# Tag Info

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 ...

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 ... 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 ... 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. 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). 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 ... 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 ... 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 ... 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 ...

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 ...

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 ...

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 ...

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

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 ...

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 ...

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 ...

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

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

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 ...

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 ...

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 ...

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 ...

1

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 ...

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 ...

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

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 ...

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 ...

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 ...

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 ...

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.

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