# Tag Info

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We could make an analogy with sound waves to understand what is the classical outcome. If we are hit by a very loud noise, we can damage our ears, no mater it has low (like an explosion) or high pitch. But being exposed to a low intensity sound for a long time is not dangerous. It is what we experience in our normal life. So I think that the picture ...

1

By assuming light is a wave and not a particle, are you assuming classical physics is correct? There are no photons? If so, your point 1) is right. That was one of the mysteries of the photoelectric effect that was hard to explain classically. A light of low intensity absorbed by a metal should add energy to the metal slowly. Eventually there should be ...

2

The refractive index of a plasma is $\sqrt{\epsilon_r}$, where the relative permittivity is given by $$\epsilon_r = 1 - \frac{n_e e^2}{\epsilon_0 m_e \omega^2}=1 - \frac{\omega_p^2}{\omega^2}\,,$$ where $\omega_p$ is known as the plasma frequency. The refractive index will become imaginary when $\omega<\omega_p$ and an EM wave ...

0

that device was probably called a solar radiometer.

1

"Maxwell's equations lead to simultaneous perpendicular oscillations of electric field as well as magnetic field with same amplitude and “wavelength”. But, they do not guarantee any propagation of these fields." Wrong. They do. "I never found a reference which provides a practical verification of existence of magnetic field component and ...

2

Warning: long answer ahead. To see pictures, scroll down. A free field is just a cavity with infinitely large boundaries, right? Right. This is how it's constructed: take a finite cavity with periodic boundary conditions, and then let its size tend to infinity. I have also heard from many sources, including colleagues, that you can see the field as ...

3

So-called "K-shell" x-rays are produced when electrons fall from outer electron orbitals to the innermost. (The "L-shell" x-rays are produced when a vacancy is filled in the second-innermost shell.) An electron falling to the hydrogen ground state emits ultraviolet radiation in a spectrum first observed by Lyman. That is, the physics of ...

1

The X-ray part of the electromagnetic spectrum covers quite a large spectrum. I assume medical imaging uses x-rays at the low-frequency end of the x-ray spectrum, and low intensity, to minimize tissue damage. (To make up for that the sensor must be very sensitive.) I expect medical imaging x-ray machines will be thwarted by aluminum foil. In industrial ...

-1

I think it is because shorter wavelengths are higher energy and more easily interact with matter. Similar to how UV is more easily absorbed than visible and visible is more easily absorbed than near IR. The same thing happens with 5.8GHz vs 2.4GHz WiFi. Longer wavelengths also diffract around objects more easily. A longer wavelength won't cast a very clean ...

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I personnally wrapped my phone with silver paper (aluminium foil to be precise) and the phone was shielded. However, the phone has to be completely covered in foil for it to work, a few mm hole will leak enough for the experiement to fail.

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Free-air laser communications can be affected by EMI if that EMI matches the of the detector (i.e. is some other matching light). This is just another way of saying stray light will interfere with your communications if it can get into your detector. But in fiber optics, the internal laser is isolated from external light sources. High frequency EMI (i.e. ...

1

It works like this. You solve Maxwells equations. Then the potential tells you how many photons you can expect on average, that is, it gives you the average of a Poisson or Gauss distribution. This distribution is the experimental prediction. We only know how it works not why it works like this.

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But even in Bohr's model, the fact that there are distinct energy levels does not change the fact that the electron is revolving around the nucleus. It's still accelerating, and therefore must radiate energy! In classical mechanics, yes, but atoms existed, instead of the electrons falling on the nucleus and neutralizing it. The data from Atomic spectra , ...

0

Bohr asserted without proof that when an electron was in an orbital, it was in what he called a "stationary state" in which it could not be visualized as orbiting at all and therefore was not compelled to radiate. He further asserted that without an energy state to occupy below the ground state, an electron in its ground state was incapable of ...

0

The Cosmic Microwave Background (aka CMB) is light radiated from hot glowing matter as it cooled from the Big Bang. It took a few hundred thousand years for the universe to cool enough that electrons could stick to protons, forming neutral H atoms. Before this, the universe was full of free charges, which absorbed light. After this, the universe was ...

3

Classically, there would be no hard limit. In a perfectly silent, classical universe, you could imagine a device to detect the wave. But practically, the strength would eventually decay below the ability of your device to discriminate it from other sources of noise (including those from within the device). In QM, any detector has a lower and lower ...

