# Tag Info

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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 ...

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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 ... 6 To properly understand what is going on you need to understand general relativity. Massless particles, like photons, travel on null geodesics and mass bends spacetime so the null geodesics are not straight lines. The problem is that neither you nor your teacher understand general relativity so this isn't a very convincing argument. But here is an argument to ... 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 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 ... 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). 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 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 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 Since the photon reflects, its momentum changes: p_{ph}'=-p_{ph}. But total momentum of the system is conserved: p_m+p_{ph}=p_m'+p_{ph}'. Thus, the mirror will change its momentum. But, if the mirror has large mass, then it'll get very small energy from the collision. For zero-mass particle (photon) falling onto the mirror with mass m_2, the energy of ... 2 You are asking about the number of photons fired from a device such as a laser at a unit time. We cannot say what the precise number of photons would be at any interval due to a version of the Heisenberg uncertainty principle:$$\Delta E \Delta t \geq \frac{\hbar}{2} It basically states we cannot suppress our uncertainty about energy fluctuations in time, ...

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The classical electromagnetic field given mathematically by Maxwell's equations can be proven to emerge from a confluence of individual photons, which photons are described by the Quantum Mechanical form of Maxwell's equations. Thus the classical wave is made up by zillions of photons with energy $h\nu$, where $\nu$ is the frequency of the classical wave. ...

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In case you "run out of photons", you must switch to probabilistic description of quantum mechanics. Let's consider an extreme case: You have an emitter of spherical waves which radiates just one single photon. You place a lot of detectors some meters apart (with same distance) from the emitter. QM says that the photon propagates as a probabilistic wave to ...

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Briefly, the formula $E=mc^2$ applies only particles at rest in an inertial frame of reference. Since there is no rest frame for a photon, no inertial reference frame in which a photon is at rest, one cannot apply the formula $E=mc^2$ to a photon. In more detail, the four-momentum of a particle has components $(\frac{E}{c}, \vec p)$ The four-momentum for ...

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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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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 ...

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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 ...

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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. ...

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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 ...

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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

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 ...

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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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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 ...

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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 ...

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

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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. ...

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What you read is correct. I am not sure if those were the exact words of your teacher but according to the general theory of relativity, sun doesn't "attract" the photon (or any other body for that matter). In fact gravity is not even a real force. Let me briefly state what the theory of relativity has to say about gravity without going into the complicated ...

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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 ...

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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 ...

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