# Tag Info

1

Phase-contrast imaging with x rays has been achieved at the LCLS. This is used to probe extreme matter states, such as the ones created by a shock launched by high-intensity laser-pulses incident on metal slabs. See this or this for instance.

0

Radio and the like are best described quantum mechanically in terms of coherent states of the electromagnetic field. These states don't have a definite "number of photons", but are rather like an infinite mush of harmonic oscillator states that behave the most like classical electromagnetism.

1

The question here really is: "When are photons not virtual?". As explained in this article: A virtual particle is one that has borrowed energy from the vacuum, briefly shimmering into existence literally from nothing. Virtual particles must pay back the borrowed energy quickly, popping out of existence on a time scale set by Werner Heisenberg's ...

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This is a perceptive question. Consider the following from the Wikipedia article "Virtual Particle": As a consequence of quantum mechanical uncertainty, any object or process that exists for a limited time or in a limited volume cannot have a precisely defined energy or momentum. This is the reason that virtual particles — which exist only ...

2

In 1995 Willis Lamb published a provocative article with the title "Anti-photon", Appl. Phys. B 60, 77-84 (1995). As Lamb was one of the great pioneers of 20th century physics it is not easy to dismiss him as an old crank. He writes in the introductory paragraph: "The photon concepts as used by a high percentage of the laser community have no scientific ...

8

Both the wave theory of light and the particle theory of light are approximations to a deeper theory called Quantum Electrodynamics (QED for short). Light is not a wave nor a particle but instead it is an excitation in a quantum field. QED is a complicated theory, so while it is possible to do calculations directly in QED we often find it simpler to use an ...

3

In this link there exists a mathematical explanation of how an ensemble of photons of frequency nu and energy E=h*nu end up building coherently the classical electromagnetic wave of frequency nu. It is not simple to follow if one does not have the mathematical background. Conceptually watching the build up of interference fringes from single photons in a ...

7

Note carefully Nick's comment. Suppose I send two plane EM waves on some collision course so they interfere. The waves will pass through the region where they meet, generating some interference pattern in that region, then they will exit that region and continue on their separate ways unchanged. In other words neither the energy nor the momentum of the waves ...

3

There is no need for high order mechanism. It is simply because a single photon can interfere with itself. If you remember the double slit experiment, they are indeed looking for a single photon passing through a slit and interfere with itself. Now if, instead we have billions of billions photons, the same single photon interference still happen ...

1

Partly this answer is just gathering together the comments above, though there are a couple of points that haven't been mentioned. Firstly, as mentioned in the comments electromagnetic waves do gravitate and the links in the comments cover this well. In the early universe (for the first 47,000 years after the Big Bang) EM radiation was the dominant ...

1

Observing the derivation of the Friss equation "http://www.antenna-theory.com/basics/friis.php", the answer to the question is because the effective" aperture of an antenna is proportional to the square of the wavelength that the antenna operates at.

2

Yes, you are basicaly right. The Poynting vector gives you the momentum of the the EM wave. At the quantum level, it is an operator of the form (see page 7 of http://www.physics.usu.edu/torre/3700_Spring_2013/What_is_a_photon.pdf) : $$\hat{\mathbf{P}} =\sum_\mathbf{k} \mathbf{k}\, \hat a^\dagger_\mathbf{k} \hat a_\mathbf{k}$$ for a given polarisation (here ...

1

In one sense you are right: the only free space "perfectly collimated" optical field is the plane wave in the sense that these are the only eigenfields of Maxwell's equations, being fields which conserve their form under propagation and only undergo scaling by an eigenvalue in such propagation. Since Maxwell's equations conserve energy in free space, ...

0

Bremsstrahlung Radiation Dipole Radiation Accelerated charges radiate. This is the golden rule. The physical causes of electro-magnetic radiation are called legion for they are many. However, the mathematical mechanism is fairly straight-forward conceptually (while being a real bear in the details). Here's the basic idea: by solving Maxwell's Equations ...

0

Bremsstrahlung is a German fancy name for a type of Electromagnetic wave (an X-Ray to be specific). According to classical electromagnetism, whenever a charged particle accelerates an electromagnetic wave is produced. The source of bremsstrahlung (or "braking radiation") is the negative acceleration of an electron in a Coolidge tube (Google it for the ...

1

I might add a few commas to that Wikipedia sentence, as "A perfectly collimated beam*,* with no divergence*,* cannot..." to show informative rather than additional parameters. To answer your question about "collimated" vs. "plane wave" , consider two point sources at th plane of focus of a lens. Each point source gives off spherical waves; the lens ...

3

A monochromatic plane wave is simply: $$x(t) = A \sin\left(\omega t + \phi\right)$$ where $A$, $\phi$, and $\omega$ are fixed, never-changing quantities. Because the properties of this wave never change, there is no way to use it to transmit information. Consider this: suppose you point a laser pointer from one building to another, so that you can see ...

2

This is an experimentalist's answer and yes, accelerated charged particles either in stable circular orbits or in linear acceleration do radiate. Classically, any charged particle which moves in a curved path or is accelerated in a straight-line path will emit electromagnetic radiation. Various names are given to this radiation in different contexts. For ...

