However, I don't really see how to exponentiate the infinitesimal form
to actually get to the finite form
In addition to the detailed answer of @LubošMotl, you may notice that conformal transformations transform light cones in light cones.
This means the following. Starting from a light cone :
(x′−a′)2=x′2−2x′.a′+a′2=0
you must find that it is also a light cone for x, that is, it exist a such as :
(x−a)2=x2−2x.a+a2=0
To do that, replace simply x′ by its value in function of x, by :
x′μ=xμ−bμx21−2x⋅b+b2x2
and, with some algebra, you will find that it is possible to exhibit a(a′,b) (a as a function of a′ and b) , which satisties (2).:
a=a′+a′2b1−2a′.b+a′2b2
This means that the transformation (3) is a conformal transformation whose infinitesimal expression is clearly:
x′μ=xμ+2(x⋅b)xμ−bμx2
(because only quadratic terms in x are allowed in infinitesimal conformal transformations)
Any other expression of the global special conformal transformation (3), having infinitesimal expression (5), will not transform light cones in light cones.
This post imported from StackExchange Physics at 2014-06-27 11:33 (UCT), posted by SE-user Trimok