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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 ...


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


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 ...


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 ...


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

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.


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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) ...



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