3

'Radioactive decays' tend to be categorised into 'alpha', 'beta' and 'gamma' decays. Alpha particles are helium nuclei, beta particles are electrons and gamma particles are electromagnetic radiation. To answer your question: It depends on the radioactive product, but gamma rays (which are produced in most radioactive decays) are electromagnetic waves.

2

Radioactivity comes in three basic types. Gamma radiation is an electromagnetic wave just like light and radio waves but of higher energy, and is described using electrodynamics. Alpha and beta radiation is charged particles (helium nuclei and electrons respectively) and again the motion of charged particles is described using electrodynamics. So ...

2

I'm not quite sure what you mean by "radioactive energies", but in general there are three types of radiation: $\alpha$ radiation: these are helium nuclei: He$^{2+}$. $\beta$ radiation: these are nothing more than electrons: e$^-$. $\gamma$ radiation: these are nothing more than photons (often denoted by $\gamma$), and are in fact traveling packets of ...

1

You have separate Fresnel equations for s- and p-polarized light. The two polarizations reflect/refract separately. You can reconstitute them on the other side to recover the new polarization vector if you want.

0

Yes, you can view a transmission line as a model of an EM wave propagating in one particular mode, in this case it is a plane wave moving in a given direction. The model can describe how the voltage and current of an antenna port would behave, especially, if you include several of these transmission lines connected to the port with ideal transformers of ...

2

Don't worry, I did research in surface plasmons and even then I was more than a year into it before I truly understood, on an intuitive level, how the light gets a 'kick' from the grating. You are correct that it is diffraction at a 90 degree angle to the normal, but there is an easier way to think about it. You say you've never taken a formal course in ...

1

The "that's how it is" answer may be given for mics with bidirectional acoustic pattern (that's just what a voice coil constrained to one axis of motion will provide), but not for cardioid type mics. The cardioid shape is apparently formed via the superposition of the omni & bidirectional shapes, implying more than one type of audio elements being ...

1

The condition comes from "phase-matching" - or in other words that the wavevector of the SPP ($\beta$ in your example) is matched to the wavevector of the in-plane component of the incident light. Now before the light hits the surface, this in-plane wavevector is given by $k \sin \theta$, but when it hits the grating, it receives a momentum "kick" of $\pm ... 5 First of all, the Wi-Fi doesn't have a defrost setting, we have that going for us. Jokes aside, I looked at the microwave and Wi-Fi that I have, and the first major difference is the power level. The microwave outputs$400W$to$2kW$where the for the Wi-Fi is just around$5W$. Also, the microwave runs on a single frequency in a very confined and shielded ... 5 Our bodies do absorb Wi-Fi energy, which causes the signal to attenuate. The thing is that the signal is so weak compared to a microwave oven that our bodies are able to get rid of the extra energy as fast as we accumulate it. Your skin just isn't going to heat up measurably. If you stand to close to a radar transmitter, for example, you will experience ... 1 If the incident radiation is unpolarized, it can be seen as the sum of two mutually incoherent terms: the first one due to a polarization perpendicular to the scattering plane and the second one contained in this plane. For each term, the scattered radiation preserves the polarization. The first term does not vary with$\theta\$, since the induced dipole is ...

1

CCD type sensors can be damaged by radiation, resulting in so-called hot pixels. This handbook says "Warm and hot pixels accumulate as a function of time on orbit. Defects responsible for elevated dark rate are created continuously as a result of the ongoing displacement damage on orbit." Now space is a pretty extreme environment, and the radiation is ...

1

It is absolutely correct that in vacuum all colors of light travel with same speed and this is why a white ray travels through the vacuum without suffering any dispersion...

2

If we assume the wave-fronts that initially enter the system to produce this image are symmetric, than the result is intuitively symmetric as all the components used to produce it are too. And anyone who looks at the images you have shown, will clearly see a symmetry about the Y-axis. Assuming they do not look for minuscule errors due to wave-front ...

6

Gamma rays are affected just like light rays, so they will be subject to a gravitational red shift and they will be bent by gravitational fields just as visible light is. It's important to be clear that in a gamma ray burst the gamma rays are not generated by the black hole. The process of forming the black hole heats the interior of the star to incredible ...

1

I think you may be misreading that the field holds no energy. All of the light can be reflected at steady state. When the light is first incident on the system, energy is stored in the plasmon part of the light-plasmon system. Once the steady state has been reached, the energy going in equals the energy coming out (less a trickle needed to make up for any ...

-1

I like thought experiments ! No need to replace those expensive melted bits !! And that is relevant to this experiment. We will wind up the RF output of the transmitter until the aerial itself is glowing a nice cherry red . Still intact (just !) as a functional aerial but emitting light (you can see its red). So assuming we have still got a good SWR and all ...

0

As other answers have mentioned light in classical electricity and magnetism is described very well by the solutions of Maxwell's equations which combine electrostatics and magnetism and describe a traveling wave of energy propagating at a speed c. This speed is not arbitrary but, as it comes out from the equations, depends on the electric and magnetic ...

0

Light travels as a wave and interacts as a particle so to be pedantic there is no way to answer the questuon.

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