# Tag Info

## Hot answers tagged electromagnetic-radiation

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 ...

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 ...

3

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

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 ...

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

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 ...

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 ...

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 ...

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 ...

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

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

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, ...

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 ...

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 $\pm ... 1 CCD type sensors can be damaged by radiation, resulting in so-called hot pixels. This handbook says "Warm and hot pixels accumulate as a function of time on orbit. Defects responsible for elevated dark rate are created continuously as a result of the ongoing displacement damage on orbit." Now space is a pretty extreme environment, and the radiation is ... 1 I think you may be misreading that the field holds no energy. All of the light can be reflected at steady state. When the light is first incident on the system, energy is stored in the plasmon part of the light-plasmon system. Once the steady state has been reached, the energy going in equals the energy coming out (less a trickle needed to make up for any ... 1 Observing the derivation of the Friss equation "http://www.antenna-theory.com/basics/friis.php", the answer to the question is because the effective" aperture of an antenna is proportional to the square of the wavelength that the antenna operates at. 1 You have separate Fresnel equations for s- and p-polarized light. The two polarizations reflect/refract separately. You can reconstitute them on the other side to recover the new polarization vector if you want. 1 If the incident radiation is unpolarized, it can be seen as the sum of two mutually incoherent terms: the first one due to a polarization perpendicular to the scattering plane and the second one contained in this plane. For each term, the scattered radiation preserves the polarization. The first term does not vary with$\theta\$, since the induced dipole is ...

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