New answers tagged

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Is scattering an interaction? Yes, scattering is an interaction between quantum mechanical entities. There are four fundamental forces that are involved in scatterings and decays. The strong which has a coupling of 1 is responsible for the nuclear forces and the periodic table of elements. The electromagnetic , which has a coupling of 1/137 The weak ...


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The way I understand scattering in classical physics is: it is underlying electromagnetic interaction plus the exclusive principle. There's no exclusion principle in classical physics. Classical physics means physics without quantum mechanics. So I had thought it would have no scattering nor interaction between dark matter particles, but it seems not a ...


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Some basic geometry reveals the following formula where $R$ is the radius of earth, $h$ is the height of the tower above sea level and $d$ is the distance traversed- $$(R+h)^2=R^2+d^2$$ The coast of Lebanon is pretty mountainous with an average altitude of $2500{\text m}$. Considering they built a tower with a decent height of $200{\text m}$. This tower will ...


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The antennae almost certainly weren’t at sea level. Find the island’s name, find its highest point, and look up the horizon distance. The other end might be trickier, unless a press release give more specifics, but you can pick a mountain in Lebanon. Maybe you’ll be able to challenge the flat-earther’s beliefs with some horizon-distance info...


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The whole point of QFT is that it is a framework that allows you to define Lorentz (co)variant scattering amplitudes. In fact, under some general hypothesis it is the only framework with that property. The expression in the OP is not manifestly Lorentz covariant, although it turns out to be, after a very cumbersome analysis. See ref.1 for a detailed ...


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Your expression for $\mathcal{M}^2$ is wrong. Inside $\mathcal{M}$ polarisation vectors are contracted with the momenta so for example $$\left|(P + P')^\mu \epsilon_\mu\right|^2 =(P + P')^\mu \epsilon_\mu \, (P + P')^\nu \epsilon_\nu =(P + P') \cdot \epsilon \, \,(P + P') \cdot \epsilon$$ It seems that you incorrectly contracted the $(P + P')$ factors with ...


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The expected probability density for Compton scattering is \begin{equation*} \langle|\mathcal{M}|^2\rangle= 2e^4\left( \frac{\omega}{\omega'}+\frac{\omega'}{\omega} +\left(\frac{m}{\omega}-\frac{m}{\omega'}+1\right)^2-1 \right) \end{equation*} where $\omega$ is the angular frequency of the incident photon, $\omega'$ is the angular frequency of the ...


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The expected probability density is \begin{equation*} \langle|\mathcal{M}|^2\rangle= 2e^4\left( \frac{\omega}{\omega'}+\frac{\omega'}{\omega} +\left(\frac{m}{\omega}-\frac{m}{\omega'}+1\right)^2-1 \right) \end{equation*} The Compton formula is \begin{equation*} \frac{1}{\omega'}-\frac{1}{\omega}=\frac{1-\cos\theta}{m} \end{equation*} It follows that \...


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1) A normalization condition for one-particle states $| \vec p \rangle$, which is Lorentz invariant is $\langle \vec p | \vec q \rangle = 2 E_p (2 \pi)^3 \delta^{(3)} (\vec p - \vec q)$ (The factor $2$ is conventional) However, if we write $\delta^{(3)} (\vec k) = \int \frac{d^3x}{(2 \pi)^3} exp (i \vec k \cdot \vec x)$ we can state $\delta^{(3)} (\vec 0) ...


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$:\phi(x_1).\phi(x_2):$ is equal to $:\phi(x_2).\phi(x_1):$ So it doesn't matter how you write it. You can check by expanding it out.


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Good question! The difference is simply the setup: In a scattering problem, one considers a wave (or particle or whatever else you can shoot at stuff) traveling towards a target. The interaction with the target causes the wave to be deformed (or the particle to bounce off in a different direction), i.e. scattered. In a source problem, one does not have the ...