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Electromagnetic waves are made of photons. These will continue to travel along geodesics unless they scatter or are absorbed. Currently the main model of cosmology has an infinite Universe so if this is correct a photon could keep travelling indefinitely if it is not absorbed.

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Electromagnetic waves in vacuum and air very accurately are subject to what is called the superposition principle, which in turn derives from the linearity of the equations of electrodynamics. The superposition principle says that waves that travel in different directions don't disturb each other waves with different frequencies/wavelengths don't disturb ...

1

Another good method is the Fourier approach. First decompose your aperture function in terms of a set of plane waves propagating in different directions, then propagate each plane wave to the observation plane, then add them up again. Thus two Fourier transforms with a simple phase factor introduced in between them. Sorry I don't have time now to write out ...

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No. There are other sources of electromagnetic radiation : Antennas, for instance: The simplest embodiment is perhaps the one used in the early Marconi experiments: Appling pulses of current from a battery to a wound (insulated) copper coil More generally, according to Maxwell’s equations, setting any electrical charged body in motion generates an ...

3

So how can we calculate the speed of light for different frequency? It depends on how much prior information you want to require of your calculation. In the simplest case you may just look up an approximate formula for the speed of light in your medium of choice as a function of wavelength or (less common) frequency. Of course, this relies on somebody ...

3

This question can be reformulated as: “Does the index of refraction of a given medium depend on wavelength” (inverse of frequency), the answer is given by Cauchy’s formula: https://en.wikipedia.org/wiki/Refractive_index

3

No. There is a Hubble flow, but no Hubble flow effect, i.e., nothing that makes objects tend to move with the Hubble flow. Matter moves in that way on large scales because some historical process (perhaps inflation) gave it that velocity distribution 13.7 billion years ago. Some of that matter later collapsed into the Sun and Earth and so on. Voyager 1 and 2 ...

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Your question is posing a characterization that's grounded in a misunderstanding and does not fit the situation at all. What you described is not what spin is, nor how it is characterized. And, in fact, some of the replies are also grounded in a misunderstanding and mischaracterization. First, photons don't have spin. They have helicity. Spin is an attribute ...

2

Yes. Any type of conductive material will absorb voltage or EMF and emit it. This includes radio, tv, mobile phone, wiFi, hair drivers, appliances, electrical lines, ECT. Any electrical wire, even unplugged, will absorb and emit EMF, its similar to a passive antenna having a "gain dB" level. The problem is that repeated grounding is needed to ...

2

Classical wave theory from Maxwell's Equations implies a rate of electro magnetic energy entering through the surface of a sphere at a rate much slower than is found in studying the photoelectric effect. About 10 minutes to launch an electron to observed velocities vs near instantaneous take off. Compare the surface integral of the Poynting Vector over the ...

8

I assume the photon moves in $z$-direction here: The two polarizations of the photon (corresponding to its two spin states) are not linear polarizations, but clockwise and counter-clockwise circular polarizations (you obtain linear polarization as a superposition of a clockwise and counter-clockwise one). These are completely rotationally invariant. The spin ...

32

If you rotate a polarized classical electromagnetic wave by 180° about its axis of propagation, you don't get the same wave back; you get the same wave phase-shifted by half a cycle. You need to make a full rotation of 360° to get the same classical wave solution back, so by your logic, the photon should indeed have spin 1. This is contrast to a ...

8

The photons are point quantum mechanical elementary particles of spin either +1 or -1, to their direction of motion. Their mass is zero and their energy equal to $hν$ where $ν$ is the frequency of the classical wave built up by thousands at least of photons. The way classical electromagnetic polarized waves are built up by their constituent photons is seen ...

2

Short instruction. In 2D case one needs instead of $\frac{e^{ikr}}{r}$ to put a Bessel function of the second kind. The 2D version of KDI is the following: $$u(r)=\frac{1}{4} \left[ \int_S u \partial_n Y - Y \partial_n u \right]\quad(1)$$ where $$Y=-Y_0(k r),$$ where $Y_0(x)$ is a Bessel function of the second kind $$\pi Y_0(x)=\int_0^\pi \sin(x \sin{\... 1 There are different variants of the Planck's distribution function: spectral density of radiance of blackbody over frequencies:$$B_\nu(\nu,T)=\frac{2h\nu^3}{c^2}\frac1{e^{\frac{h\nu}{kT}}-1},\tag1$$the same density, but over wavelengths:$$B_\lambda(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac1{e^{\frac{hc}{\lambda kT}}-1}.\tag2 These are different ...

