Scale factor of special conformal transformation

[stegosaurus]

Homework Statement

(From Di Francesco et al, Conformal Field Theory, ex .2)
Derive the scale factor Λ of a special conformal transformation.

Homework Equations

The special conformal transformation can be written as

x'^μ = (x^μ-b^μ x^2)/(1-2 b.x + b^2 x^2)

and I need to show that the metric transforms as

g'_μν = Λ(x) g_μν

The Attempt at a Solution

My attempt was to differentiate the transformation law in order to then use the chain rule (the derivatives are intended as partial):

g_σλ=dx'^μ/dx^σ dx'^ν/dx^λ g'_μν

For a particular partial derivative I get:
dx'^μ/dx^ν = (δ^μ_ν-2b^μx_ν)/(1-2 b.x + b^2 x^2)- (x^μ-b^μ x^2)(-2 b_ν+2b^2 x _ν)/(1-2 b.x + b^2 x^2)^2

however plugging this and the other similar term in the chain rule gives rise to a very long expression which does not appear to simplify (I've checked it does in 1D, if all quantities were scalars).
Am I doing something wrong, or am I just missing something?

 

[strangerep]

[...] plugging this and the other similar term in the chain rule gives rise to a very long expression which does not appear to simplify

Oh c'mon, it's not all that "long". (It might seem less intimidating if you used latex and \frac ...)

The combined numerator ends up as a sum of 4 terms, and you can contract some of indices with $g$.

You'll have to post the whole expression before we can help you figure out what's wrong.

 

[stegosaurus]

You're right, I made a dumb mistake early on and that prevented me from getting to the final answer! Now I get it.

 

[metalvaro18]

$$ \frac{\partial x'^\mu}{\partial x^\nu}=\frac{\delta^\mu_\nu-2 b^\mu x_\nu}{1-2(b\cdot x)+b^2x^2}-\frac{(x^\mu-b^\mu x^2)(-2 b_\nu+2 b^2 x_\nu)}{\left ( 1-2(b\cdot x)+b^2x^2\right )^2} $$
is this expression correct?

 

[strangerep]

$$ \frac{\partial x'^\mu}{\partial x^\nu}=\frac{\delta^\mu_\nu-2 b^\mu x_\nu}{1-2(b\cdot x)+b^2x^2}-\frac{(x^\mu-b^\mu x^2)(-2 b_\nu+2 b^2 x_\nu)}{\left ( 1-2(b\cdot x)+b^2x^2\right )^2} $$
is this expression correct?

Afaict, it looks ok to me.