New answers tagged

1

Yes it does. For example a (perfectly) black surface would absorb all photons, and hence the radiation pressure would be given by: $P_R=\frac{E_f}{c}$ Where $E_f$ is the photon energy flux and $c$ is the speed of light. A (perfect) mirror reflects all the light, and so you would get double the momentum transfer, and hence double the radiation pressure. ...


0

Yes. A totally reflective surface will bounce the photon back, resulting in twice the momentum change in merely absorbing it. However, it gets a little more complex because an absorbing surface heats up and then itself radiates. If there is perfect thermal transfer it will radiate isotropically. So shape, and materials, matter with absorbers.


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According to the Stefan-Boltzmann law, radiated power is proportional to the fourth power of the absolute temperature of the body. With T=0 the radiated power would be zero.


-1

Compton scattering occurs on free electrons, i.e. not in a bound state. The equivalent to a Compton scatter would be a scattering of a photon off the field of a solid, momentum and energy balance happening collectively with the total mass of the solid. In the best case it would be a whole atom that the photon would scatter inelastically off, the atom taking ...


-1

In some cases it does eject a photon with a lower wavelength, if it did not do this then the laws of conservation of momentum would not be supported thus disproving many aspects of modern physics. The problem with this is without the experimental evidence or data, it is hard for someone to calculate or predict the new photons wavelength, let alone detect it. ...


2

Electromagnetic (EM) radiation between 400nm and 700nm in wavelength is the same thing as light. There is no difference. Neither is there a distinct difference between light, ultraviolet, x-rays, gamma rays, infrared, microwaves, or radio waves. Those names are all just human convention to specify EM radiation in certain frequency ranges. And regardless ...


1

In the particular setup that you show, where both the car with the photon source and the surface with the detector move in the same direction at the same speed, the result is the same regardless of the emitted object is a ball or a photon: it will hit the detector. This is best understood by a transformation of reference frames: Instead of looking at the ...


1

Yes it will be affected by the car's movement (if you are outside the car), since momentum is conserved. If you are moving with the car, the source is not moving relative to you, so the photon is moving down straight. PS from the outside point of view, it will seem like the photon traveling diagonally covers more distance. But since the speed of light ...


0

Photon energy density does affect the expansion of the universe, similar to the effect of dark matter and dark energy. At the present time, however, the photon energy density has a much smaller effect than either dark energy or dark matter. This is due to the different ways in which the energy densities of the various components scale with the size of the ...


0

Usually a higher-energy photon hits a molecule, excites it, and in some time afterwards (from picoseconds to days) the molecule re-emits another photon with lower energy. Using two monochromators, you can obtain two-dimensional fluorescence spectra, identifying that some spectral features depend on the excitation wavelength, while others do not. In some ...


3

Absolutely it can - and it happens all the time. If you excite an atom, it can go through various "stages" of decay back to the ground state - with each drop in energy resulting in an emission of radiation. This happens during photosynthesis: see this page from which I copy this image: As you can see, there are multiple paths for the energy to be lost by ...


7

Yes, but you'll have to go really, really fast. And even then, don't worry about the photons. The relation between velocity $v$ and the observed and "true" wavelength $\lambda_\mathrm{obs}$ and $\lambda$ of the light is $$ \lambda_\mathrm{obs} = \sqrt{\frac{1-v/c}{1+v/c}} \lambda. $$ If you consider optical (i.e. visible to humans) light with a wavelength ...


0

So, I'm not sure how much relativity you know, but usually we write the 'Proper time' for a particle moving through the x direction and time as $$ (c \Delta \tau)^2 = (c \Delta t)^2 - (\Delta x)^2 $$ Where proper time $\tau$ is just the time that the particle would measure in it's own internal frame. So if you are travelling fast on a rocket ship, someone ...


2

What we call reflection is in reality a more complicated process than bouncing a ball to a wall. For the part of the electromagnetic radiation that we call visual light and for low densities of this light the surface electrons are responsible for the absorption and re-emission of this photons. So yes, mirror will gain momentum and the photons will lose ...


1

The atom absorbs the photon that kicks up an electron to an excited state, and it is the atom that will emit a photon when it de-excites. Not the electron. Is the invariant mass of an atom higher when the electron is in an excited state? Take the hydrogen atom. The ground state energy is at -13.6eV. This means that the mass of hydrogen is less than the ...


0

In classical physics, a system can be described by a set of numbers whose values can all be measured using a single instance of that system. In quantum mechanics, a system is characterised by the values of observables where those values are represented by Hermitian matrices. To describe how information is transferred between quantum systems you have to ...


1

let's assume we have 2 rays moving at 180°. If the container ( and the background ) move instead of the photons, the container would have to move in the direction of both rays. But, it's impossible since we assumed 2 antiparallel rays. How can container move forward and backward simultaneously.