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A probability density can be obtained by integrating the differential cross section over $d\Omega$. For Moller scattering we have $$ \frac{d\sigma}{d\Omega} =\frac{\alpha^2}{8E^2} \left( \frac{1+\cos^4(\theta/2)}{\sin^4(\theta/2)} +\frac{8}{\sin^2\theta} +\frac{1+\sin^4(\theta/2)}{\cos^4(\theta/2)} \right) $$ Let $$ I(\xi)=2\pi\int_\alpha^\xi\frac{d\sigma}{...


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Just to fill in a few of the blanks of TwoBs answer: Using \begin{align*} \sum_{s} \epsilon^s(k)^{\mu}\epsilon^s(k)^{\nu *} = -\eta^{\mu\nu} + \frac{1}{2E^2}(k^{\mu}\bar{k}^{\nu}+k^{\nu}\bar{k}^{\mu}), \end{align*} from TwoBs notes, we would like to calculate \begin{align*} \sum_{s,s'} |\mathcal{M}|^2 = 4e^4 \left( \frac{p_{\mu}p'_{\nu}}{p\cdot k}-\frac{p'...


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So let us first define what is a bound state, in particular with respect to scattering theory. Scattering theory studies asymptotic states coming out of interactions in the limit of infinite times. That means that if we want the bound state to appear in our out states, it must be stable, or not spontaneously decaying after a finite amount of time. If it does ...


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The integral should be done over $d\Omega$. Here is the differential cross section for the process. $$ \frac{d\sigma}{d\Omega} =\frac{\alpha^2}{16E^2} \left(1+\cos^2\theta+\frac{m^2+M^2}{E^2}\sin^2\theta+\frac{m^2M^2}{E^4}\cos^2\theta\right) $$ For $m=0$ we have $$ \frac{d\sigma}{d\Omega} =\frac{\alpha^2}{16E^2} \left(1+\cos^2\theta+\frac{M^2}{E^2}\sin^2\...


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Another reasoning leading to the conclusion that one cannot just simplify the expression for $\mathcal{M}$ (first equation) using the Lorentz gauge condition $\epsilon . k=0$ and subsequently replace in the simplified expression (second equation) $\sum \epsilon_\mu \epsilon_\nu^*$ by $ -\eta_{\mu \nu}$ is the following. The matrix element of the first ...


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One way to reduce the number of terms is to set $m=0$. For $E\gg m$ and using the approximation $m=0$ the result for the trace calculation is $$ -8s^2 $$ This result (obtained by computer) indicates the magnitude of simplification that occurs. Complete derivation follows. In center of mass coordinates the momentum vectors for Moller scattering are $$ ...


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These identities should work: \begin{align*} s&=4E^2 \\ t&=-2E^2(1-\cos\theta)=-4E^2\sin^2(\theta/2) \\ u&=-2E^2(1+\cos\theta)=-4E^2\cos^2(\theta/2) \end{align*} Derivation of probability density for Bhabha scattering: In a typical collider experiment the momentum vectors are \begin{equation*} p_1=\begin{pmatrix}E\\0\\0\\p\end{pmatrix}\qquad ...


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That depends strongly on many factors. The reason for the decrease in thermal conductivity at high temperature is that the mean free path goes down because of phonon-phonon scattering (interaction with the lattice). At some point the phonons do not propagate much further than a lattice period and the lattice melts. So those are the shortest mean free paths. ...


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The Rayleigh scattering cross-section from nitrogen molecules is around $1.6\times 10^{-30}$ m$^2$ at 400 nm (blue light). At sea level, there are about $2\times 10^{25}$ molecules per cubic metre. The mean free path of a blue photon is therefore 31 km. Since the Earth's atmosphere has a scale height of around 8 km, then the vast majority of photons ...


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Wikipedia states very strongly that explanations of this in terms of Rayleigh scattering are wrong, and that the real explanation is the absorption of blue light by ozone. Ozone doesn't absorb blue light (much): on the contrary, it absorbs red light much more, thus making the sky look blue. See in particular my answer for the question at Chemistry.SE: What ...


