Question

Derive the finite special conformal transformation

Answer 1

Answer:

numerous discussion of the special conformal transformation, they cite the finite transformation as

xμ′=xμ−bμx21−2x⋅b+b2x2xμ′=xμ−bμx21−2x⋅b+b2x2

This can be found from integrating the infinitesimal conformal transformation

δxμ=2(b⋅x)xμ−x⋅xbμδxμ=2(b⋅x)xμ−x⋅xbμ

I found the derivation given as an answer on this site. I completely understand what they did, but at the end of the day they get the answer to be

x(t)=x0−x20(tb)1−2x0(tb)+x20(tb)2x(t)=x0−x02(tb)1−2x0(tb)+x02(tb)2

Their starting point was the differential equation x˙=2(b⋅x)x−x2bx˙=2(b⋅x)x−x2b.

Why, exactly, is there a tt in the second equation but no tt in the first? Is tb=b