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Space is mostly black, because most of space doesn't have light coming from it for you to see. The moon, being basically a big rock, unsurprisingly does not change this: You would expect that standing on a rock would not affect what you see. Earth is unusual in having a sky because it has an atmosphere. The atmosphere is a bunch of gas that changes the ...


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In addition to the lack of an atmosphere (the primary reason, already addressed in other answers), something that compounds the effect even further is that the lunar surface is quite reflective, and because the people who are taking the images of lunar surfaces often want to have Earth and/or astronauts in the shots and also be able to make out detail in the ...


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With the atmosphere of the moon being $10^{14}$ times less dense than that of Earth, there is negligible scattering, so whereas on Earth, approximately 25% of direct solar radiation is scattered around (making the sky light up and appear blue), there is no mechanism for this on the moon, and all light from the sun travels (essentially) unaffected to the ...


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The moon does have a night and a day, but this isn't as fully connected to your question as you might think. The moon is tidally locked with the earth, meaning that the same side always faces earth. Since the moon also orbits around the earth (with a period of a lunar month), this means each side changes, over the course of a lunar month, between facing ...


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Good question! I can maybe guide you in the right direction, although I only found this post because I wanted clarity myself. The reduced width idea comes from the R-matrix formalism (a good paper is by Descouvemont and Baye here). The most basic understanding of it is that the most general cross-section for an interaction of two nuclei (in which a ...


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The integral can be easily evaluated $$ \int _0^\infty x e^{-ax}\text{sin }bx \,dx = \frac{2ab}{(a^2 + b^2)^2} $$


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The central idea is that you can translate probabilities from a single-particle picture into fractions of particles in a many-particle picture (assuming no interaction). Consider $T$ for a moment for a single particle. Let's just, for ease of writing, say $T=.9$. So if you send in a single particle from the far left, 90% of the probability density ...


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I'm going to try to give a very concise answer. Your illustration and presumably your intuitive picture are predicated on the superposition principle. You're thinking that if each wave packet is separately a solution of the wave equation, then their sum will be as well. That is not the case. The superposition principle is a property of linear wave ...


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From the Outside Your intuition is right. Rayleigh Scattering would come into effect, making the insides red and the outsides more blue. If the sun is at your back, or if the angle of sun-air-you is a right angle, you will still see light coming off of the air. However, this reflected light may be faint enough in places that you get the color of the night ...


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I realize this is an old question and you have probably moved on... Flicking through the paper, the thing that strikes me is that the wave is normally incident. Unless that is the geometry you are using, the equations are not going to work for you. While an arbitrary wave front can be thought of as being composed of a smarties of plane waves, those wave ...


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Your main question was "Is there an example where the overlap between a bound state and the scattering states makes a measurable contribution to the energy in the perturbative regime?" Actually, I disagree with the statement of the other answer that the scattering states must be included in the perturbative calculations only if the result is to be highly ...


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The normal nonresonant Raman scattering happens when a photon interacts with a molecule; the molecule absorbs the photon momentarily and re-emits it with slightly less energy. In an energy diagram, that looks like this. The frequency of the incoming photon is $\omega_i$, and the frequency of the scattered photon is $\omega_s$. The thick lined level is ...


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This is the relation between the scattering matrix and the Green's function (notice that $(H_M-E-i\pi WW^\dagger)^{-1}$ is basically the Green's function, where $i\pi WW^\dagger$ is the self-energy correction due to coupling to the leads). For a pedagogical account, a good reference is Datta's "Electronic transport in mesoscopic systems", which is in general ...


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Sigma provides plethora of information on silver and gold nano-particles. Silver absorbs around 400nm and gold around 500nm. You can see that shapes of spectra are very complex. This figure shows dependence of peak wavelength from particle size. It seems like Mie is used in description of metal surface plasmon resonance. Nature review Light scattering and ...


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Amplitude coefficients are complex. The reflection and transmission coefficients must account for both amplitude change and phase change. In order to account for both of these, complex coefficients are required. These are the most general, and are needed for a complete description. In some special (and simple) cases, the phase shift is $0^\circ$ or ...



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