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1

Binocular vision has already been discussed, but it left out an important aspect. A single eye is sensitive to distance. The shape of the lens changes to focus on near/far objects. The reason this is needed is that our pupil has finite size and cannot be modeled as a pinhole. The same physics is going on here as in a lens of a camera focusing on an ...


3

To have depth perception two eyes are needed. Our two eyes are some distance apart which causes the photons from an object to arrive at slightly different angles. The brain then reconstructs the depth field from these differences. Similarly, we can figure out how far nearby stars are by using images made by a telescope at two different times of the year, ...


2

There is no edge of the universe. The standard model for cosmology is based on the FLRW metric. It is what happens when you assume the universe is homogeneous and apply general relativity. In this model the universe could have a finite volume that's not growing too fast, where paths just loop back on themselves. A photon in such a universe would keep ...


0

Will an unhindered (un-scattered) photon go to the edge of the universe? To add to the answer of @RedAct : If you are thinking of the three dimensional part of the universe we observe now, it is an instantaneous universe, i.e. time t has one value. In this sense we are at the edge of the expanding universe, the right cutoff at 13.8 billion years in the ...


1

The answer probably depends on how that question is interpreted. The universe is expanding. The ultimate fate of the universe isn't known for sure, but the growing consensus among cosmologists is that the universe will probably continue to expand forever. If that's the case, then a photon that leaves the Earth now will never catch up to what is currently ...


1

Lets attempt some answers: Both can happen, a quantum transtion can be associated with a photon exchange or a photon exchange can be associated with a quantum transition (this is just 2 ways to state the conservation of energy in these cases) Photons do not have mass but they do have momentum. There are some approaches in physics which associate a virtual ...


0

I have personally measured many photons, certainly more than one. Not once did a photon tell me that it was the same one that I measured an hour earlier. :-) That time stands still for a photon is not true, by the way. Photons simply make a full rotation from the time-like to the space-like coordinate axis. To a photon "when" becomes "where". You can see ...


3

You are using the concept of time in a mixture between Newtonian ideas and Relativistic. It is true, that for the photon time is slowed to a standstill, however, for an observer who has mass, time still flows, one can measure simultaneously two separate photons, that are not causally connected and know these are two separate entities since having measured ...


1

Yours is a subtle question with a rather subtle answer. From the way you ask the question, you seem to be thinking of the photon as a little billiard ball. It is not. It is an excitation of a quantum field, which is described very differently from a "particle" in the classical sense. Even so, the short answer, in some very subtle ways, is that indeed there ...


4

It's not necessarily true that most of the photons that strike a wall will be absorbed and turned into heat. The whitest white paints can have a light reflectance value of up to about 85%. There isn't a "wavelength corresponding to white color". An ideal white surface reflects as much as possible of all wavelengths in the visible spectrum. That sounds ...


0

1) No, substances almost never completely absorb photons. Otherwise you could not see them. In case a substance would absorb all photons (which is quite hard to achive intentionally) it would be pitch black even if you shine arbitrarily strong light on it (-> black-body). 2) It will be reflected back and forth, but only a finite amount of time. This is ...


1

Almost always, when photons hit matter or interact with it, they are not reflected in the way a billiard ball bounces off a billiard table edge. Rather, they are absorbed, the absorber rises into a metastable state, and then a new photon is emitted on the decay of the metastable state. Sometimes, though, when photons undergo an interaction with a lone ...


3

Almost but not quite. Qualitatively the spectrum is the same with the $1/n^2$ spacing, but the scale of the spectrum is set by the reduced mass $\mu$, $$\mu = \frac{1}{\frac{1}{m_l}+ \frac{1}{m_p}}$$ where $m_p$ is the proton mass and $m_l$ is the lepton (muon or electron) mass. Since $m_p \approx 2000 m_e$, it is not a large error to take $\mu = m_e$ for an ...


1

Photon photon interaction, which is what a collision will mean, is practically non existent as it is higher order and in the context of this question, light sources, non existent in reality. If we reach gamma ray energies then particles will be produced but this has nothing to do with this question. There will be interference patterns as whenever coherent ...


2

I think what you say is correct. By saying it should be unitary, he means it should be time-reversible. It is indeed time-reversible if before reaching the detector 2 it is in the same state as the initial state. It starts off with the superposition, and then only vertical polarization in the top part and horizontal in the lower part of the arms, which ...


0

"If a photon hits an electron". If only it were that simple! The electrons (in this case) are bound in atoms (or ions). If the photon has the "right" energy, then the probability of the atom making a transition from one state to another may become appreciable and the transition may occur. If the photon does not have the "right" energy - and you can think of ...


