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48

(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 ...


34

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 ...


27

This was demonstrated by "Experiment 144" at SLAC in 1997. Here is a list of publications from that project, for instance "Positron Production in Multiphoton Light-by-Light Scattering", whose abstract reads: A signal of 106±14 positrons above background has been observed in collisions of a low-emittance 46.6 GeV electron beam with terawatt pulses from a ...


26

The keywords here are Rayleigh scattering. See also diffuse sky radiation. But much more simply, it has to do with the way that sunlight interacts with air molecules. Blue light is scattered more than red light, so during the day when we look at parts of the sky that are away from the sun, we see more blue than red. During sunset or sunrise, most of the ...


16

The two effects are not related. The size appearing larger is a matter of some speculation to this day, but it is purely a psychological effect. If you want to prove this, take a look a the moon while standing up and looking between your legs. It won't look nearly as large. The red/orange color is related to the sunset being red. In fact, it's the same ...


14

(Source, Wikipedia Commons) The moon is generally called a "Harvest Moon" when it appears that way (i.e. large and red) in autumn, amongst a few other names. There are other names that are associated with specific timeframes as well. The colour is due to atmospheric scattering (Also known as Rayleigh scattering): may have noticed that they always ...


13

This was something that confused me for awhile as well until I found this great set of notes: homepages.physik.uni-muenchen.de/~helling/classical_fields.pdf Let me just briefly summarize what's in there. The free Klein-Gordon field satisfies the field equation $(\partial_{\mu} \partial^{\mu} +m^2) \phi(x) = 0$ the most general solution to this equation ...


12

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 ...


11

I guess you were asking about elastic photon-photon interaction. If that is the case, I remember reading some proposals (and argumentation of feasibility) of experiments using existing laser facilities in: E. Lundström et al. Using High-Power Lasers for Detection of Elastic Photon-Photon Scattering. Phys. Rev. Lett. 96 no. 8, 083602 (2006). ...


11

As has been said by many answers; all gases aren't colourless, for example chlorine gas is a pale yellow; which is a good things as its 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 ...


9

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 ...


9

Conservation of particle current is nothing but the statement that a theory has to be unitary. In other words the scattering matrix $S$ has to obey $SS^\dagger=1$ Defining $S=1+iT$ i.e. rewriting the scattering matrix as a trivial part plus interactions (encoded in $T$ which corresponds to your $f$) one finds from the unitarity condition: ...


9

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 ...


8

Suppose you treat scattering of a particle in a central potential. This means that the Hamiltonian $H$ commutes with the angular momentum operators $L^2$ and $L_z$. Hence, you can find simultaneous eigenfunctions $\psi_{k,l,m}$. You might know, for example from the solution of the hydrogen atom, that these functions can be expressed in terms of the ...


8

These terms apply when you're solving the Schrodinger equation with a potential that goes to zero at large distances. In this situation, the solutions with $E<0$ have the property that $\psi$ dies away to zero for large distance. So the particle is, with high probability, guaranteed to be in a confined region (not at large distance). So those are bound ...


8

P&S almost seem to argue that they need to be included for the simple reason that, if they didn't include them, they would get a nonsensical (i.e. infinite) result. Well, I am confident that Peskin and Schroeder not only "seem" to argue in this way but they explicitly and comprehensibly enough write this fact because it is both true and important. ...


8

Short version: In the infinite potential well, $E \geq 0$ (because $V_{min}=0$, and $E \geq V_{min}$). In your finite potential well, it sounds like you are looking for bound states, in which case $E < 0$, so you absorb the negative into the square root. Long version: When you are tackling a QM problem, first you should figure out the admissibility of ...


7

Surface coating of an integrating sphere is optimized for low losses. This white coating (barium sulfate or PTFE) acts like an ideal lambertian scatterer. all light is scattered (Ok, not 100%, but a very high percentage like 99,5%. See ressources) it is emitted in the hemisphere following the cosine law: perpendicular to the surface it's highest. ...


6

The main problem about a rigorous solution to such a scattering proplem is that computations are extremely demanding. Just imagine you have a wavelength $\lambda$ of some $400$nm to $700$nm for visible light (from here): Now, to do physically meaningful simulations, you will need a sub-wavelength lattice which makes any computational cell above, say ...


6

There is a very easy way to see this and it is through an $\hbar$ series. This claim can be traced back to Sydney Coleman and states that in the ultraviolet one is doing an expansion with $\hbar$ going to zero. A previous answer cited these lectures on classical fields but I would like to start from the generating functional of the scalar field theory and ...


6

I've a vague memory that light-on-light scattering puts an upper limit on the energy of photons that will propagate over cosmic distance scales. So one probe is very high energy gamma astronomy.


6

There is a ward identity that links the charge renormalization to the photon's wave function renormalization. Ward identities are relationships between correlation functions that follow from the quantum theory having a symmetry. In this case the gauge invariance of QED relates (among other things) the electron's two point function (propagator) to the ...


6

The fact that only connected Feynman diagrams contribute to the scattering amplitude can be interpreted in terms of the vacuum of the theory. Omitting disconnected diagrams amounts to shifting the vacuum: the vacuum of the interacting theory differs from that of the free theory. Regarding your second question: strongly connected (also called one-particle ...


6

Forward scattering need not be equivalent to "no scattering" - and, indeed, will only rarely be indistinguishable from it. In the usual scattering-theory setup, you have an electron coming in in a plane wave $$\psi(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}=e^{ikz}$$ and impinging on some short-range potential. This will add to the wavefunction a scattered ...


6

Rayleigh scattering is also present in water, but the main reason of the blue color of the sea is absortion, due to vibrational transitions. See: http://www.dartmouth.edu/~etrnsfer/water.htm


5

Here: ...the assumption that the time of the measurement is much larger than the time needed for the transition...


5

We do all the "cross section business" because we want to predict results of experiments. Let's take for example some particle with two polarizations states: "+" and "-". You know that experimentalists will collide 1 000 000 pairs of particles, with polarisation of initial particles being unknown. Best thing you can do is to hope that in experiment ...


5

$S^{-1}=S^*$ is just the condition for unitarity. It is usually written as $S^*S=1$ (together with invertibility) and means that $\psi^*\psi$ doesn't change when $\psi$ is replaced by $S\psi$: $(S\psi)^*(S\psi)=\psi^*S^*S\psi=\psi^*\psi$ Therefore probability is conserved, a must for a good scattering matrix. In general, unitarity of the S-matrix is a ...



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