mainly produces X-ray photons. The kinetic energy is partly converted into photons. The annihilation between a particle and its antiparticle leads to ... 0 I tend to think particle-wave duality this way: The actually physical state of the system is some abstract mathematical object. In old fashion quantum mechanics, it's a wave function (or ray or whatever). In quantum field theory, it's the quantum field (super position of many classical field). What do we mean by a free particle? One way to informally "... 1 There are no different types of light in the elementary sense. Whys should we distinguish types of light based on frequency or spin, as we don't do this for any other particle. 0 It's up to you how you think about electromagnetic radiation. The more EM phenomena you observe, the more you will relate those phenomena to light as particles or aa waves. In order to explain photoelectric effect, Einstein suggested that light is made up of photons The photoelectric effect is one of these phenomena which where not explainable with light ... 3 Your question is a little confused at the end, but I think the answer to what you're trying to ask is "yes". Names like infrared and gamma apply here to ranges that have been divided up for historical and practical reasons, but they do not denote something other then an electromagnetic wave within certain frequency ranges. If you had the hypothetical ... 0 [F]or some reason, it seems too simple in my head that anyone at home could theoretically create anything from radio waves to gamma waves by generating electrical signals at different frequencies. I don't see how to easily create a device that vibrates at 10¹² Hz, not even 10⁶ Hz. Technologically, this is not as simple as it sounds. As mentioned by @Brick, ... 1 As my2cts mentions in the comments, the magnetic field in the presence of currents is only partially determined by the electric field. However, in vacuum they do determine each other fully. Remember that the magnetic field is not only governed by Farday's law of induction$$ \epsilon_0^{-1}\mathrm{rot} D = -\mu_0\partial_t B  but also by Ampère's law \$...

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