Integrating infinitesimal conformal transformations

[Rocky Raccoon]

While it's pretty easy to derive the infinitesimal version of the special conformal transformation of the coordinates:

x'^{\mu}=x^{\mu}+c_{\nu}(x^{\mu} x^{\nu}-g^{\mu \nu} x^2)

with c infinitesimal,

how does one integrate it to obtain the finite version transformation:

x'^{\mu}=\frac{x^{\mu}-x^2 c^{\mu}}{1-2 x^{\nu} c_{\nu} + x^2 c^2}

with c finite?


I've never delt with nonlinear transformations that don't exponentiate trivially. Also, how does one integrate over a parameter that is Lorentz contracted with another vector?

Thanks for your help.

 

[Bill_K]

(First of all, I think you're missing a 2, it should be (2x^μx^ν - g^μνx²)).

In general, the way to exponentiate a transformation that acts nonlinearly is to find a different set of variables on which it is linear. In the present case, the conformal group can be generated by a) Lorentz transformations Λ, b) translations T, and c) inversion R. The inversion is R: x^μ → x^μ/x^νx_ν. The transformation you're looking at is called a transversion, and can be written in terms of the generators as V = RTR^-1. This gives you a clue that what you need to do to linearize V is to look at its effect on the inverted position vector, y^μ ≡ x^μ/x^νx_ν. The action on y^μ will be linear and easily exponentiated, and then you can go back and express the result in terms of x.

 

[Rocky Raccoon]

Yes, you're absolutely right, I'm missing a factor of 2.

So, you're saying that a simple change in variables will do the trick. I'll certainly try to do it that way. But, I was more thinking along the lines of solving the equation: dx^{\mu}=dc_{\nu}(2x^{\mu} x^{\nu} - g^{\mu \nu} x^2)? Can it be integrated?