# Tag Info

## New answers tagged photons

1

Great question -- it opens up a lot of physics. My favorite example is "why do some conducting metals look gold/copper/etc. rather than grey?" Turns out it's due to relativistic effects acting on the various electrons' orbitals. Then there's the blue jay, whose blue color is entirely interferometric rather than absorptive/reflective. and more :-)

0

A material displays a color when light is reflected off a surface. When a certain wavelength reaches a surface, if the energy $E=\frac{hc}{\lambda}$ of the photon correspond to the difference between two electronic states then it has a certain probability to be absorbed. The probability of being absorbed depends on the density of electronic states of ...

0

Interesting question. I believe it has to do the energy levels of the atoms of the surface. I assume you understand the concept of atomic orbitals - When light encounters a surface, the wavelengths absorbed are those that contain the right amount of energy to move electrons up to the next orbital. ... Upon a google search: If the energy difference between ...

2

I'd like to add to Kostya's Excellent Answer and also Marek's. Kostya is actually describing a quantum superposition of free photon and excited matter states. Often in this scenario, the refractive index is described as arising from the repeated absorption and re-emission of the vacuum photons by the atoms/molecules of the medium. This is a good first ...

0

Clear as mud. Glass does not absorb or reflect photon/light waves. It allows the energy to be transmitted through because of its molecular structure.

-1

All that upon are in my point of view (Information Science) no definitive answers. Only some mathematical relations can't answer this question: Are electrons and photons excites of the same electric field? Photons are said to be that particles which interact in the electric field. Therefore they are also the massless kernel elements of electrons, but in ...

4

Electrons don't gain mass when they hop up an energy level--we know this because we calculate the electron levels assuming a particular electronic mass, so if the electron mass changed, it would also have a different energy level. Photons are massless. Nothing that has mass can travel at the speed of light. Now, $E = mc^2$ is not a complete formula. The ...

3

In a curved spacetime, the wave four-vector $k^a$ for a solution to Maxwell's equation (in the appropriate geometric-optics limit) satisfies the geodesic equation: $$(k^a \nabla_a )k^b = 0.$$ Effectively, this means that "the derivative of $k^a$ in the direction $k^a$ is zero". (Note that the first term is just a directional derivative in the direction of ...

3

Gravitational red/blue-shifting conserves energy. It merely swaps potential energy for kinetic energy. A photon outside a gravity well has zero gravitational potential energy. As it falls into the well, it gains momentum and loses potential energy, similar to anything else falling into a gravity well. At least, that's a simplistic way of looking at what's ...

0

(I am guessing that you by "decay" mean, "is the material loosing mass slowly". As the other answer suggests, decay is a word reserved for something else in physics.) "Knocs electrons off" is in this sense quite a simplistic way to describe it. Rather see it like this: The photoelectric effect is when a photon (light) excites an electron. "Excites" ...

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No decay within atoms is where the neutrons or protons change such as beta which is a neutron to proton. Removal of electrons doesn't alter the element or nucleus structure.

1

My answer to the question is first a question. How do you photograph a photon? You can measure a photon by detecting it's presence on photographic film, or by using some sort of photomultiplier and digital detector, or by a handful of other ways. My point is that these all require the photon to be absorbed by the detector, thus they must be localized, and ...

1

This is true. If for example you subject Hydrogen gas to a perfectly monochromatic 121.57 nm laser, then all that will happen is that the gas will scatter the light in all directions, glowing without increasing the temperature. Otherwise there are many different phenomena that are involved in the heat transfer of energy by radiation. For example in solids, ...

0

I found a nice discussion of this from several years ago at physicsforums. Inside, Ben Crowell makes an argument that does in fact use time dilation: if you're willing to accept that the properties of a massless photon are the same as the $m\rightarrow 0$ limit of a massive photon, it's easy to see the decay rate has to be zero. One way to see this is that ...

0

It is not possible to find a frame of reference where a photon is at rest. I will argument in two different ways: 1. Maxwell equations and electromagnetic argument: From Maxwell it is expected that electromagnetic disturbances propagate in vacuum at a constant speed c~299792458 m/s which is the maximum speed for the propagation of electromagnetic ...

3

The answer is "no", not as a straight 1-body -> 2-body decay. Proof: let $E_\gamma, E_1$, and $E_2$ denote the energies of the initial photon and the two daughters, and define $\vec{p}_\gamma$, $\vec{p}_1$, and $\vec{p}_2$ similarly for their momenta. By conservation of energy and momentum, we have $$E_\gamma = E_1 + E_2,$$  \vec{p}_\gamma = \vec{p}_1 ...

5

The selection rules for atomic transitions are entirely governed by how much angular momentum the photon can carry away. A dipole transition (in the nomenclature, E1 or M1, depending on its parity) carries away angular momentum $\hbar$, so the atom's initial and final state must have angular momentum quantum numbers $J$ differing by zero or one. A ...

-1

You should remember that photons don't have any mass, so relation is $E=p/c$ for photons. Moreover, you have to count electrons mass when you apply energy equations. Electron mass will vanish, but it's important to take account of it. I mean, before the square root, you don't have the electron mass, you have the kinetic energy.

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No, the virtual photon is not a particle, since a virtual particle is what one calls the internal lines in a Feynman diagram, and there are no asymptotic particle states associated to these lines, so a virtual particle is not a particle in the usual (or any other rigorous) sense. Therefore, the question is non-sensical because it is not clear what an ...

