# I Conformal transformation of the line element

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1. Jul 22, 2016

### physicality

Let us see how the line element transforms under conformal transformations. Consider the Minkovski metric gij, a line element ds2=dxigijdxj, and a conformal transformation

δk(x)=ak + λ xk + Λklxl + x2sk - 2xkx⋅s

We have δ(dxk)=dδ(x)k=λ dxk + Λkldxl + 2 x⋅dx sk - 2dxkx⋅s - 2xkdx⋅s

And so the line element transforms by δds2=δ(dxi)gijδ(dxj)=
(λ dxi + Λildxl + 2 x⋅dx si - 2dxix⋅s - 2xidx⋅s) gij (λ dxj + Λjrdxr + 2 x⋅dx sj - 2dxjx⋅s - 2xjdx⋅s)

How can we see that δds2=(2λ-2x⋅s)ds2

2. Jul 22, 2016

### samalkhaiat

The variation symbol $\delta$ is a derivation. So, you should consider

$$\delta \left(ds^{2}\right) = \eta_{\mu\nu} \ \delta \left(dx^{\mu}\right) \ dx^{\nu} + \eta_{\mu\nu} \ dx^{\mu} \ \delta \left(dx^{\nu}\right) . \ \ \ (1)$$

Now, for the infinitesimal conformal transformation

$$\delta x^{\mu} = a^{\mu} + \lambda x^{\mu} + \omega^{\mu}{}_{\nu}x^{\nu} + c^{\mu}x^{2} - 2 (c \cdot x ) x^{\mu} ,$$

if we take the partial derivative with respect to $x^{\sigma}$, we get

$$\partial_{\sigma} (\delta x^{\mu}) = \delta^{\mu}_{\sigma} \left( \lambda - 2 c \cdot x \right) + \eta_{\sigma \tau} \left( \omega^{\mu \tau} + 2 ( c^{\mu}x^{\tau} - c^{\tau}x^{\mu}) \right) . \ \ (2)$$

In terms of the following local parameters

$$\Lambda (x) = \lambda - 2 c \cdot x ,$$ $$\Omega^{\mu \tau}(x) = - \Omega^{\tau \mu}(x) = \omega^{\mu \tau} + 2 (c^{\mu}x^{\tau} - c^{\tau}x^{\mu}) ,$$ equation (2) becomes

$$\partial_{\sigma} (\delta x^{\mu}) = \delta^{\mu}_{\sigma} \ \Lambda (x) + \eta_{\sigma \tau} \ \Omega^{\mu \tau}(x) .$$

From this, you get

$$d (\delta x^{\mu}) = \Lambda (x) \ dx^{\mu} + \eta_{\sigma \tau} \ \Omega^{\mu \tau} \ dx^{\sigma} . \ \ \ \ \ (3)$$

Substituting (3) in (1), we find

$$\delta \left(ds^{2}\right) = 2 \Lambda (x) \ \eta_{\mu\nu} \ dx^{\mu} dx^{\nu} + \Omega_{\mu \nu}(x) \ dx^{\mu} dx^{\nu} + \Omega_{\nu \mu} (x) \ dx^{\mu}dx^{\nu} .$$

The last two terms vanish because $\Omega_{\mu\nu} = - \Omega_{\nu\mu}$. So you are left with

$$\delta \left(ds^{2}\right) = 2 \Lambda (x) \ ds^{2} .$$

3. Jul 25, 2016

### physicality

Right, variations satisfy the Leibnitz rule. Thank you very much, sir.