3

If that happened, we would be able to detect it by looking at correlations between successive photons' "decisions." That is, suppose you represent each pair of consecutive photons (1 and 2, 2 and 3, 3 and 4, etc.) with $+1$ if they both made the same "decision" or $-1$ if one went through the polarizer and the other didn't. Take the average of these numbers ...


5

There is no such mechanism. The probability for a photon to pass through a polarizer at an angle $\theta$ is $\cos^2(\theta)$, regardless of what has happened before, and regardless of how many photons "at once" try to pass through it. As Bell's theorem tells us, the quantum world is really random (or non-local).


0

A perhaps not completely rigorous, but easy to understand derivation: It can be shown that the magnitude of angular momentum of circularly polarized classical radiation of frequency $\omega$ and energy $E$ is given by $E/\omega$. If we now assume that radiation is quantized in packets of energy $E = \hbar \omega$, we arrive at the conclusion that the ...


-2

Perhaps this helps: https://de.wikipedia.org/wiki/Benutzer:HolgerFiedler/Strukturen/Summary It explains how the photons electric field changes to the magnetic and back and there are no points of zero energy. But what you have to keep in mind is that radio waves are cliffs rent waves, they are omposed from a lot of photons and the radio wave is zero at all ...


1

As was pointed out in the comments most photons are produced individually from deexcitation of electrons from higher energy orbitals, where they had been pushed by heat ( as with heat filament lamps) to lower ones. Since I am a little new to this topic, a little background of entanglement would be appreciated as I might be wrong in my conceptualization. ...


1

Just consider the gauge transformation after Fourier transforming everything. A Fourier transform turns derivatives into momenta, such that we get \begin{equation} \tilde A_\mu \rightarrow \tilde A_\mu - \frac1e k_\mu \tilde\alpha \;. \end{equation} This mean that only the component parallel to $k_\mu$ (the longitudinal one) will change, while the ...


5

An experimentalists answer: If you divide the energy of the electromagnetic wave by hv you will have the number of photons that are building up the electromagnetic wave.


11

You only need to rewrite $\mathbf B$ and $\mathbf E$ in terms of field $A_{\mu}$ (here $\hbar = c = 1$), $$ \tag 1 \hat{\mathbf B} = [\nabla \times \hat{\mathbf A}], \quad \hat{\mathbf E} = -\frac{\partial \hat{\mathbf A}}{\partial t} - \nabla \hat{A}_{0}, $$ which is written as infinite "sum" of photons: $$ \tag 2 A_{\mu} = \sum_{\lambda} \int ...


1

In most quantum processes you cannot get just particles (without antiparticles) as products. That would violate some conservation laws (charge conservation mainly). So the quark-gluon plasma was a mixture of quarks and antiquarks. As a consequence, after the QCD cooled, you get both hadrons and antihadrons. These annihilated, but there was certain assymetry ...


0

First clearing up a misconception. I understand that protons and neutrons (which are baryons, which are hadrons) are made out of quarks, and quarks are held together by gluons (at a high level). So those are all the ingredients listed in the title "quark-gluon plasma". Nope. There are no hadrons in the QGP phase (regarded as a pure phase, I'm ...


5

Gluons and quark-antiquark pairs, like photons and lepton-antilepton pairs, are excitations of the vacuum. Any volume of space with enough energy density to contain a quark-gluon plasma also has, by definition, enough energy density to contain a gas of photons and electron-positron pairs. The difference is that the quark-gluon plasma is governed by the ...


2

The chronology in Wikipedia is emphasizing what is changing in each epoch. Particles that are lighter than the ambient temperature are presumed to be in thermal equilibrium, but not doing much of interest at that time. In particular, the light leptons and photons are created and annihilated all the time, so there is a sea of electrons, positrons, and ...


1

You can think of light as the carrier of the electromagnetic interaction. The particles interact with light, not directly with each other. It is an experimental fact that light does not interact with itself. Note that this is not the case with quantum chromodynamics (the theory of nuclear matter). This theory is built along the same lines as quantum ...


1

Light waves cannot be longitudnal in free space. Transverse EM waves do exert pressure, but it's too small to be perceived by humans.


0

Martin wrote: ", hence they can only interact via the weak force, which is, as the name says, weak." Makes me wonder, is "weak" a good name? Or is this interaction better described as short, and still full of surprises? I'm thinking of {Dirac, Weyl, Majorana} spinors, rates of chiral oscillation, inertial-mass... ... which makes me wonder about the ...


1

Unfortunately, in quantum mechanics "ordinary" reasoning does not get you anywhere. The photon, like any other particle, is neither a particle nor a wave; it is an entity that we can only describe mathematically. It's only when we observe it that it shows up as either particle or wave. Or senses, and hence our logic, evolved to make sense of the real ...


