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Relativistic Coulomb Scattering

  • #1
Hi. Landau and Lifgarbages give an equation describing the angle of deflection of a charged particle of a given initial velocity and impact parameter. It's given on page 102 of their Classical Theory of Fields, available here: http://books.google.com/books?id=QIxD3Z82AagC&dq=classical+theory+of+fields&printsec=frontcover&source=bl&ots=gMlrhO83op&sig=1_vW5i6UAgI_prJ788P-eb4NR1w&hl=en&ei=0-MAStCBKZjcMYGugdoH&sa=X&oi=book_result&ct=result&resnum=5#PPA102,M1. It's in the solution to Problem 1, at the bottom of the page.

I'm just curious where the equation in Problem 1 came from, in particular where the arctan came from. It's simple to solve 39.4 for the angle letting r-->infinity, but that gives an arccos, not an arctan. L-L just state that as the answer without working through it, as if it's obvious. Does anyone see how to derive it?

Thanks!
 

Answers and Replies

  • #2
gabbagabbahey
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If you have an equation of the form [tex]\cos\theta=\frac{a}{b}[/tex], then you can think of [itex]a[/itex] being the adjacent side of a right triangle, [itex]b[/itex] being the hypotenuse and hence [itex]\sqrt{b^2-a^2}[/itex] as the opposite side [tex]\implies \tan\theta=\frac{\sqrt{b^2-a^2}}{a}[/tex]
 

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