3

Your question is the one that Einstein pondered for long time and from which Special Theory of Relativity was born. He wondered what could happen if you travel at the speed of light how would you see a ray light. The problem was that according to Maxwell's Electrodynamics, explained light as oscillating $E$ and $B$ vectors along space and time, so as a ...


1

Elementary particles do not have consciousness, individuality or volition. They follow the rules of the boundary value solutions of the quantum mechanical equations they obey. The relativistic quantum mechanical mathematics have zero mass particles moving at velocity c, and in all valid frames massive particles move at velocities less than c. It is the ...


-1

intensity is defined as power/area, here, more energy that reaches your eyes, more light you observe, By symmetry if we consider a isotropic (symmetrical) point source with total power being emitted P, intesity at any d(A) {small area} at the surface of imaginary sphere will be P/4π(r)^2 so, Intesity falls by second power of radial distance


2

There is light all around the lamp, and all around you (except of course for the points that light cannot reach because of walls or whatever). You just only perceive the light that comes into your eyes. The light that comes directly from the lamp is generally more intense than that reflected by the objects around you, so that you tend to ignore the latter. ...


1

You are describing a spherical wavefront i.e. light radiating outwards from a point source with spherical symmetry. But suppose you have two such point sources near enough to each other that their wavefronts overlap. Now your expansion model has to have space expanding simultaneously in opposite directions. Consider also that a wavefront can be any shape. ...


1

Neither image displays diffraction. They both illustrate shadowing. The second photo was apparently shot on a clear sunny day. All of the rays come from the sun. Some are blocked, those that aren't go straight down to the floor. The second was probably taken on a hazy day. Rays originate from the haze; they come from all directions, and make their ...


0

Current is moving electrons. A photon with a frequenzy will give its energy to the electrons in the antenna. The energized electrons then travel thru the wire to the radio at the same frequenzy as the photons will give of its energy. When the photon field, or rather the electromagnetic field oscillates from positive+ to negative-, as in the AC used to ...


-3

If I understand QFT correctly, you can think of the electron field as an all pervading bubbling cauldron (called vacuum fluctuations) of premature, not-ready-for-prime-time (potential?) electron energy balls, having energy below the minimum (but not zero) quantum energy required for an adult electron. If they get whacked with enough energy they can start ...


1

Your overall efficiency will be well less than 50%. There are several factors at play here: To generate a positron with photons, you also have to generate an electron in order for the quantum symmetry to be preserved. When you generate a pair this way, they end up with half the energy each. If you are only using one of those for your transmission vessel, ...


0

The general theory For each irreducible representation of the Poincare group with a given helicity $\pm s$, which is realized by relativistic creation-destruction field, $$ \hat{\psi}_{l}(x) = \sum_{\sigma = \pm s}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3}2\epsilon_{p}}}\left(u_{l}(\sigma , \mathbf p)\hat{a}(\sigma, \mathbf p)e^{-ipx} + v_{l}(\sigma ...


2

The trick is that because the "slit" is infinitely wide, you shouldn't work in the far-field approximation (Fraunhofer difraction integral which leads to fourier optics), but with distances computed to the square order (Fresnel diffraction integral). The results are appropriately named Fresnel integrals (this time this is a name of a special analytical ...


0

It all depends on definitions used of course, but overwhelmingly I have found that when people talk about "unpolarized" or "depolarized" they mean a classical mixture of pure quantum states. So they definitely do not mean a superposition state. Another way to understand that a superposition is not "unpolarized", or at least that this is not a good ...


0

Let us try to clear up what light means and what a photon means. Light is a solution of classical Maxwell's equations , an electromagnetic wave. As a wave it can have ploarizations that go through filters etc. A photon is an elementary particle which has as energy h*nu, nu the frequency of an electromagnetic wave solution, and is described by a wave ...


0

First, you are right about the unpolarized light. Second, whether a photon will be absorbed or not is according to the so-called selection rule which is determined by the conservation law (In this case it is the conservation of angular momentum). Thus you may think that the polaroid is "trying to" absorb each photon. However, the possibility for the ...


3

The electron-positron pair has a center-of-mass reference frame where the momentum is 0. Obviously, there exists no one-photon system with positive energy which has 0 momentum, as the energy-momentum relation for a photon is $E = p c$.


1

The maximum kinetic energy of the electron occurs when the change in wavelength of the photon is a maximum. The formula for wavelength change is usually given in terms of the cosine of the scattering angle of the photon. So you can decide which scattering angle gives you the maximum change in wavelength and hence the maximum KE gain of the electron. Once ...


