Tag Info

Hot answers tagged

19

The key phrase is "may be considered". That is, Huygens principle, as a mathematical procedure, gives (approximately) the right answer. The principle does not say that light is actually being generated at each point on the wavefront. Another way to look at it: the light at a phase front already exists there. There is no need to generate it all over ...


17

The definition of a wave is not that it is the oscillation of a medium. Waves are called waves because they are solutions to a wave equation, which is, for a generic "excitation" $A(t,x)$ depending on the time $t$ and some spatial coordinate $x\in\mathbb{R}^n$, of the general form $$ \frac{\partial^2 A}{\partial t^2} = c^2\Delta A$$ where $\Delta$ is the ...


14

Some numbers come from a review paper by Cullers (2000), who discusses the SETI Phoenix project. There, it is claimed that the Arecibo dish is capable of detecting a narrow band, coherent signal of $f=10^{-27}$ W/m$^2$ given a 1000 second observation. Assuming that this is an isotropic signal, then the implied power at distance $d$ is $p=4\pi d^2 f$, which ...


10

In addition to the other answers, back in the olden days they were thought of as oscillations in the ether. As a result of the Michelson-Morley experiment back in 1887, physicists began to think that there was no ether. But the term didn't change.


7

It's like saying sound produces sound: actually, particles move other particles, and this entire process is sound. Huygens' principle talks about propagation, not "emission" in the conventional sense. The light is a wave in the electromagnetic field and the changing field at a point in space induces field in the surrounding space when time moves forward. ...


6

The red, orange, yellow, and white parts of a candle flame results from glowing soot. The color in this part of the flame is indicative of the temperature. The spectrum in this part of the flame is fairly close to that of a black body. The blue part of the candle flame at the bottom of the flame results from chemiluminescence. Chemiluminescence is not black ...


5

Huygens assumed the existence of "ether particles". And every point in wave front, which your statement of Huygens principle is saying to be the source of secondary wavelets, are nothing but the ether particles, which Huygens assumed to give rise to its own individual wave. Huygens said that light waves were longitudinal as they passed through a ...


4

Electromagnetic waves are called waves because there are waves (propagating disturbances), waves in the electromagnetic field. These electromagnetic waves, like material waves, transport energy. According to the Wikipedia article "Wave" In physics, a wave is disturbance or oscillation (of a physical quantity), that travels through matter or space, ...


4

This paper contains an important analysis of the different trade-off between bandwidth and energy efficiency. The interesting conclusion from that paper is that the most energy-efficient way to send and receive interstellar messages (over flat spacetime) that maximise the bit-rate requires making the bandwidth of transmission very large. In particular, this ...


3

An oscillating magnetic field is always accompanied by an electric field. This is because Maxwell's equations tells us that (amongst other things): $$ \nabla \times {\bf E} = - \frac{\partial{\bf B}}{\partial t} $$ On the right hand side of this equation the symbol $\partial{\bf B}/\partial t$ means the rate of change of the magnetic field with time, and ...


3

In the book "Physics of the Plasma Universe" Dr. Anthony Peratt puts candle flames near the bottom of "energy in electronvolts" portion of the 'plasma spectrum'. If you look at the chart below, you'll see candles flames about midway (ok, cosmologically) between the ends: solar bodies and laser radiation terrestrial flames interstellar charged gases ...


3

The optical isolator component is active. It consumes energy and so is no different (thermodynamically speaking) from the heat-pump in a refrigerator. If you are talking about a passive component that draws no power then, a surface that allowed light pass in one direction only would violate the second law of thermodynamics. To see why, imagine two rooms, ...


3

Searching that book for "lineshape function" will return this page, which explains what that means. Essentially, the atoms in the gain medium are usually able to respond to frequencies $\omega$ which are close to, but not necessarily exactly equal to, the central frequency $\omega_0$. The response is strongest at $\omega_0$ and then it tapers off over some ...


2

The electromagnetic waves satisfy the Maxwell equations for waves. They don't need a medium for propagating, because these waves are their own energy-carriers, the photons. By that, they differ from water waves whose energy is propagated by the intermediation of the water molecules, or sound waves whose energy is propagated through the molecules of the ...


