# [Physics] Finite Transformation of the Special Conformal Transformation

conformal-field-theorygroup-theory

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

$${x^\mu}'=\frac{x^\mu-b^\mu x^2}{1-2x\cdot b+b^2 x^2}$$

This can be found from integrating the infinitesimal conformal transformation

$$\delta x^\mu =2(b\cdot x)x^\mu-x\cdot x b^\mu$$

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

$$x(t)=\frac{x_0-x^2_0 (tb)}{1-2x_0(tb)+x_0^2(tb)^2}$$

Their starting point was the differential equation $\dot{x}=2(b\cdot x)x-x^2 b$.

Why, exactly, is there a $t$ in the second equation but no $t$ in the first? Is $tb=b^\mu$? Also, where in the derivation I've cited do they only consider the case of $\mu=0$ (i.e., time)? Any clarification would be greatly appreciated.

@Trimok solved the problem most elegantly in his comment to the question cited, and since you are troubled by @Josh's simplifying changes of variables, $b^\mu\equiv \hat{b}~ t$, I'll avoid them to merely integrate the variation $$\delta x^\mu =2(b\cdot x)x^\mu-x\cdot x~ b^\mu$$ directly. It immediately implies $$\delta \left (\frac{ x^\mu}{x^2}\right) = \frac{ \delta x^\mu}{x^2} -2 x^\mu \frac{ x\cdot \delta x }{x^4} = -b^\mu ~.$$
That is, the vector $x^\mu/x^2$ evolves by shifting along $-\hat{b}^\mu$, linearly in the magnitude $|b|$ of $b^\mu$, so with constant unit speed in this "pseudotime" $|b|$. Integrating this simplest of advections for finite pseudotime, we immediately get $$\frac{ x'^\mu}{x'^2}= \frac{ x^\mu}{x^2} -b^\mu .$$ Square both sides, to get the normalization, $$\frac{ 1}{x'^2}= \frac{1}{x^2} +b^2 -2\frac{ x\cdot b}{x^2}= \frac{1-2x\cdot b+b^2 x^2 }{x^2},$$ which divides the above vector equation to yield your conventional form for it, ∴ $${x^\mu}'=\frac{x^\mu-b^\mu x^2}{1-2x\cdot b+b^2 x^2}~.$$