Hot answers tagged scattering
23
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 ...
18
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 ...
14
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 ...
13
(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 ...
9
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). ...
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:
...
8
This is fundamentally no more difficult than understanding how quantum mechanics describes particle motion using plane waves. If you have a delocalized wavefunction $\exp(ipx)$ it describes a particle moving to the right with velocity p/m. But such a particle is already everywhere at once, and only superpositions of such states are actually moving in time.
...
7
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 ...
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
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 ...
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
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. ...
5
The summing over final states and the averaging over initial states is a good observation that I always emphasize as the origin of the arrow of time. As soon as one considers mathematical logic, this asymmetry has to arise.
Why are we summing over final states? Because "we don't care" about which of them occurs (and no one knows). We're calculating the ...
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 ...
5
In principle, bound states are possible in a QFT. In this case, their states must be part of the S-matrix in- and out- state space in order that the S-matrix is unitary. (Weinberg, QFT I, p.110)
However, for QED proper (i.e., without any other species of particles apart from photon, electron, and positron) it happens that there are no bound states; electron ...
5
In general, the scattering of light from some object depends on the how close the wavelength of light is to the size of the object.
To make an analogy, if a tidal wave with a wavelength of several kilometers hits a telegraph pole with a radius of 15 cm it isn't going to scatter very much. On the other hand, waves with a wavelength of a few cm, e.g. ...
5
Sorry, a solid angle is something different than an ordinary angle, see
http://en.wikipedia.org/wiki/Solid_angle
so it is not measured "with respect to anything". Solid angle $\Omega$ measures the size of a set of directions in the 3-dimensional space via the formula
$$ \Omega = \frac{A}{R^2} $$
where $A$ is the area of the intersection of all these ...
4
I had the same problem - it was one of just two main problems with the motivation of this whole "twistor uprising" (the other, surviving problem for me is the relevance of the whole formalism for the off-shell gauge theory which I find important, especially for AdS/CFT etc.) - but it has been fully answered for me. The solution is as follows:
In general, ...
4
The Rayleigh scattering is one element of the solution, as j.c. explained.
However, when there is dust in the air (for instance after a volcano sends huge quantities of tiny rock particles into the sky, like what happened with Eyjafjallajökull, or because of sand or smoke particles), you might have noticed that the sunsets happen to be more colourful.
This ...
4
(Edited to reflect updated question).
The solution has a single parameter, which we can take to be the angle the impacting particle goes after the collision. A particular solution is determined only after this parameter is given. Usually we would talk about the collision of, for example, two spheres, in which case this parameter could be related to ...
4
At David Zaslavsky's suggestion I'll transfer this from the comments to the answers (I was a bit hesitant because I don't know how reliable youtube videos are to still be around in, say, 6 months time!):
This little youtube video might help. You can only resolve the objects by looking at the reflected waves. The amount of detail you can get in the ...
4
There is only one S-matrix, but it is frequently introduced in sloppy ways.
The S-matrix is a unitary matrix between two isomorphic Fock spaces whose 1-particle sector contains precisely one particle for each bound state of the system. It can be constructed by the usual adiabatic textbook procedure if and only if there are no bound states (which is a ...
4
The scattering states must be included in the perturbative calculations if the result is to be highly accurate. In particular, it is not justified to ignore the continuous spectrum at energies close to the dissociation threshold.
The Hilbert space in the position representation is the space of square integrable functions on $R^3\setminus\{0\}$ with respect ...
4
I tracked down a spectrum of the sky at an altitude somewhere below 51° and overlaid it on the colors of the spectrum:
From this diagram, it appears that the intensity of the light admitted through the atmosphere diminishes significantly before reaching violet. Unless the perceiving retina was overpoweringly tuned to ...
4
The blackbody spectrum of the sun is the following, given $T=5778 K$. I admit I'm just copying from Wikipedia.
$$I(\nu,T) =\frac{ 2 h\nu^{3}}{c^2}\frac{1}{ e^{\frac{h\nu}{kT}}-1}$$
The comic suggests that the reflection from scattering transforms the above spectrum by $1/\lambda^4$ (as in, it is multiplied by this). Light is a wave, so $\nu \lambda=c$. ...
4
It is simply a matter of notation. The $p_1$ (and hence $E_1$ and $E_2$) in
$$\int d\Pi_2=\int d\Omega\frac{p_1^2}{16\pi^2E_1E_2}(\frac{p_1}{E_1}+\frac{p_1}{E_2})^{-1}$$
is no longer an integration variable; it has the fixed value that satisfies the delta function $\delta(E_{cm}-E_1-E_2)$ in the previous integral. The factor ...
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