New answers tagged

1

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 ...


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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 ...


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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.


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If you have a system of independently oscillating point charges as radiators and they do not have a coherence among themselves. Then, If you take single dipole it radiate in a dumbbell shape. If you orient these radiators randomly oriented in space the radiation will propagate as a spherical wave. If you let this wave pass through a slit then you will see ...


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 ...


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Most objects are porous and allow the elector-magnetic signal through. Little electrons can pass through what appears to be solid material like photons through glass. Some are still trapped in the atoms. Denser metals like lead do not have enough space between the atoms to allow the electromagnetic waves through. Certain wavelengths can pass through certain ...


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Does light carry any information from the source? Definitely. Light has properties that can be used to determine material of the source, as well as material through which the light ray has passed. Spectrum of light, is a great source of information which is actually equivalent to timber for sound. This is how astronomers find out what elements make up a ...


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 ...


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 ...


0

In a sense, the size of the universe limits the wavelength of a photon: any photon that has larger wavelength than the size of the universe, cannot exist entirely within this universe. It is not clear that this can ever be tested, however. In high energy (short wavelength) the lack of a limit to thermal radiated light was an important reason for the ...


0

Brian Dodson has posted a brief explanation to this at Quora. Basically it is suggested that "the largest energy that is sustainable as an electromagnetic wave is approximately 1 MeV, or a wavelength of 0.01 Angstrom." https://www.quora.com/Whats-the-longest-possible-wavelength-lowest-possible-frequency-lowest-possible-energy-of-electromagnetic-radiation ...


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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.


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If your walls are bonded to floors and ceilings with similar conductive material, and if doors are conductive and gasketed... and if windows and ports such as air ducts are treated with screen or mesh conductors over their apertures, then it's a Faraday cage, all right. For safety, as required by most building codes, it must be grounded (bonded to the ...


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Noorism philosophy of universe interprets this phenomena as one example to conversion of '' vacuum energy in to EM energy '' that is vacuum energy and EM are inter-convertible.An accelerating or decelerating electrons causes the fluctuation in the vacuum where C N F space time field is present. Charge and spin of electron produces electromagnetic version of ...


0

1)Consider it rather as a mirrors than a Faraday cage. You protect yourself from the high frequency waves, not just electrostatic field. Faraday cage stops only waves with wavelength bigger than the holes in this cage. The rest i don't know.


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You want to read the classic paper by Richard Beth, Mechanical detection and measurement of the angular momentum of Light, Physical Review 50 115 (1936). Beth used bright circularly polarized light to drive a torsion pendulum in a vacuum chamber, and was able to observe torques due to circular polarization of order $10^{-16}\rm\,N\,m$. This was with a ...


0

   The metal mesh, or 'cage' around a microwave's oven cavity acts as a Faraday cage (see Wikipedia article on Faraday cage here), although a 'true' Faraday cage is grounded, and a microwave cage is not.    A cellular phone inside a Faraday cage will be protected from outside EM transmissions, just as conversely, the ...


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 ...


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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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The actual reason that radiowaves do not penetrate the Earth, which is that the Earth is partly or very conductive and either dissipates or reflects the waves. The dissipation length or "skin depth" at most radio wave frequencies is a few to tens of metres at most. However, this conductivity, especially seawater, combined with the conductivity in the ...


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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 ...


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Radio waves have frequencies between 3 kHz and 300 GHz which means wavelengths between 100 km and 1 mm because $c = \lambda \nu$ where $c$ is the speed of light, $\lambda$ is the wavelength, and $\nu$ is the frequency. Sometimes, a factor of $2 \pi$ appears in that equation depending on how frequency is defined. As a rule of thumb, if the wavelength is much ...


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. ...


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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 ...


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Kinetic energy and photon absorption Particles - when moving into a magnetic field - have gained before kinetic energy. As GRB wrote: Every time a charged particle has to be accelerated, a photon has to be involved. If you want to linearly accelerate a charged particle, you have to shoot photons at its back. To be precise this one of the possibilities ...


0

If you assume that you can't change the signal faster than the frequency of the wave then the frequency sets an upper limit on the number of changes in the signal per second. In fact you can send information faster than this with some clever coding schemes if you are able to measure the amplitude/phase of the signal accurately.


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When it comes to gravitational waves, no we have not determined them to be EM waves. Although we can't completely rule it out. For example, there is a theory that reconciles EM force with weak nuclear, and it is called Electroweak. Electroweak interactions happen above 100 GeV, when EM and weak nuclear become the same force. Edit: Maybe someone will develop ...


