64

If the light has nothing to scatter off of to reach your eyes you won't see anything.


15

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 ...


14

No. Light has to physically interact with the sensors in your eyes for you to be able to see it, and likewise for cameras, too. The reason you can see "light beams" in a terrestrial environment is that in the atmosphere, some of the light can be scattered so it does get into your eyes. In vacuum, this does not happen.


12

The assumption is wrong. Rayleigh-scattering does happen independent of altitude. That it happens at sea level, too, becomes clear when you look at the other bank of a large river mouth or a nearby coast over a bay: they loose much contrast and everything is blue-ish; the same effect happens in the mountains looking from one peak to the other. In the sky ...


11

In a perfect vacuum, not only can you not see light that isn't traveling toward you, you can't even see light that is traveling toward you until it actually reaches your eyes.


10

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 ...


10

Both are right in their respective situations. Optical refraction is a relatively low-energy (non-ionizing) effect which does not exchange energy with the medium. The atoms have electromagnetic fields which interact with that of the photon, increasing the electrical permittivity $\epsilon$ of the medium over that of free space, $\epsilon_0$ (I am not sure ...


10

One would think that those molecules would have to possess some degree of common alignment in order to produce light that possesses nonrandom polarization. Molecules are quantum mechanical entities and light interacting with individual molecules should be thought of in photons. Nevertheless, the classical elecromagnetic light with its description by the ...


10

If we speak of electrons, then it is appropriate to speak of "wind* as a flow of atoms. Electrons are usually confined to atoms or materials, inside of which the wind atoms do not penetrate, so the collision of electrons with wind are very unlikely. One place where one does encounter a stream of unbound electrons is electron microscope. In this case one ...


7

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 ...


7

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 ...


6

I think that there is somewhat of a biophysics question here but it is perhaps buried. In theory there is no contrast difference, but in practice... If you are designing for contrast in particular, there is almost no sense in which light-on-dark and dark-on-light can be properly distinguished in the abstract. Like, if "black" and "white" are consistently ...


6

The key difference between treatments of quantum mechanical scattering and classical scattering is the nature of the incident particle. In the quantum mechanical case, the incident particle is typically treated as a delocalized plane wave, whereas in the classical case the particle is treated as a point particle. The delocalized nature of the plane wave is ...


6

The root cause is that you see no radiation from an oscillating electric dipole when looking along the axis of oscillation. Unpolarized light can be considered an equal mixture of two perpendicular linear polarisation states (with random phases). Such light causes electric dipoles in the atmosphere (i.e. molecules) to oscillate in those two perpendicular ...


6

I think your question is really based on isotropy and anisotropy: How can an isotropic medium produce an anisotropc effect? The answer is that the direction of the sunlight provides a preferred direction. Given the fact that the atmosphere is isotropic, polarization (if it exists) of scattered sunlight must have radial symmetry around the axis defined by ...


6

If one imagines a macroscopic object moving through air, one can very well approximate air resistance as the action of a continuous fluid. At low speeds, this situation can be described by viscous friction, which is a force proportional to the speed of the object, and with direction opposite to its direction of motion (I neglect here the effect of turbulence)...


5

Reflection is a form of scattering, but it is coherent scattering. When scattering occurs from a lot of identical particles rigidly fixed in a plane, then all the possible ways a single photon can be scattered from the plane of particles are fixed in phase. That is coherent scattering, and it is reflection. If the particles instead are, e.g., in a gas, ...


5

Light must reach your eyes/detectors one way or another - no matter vacuum or not. It may be directed there in the first place or scattered somehow. Then again, scattering is a complex matter. Both your eyes and the light source have some edge diffraction so SOME light will get into your eyes (given enough time to propagate) no matter where everything is ...


5

Light passing through a fluid can be polarized for two reasons. We call this interaction, where different polarizations of light scatter differently "birefringence." First, if the components of the fluid have some preferred direction, then the light will be polarized in the corresponding orientation (e.g. chiral molecules). Second, if the fluid ...


4

The comments by KF Gauss are on the right track: non-relativistic quantum electrodynamics (NRQED) is a good foundation. The question is then how to recover the "effective potential" description of X-ray diffraction from NRQED. This isn't my specialty, but I reviewed the NRQED derivation of elastic X-ray scattering from a crystal, using the references listed ...


4

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 ...


4

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) ...


4

There is no contradiction. The division you make into "spherical scattering" and "pancake scattering" is not quite there, because you fail to consider the other particle involved in the collision: If you collide two protons together at high relative speed, they will both be pancake-shaped in the centre-of-momentum frame (the lab frame). In the rest-frame of ...


4

The first equation in the question is from equations (6.6)-(6.7) on page 179 in Peskin and Schroeder's An Introduction to Quantum Field Theory. According to page 177, the kick occurs at $$ \newcommand{\bfA}{\mathbf{A}} \newcommand{\bfk}{\mathbf{k}} \newcommand{\bfp}{\mathbf{p}} \newcommand{\bfv}{\mathbf{v}} \newcommand{\bfx}{\mathbf{x}} t=0 \hskip2cm \bfx=\...


4

Kinematics, mostly. More specifically, the final-state particles generally are moving away from each other, and so are unlikely to collide, whereas the initial-state particles generally are moving toward each other, so they are likely to collide. Note that there are, in fact, situations in which you do see the final-state particles return to the initial ...


3

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 ...


3

In the comments on the question I mentioned that the plane-wave solution "gets the right answers". The reason for this is related to the notion of time-independence. Any experiment where you shoot a single particle at a barrier/scattering-site and then measure where the outgoing particle lands in intrinsically a time-dependent experiment and it corresponds ...


3

From below you are looking up at the atmosphere with a black background beyond it. From above the background is reflecting significant amounts of light and so you see the background. This picture from a satellite shows the background visible but getting less so as the light passes through more and more atmosphere towards the rim. Here is the effect at ...


3

But none of the radiation is scattered exactly at any particular angle. You always have to integrate over a range of solid angle. Therefore none of the scattered radiation precisely contributes to $I(s)$. Of course you can modify the Beer-Lambert law to add a source term to the RHS of your equation to account for the contribution of scattered light to the ...


3

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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