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What is forward scattering? If it is equivalent to no scattering, then why not call it "no scattering"?

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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 wave $$\psi_\text{scattered}(\mathbf{r})=F(\theta,\phi)\frac 1r e^{ikr}.$$ The form factor $F(\theta,\phi)$ governs the angular structure of the scattered wave, and the case where $\theta=0$ is called forward scattering.

Note that:

  • The forward-scattered wave is part of a spherical wave and its amplitude decays with the distance from the scattering centre in a different way to the incoming wave. In practice, the incoming beam will also suffer from wavepacket spreading, but in general the forward-scattered wave will be weaker unless special scattering conditions are at play.

  • The form factor in general includes a phase. This means that the forward-scattered wave will interfere nontrivially with the incoming beam, providing a delay in the phase of the final wave.

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    $\begingroup$ On the practical side experimenters sometimes use "forward scattering" to mean all $\theta$ smaller than the smallest scattering angle which can be detected because to go closer would cause the halo of beam to fall on the detector. As usual this definition is convolution of the precise one with an experimental resolution function. $\endgroup$ – dmckee --- ex-moderator kitten Dec 10 '13 at 17:34
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If you want an even more everyday example than Emilio Pisanty's example: "no scattering" would mean that the would be scattering object in question (modelled by the short range potential in Emilio's answer) would beget no change the the forward travelling wave. Otherwise put, an observer sensing the incoming plane wave could not tell whether or not the object were in between them and the source of (wontedly plane for these scenarios) waves. Forward scattering means that the scattering object leaves signs of its presence in the forward travelling wave.

As an illustration, I don't know whether they still do this in Britain, but it used to be that one must by law buy a licence to have a television set. To enforce this law, the state had people driving around in vans (see them in action here) detecting unlicensed televisions from the scattered wave that must always arise from the television's antenna system whenever it interacts with the electromagnetic field. This scattered wave can be sensed at almost any orientation relative to the receiver and transmitter. In this scenario, "no scattering" would mean that you would be safe from the TV detector man.

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  • $\begingroup$ You still need to buy a TV license, and they still have door-to-door inspections. I'm not sure whether they still have the detector vans, though; they are possibly not that useful since you need a license for any device that shows TV content "as live", including streaming from the BBC website. $\endgroup$ – Emilio Pisanty Dec 11 '13 at 15:48
  • $\begingroup$ ... and apparently they still have the vans, plus handheld detectors. More in Wikipedia. $\endgroup$ – Emilio Pisanty Dec 11 '13 at 15:50
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The optical theorem (https://en.wikipedia.org/wiki/Optical_theorem) relates the imaginary part of the forward scattering amplitude to the total scattering cross-section, $ \sigma_\mathrm{tot}=\frac{4\pi}{k}~\mathrm{Im}\,f(0)$, so if $ \sigma_\mathrm{tot}$ is non-zero, so is the forward scattering amplitude.

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In QFT, forward scattering means $\theta\approx0$. So it is basically some kind of small angle approximation. This is the answer to my question. ($\theta$ is the usual angle in spherical coordinates)

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