Conformal transformation of the line element

[physicality]

Let us see how the line element transforms under conformal transformations. Consider the Minkovski metric g_ij, a line element ds²=dxⁱg_ijdx^j, and a conformal transformation

δ^k(x)=a^k + λ x^k + Λ^klx_l + x²s^k - 2x^kx⋅s

We have δ(dx^k)=dδ(x)^k=λ dx^k + Λ^kldx_l + 2 x⋅dx s^k - 2dx^kx⋅s - 2x^kdx⋅s

And so the line element transforms by δds²=δ(dxⁱ)g_ijδ(dx^j)=
(λ dxⁱ + Λ^ildx_l + 2 x⋅dx sⁱ - 2dxⁱx⋅s - 2xⁱdx⋅s) g_ij (λ dx^j + Λ^jrdx_r + 2 x⋅dx s^j - 2dx^jx⋅s - 2x^jdx⋅s)

How can we see that δds²=(2λ-2x⋅s)ds²

 

[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} .

 

[physicality]

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