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It turns out that this structure is the a doppler broadened line coming from the $^{10}$B(n,$\alpha$) reaction which populates the 477.6 keV excited state in $^7$Li.


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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.


1

There is no physical principle that á 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 ...


0

I like the following explanation. Although mostly mathematical, it illustrates a point Feynman tried to convey in his lectures. Energy is always conserved, and whenever it seems like it isn't you just need to look harder. Let's recall that in the context of classical particle mechanics one defines the energy of the system of $n$ particles as the sum of 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 \...


0

Wow, Your question has many aspects. First, We are surrounded by electromagnetic fields. We can divide these fields in two parts 1). Ionizing fields (having wavelength ~200 nm and below) and 2) Non ionizing fields (wavelength above ~300 nm). Ionizing radiation (such as x rays and UV rays) can affect our body even when their energy density is small, because ...


0

The best intuitive explanation I can come up with is that the Fresnel equations, which perfectly describe the reflexion of light from a plane, optical quality interface, are an expression of the requirement that the field vectors should be continuous across the interface. This requirement is itself an expression of the absence of either static or flowing ...


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 ...


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 ...


0

You cannot immobilize light, but you can absorb it, convert it to heat. A first step in measuring light intensity (watts/square meter) is to make a black body to absorb the light. Then, by conservation of energy, the heating of that black body tells you how much light is being absorbed, in watts. The illuminated area of the black body is the "square ...


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

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\...


0

A light wave is a traveling disturbance in the electric and magnetic fields. In the far-field these two components are coupled and in-phase so that one can define the amplitude entirely in terms of either the electric field strength or the magnetic field strength; by convention we give the electric field strength. So, in SI, the amplitude of a light wave is ...


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.


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If I may simplify this a bit.........I will compare to a gong. When you stike that gong sound propagates though air. Photons are the end result of gamma radiation. When you 'bang a gong' that energy is released. 'It' must radiate out. Same as with light energy or photons.


-1

So an amplitude in terms of an electromagnetic wave is related to its intensity. So technically, the more intense a light wave is at a point, the more its amplitude. What is the intensity Intensity is inversely proportional to the A^2 of a wave. If you know the intensity, then the amplitude can be found using this formula I=1/2*pvw*A^2 where p is density, v,...


0

Direct measurement of the amplitude of the optical field requires interferometric techniques. One that works is the FROG - Frequency Resolved Optical Grating. There are many variations on it today, including from the original developers: Trebino Research Group. These devices were designed for ultrafast pulses. For CW one usually just measures intensity ...


1

In general, gaussian beams are understood to be taken as linearly polarized, but this is only approximately true. The full polarization is a complicated object to handle, and it is not necessarily available in a closed elementary form. It is described, for example, in Analyses of vector Gaussian beam propagation and the validity of paraxial and spherical ...


5

A comment about coherence in general I find the definition of coherence as some sort of "unrelated phase" problematic for a couple of reasons: This formulation somewhat implies that coherence is discrete, i.e. there is incoherent and coherent. Of course that is not true, you can have a continuum partially coherent states. But what quantity are you going ...


2

Apart from the heating due to sound absorption, as per the comment by HolgerFiedler, I don't think you will find a mechanism that can radiate due to polarization effects in the medium. Any EM radiation would be at the frequency of your acoustic waves. With the difference between the speed of sound and the speed of light, that would be very-long-wave ...


3

This is essentially the same question as "how does a tiny dipole antenna transmit a wave of several kilometers wavelength"? If you're designing an antenna to transmit a wave, you can indeed get a dipole a few centimeters long to transmit low frequency RF, say less than one MHz. Many commercial transceivers do this; it's not ideal, and you need to drive the ...


2

It's like this: The speed of light is constant c. Now, during the time when one electron 'jumps' from a higher state $\psi_2$ to lower state $\psi_1$, it oscillates from those states at a frequency equal to $\nu$ in $E = h\nu$, the energy of the photon released. The oscillation induces the formation of electromagnetic waves. So, the frequency of a light ...


0

There are lots of emission and absorption lines – at different energy scales. Atomic spectra, vibrational spectra, rotational spectra, and so on. For each of them, one may discuss millions of molecules in principle (most of the large number are organic molecules). The amount of data to cover "everything" is clearly unrealistically huge. But any subset of ...



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