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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 ...


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 ...


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 ...


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

Any physical phenomenon is potentially capable to cause some change to any other phenomenon, more or less directly. If it was not the case, the physical world could be divided into completely independent realms; there would not be the one single world we call Nature. Practically though, many if not most of the actually existing interactions between systems ...


5

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 ...


5

I can answer half your question in that a sound can change the path of light. A change in the density of the air produces a change in the refractive index of the air and so a Schlieren photograph can make this visible. Here is a YouTube video to show a sound wave produced by clapping.


5

Amateur "ham" radio operators who communicate with HF (3.0 to 30 MHz) frequencies can hear their own signal as it has circumnavigated the globe. This almost only happens with operators using Morse Code (CW) where the distinct signal can be heard and detected with sub-second intervals. Also, ham operators make a distinction between short path (SP) and long ...


4

If we assume that velocities are well below the relativistic region then in the absence of radiation the equation of motion of the electron is simply: $$ \frac{d^2x}{dt^2} = \frac{Eq}{m} \tag{1} $$ where $E$ is the field strength and $q$ and $m$ are the electron charge and mass. As Lawrence says, the power emitted as radiation is given by the Larmour ...


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

Possibly you are mixing different concepts together. A short answer is the following: indeed, for a single photon, the energy is proportional to its frequency. That is \begin{align} E_{singlephoton}=h\nu. \end{align} However, for a wave package or an electromagnetic wave, the total energy is proportional to the photon number, $n_{photon}$, as well. That is \...


3

All of your claims are essentially true. The angular momentum of light, in both its orbital and spin varieties, is indeed angular momentum that can be transferred to matter to make it spin and give it the garden variety of mechanical angular momentum. This is well explained in the relevant Wikipedia section, with good references for experiments that show it. ...


3

No, radio signals do not go through the Earth. Not for any one reason, but for a series of reasons, many of which have been talked about here so I will make a comprehensive list, and there might be even more to add on! The Earth is quite dense: As @Renan pointed out, x-rays can't even penetrate bone, and the photons in an x-ray are significantly more ...


3

The Larmor formula comes from the Poynting vector $$ \vec S = \frac{c}{4\pi}\vec E\times\vec B $$ The electric field in a relativistic content is derived with the Lienard-Weichert potential $$ \vec E = q\frac{\hat n - (\vec v/c)}{\gamma^2(1 - \vec v\cdot\hat n)^3r^2} + \frac{q}{c}\frac{\hat n \times((\hat n - (\vec v/c)))\times\vec v_t/c}{\gamma^2(1 - \vec v\...


2

The technique is to sight in on known frequencies of sources in the Milky Way and other galaxies. Any signal bearing the multiband set of data is subtracted.


2

How come in all my EM courses...I never had to account for this loss of energy in the form of EM waves? In the simple examples shown in introductory courses, the loss of energy of charged particle moving in external field via radiation is neglected, because radiation of EM waves is a difficult topic and even if it is not neglected, it can be shown its ...


2

The best way to look at this is with Larmor's formula, which gives the power radiated by an accelerated charged particle. $$ P = \frac{q^2 a^2}{6\pi \epsilon_0 c^3},$$ where $a$ is the magnitude of the acceleration. So the short answer is yes, it is true.


2

Light coming from some source usually consists of many frequencies. Distribution of them is called "spectrum of light". Each frequency (in the visible to a human's eye range) corresponds to some colour. If we limit the spectrum of light to a single frequency, we will see this colour. However, not all colours we perceive can be obtained in this way and also ...


2

Hint: Look at the Poynting-vector associated with an electric charge Q and a bar magnet, and then consider $\int\int {\bf E}\times {\bf B} . d{\bf S}$ around a closed surface.


2

Yes, sound waves in a gas, liquid or solid can affect the light passing through it, as the motion of the atoms due to sound waves changes the atomic spacing, and this changes the index of refraction slightly. So the light would be diffracted and some amount of the light would experience a frequency shift up and a frequency shift down by the sound wave ...


2

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 ...


1

As you said waves have particle nature. The correct statement will be to say that the quantized fluctuations of the field can be visualized as particles. An intuitive example would be to imagine yourself in a pond with no ripples. In such a situation you will not feel any thing but if the pond has ripples then you would feel as if something is hitting you i....


1

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 ...


1

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 ...


1

Your question is really broad. As correctly pointed out by @John Rennie there can be so many signals that can pass from the solids. It must be noted that even if the signal can pass through solid it will experience certain losses. Mechanical waves such as sound took advantage of elasticity of the material. The sound oscillations are transferred through ...


1

Timbre is a consequence of harmonic content which is a consequence of a sound being made of multiple pure sounds. Thus a 256Hz square wave has pure sinusoidal components at 256, 768, 1280,... Hz. An equivalent in light terms would be the spectral content of the light. For instance, take the bright yellow D line(s) of sodium on the one hand and a visually ...


1

There is no physical principle that รก priori restricts the range of possible wavelengths of electromagnetic radiation. However, the higher the frequency the more energetic the radiation will be, so it will be harder to make that kind of radiation. As for low frequency you can also have as long a wavelength as you wish. In fact, you can think of a static ...


1

The spectrum of emitted rays is defined by the energy output of the reactions happening in the device and its surroundings. Basically, after an exothermic nuclear reaction, the produced particles smash into surrounding material and bounce for some time until their energy is dispersed. This energy heats up the medium which begins to emit black-body radiation ...


1

It is incomprehensible to me that the most up-voted answers, and the ones posted by the most seeminglly knowledgeable people, all attempt to treat this question in terms of individual photons striking individual electrons. In fact, the phenomenon of transparency is all but incomprehensible in terms of such quasi-QM explanations. The natural, sensible ...


1

This answer is a little circular and like Burley's. Transparent materials have uniform electromagnetic coupling between its molecules. Think of glass as a uniform array of tiny capacitors.



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