# Tag Info

1

This is just a complement to the previous answers which give the correct response. If you want to think about it in an intuitive way, imagine that the interaction between electrons and photons becomes weaker. In the limit when it becomes nearly zero, the light will be almost not scattered at all and will continue in a straight path.

2

It is momentum that defines the incoming direction and momentum transfer the outgoing one. The photons, quantum mechanically carry momentum equal to p=h*nu/c . Momentum is a vector and defines directions. An electromagnetic field is an emergent classical quantity built up by innumerable photons. There exists also a momentum defined for the classical field ...

1

Let me offer you a slightly modified version of your question to illustrate a way of re-formulating it your thought process. How does a pool ball know from which direction the cue ball hit it? The answer is the same in the sense that "the particle" does not know all by itself, "the system"1 has certain invariant quantities (like momentum and energy) ...

0

Certainly late for your homework problem set, but for future readers: I think it best to go to the source, Ernest Rutherford, "The Scattering of $\alpha$ and $\beta$ Particles by Matter and the Structure of the Atom", London, Edinburgh and Dublin Philosophical Magazine and Journal of Science, Volume 21, Issue 125, pages 669-688 (1911). This is the paper in ...

0

So one detail I omitted from the question was that: $$\psi_{sc}(k)=\frac{g+g I}{2\pi(k^2-p^2)}$$ Where: $$I=\int^{\infty}_{-\infty}\psi(q)dq \space\space\space\space (1)$$ (I had used in arbitrary prescription in the original description of the problem, this is what I obtain before solving for $I$)$$\\$$ Using equation (1) we can solve for I, obtaining: ...

5

SSC: synchrotron self-Compton BBC: Compton upscattered blackbody radiation; Compton upscattering of stellar blackbody photons XC: upscattering of photons emitted by the accretion flow; accretion flow photons From: http://arxiv.org/abs/1307.1309 and http://arxiv.org/abs/1403.4768

0

I) Before we start let us briefly recall certain aspects of the formalism from Ref. 1. The Minkowski sign convention is $(+,-,-,-)$. The 'phase space' measure for a particle $A$ is $$d\tilde{k}~:=~ \frac{d^3k}{(2\pi)^3 2\omega_{k,A}} ~=~ \frac{d^4k}{(2\pi)^3} \delta(k^2-m^2_A)\theta(k^0),$$ $$\omega_{k,A}~:=~\sqrt{{\bf k}^2+m^2_A}~>0~. \tag{3-35}$$ ...

3

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

0

You obtain this by Wick's Theorem, which can be stated as $$T\{\phi_1\phi_2...\phi_n\}=N\{\phi_1\phi_2...\phi_n+\sum\text{all possible contractions of }\phi_1\phi_2...\phi_n\}$$ where N is the normal ordering operator which puts all the daggered fields on the left ( for example ...

2

Here's a parallel answer to Luboš's but purely classical. Start by noting that the momentum vector of a plane wave with wavelength $\lambda$ is: $$\vec{p} = \frac{2\pi}{\lambda}$$ In some elastic scattering experiment, e.g. X-ray or some other diffraction measurement, we have something like: where $\vec{p}_{in}$ is the momentum of the incoming wave ...

6

Microscopically, i.e. in the quantum theory the scattering with radiation is a collision of particles with photons such as $$e^- + \gamma \to e^- + \gamma$$ The momentum vectors of the particles above are $$\vec p_1+\vec p_2= \vec p_3 + \vec p_4$$ where the identity holds due to momentum conservation. But in general $\vec p_1\neq \vec p_3$ and \$\vec ...

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