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1

Yes, this is actually often used in a spectroscopic technique called REMPI -- see the image on this wikipedia page https://en.wikipedia.org/wiki/Resonance-enhanced_multiphoton_ionization There are some important physics techniques that rely on interaction with two photons -- two photon spectroscopy (http://cua.mit.edu/8.421_S06/Chapter9.pdf). Some other ...


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The key to the effect is the idea - really just a conjecture, and as the paper states, "these ideas have not been met without controversy" - that there are materials that are opaque to gravity waves in the same way that conductors are opaque to EM waves. If that's true, then the cavity between parallel plates will contain a reduced set of modes of the ...


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charge is an intrinsic property of any particle. we in principle cannot change the intrinsic property of any particle. photons are the carriers of electromagnetic interaction(action at a distance).


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I have read the paper and the gravitational explanation of the Casimir effect is not the main point, it's the experiment that provides evidence of the graviton's existence. The paper starts by drawing a macroscopic analogy between Maxwell's equations and Einstein's linearized field equations (known as gravitoelectromagnetism). The author uses GEM because of ...


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The point is that a Dirac field isn't really a solution for an electronand positron but instead an electron and the conjugate of the positron. This means that the Dirac electron field does indeed obey, \begin{equation} e ^{ i \alpha (x) Q } \psi = e ^{ - i \alpha (x) } \psi \end{equation} Such misconceptions are dissolved if one works in the Weyl ...


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If yes, how I can do it? Try the Feynman Parameter trick. It allows you to take the product of terms in the denominator and convert it into a sum. Then the sum looks like a standard integral. See, for example, Google or Peskin & Schroder at ~page 250.


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I cannot answer to all the questions but would like to stress something regarding what the Casimir effect tells us and what it doesn't. If you look at how it is derived for the usual EM interaction, an experimental verification of the standard Casimir effect tells us that: the EM field can have standing waves between two plates and outside them There ...


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The standard classical description due to retarded Lienard-Wiechert potentials does not describe this reaction force back on the source particle. There is force back on the first particle, but it occurs after a delay necessary for the retarded field of the second particle to get back to the first particle. This delay is due to the choice of retarded ...


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For what it's worth, I showed in my recent article http://link.springer.com/content/pdf/10.1140%2Fepjc%2Fs10052-013-2371-4.pdf (published in European Phys. J. C) that one can eliminate the Dirac field from the Dirac-Maxwell electrodynamics after introduction of a complex electromagnetic 4-potential (producing the same electromagnetic field as the real ...


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In our modern understanding, every electron is thought to be a localized excitation of the electron (or Dirac) (spinor) field $\Psi(x^\mu)$, while every photon is considered to be an excitation of the photon (vector) field $A^\nu(x^\mu)$, which is the quantum field-theoretic counterpart of the classical four-potential. Thus, the answer to your questions ...


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Basically recoil, since photons can also carry momentum. The reason why this recoil can also lead to attraction instead of plain repulsion is that the exchanged virtual photon has a definite momentum, so it can not have a definite position at the same time (the position wave of a virtual photon is stretched out to infinity). The momentum can be transfered ...


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To find the lifetime you can use a mixture between QED and non relativistic QM. You use QED to find the cross-section of $e^+ e^- \rightarrow \gamma\gamma$ which is to first order in perturbation theory for the singlet state $\frac{4\pi}{cv}\left(\frac{\alpha}{m}\right)^2$. Then you must multiply this by the electron luminosity, this is where you use the QM ...


3

Quantum mechanics and classical E&M have created a whole bunch of fruit. In no particular order: Light is an electromagnetic wave. This means that we are routinely making light with LEDs and the like. It's pretty cool that you can design better antennas, or waveguides, or what have you. While we're at it, knowing why Stefan-Boltzmann radiation goes ...


1

The Cosmic Microwave Background includes photons that will not be absorbed before the universe inflates to the point where there is nothing to hit ever again. If parts of the last scattering were somehow barred from releasing that energy, I think we would notice. The photon carries energy. Particles do that. How is it any different from the electron, which ...


2

It is pretty simple just use the following formula, $$ \int d^3 x e^{i(p+q)x} = (2\pi)^3\delta(p+q)$$ and thus on integrating $d^3 q$ you will have $\sqrt{2E(p)2E(q)} = E(p)$ in the downstairs, and then it's pretty straightforward.


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As an outcome of his PhD thesis work, Richard Feynman and John Wheeler wrote a series of papers on how the kickback on an electron as it emits a photon can be modeled accurately as the result of an "advanced photon" traveling backwards in time and striking the electron. No, the Feynman-Wheeler theory considers model of classical electromagnetic ...


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Perhaps this almost ventures into the realm of philosophy, but the original premise oddly treats the photon differently from the electron. The only reason we think of it this way is because it is more common to observe the photon being created, since it is a boson. But quantum electrodynamics dictates that as a photon is traveling along at a great ...


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Quantum electrodynamics is needed to describe Nature instead of classical electrodynamics because quantum phenomena are observed – and have been observed at least since 1900 – which prove that classical physics in general and classical electrodynamics in particular is incorrect as a description of Nature and a better theory is needed. The quantum phenomena ...


1

In terms of Feynman diagrams, a "coupling" translates to a vertex factor. The Lagrangian for a free electromagnetic field is $$\mathcal{L}=-\frac{1}{4}F^2$$ as you well know. Now suppose we have an electron field $\psi$ too. We want this electron field to "interact", or couple, with (to) the photon field. The free Dirac Lagrangian is ...


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Light, the classical electromagnetic field , is built up/emerges from a large number of photons, not in a simple manner. Photons are elementary particles and therefore can only be described within a quantum mechanical framework. They have a wave-function that obeys the potential form of Maxwell's equations turned into operators which operate on the ...


0

In quantum mechanics, the magnetic moment operator is related to the spin operator by: $\vec\mu = -\left(\frac{e}{mc}\right)\vec{S}$ In other words, they are directly proportional up to some known physical constants. This means that measuring the spin of an electron is exactly equivalent to measuring its magnetic moment: if you obtain either quantity, you ...


1

In the photoelectric effect however the photon seems to give all it's momentum and energy to the electron? No, the photoelectrons are emitted with a range of energies. The well known expression: $$ E = h\nu - \phi $$ gives the maximum energy, but photoelectrons are emitted with energies ranging from zero to this maximum value. The emission of ...


1

A short, mathematical answer to the question is found in the properties of Fourier transforms. The temporal response of the environment to a perturbation is given by the Fourier transform of its frequency response to the same perturbation. Therefore, if a broad range of frequencies in the bath are perturbed, the response occurs over a narrow range of times. ...



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