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From one of your comments to other answers: I think the sunlight doesn't lose its blue that much on the way to the point where red scattering dominates. Why wouldn't it be able to? Rayleigh scattering has cross section proportional to $\lambda^{-4}$. The spectrum of light illuminating the volume scattering it gets multiplied by $\lambda^{-4}$, which ...


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The reason is the same as in rainbow; different colours (wavelenghts)bends differently (sorry my language). This Blue hour is the sunlight which arrives below the horizon. The reason why other light doesn't is on their wavelength. The Blue light can bend the most, and thus it can still reaches the Earth surface an hour after sunset, while green, yellow and ...


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When you look at the sun overhead (not advisable) you see the white light from the sun with a little bit of the blue scattered horizontally. The blue sky is blue light scattered from sunbeams going elsewhere. When the sun is near your horizon, its light passes through a much greater distance of dense atmosphere, and most of the blue is lost from the beam ...


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Considering 11 km as the thickness of the atmosphere that scatters the blue light, when the sun is just at the horizon, the length travelled by the sun light is: $$l = \sqrt{(6371+11)^2 - 6371^2} = 374 km.$$ When that rays come to the place where we are looking at the horizon, much of the blue components were already scattered along the long journey ...


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Phonons are lattice vibrations. The distance between two consecutive phonons is of the order of 1/N where N is the number of atoms in the lattice. At room temperature a phonon travels approx 10 to 100 lattice constants before scattering. In this article https://www.nature.com/articles/srep17131 they say that a phonon travels $< 1 \mu m$ before ...


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The linked video shows a TEM image. In transmission electron microscopy, you have a parallel beam of electrons incident on the material, which gets scattered by the atoms in the material. Note that, it is not just the electron cloud, but the potential formed by the atomic core (nucleus+ core electrons) and the valence electrons together, that scatter the ...


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Because the oscillating scattering electron behaves like an oscillating electric dipole in the sense that both can be represented as a small oscillating source of current. The radiation fields due to such a system are described in any Electromagnetism textbook. The oscillating charge acts like an oscillating current, backwards and forwards in the direction ...


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In the first expression that you write for $i\mathcal{M}$, you have simplified to the second line using a gauge condition (Lorentz gauge) $\epsilon_\mu k^\mu=0$. This is a legal choice but it is incompatible with replacing both polarizations sums with $-\eta_{\mu\nu}$, since you have partially fixed the gauge in this way. It's simple to check that squaring ...


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I cannot speak to this specific case, but, in general, the poles of the reflection coefficient correspond to bound states. It is helpful to think about a similar classical problem. In electrodynamics, the bound states of a planar system (polaritons) are at the poles of the reflection coefficient. Consider the definition of reflectance. $$R = |r|^2 = \left|...


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Any superposition/linear combination of monochromatic left- and right-moving exponential waves is of course equivalent to a superposition/linear combination of sine and cosine waves, as OP demonstrated above. However from the point of view of a 1D scattering experiment where wavepackets enter from (and leave to) spatial infinity, the travelling exponential ...


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If the spatial part were to be multiplied by (for example) $\cos \left( \dfrac{\hbar k^2}{2m}t \right)$ this would be true. In this case you get a standing wave. But multiplying by $e^{-i\frac{\hbar k^2}{2m}t}$ yields a travelling wave.


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The Chappuis absorption bands occur at wavelengths between 400 and 650 nm. Within this range are two absorption maxima of similar height at 575 and 603 nm wavelengths As seen here in the optical wavelengths , most of the visible light on the left of 500nm is absorbed, due to the large absorption lines leaving dominant the blue sector. In ancient times in ...


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No, of course you can't do this. For starters, you can't just conjure up these two new position variables. Also, the loop diagrams would involve integration with respect to the loop momentum, leading you to a completely different class of functions. Finally, if you include the coupling constant, you realise that these loop diagrams are higher order ...


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