How Is the Arctan Derived in Relativistic Coulomb Scattering?

[Trickster2004]

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=QI...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!

 

[gabbagabbahey]

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