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6

Nonlinear optical elements are called nonlinear precisely because of the behaviour you note: because the optical response of the material does not depend linearly on the driving fields. The response may then have a quadratic or higher dependence on the driver, which is usually written in the form  \mathbf P =\varepsilon_0 \chi^{(1)} \mathbf E + ...

5

Reflecting energy back into the photosphere using mirrors is equivalent to asking what happens if we artificially increase the opacity of the photosphere - akin to covering the star with large starspots - because by reflecting energy back, you are limiting how much (net) flux can actually escape from the photosphere The phenomenon could be treated in a ...

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More than one photon can be absorbed, but the probability is minute for usual intensities. As a scale for "usual intensities" note that sunlight on earth has an intensity of about $1000\,\mathrm{W/m^2} = 10^{-1}\mathrm{W/cm^2}$. The intuitive reason is, that the linear process (an electron absorbs one photon) is more or less "unlikely" (as the coupling ...

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There are very strict limits on the mass of the photon already, so it would only affect our understanding of physics on the largest scales. The cosmologists would have some hard thinking to do, for instance. However, contrary to a comment, it would not affect relativity beyond requiring us to reconsider the usual name for $c$: not "the speed of light" but ...

4

Let's start with your assumptions: 1) Photons travel at the speed of light. Right. No problem with that. 2) From the photon's reference spacetime is contracted to 0 length in the direction of photon travel. Wrong. The photon doesn't have any kind of reference frame. To appreciate why, imagine you're travelling at the speed of light. We know you can't ...

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Yes it's possible, and it's called (unsurprisingly) two photon emission. However the probability of two or multi photon emission is generally lower than for single photon emission by several orders of magnitude, so it's hard to observe.

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I think there is some confusion here. Photons are always massless. They also always move at the speed of light. Therefore every example of a photon in nature has zero mass. Perhaps you are thinking about a photon moving through a medium other than a vacuum. In this case, we can view the photon plus the interactions with the medium as a quasiparticle with a ...

3

Does the invariant helicity property contribute to the the concept of a photon and an "anti-photon" being the same entity? Not really, and I think you're getting mixed up between helicity and chirality here. Take a look at this deep-water wave image by Kraaieniest. See how the red-dot test particles move in a helical-like fashion? They can't move "the other ...

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Please recall how the concept of a photon came to life. 1.You have Maxwell's equations, 2. you look for wavy solutions (all classically),3. you get a classical Lagrangian for EM fields and the classical action accordingly. You quantize this classical action. Frequency in this case is an energy (Planck constant is unity in selected system of units). The ...

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No, the photons do not travel in a helix, they travel in a straight line but with a phase delay that is dependent on position. Looking across the beam's wavefront there is a phase delay that is dependant on the polar angle $\theta$ around the beam axis. If we take a simple helical mode's complex amplitude as $\zeta(r,\theta,z) = u(r,z) e^{-ikz} e^{il ... 2 The key difference is that Compton scattering occurs on a free electron. It's all about making energy conservation work out. In an atomic process, the atom goes form having energy$E_1$to$E_2$, and its energy is fixed by the atomic state it goes to. It can't be a$2 p ^1$electron and also have extra energy it hangs onto. (Things are slightly more ... 2 Laser modes are the eigen-modes of a laser resonator: only specific distributions of electro-magnetic field can "resonate" in each particular resonator. Due to the 3D nature of our space, each mode is described by 3 numbers, or indices,$m$,$n$,$q$. The latter is the longitudinal mode number, and is easy to understand: to form a standing wave, the ... 2 You are correct that the reason the photon is not spin 0 is because there are multiple polarizations. Saying that a particle is spin 0 means that under rotations of your spatial coordinate system the mathematical description of the internal state of the particle does not change. For the massive photon case there are three polarizations and when you rotate ... 2 People are addressing the speed question, but just to be clear: a photon can be very low energy. For instance, radio waves are much lower energy than gamma rays, even though both are made of photons (and, in vacuum, both travel at the speed$c$). What determines the energy of a photon is the frequency of the excitation (frequency of the corresponding light ... 2 Non-Linearity means that the dispersion relation becomes non-linear. Linearity is an assumption which only holds true for low intensities. Almost every material has some non-linear effects if the light source is only powerful enough. The polarization vector for example becomes:$P = P_0 + \varepsilon_0 \chi^{(1)} E + \varepsilon_0 \chi^{(2)} E^2 + ...

1

Although it has been said in other comments and answers, it bears repeating succinctly: photons (as far as any experiment can tell) are massless and therefore always move at the universal, invariant speed of light. There is NO non-relativistic description of the photon. Even the "classical" description of light - Maxwell's equations - can be interpreted as ...

1

In quantum treatment, phonons are quantum harmonic oscillators, which is studied in any quantum mechanics textbooks. The energy spectrum is readily studied there. https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator The $1/2\hbar\nu_E$ comes from uncertainty principle. (A quantum harmonic oscillator is confined in space so its momentum cannot be zero) ...

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Polarization and gauge symmetry In QFT, the dynamical varible is the four-potential $A_\mu$. The electromagnetic field is defined by $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$, an antysymmetric tensor wich six independent components: 3 for the electric field and 3 for the magnetic field $E^i = - F^{0i}$, $B^i = ... 1 Refractive Index is when light travels more slowly in a medium. Here is an example of light being slowed down to 38 miles per hour. The speed of a photon does not affect its energy. It has zero mass, therefore zero kinetic energy. The energy it has is due to its frequency (color), and nothing else. (However, it does have momentum!) 1 This question is too broad. It involves ALL the objects in the universe which have a surface, i.e., everything. I'm going to avoid giving a lecture here. In some liquids and most gases the electronic structure of each individual atom or molecule is enough to describe their spectra. The "property" you are looking for in the case of solids is the band ... 1 There are photons traveling in all directions, not just the dozen or so you show. The further from the source the telescope is, the smaller the amount of solid angle it covers and the fewer photons it will gather. A$1 m^2$telescope pointed at the sun will receive about$1.4 kW$. Taking a typical photon energy of$2 eV$that is about$4.2E21$... 1 As John Duffield pointed out in his answer, there is no reference frame in which the photon is at rest. This is the simple reason why most parts of your statements do not make sense. Still, there is an interesting perspective to the issue you raise: Normally one considers the events or states "photon emitted", "photon is traveling", "photon absorbed" as ... 1 "Spin" is dictated by the representation of the Loretz group the field the particle is a quantum of transforms in. Half-integer spins are fermions, integer spins are bosons by the spin-statistics theorem, where representations of the Lorentz group are labeled by two numbers$(s_1,s_2)$, whose sum is what we call spin in this context. The significance of the ... 1 The photon does not have an rest frame because that would imply it's velocity is 0, when by postulate it must be moving at C in all frames. Rather pointless to wax on about not experiencing any time or what not. 1 Remember that if the electric and magnetic field components are perpendicular and in phase, then you can always calculate the magnetic induction component as$|\vec{B}| = \frac{1}{v}|\vec{E}|$; where$v$is the speed of the wave, and is given by$v = \frac{1}{\sqrt{\epsilon\mu}}$, where$\epsilon$and$\mu$are the relative permittivity and permeability of ... 1 For a non rotating spherical black hole the event horizon is at a circle of circumference$4\pi MG/c^2\$. And gravity gets weaker the farther away you get. At that location even moving straight outwards at light speed won't let you travel to someplace where gravity is weaker. But if you aren't moving straight away then you will get trapped even farther out. ...

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