2

What constituent of internal energy does an electron excitation represent? You can think of electrons as just like planets orbiting the sun and get the correct answer to this question. An electron in a higher energy level has less kinetic energy, but more potential energy as it is (generally) farther from the nucleus. The net result is more energy. ...


2

The change in electronic excitation represents both a potential and a kinetic energy term in classical physics, but there is no simple correspondence to classical physics terms, when you are looking at quantum systems. All we really care about is the total energy difference between electronic states. Those energy differences correspond to the energies at ...


2

Not merely can it transfer its momentum as well as its energy when it interacts, but it must. If the target atom is in a fluid context (liquid, gas, plasma), then that energy and momentum must end up in the target or some other reaction products(s). In a solid context the Mössbauer effect can be an issue, allowing the transfer of that momentum to a much ...


3

It is not very clear to me if you are asking about energy or momentum. You should also ask about a specific interaction process as there are many, this is required especially to answer your last, quantitative, question. However, generally speaking, a $\gamma$ photon cannot give some of its energy to anything else: it is all or nothing. Even in the Compton ...


2

This is an instructive video on the double slit experiment with photons which is experimenting with single photons. At about 2.5 minutes it explains how the experiment is done with single photons. In a nutshell, by lowering the intensity of a light source to the point of zero emitted by the source and then slowly increasing it. The detection hinges on the ...


3

Generally speaking, the emitted frequency is not necessarily the same as that absorbed but let's suppose that it is. As the white light (composed of all the different frequencies) encounters the material (the red box, below), certain frequencies are absorbed and then re-emitted but in random directions. So they're scattered. Therefore, far fewer photons of ...


1

The "slowing" of light in a medium can be entirely explained using a classical wave-based approach. An incoming EM wave wiggles the electron clouds around the atoms in the material. These electrons clouds re-emit a much weaker EM wave having a very small amplitude. This re-emitted wave is 90-degree phase shifted from the original wave but superposes with ...


0

To make one thing perfectly clear: photons do interact with each other. That interaction is simply very weak for low energy photons, but non-trivial at very high energies. Photon-photon interaction does, of course, not play any measurable role in optical experiments with visible light. As for the Hanbury and Twiss effect... IMHO it's trivial and classic. ...


1

There's no contradiction. Without considering high-energy physics, photons don't interact in free space in the sense that they pass through each other intact. Here's a very recent review article: http://www.nature.com/nphoton/journal/v8/n9/full/nphoton.2014.192.html Free space photons can of course lead to interference patterns and the HBT effect is still ...


0

We know four fundamental forces, three of them being included in the Standard Model. At low energies (compared to the mass of $W^\pm$) the forces have strengths as follows: strong force > EM force > weak force > gravity Also, photon and neutrino are really quite different. Photon is a force carrier, while neutrino is a matter particle. Now, matter ...


1

The problem with the picture (and probably with your understanding of the physical process) is that it assumes photons as classical particles on well-defined trajectories. If this were a true picture of reality, your objection would be justified. This, however, is not so. In order to describe the process properly, one has to acknowledge the quantum nature ...


1

The photons do not have a well defined trajectory. The diagram shows them as if they were little balls travelling along a well defined path, however the photons are delocalised and don't have a specific position or direction of motion. The photon is basically a fuzzy sphere expanding away from the source and overlapping both slits. That's why it goes through ...


11

The photon does couple directly to charged stuff, e.g. via Compton scattering. This is indirectly related to the spin, as direct interactions between fermions are hard to construct. The neutrino on the other hand does not couple immediately to any other matter particle. It requires a force-carrier. Now as it turns out the only force carriers that care about ...


8

Neutrinos having no charge means they don't participate in electromagnetic interactions, which are the strongest (at least long range). Them being leptons means they don't interact with the strong force (which is, as the name says, strong), hence they can only interact via the weak force, which is, as the name says, weak. Photons on the other hand do not ...


0

It's a matter of definition. For example, by photon, do you mean first or second quantization (the latter being the canonical treatment)? Although there's no ambiguity in theoretical physics, the nature of photon has generated much debate in the community of applied physics. See, for example, these articles: http://arxiv.org/pdf/quant-ph/0605102.pdf ...


0

You have really figured out the answer yourself. On one hand, you have a natural phenomenon (light) and on the other you have our models (wave description, photon description). When speaking of the reflection of rarefaction of photons, the author implicitly assumes the reader to know about the wave-particle duality (see links in the comments). I think you ...


2

Vacuum magnetic birefringence basically involves the same loop diagram as light-light elastic scattering except that two of the four photons come from a magnet. Detecting this effect is the aim of the PVLAS experiment in Ferrara, Italy. See arXiv:1406.6518 and references within. The experiment is running at the moment but the sensitivity is not good enough ...



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