2

The issue here is how much the refractive index $n$ tells you about dissipation. As you rightly said, the imaginary part of $n$, which depends on both real and imaginary parts of $\epsilon$, leads to an imaginary part in k which describes an exponentially decaying electric field. However, this doesn't necessarily correspond to dissipation (i.e. a drop in ...


2

If you have a good grasp on the relative scales of other things in physics, you may be able to relate the wavelengths to those. Otherwise, your best bet is just to memorize the wavelengths (or frequencies). Since $f = \frac{c}{\lambda}$ for light, you'll be able to figure out the frequencies if you know the wavelengths. Here are some things I use to help ...


2

The device on the armature is a compact, commercial, bremsstrahlung-xray source. These have been used in dental officess for decades with both film and solid-state detector system Presumably the object you placed inside your mouth was a battery powered, pixelated solid-state detector of some kind. They can be as simple as CCDs like those in a digital ...


2

So, this is an old post that I came across when I had a similar question. Here's a paper where they dissolve different amounts of ions in the water and found that the ability for the microwave oven to heat the water actually reduces as more ions are introduced.


1

The Cosmic Microwave Background includes photons that will not be absorbed before the universe inflates to the point where there is nothing to hit ever again. If parts of the last scattering were somehow barred from releasing that energy, I think we would notice. The photon carries energy. Particles do that. How is it any different from the electron, which ...


1

Would the human start emitting photons and die? EDIT: This answer is an answer to the original question regarding "a billion" positions. The question was subsequently edited to now read "2.3*10^28" positrons. That is not cool. The human would start emitting photons. This is exactly what happens during a PET (Positron Emission Tomography) scan at your ...


1

The rest mass of an electron is 0.511 MeV. When an electron and a positron annihilate their mass turns to energy (two 0.511 MeV photons) so for each annihilation an energy of 1.022 MeV is released. One electron volt is $1.602 \times 10^{-19}$ joules, so in joules the energy released is $1.637 \times 10^{-13}$ J. You ask what happens if $2.3 \times 10^{28}$ ...


1

By definition, an electromagnetic wave is a solution to Maxwell's equations in vacuum. The electric field of a EM wave solution is always perpendicular to the direction of propagation. Let me denote this electric field by $\vec{E}_{EM}$. If $\vec{v}$ is the velocity of the wave, then we must have $\vec{E}_{EM} \cdot \vec{v} = 0 $. However, Maxwell's ...


1

The pot may have been steel (not that good a conductor) and thin. The lid may not have fit well, leaving gaps. Cell phone signals are short wavelength, meaning a small gap will not completely block them. Cell phones are good at picking up weak signals. Next time, just turn it off? Or answer it and tell the poor guy where his phone is?


1

Since no one else has mentioned it ... If you want to have a better conceptual understanding of the apparent slowing of light (and other electromagnetic waves) in materials, I strongly suggest reading Richard Feynman's lectures, especially Chapter 31 of volume I. That will give you much more explanation than is possible in this forum. All the Feynman ...


1

The efficiency of the pumping source is $x$ means that $x$ amount of electrical power is converted to energy which is useful for pumping the laser medium. The absorption of the pump is $y$ means that $y$ amount of the energy from the pump source is actually pumped into the medium to generate the population inversion necessary for lasing. The total amount ...


1

Classically speaking, electromagnetic radiation is an oscillation of electric and magnetic fields which propagates. The movement of charged particles causes those oscillations - the motion of charges and the electric and magnetic fields are coupled - see Maxwell's equations. For example, an antenna broadcasts electromagnetic radiation when an alternating ...


1

There are two possible factors that are involved here: The plasma emits electromagnetic waves, which means there will be an oscillating electromagnetic field spreading around the plasma ball, which would explain why the voltage detector beeps 3 feet away. In fact, it is an interesting experiment to study the electromagnetic waves propagating from the ...



Only top voted, non community-wiki answers of a minimum length are eligible