-1

I believe that the photons of EMR "escape" the source of them (excited electrons), whereas somehow the photons in a magnetic field just circle around.

1

Pair production can also happen in extremely strong electric fields via the so-called "Schwinger mechanism". Unlike most other results in quantum field theory, in this case one can exactly compute the probability of creating pairs per unit volume and time, see e.g. here.

1

Gamma radiation is the highest in energy on the electromagnetic spectrum. Any photons with energy over $100keV$ are classified as gamma radiation. As stated in other answers, for a photon to produce a particle-antiparticle pair, the energy of the photon must be at least twice the mass-energy of the type of particle produced. There are 2 particles of equal ...

1

Pair producing an electron and a positron requires an energy at least equal to their masses, $2m_ec^2$. This would just create them stationary, and you'd need an even greater energy to give them some momentum. Since $m_e = 0.511$ MeV/c$^2$, the minimum energy required for pair production to occur is $\sim 1$ MeV. What photons have an energy close to that ...

2

"Pair production is when a particle-antiparticle pair is produced from a single gamma photon. The gamma photon must have enough energy to produce that much mass. Pair-production usually happens near a nuclear, which helps to conserve momentum." It is a hand waving way of describing pair production. Gammas are photons of high energy by definition. ...

1

It's not really worthwhile in this type of situation. (It makes sense in other situations however ... like transferring power from the ground to an airplane or satellite.) The two most plausible system types are: (A) Microwaves / radiowaves: Emitted by an antenna, collected by a rectenna (B) Visible / infrared: Emitted by a laser, collected by a ...

0

No, because you need an infinite amount of energy to make a massive particle go at light speed. Photons are massless, therefore there is no relation.

2

Short answer: yes. Longer answer: unlikely to be distinguishable from noise. Here's what happens: some photons may be reflected to various items in the camera, e.g., if the autofocus has its own sensor. The photons which follow the optical path to the CCD or CMOS detector have a certain probability of generating a photoelectron. This probability is called ...

3

It seems to me that you just cannot tell the difference between a Bose condensate and nothing in this case. What will change if you add some photons or phonons with zero energy to the system? No characteristics of the system will change. So it seems to me we have no criterion to decide if there is a Bose condensate in this case, and what's more important, it ...

0

I could give a shorter then the other answers because of a different understanding how a double slit works. See my questions and answers in this forum. Every photon with its E and B field components get into interaction with the surface electrons from the slits material. Their field is quantized and this we see as fringes. Diffraction we not only get behind ...

2

Will the pattern the 1 million entangled photons (An) created on the right screen in case 1), be discernible from the pattern in case 2), in the sense of being able to state with high probability which of the 2 cases applies, by analyzing the distribution of the photons (An), which hit the screen after passing through the right slit? No, the pattern of the ...

4

Your question is muddled and unclear (in my professional opinion - you may think otherwise and that's OK), but I think I can make one thing clearer. If you have two entangled systems, $A$ and $B$, and perform independent experiments on either side, then nothing about the choice of experiment you perform on $B$ will have any effect on the local results from ...

2

There is also, a gas of photons in the right side. It was trapped there when you assembled your box, and since you are assuming a perfectly zero emissivity, these photons must be perfectly reflected from all surfaces. That means they are blue shifted if the wall moves toward the right and red-shifted if the wall moves toward the left. Result: If you ...

1

I'm going to have to give an answer that's very different to Jimmy360's. Apologies. How does the potential and kinetic energy of a photon relate? They don't. The photon is all kinetic energy. Do they mean the same thing? No. When you drop a brick, its gravitational potential energy is converted into kinetic energy. When you dissipate this kinetic ...

0

Quote from a webpage a bit over my head :-) greatians.com Photon has linear momentum. Photon travels in vacuum space at the ultimate speed of light. Photon has the quantized energy of hf as given by eq. WD.1.2. E = hf … eq. WD.1.1 where h = Plank’s constant ...

0

Imagine a beam of light, going towards a massive object. It has potential energy in the gravitational field. Of course, the potential energy has to become kinetic energy. This is done by shifting frequency. The energy of a photon is given by $E = hf$ so to increase kinetic energy we must increase frequency. If the beam of light was red, it will be a higher ...

1

Is light a particle which has a electromagnetic field around it OR does the particle itself travels in a wave like motion? The latter. Light consists of electromagnetic waves which have a quantum nature, wherein we say the photon is a quantum of light. This tends to get converted into a "particle" of light wherein people think of the photon as some ...

1

Is light a particle which has a electromagnetic field around it OR does the particle itself travels in a wave like motion? This depends of what you want to calculate. For atomic processes one use QED and photons are disturbances in EM field existing everywhere. But really one calculate processes in atomic and subatomic dimensions. If one want to ...

1

This is I guess a common problem that stems from the way we use the word "intensity". Normally, we would use intensity to mean something like the energy going through a unit area per unit time. In the case of the photoelectric effect, we instead generally mean the number of photons (per unit time), as it is this that decides how many photo-electrons are ...

1

a) What would happen if we did NOT detect an interference at D0 8ns prior to the entangled idler photons reaching D3 or D4 and decide to remove the BSa and BSb beam splitters really really fast such that the idler photons would travel to either D1 or D2 instead, hence the "which path" is not known and therefore we should actually see an interference pattern ...

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