4

The values being plotted are "intensity per unit wavelength" or "intensity per unit frequency", which are different things. So it's not just a transformation on the x-axis; the actual quantity being plotted is different. It's scaled by the derivative of wavelength over frequency (which is not a constant and therefore affects the maximum). ...

2

I see two questions. Q1. Does every phenomenon of EM radiation have to be explainable at the level of photons? All electromagnetic radiation involves photons. However, often it is enough to just consider the classical em field. This is the case when the em field intensity is high so that many photons are involved and no shot noise occurs. For low frequencies ...

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Does every phenomenon of EM radiation have to be explainable at the level of photons? No. In quantum field theory, the electromagnetic field is still a field, albeit a quantum one. We call it "quantum" because it uses non-commuting operators. It's not made of particles. Particles (photons) are one of its many possible manifestations, but most of ...

3

Yes, several quantities vary with distance from the source. One useful quantity is the radiant flux (the amount of energy passing through a given area in a given amount of time -- basically how 'bright' it looks from your position). In fact, for a source that is spherically symmetric or far enough away that we can regard it as spherically symmetric, the ...

0

Intensity changes because energy of the wave is absorbed by the medium in which it propagates. Also the wavefront appear different. Far away from a source, waves have plane parallel wavefront, like light from the stars.

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Electric potential depends on distribution of electric charge in space, both on the ring and outside it. Induced EMF can be determined from the magnetic field rate of change alone, but electric potential cannot, because the distribution of charge on the magnetic field source and other nearby bodies is unknown. In a carefully prepared situation where the ...

2

The most important insight to this is that in communications we do not really encode "bits" instead we encode waveforms, in other words messages are encoded into waveforms not into "bits". Now waveform communications aimed at a specific spatial direction have several "dimensions" that can be used to carry the message. time ...

4

Any form of digital radio communication already employs error correction. It's not so much a question of whether or not to use error-correction, but an assessment of the amount of overhead to be allocated to provide support for error correction. Such overhead reduces the bitrate of the transmission. Given the state of technology of error correction I would ...

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At the moment I'm writing this, there are already five answers, so what could I add more? Let us see: "Classical electrodynamics (CED)" is to be used as an opposite to "Quantum electrodynamics (QED)" and means "the theory of electromagnetic interaction between matter (sources for fields) and electromagnetic fields (i.e. time-variable ...

1

I think you may be looking at this from the wrong perspective. Light doesn't exist as some fundamental entity and we have discovered that it produces oscillating fields. Instead, there exists oscillating electromagnetic fields, and we call that phenomenon light. Light doesn't create those oscillating field, it IS those oscillating fields.

3

The EM waves act on charges, and the frequency of the movement of that charges is selected by the LC reception circuit of the radio. But without enough carries present in a good antenna, there is no much signal to select. The effect of the human body is to provide movable charges for the EM waves. When touching the antenna, it increases its role. But even ...

14

Small pocket radios usually have fairly poor antennas in them, because of their small size. But by placing your hand nearby the antenna of that small radio, you are creating a capacitor in which one terminal is the radio chassis and the other is your hand. Any radio frequency signal induced in your skin by radio broadcasts will be capacitively coupled from ...

1

In addition to using the relativistic expressions and the formalism involving the electromagnetic fields to write down Maxwell's Equations, I think the adjective "relativistic" means that one needs to be more careful to treat the whole system of fields and sources [and media ] and measuring apparatuses relativistically (with relative speeds ...

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I partially agree with @DescheleSchilder, but I also think that by relativistic electrodynamics one usually means quantum electrodynamics (QED). The reason is that in QED all particles are treated in relativistic way, on equal footing - this is how the term relativistics, which referred to the teratment of electrons, protons, etc. has become attached to ...

1

Electrodynamics could be considered the study of time-variable electric and magnetic fields via Maxwell's equations. The distinction I would draw with Relativistic Electrodynamics is when one starts to consider how those fields would appear/behave in other frames of reference, although some would offer the valid argument that the distinction is when you ...

0

Take a look at single edge Diffraction patterns: https://images.app.goo.gl/RSYzmToh96thgAfM6 A single slit has two edges with diffraction patterns going opposite directions. A double slit has four edges with diffraction patterns going opposite directions and overlapping each other.

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