0

The gas in a star is emitting light by blackbody radiation at the same rate as it's absorbing emitted radiation so it's temperature varies extremely slowly with time. The blackbody radiation increases the rate of diffusion of heat in a star. See Does nature really follow the heat equation?. That diffusion of heat is what people call photon random walk. No ...


2

From the theory of Thompson scattering (see http://farside.ph.utexas.edu/teaching/em/lectures/node96.html ) we know that a charged particle of mass m interacting with a plane wave electromagnetic field of Strength $E_0$ and frequency $\omega$ has an effective dipole moment of magnitude $$d=\frac{e^2E_0}{m\omega^2}$$ Note that the dipole moment scales ...


1

It relies on conservation of energy and momentum and the equation for energy in special relativity: $E^2 = (pc)^2 + (mc^2)^2$. Here you go. Energy of photon: $E_\gamma = \hbar\omega = p_\gamma c$, where $p_\gamma$ is the momentum of the photon. Assume the electron is initially at rest, so it's energy is simply $m_ec^2$. By conservation of energy, the ...


2

The formula E=h*f is a quantum mechanical formula for the elementary particle called a photon. The photon travels with velocity c in vacuum. If it is not a vacuum it will interact with appropriate quantum mechanical interactions, and in the interval between interactions it will be traveling with velocity c. If its interaction is elastic, as happens in ...


1

What bugs you here is that the word "speed" isn't precise at all when we talk about a wave in a dispersive medium, such as water. What we usually refer to when we say the "speed of light" is the constant we call $c = 299792458$ m/s, which is equal to the velocity of light in the vacuum. Vacuum is non-dispersive, so the group and phase velocity are the same, ...


0

Only in vacuum light travels without disturbances. To be precise, really without any disturbance light would travel only in a space with constant gravitational potential (and such a potential distribution does not exist). Every source of gravitation deflects light. But this does not has an influence on different trajectories (geodesic lines) nor leads to ...


1

In one word the answer is interference. Now more details. Assume you are traveling through something which has refractive index different then 1, like water. It actually means that for "slightly" different frequency of light water molecules will start to absorb the photons, i.e. shake together with electric field. However if you stay away from that ...


3

When people say that photons always travel the speed of light, they meant in terms of reference frames. And in fact, even through water, photons do travel the speed of light. The only reason why they appear to travel more slowly is because they "run into" molecules, and the path is not straight. In a vacuum, there are no molecules to hinder the light, and it ...


2

Light does not travel at the same speed in all mediums. In water the speed of light is less than that in vacuum.


16

The problem is that you are confusing light intensity with energy of a single photon. The photoelectric effect requires a certain energy per photon to work. But low light intensity just means fewer photons come - you can actually see the grain if the conditions are too dark: every pixel can get ~10 photons or less... and yet still, each photon that comes has ...


3

The electromagnetic potential $A^\mu$ is a four-vector, and hence transforms in the fundamental representation of $\mathrm{SO}(1,3)$, i.e. $A^\mu\mapsto \Lambda^\mu_\nu A^\nu$ where $\Lambda$ is the usual 4x4 matrix associated to a Lorentz transformation. Your question seems fundamentally confused about the difference between the field and the particle. The ...


2

Think about it this way. If you had a radioactive atom it might decay and emit a photon. And you don't know when. So there is a whole range of times for when a nearby photon detector might go off. But there is also an issue of coherence. Which is a specific technical term, not a word people use for emotional purposes. If you have a short coherence length ...


1

We don't harvest waste artificial light, because it would be ridiculously expensive to do so. The energy in sunlight is, at full sun, $1000W/m^2$. That's way higher than any artificial light in normal circumstances. So it's far more economic to position solar panels to optimise collection of daylight, rather than to capture artificial light at night. Any ...


2

Building on a comment by CuriousOne (who, honestly, should leave off commenting since he only ever writes answers in then anyway): A photon is not an object in and of itself. A photon is an excitation in a quantum field, which is not localized but fills space. In the double slit experiment you have an emitting source, a mask with two slits, and a ...


1

As I understood two photons will not interact with each other to produce interference pattern rather one photon behaves like a wave near two slits and go through both holes at same time. It looks like particle is spread in space and behave like a wave and will go through both holes at the same time to produce interference pattern.


3

What happens to photons when they hit our eye? Some of them pass through the iris and are focussed by the lens onto the retina where they are absorbed by rods or cones. where do they end up? Some of them end up absorbed by rods/cones, some by other tissues, some are reflected (c.f. red-eye in photography). why our eye don't get overheated? ...


1

For the theoretical background with an occasional nod to experiments, my bible is "Optical Coherence and Quantum Optics" by Mandel and Wolf. This book is both a text book and a reference for researchers. It covers the basics of random signals, quantum mechanics, the quantum theory of radiation, quantum optics, a bit of nonlinear optics, a bit of laser ...



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