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29

You have your "prove" in the wrong place. The way to prove that ground-state electrons in hydrogen atoms don't emit radiation is the following: Construct a sample of ground-state neutral hydrogen atoms. Place this sample near a detector which is sensitive to the sort of EM radiation you expect. Die of old age waiting for a signal, because ground-state ...


21

At tree level, a matter-antimatter annihilation reaction doesn't just produce gamma rays, nor can you exclude neutrinos in the final state. Even the simplest such reaction can—given enough energy—produce a variety of particle-pairs. However, those pairs are subject to two subsequent processes: If the particles are not stable, they will decay towards ...


18

The short answer is that there is no known theoretical strictly positive lower bound to the speed of light. Any positive number, no matter how small, is possible, although limits are set for each candidate material, as I explain at the end of my answer. One has to be pedantic to understand the lack of limitation to a more generalized "speed of light". From ...


18

Even though there is a single photon in a volume of your choice the light is still a wave. An experiment was performed which proved this. In this experiment a Michelson interferometer was set up and the incident light is so weak that only one photon was in the whole setup at a time. A photographic plate was used to detect the interference pattern. Now just ...


16

How the light slows down in a matter, it depends on the recipe of its refractive index. For common, ordinary materials it is in the range of 1-3. Bose-Einstein condensates have an extreme refractional index, even millions or billions. In a BEC with a refractive index of $10^9$, the speed of light is only $30~\mathrm{cm/s}$. Here is a relative old article ...


15

The existence of hydrogen atoms is enough to demonstrate that the electrons don't emit radiation. If they did, that energy would have to come from somewhere. The only place it could come from would be a reduction of orbital radius until the electron finally reaches the nucleus. If you accept that electrodynamics applies, then you have to accept that atoms ...


13

Looking into this, the first thing that came up was a technology that used infrared light to determine the curvature of the pages of a book so it could be scanned non-destructively (in normal scanning, the pages need to be flat - before google's idea this was only possible with glass plates, which was inefficient, or dis-binding the book, which destroyed the ...


13

The proposal probably refers to terahertz (THz) imaging; THz is the band that is above microwave, but below infrared, between 100 um and 1 mm wavelength; often referred to as T-Rays. Because of their relatively long wavelength the T-Ray penetrates well into non-conductive materials, and is far, far below the ionization threshold. Thus it is safe to work ...


11

The power radiated by a charged particle moving at non relativistic speeds, whose acceleration is $a$ and charge is $q$, is given by the Larmor Formula $$P =\frac{2q^2a^2K}{3c^3}.$$ The product above is nowhere near $c^3$ for planetary motion and hence the radiation emitted is negligibly small.


11

In the field of PET (positron emission tomography) people tend NOT to call the product of the annihilation "gamma rays", but rather "annihiliation photons". While that may be a subtle distinction, the view is that a "gamma ray" is emitted by a nucleus, while a "photon" is a more general term used for a quantum of electromagnetic energy. But according to ...


9

Because of its wave nature, the electron in its ground state is actually smeared symmetrically about the proton (ignoring spin-spin effects), and spherically symmetric charge distributions do not radiate (there's no special direction). Accelerated charges do not always radiate em radiation. See also How to find the magnetic field due to a revolving electron ...


9

Light does not slow down during a reflection. Light is a signal disturbance in electric and magnetic fields. These disturbances propagate through space at a fixed speed $c$ in vacuum. The situation is completely analogous, in a mathematical sense, to a wave pulse that is sent along a string. When the pulse encounters a boundary, it flips direction, and may ...


8

I believe some of the answer in the links are correct, others are less obvious and might even be confusing. I am not gonna repeat the arguments there, but to stress the following idea. You cannot demonstrate that using classical electrodynamics. The theory as is does not apply to quantum objects and thus it was modified. The equations are the same, they are ...


6

In this case, it would be useful to not consider light in its particle form as photons, but instead to consider it as a wave - see this wikipedia page. Then, the wave is simply reflected from the surface, without us having to consider the kinetics of any particle. The wave, in a vacuum, would continue to propagate at the speed of light, regardless of the ...


6

Correction, a single photon does not have a circular polarization. It has spin +1 or -1 to the direction of its motion. Qualitatively Left and right handed circular polarization, and their associate angular momenta. The way the classical wave emerges from the quantum mechanical level of photons is given in this blog entry, and it needs quantum ...


