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Increasing the contrast on the projected object might help some, as well as changing the lens, but probably the best (easiest and cheapest) way to increase the contrast on the final image is to edit it after the fact, using some image/video manipulation software.


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A function can be interpretable as a scattering amplitude if that function satisfies the axioms of relativistic S-matrix theory [1]: Lorentz invariance Unitarity (Not realized by the beta function, but may be dropped if the function is interpreted as a Born approximation to the exact amplitude) T, C, P invariance (only for strong nucl. interactions) ...


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Actually, when I was studying quantum mechanics, I was surprised that some everywhere nonzero potentials produce only a finite total cross section. It is intriguing because classically, for such potentials, any particle injected will be scattered, although might with a very small deflection angle. I mean, it is not surprising to have an infinite total ...


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I am not sure the parameters you mentioned are enough. For example, in some viscous liquids, one can observe the Mössbauer effect, where gamma-ray absorption differs dramatically for the liquid and a single particle; therefore, even details of the nuclear spectra can be important.


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The BdG equation reads, in full spin-space, and for a s-wave superconductors $$\left(\begin{array}{cccc} H_{0}\left(p\right) & 0 & 0 & \Delta\\ 0 & H_{0}\left(p\right) & -\Delta & 0\\ 0 & \Delta^{\ast} & -H_{0}^{\ast}\left(-p\right) & 0\\ -\Delta^{\ast} & 0 & 0 & -H_{0}^{\ast}\left(-p\right) ...


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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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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 ...


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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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