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Hot answers tagged scattering

79

This isn't quite how rainbows work. The standard explanation is that light bounces around inside each droplet, and getting reflected once, and exiting at an angle: Image source However, the real picture is a little bit more complicated. When sunlight hits a water droplet, the rays will refract when they come in, (partially) reflect back when ...

64

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

55

(photograph credit: Efram Goldberg) [Note: left-most ampule is cooled to -196°C and covered by a white layer of frost.] $NO_2$ is a good example of a colorful gas. $N_2O_4$ (colorless) exists in equillibrium with $NO_2$. At lower temperature (left in Wikipedia photo), $N_2O_4$ is favored, while at higher temperature $NO_2$ is favored. For a gas to have ...

41

Your picture shows that an observer at X will see both the strongly scattered red light and the strongly scattered blue light, but he will see it coming from different directions. That's the same way you usually see things: different amounts or colors of light reaching your eyes from different directions, and thereby creating an image on your retina.

40

First of all, gas molecules are not invisible. There are plenty of elements whose gaseous state is quite colored, but these (iodine, e.g.) are in such rare amounts in the atmosphere that the net effect is not discernable to the eye. Next, if you Google for "atmospheric transmission curves," you'll see all sorts of spectral absorption going on, again at ...

40

This is a good example of how Science works. Geiger and Marsden observed that some of the alpha particles were being backscattered. This is inconceivable if the alpha particle is scattered by a lighter particle. If one considers a particle of mass $m$ and initial velocity $v_1$ striking a target of mass $m'$ at rest, without changing its direction, then ...

36

Wikipedia explains this rather well but I'll pick out the relevant stuff for you. Before the Geiger–Marsden experiment, the general idea was that atoms were built of some permeable positive substrate in which some negative particles were floating around; the so called plum-pudding model. If we shoot $\alpha$ particles on this setup they should all pass ...

30

If I understand you right, you're referring to the phenomenon seen in this picture (from the first Google hit), that near the horison the color of the sky is more light-blue (not exactly white): Rayleigh scattering The scattering in the atmosphere is for a large part Rayleigh scattering off of nitrogen and oxygen molecules, which are much smaller than the ...

26

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 ...

25

OP has discovered on their own a primitive application of the Schwinger-Dyson equations. Congratulations! A very gentle introduction to the Schwinger-Dyson equations. ... or how to calculate correlation functions without Feynman diagrams, path integrals, operators, canonical quantisation, the interaction picture, field contractions, etc. Note: we will ...

21

I think that it is important to recognize the practical difference between Raman scattering and fluorescence. If the energy of the photon is resonant with some molecular transition (meaning that it is equal to the energy difference between ground energy state and one of the excited states of the molecule), that the molecule can absorb this photon undergoing ...

20

The S-matrix (scattering matrix) is the unitary operator $S$ that determines the evolution of the initial state at $t=-\infty$ to the final state at $t=+\infty$. $$|\psi(t=+\infty)\rangle = S |\psi(t=-\infty)\rangle$$ This matrix/operator is therefore a collection of complex numbers that are ready to calculate the probabilities of various scattering ...

20

The first question we have to ask is: what is a one particle state in an interacting theory? It is reasonable to require that they are states that are both momentum eigenstates and energy eigenstates. (In fact, as the Hamiltonian and the momentum operator commute, these are not two different conditions.) Weinberg, in his famous textbook, says that particle ...

20

This is the illustration of the Chadwick experiment (from Wikipedia): The key point is that the scattering is close to the forward direction. So if we draw the diagram for the Compton scattering as: The angle $\theta$ is small. This diagram shows the hypothetical incoming gamma ray hitting a proton in the paraffin and scattering it. The equation for the ...

19

The Hankel functions are not really the most natural way to get a cylindrical coordinates expansion for a plane wave, which is in terms of Bessel functions. (Why is this? it's because the Hankel functions are singular at the origin and plane waves are not.) You can then rephrase it in terms of Hankel functions if necessary. While this is of course an example ...

19

As has been said by many answers; all gases aren't colourless, for example chlorine gas is a pale yellow; which is a good thing as it's very dangerous. So the gases in our atmosphere are colourless. But this is completely the wrong way round to look at it. If our eyes operated at frequencies that were blocked by gases in the atmosphere they wouldn't work ...

16

Neuneck's answer is the pithiest description of how you get normalisable states as superpositions of non-normalisable states, but the following is more of a "why" these states happen. Hopefully, you should see that this discussion is independent of the number of dimensions. Practically speaking, the reason why there are always such states it is because ...

16

Good question; I remember spending hours trying to understand this when I first learned QFT. Let's address your two main points in turn. First, you say I don't understand how rhyme these two different pictures. Let's outline how to connect the two pictures in steps. It's a good exercise to try and work through all of the gory details yourself, so I ...

15

Some gases actually are visible (nitrogen dioxide for instance). The air is invisible, because its molecules don't absorb the visible light. These molecules simply don't have useful vibration modes available to absorb these wavelengths, or the electrons in their orbitals can't utilize the frequencies of visible light to move to higher orbital (the energy ...

15

The reason the sky is blue on Earth is because of the Earth's atmosphere. The molecules and gas in the atmosphere interact with solar light via Rayleigh scattering, which allows for blue light to be scattered more efficiently than lower frequencies. This results in an abundance of blue light, which makes the sky look blue. Actually it should be said that ...

15

The picture in your question represents a halo rather than a rainbow: the rainbow is seen when the Sun is behind you, while halos appear when the Sun is in front of you. The actual mechanisms producing colours vary between the phenomena, but the basic idea is the same: if light of certain colours comes to you from different directions, your eye will ...

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

*1/2. In and out states (of massive theories) are joint energy-momentum eigenstates (spanning asymptotic in and out Fock spaces) of asymptotic, free Hamiltonians (and momentum operators) associated with the bound states of a theory. These Hamiltonians are not identical with the Hamiltonian defining the finite-time dynamics of the theory; in simple cases (...

14

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 ...

14

In 1930 it was discovered that Beryllium, when bombarded by alpha particles, emitted a very energetic stream of radiation. This stream was originally thought to be gamma radiation. However, further investigations into the properties of the radiation revealed contradictory results. Like gamma rays, these rays were extremely penetrating and since they were ...

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.

13

Rayleigh scattering is scattering from polarizable entities. The incident light induces a dipole moment, which re-radiates. Thomson scattering is scattering from free unbound charged "unpolarizable" particles. The cross section for Rayleigh scattering decreases with the fourth power of wavelength. That for Thomson scattering is independent of wavelength. ...

13

The reference frame of the center of mass is, by definition, the one where the total $3$-momentum vanishes. It exists almost always also for massless particles as I go to discuss. The total $4$-momentum $P$ of a system of $N$ free particles is the sum of their $4$-momenta of the particles, i.e., $$P = \sum_{a=1}^N P_{(a)}\:,$$ where each $P_{(a)}$ is a non-...

13

Absolutely. If Compton scattering occurred in first order in $e$, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the $z$ direction. Then the electron 4-momentum is $$p^\mu_{\text{in}} = (m,0,0,0)$$ while the photon 4-momentum is k^\mu_{\text{in}} = (\omega,0,0,\...

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