Tag Info

0

What Maxwell derived was from radio waves. Radio waves are modulated photon radiation. Electrons in an antenna rod get accelerated at once and emit photons. The density of this photons distributes in space. Detecting this radiation with a receiver one get a sinus wave form. The frequency of this wave has to do with the frequency of the antenna generator and ...

0

Fausto, I'm sorry, but there's a lot wrong here. You need to backtrack, and take one thing at once. For example, gravity is not a force in the Newtonian sense. No work is done on the descending photon. Throw a 511 keV photon into a black hole, and the black hole mass increase is 511keV/c². Conservation of energy applies. The descending photon doesn't gain ...

0

It turns out that one photon states of the electromagnetic field can be written in a way such that the state "propagates" fulfilling Maxwell's equations. This is an exact model as I discuss this in more detail in my answer here. So we begin with a one-photon Fock state of the quantized electromagnetic field. Let's keep our discussion to one mode, so one ...

4

No, it is sufficient for the photon energy to exceed the band gap. Any excess energy is transformed into kinetic energy for the electron in the new band. You get exactly the same effect when ionizing an atom - the excess energy simply powers the electron into a faster continuum state. You should also take into account that photon energies are never exactly ...

2

When a wave travels through a rope, the rope goes up and down, the position of all the 'rope-particles' changes, they oscillate and this makes up the wave. With light, it is indeed the electromagnetic field oscillating, but you shouldn't think of the arrows that represent that field in your first picture of light as 'extending into the rest of the space'. ...

0

It is very important to keep in mind that the double slit interference is a quantum mechanical phenomenon. The ideal would be to solve the quantum mechanical problem and get the wavefunction , i.e. photon , two slit field pattern, screen , detector and impose the boundary conditions to get the system wave function and see the probability of the photon ...

1

The key point here is the notion of distinguishability. If the two paths are made distinguishable, by associating different polarization to each path, then there is no way for the two amplitudes to interfere when they meet again because the "which-path" information is present. It doesn't matter whether or not the observer decide to actually look at the ...

3

You haven't said precisely what instrument you are using, but what you are seeing is normal for grating monochromators. Suppose we have light of one and only wavelength $\lambda$ incident normally on a grating. Scattered light will be seen at the following angles, $\theta$: $$d \sin \theta = m \lambda \qquad \text{for } m=0, \pm 1, \pm 2, \pm 3, \ldots$$ ...

1

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 ...

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 ...

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!)

0

Intensity is an objectively measurable attribute of light. It is the rate at which energy is delivered to a surface. Intensity is energy delivered per unit time per unit area. The intensity of light is a measurement of photon irradiance, which is the number of photons delivered per square meter per second. You can measure intensity with a photoelement, ...

0

My question is why does light move in such a way rather than vibrating or rotating, or any other actions, in-situ? Is it due to absence of Higgs boson or perhaps I should ask why does not all elementary particles move in a straight line? You have managed to mix two different frames of reference in the above paragraph. Light belongs to the classical ...

-2

Most likely is its always hotter than the outside. the frames absorb heat up into the 160 plus mark depends on where you live. 100 degrees usually about 150 degrees gained we call it solar gain.

5

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 ...

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

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 ...

0

Several answers here already talk in great detail about how electron orbitals affect if a photon will be absorbed or not, but this is not the whole story. The color from reflected radiation is indeed the only factor if the surface is completely flat and perfectly reflective, excluding the black-body radiation, but most surfaces are not. Take for example all ...

0

No, a laser pointer doesn't create entangled photon pair, there is no process involved that would cause it. Until now I haven't heard of entangled more-than-two particles. However now that you've asked I stumbled over a recent publication mentioning entangled photon triplets.

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 ...

3

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.

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 ...

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 + ... 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 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 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) ... 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 ... 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 "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 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 ... 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 ... 0 Your question has nothing to do with the (hypothetical) photon's reference frame. Even if the spacetime interval of lightlike movements is reduced to zero, the order of events is persisting - In the same way as when you cover one sheet of paper by another, you cannot reach the hidden sheet even if their respective distance approaches zero. Edit: By the ... -3 Photons have a varying electric and a varying magnetic field, perpendicular to each other and perpendicular to the photons propagation (in vacuum). This fields have a direction, for example from plus to minus and from north to south (this is a convention). What ever you try, you will end up with exact two combinations of the fields directions: -2 As often, hasn't something been overlooked. Energy & mass cannot be interchangeable with regards having the ability to generate gravitational force, because otherwise 'binding energy' would also contribute, obviously it does not. Do we have an impasse, maybe not, if photons actually possess mass after all. I have calculated algebraically that when matter ... 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 ... 0 A photon should be discussed as a wave, instead of a particle (as it ceases to exist in rest state). As such two waves of the same wavelength and frequency may or may not be "negative" to each other, depending on their phase difference. When their phase difference is an odd multiple of pi, they will invariably cancel out each other upon interference. 0 The speed of light it is medium-dependent, note c/v= \mu. But in this case, it is correct to say that a photon has a KE given by E = hc/ \lambda in vacuum. At the interfaces between mediums, however, using Huygen's principle, \lambda1/ \lambda2= v1/ v2. So light has a definite speed in a given medium, with its energy (note it only has kinetic ... 0 A high level description. In the case of reflection at a metallic interface the electric field of the photon forces the electrons in the metal to oscillate. The oscillation means that the electrons are being forced at the same frequency as the photons. Thus as the electrons in the metal oscillate they begin to emit light in response. The frequency, ... 3 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 ... 2 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 ... 0 Your statement "Maxwell's equations imply that magnetic fields are due to changes in electric fields." is not complete. A corrected statement is that Maxwell's equations imply that magnetic fields are due to changes in electric fields AND due to currents (which can be stationary):$$ \nabla\times\mathbf{H} = \mathbf{J}+\partial\mathbf{D}/\partial t  As ...

0

Photons indeed have phase (or phases, in the plural) as folloows: one can turn Hossein's answer on its head in the following way. A "lone photon" is simply a particular pure state - a so called Fock State - of the quantum electromagnetic field (see the "Quantization of the Electromagnetic Field" Wikipedia article) where the number observable is certain to ...

2

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 ...

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. ...

0

Light is only electromagnetic radiation similar to what radio waves are. Do you see radio waves? No. A radio receiver is needed to translate these waves into music, speech etc from a radio station. In the case of light the receiver is you who is "designed" to translate the waves of light and/or photons into what you see.

Top 50 recent answers are included