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Physics
Classical Physics
Electromagnetism
Rutherford scattering with Coulomb cutoff
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[QUOTE="Reverend Shabazz, post: 6022975, member: 604901"] I thought so too, but I have doubt since he prefaces that formula by saying: [ATTACH=full]227752[/ATTACH] or, in an older version of his book: [ATTACH=full]227754[/ATTACH] The wording sounds like he's doing more than approximating ##\theta##. But I could be wrong.. For the record, the question with solution for 13.1 is found here: [URL]http://www-personal.umich.edu/~pran/jackson/P506/hw11a.pdf[/URL] It is just the simple scenario where the incident particle remains on a straight line trajectory and the Coulomb potential has no cutoff. As such, the only momentum transferred is perpendicular to the incident's velocity (see section 1.1 here for details: [URL]http://www.phys.lsu.edu/~jarrell/COURSES/ELECTRODYNAMICS/Chap13/chap13.pdf[/URL]). In the cutoff case, though, my thought is that the momentum transferred to the target in the direction parallel to the velocity will be non_zero in contrast to the no-cutoff case, and thus, when combined with the perpendicular momentum, will scatter the target at an angle. I would anticipate that, in calculating the resulting differential cross section, it would produce Jackson's result. Admittedly, I haven't tried slugging through the calculation yet. [/QUOTE]
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Electromagnetism
Rutherford scattering with Coulomb cutoff
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