5

Let me explain in purely classical terms (not the description of reality, but easy to imagine). You realized that when a ball bounces off the wall, at a certain point, it has no momentum. However, it must still have all the energy of the movement (neglecting losses to environment) - where did that energy go? The ball is formed of many discrete atoms, bound ...


4

In addition to the answers already given, which answer the question pretty-well, I'll say that, historically, this exact question was the one which puzzled Niels Bohr enough to inspire him to advance his famous theoretical-explanation for the several observed frequencies of the radiations emitted from hydrogen-atoms ... in general, the fact that electrons in ...


4

The battery on your smart phone only provides approximately 3 W of power to the phone's transmitter. This is enough power to reach a nearby cell tower, but totally insufficient to send a signal for several hundred miles. Due to this, the nearest cell tower is used to relay your signal to another cell tower that is located close to the party that you are ...


4

Field interaction does not "give rise to particles". In fact, field interaction makes it particularly difficult to speak of particles. To understand how particle states and fields are interrelated, we must employ quantum field theory. This answer of mine roughly sketches how a particle state is defined in QFT - and the fact is that such particle states are ...


4

Why 2 distributed lines is represented with series inductor and resistor along with parallel capacitor and resistor? What is the motivation for that? Current through the wires is associated with (let's not discuss which causes which) a magnetic field, and we can represent that with an inductor. Potential difference between the conductors of the line is ...


4

The answers given here make me wonder, because I sense in here perhaps a misunderstanding. Or maybe I'm wrong, which might be more likely. :-) The answers here refer to distances light travels. But as far as I understood, light is never slower than 299 792 458 m/s. I guess it may "look" like from a point of reference that light has slowed down, when a event ...


4

To expand on a @ACuriousMind's comment, consider an electron-positron pair such that their center of mass is at rest. Since energy and momentum are conserved in the annihilation process, the resulting photons will be emitted in opposite directions with an energy $E = m_e = 511\ \text{keV}$ each (units such that $c = 1$). If the pair is instead in motion, one ...


4

The amount of energy to emit is enormous ($2mc^2$ plus kinetic energy), so the emitted energies per photon will be at least the rest mass of the electron, which is half GeV (not counting some extreme and unlikely redshifts due to the collision point moving with respect to the observer). And that's well in the gamma spectrum.


4

The electromagnetic field in a medium gets attenuated exponentially. $$ \mathbf E = \mathbf E_0e^{-x/\delta} $$ Where $x$ is the distance the signal has traveled. Since the power of the signal is proportional to the square of the field, then the power will be attenuated by $P = P_0 e^{-2x/\delta}$. The quantity $\delta$ is called skin depth. Its the ...


3

Dielectric materials have a loss tangent : at a given frequency, the response to EM radiation will have a phase lag relative to the incident wave. For pure dielectric materials, the phase angle is zero; when there are losses in the material, the loss angle will be non-zero. For a given loss angle (usually given as the tangent of the angle) the amplitude of ...


3

Range of electromagnetic radiation is a function of several factors: Inverse square law: if you can consider your source to be a point (good approximation when you are many times further than the size of the source), then intensity (energy per unit area) drops with the inverse square of the distance. In the setting of a home WiFi, that makes the signal ...


3

Many ways to detect single particle radiation, charged or not. Scintillation counters can use photomultiplier tubes to detect single particles, single protons, neutrons, positrons, uncharged gamma rays etc. It depends on them having enough energy to ionize the material used for the scintillation. Plenty other ways, for instance a modified version of MRI ...


3

In general, the interaction of matter with light is dominated by electrons because they are lighter and this makes their coupling with light stronger than that of nuclei. However, this doesn't mean that nuclei cannot interact with light: they are still electrically charged particles, and they do interact with electromagnetic radiation to some extent. On the ...


3

In a comment elsewhere you write that you're interested in understanding how quantum-mechanical theory describes the radiation that a hydrogen atom does and does not emit. In your question you ask about another answer that suggests some significance to the electron having zero total momentum; I think that's a feature of the coordinate system choice rather ...


3

The blue end of the spectrum is higher frequency, meaning shorter wavelength. A cage that can block blue is going to block (technically attenuate) the given wavelength and any that are longer (lower frequency). If you start blocking at blue, you'll be blocking the entire visible spectrum and on down into the IR, microwave, and radio. To filter out UV, the ...



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