- #1

Frostman

- 115

- 17

- Homework Statement
- Following the non-linear transformation of coordinates (conformal transformation)

##x'^{\mu}=\frac{x^{\mu}-x^2a^{\mu}}{1-2a_{\nu}x^{\nu}+a^2x^2}##

with ##a^{\mu}## constant four-vector, the relativistic line element is transformed (obviously) in the following way

##g_{\mu \nu}dx'^{\mu}dx'^{\nu}=K_{\mu \nu}dx^{\mu}dx^{\nu}##

- Relevant Equations
- Tensor calculus

I got stuck in this calculation, I can't collect everything in terms of ##dx^{\mu}##.

##x'^{\mu}=\frac{x^{\mu}-x^2a^{\mu}}{1-2a_{\nu}x^{\nu}+a^2x^2}##

##x'^{\mu}=\frac{x^{\mu}-g_{\alpha \beta}x^{\alpha}x^{\beta}a^{\mu}}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}##

##dx'^{\mu}=\frac{dx^{\mu}-a^{\mu}g_{\alpha \beta}(dx^{\alpha}x^{\beta}+x^{\alpha}dx^{\beta})}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}-\frac{(x^{\mu}-x^2a^{\mu})(1-2a_{\nu}dx^{\nu}+a^2g_{\alpha \beta}(dx^{\alpha}x^{\beta}+x^{\alpha}dx^{\beta})}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2}##

##dx'^{\mu}=\frac{dx^{\mu}-2dx\cdot xa^{\mu}}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}-\frac{(x^{\mu}-x^2a^{\mu})(1-2a\cdot dx+2a^2dx \cdot x)}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2}##

##dx'^{\mu}=\frac{(1-2x_{\mu}a^{\mu})dx^{\mu}}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}-\frac{(x^{\mu}-x^2a^{\mu})(1-(2a_{\mu} -2a^2 x_{\mu})dx^{\mu})}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2}##

I stop here, because in the second fraction there is that 1 in the numerator that I don't know how to collect with respect to ##dx^{\mu}##.

I suppose to express ##1## as ##\frac{dx_{\mu}dx^{\mu}}{\sqrt{dx^2}}## but I don't know if is it correct because I leave inside the ##K_{\mu \nu}## this term ##\frac{dx_{\mu}}{\sqrt{dx^2}}##

##x'^{\mu}=\frac{x^{\mu}-x^2a^{\mu}}{1-2a_{\nu}x^{\nu}+a^2x^2}##

##x'^{\mu}=\frac{x^{\mu}-g_{\alpha \beta}x^{\alpha}x^{\beta}a^{\mu}}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}##

##dx'^{\mu}=\frac{dx^{\mu}-a^{\mu}g_{\alpha \beta}(dx^{\alpha}x^{\beta}+x^{\alpha}dx^{\beta})}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}-\frac{(x^{\mu}-x^2a^{\mu})(1-2a_{\nu}dx^{\nu}+a^2g_{\alpha \beta}(dx^{\alpha}x^{\beta}+x^{\alpha}dx^{\beta})}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2}##

##dx'^{\mu}=\frac{dx^{\mu}-2dx\cdot xa^{\mu}}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}-\frac{(x^{\mu}-x^2a^{\mu})(1-2a\cdot dx+2a^2dx \cdot x)}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2}##

##dx'^{\mu}=\frac{(1-2x_{\mu}a^{\mu})dx^{\mu}}{1-2a_{\nu}x^{\nu}+a^2g_{\alpha \beta}x^{\alpha}x^{\beta}}-\frac{(x^{\mu}-x^2a^{\mu})(1-(2a_{\mu} -2a^2 x_{\mu})dx^{\mu})}{(1-2a_{\nu}x^{\nu}+a^2x^2)^2}##

I stop here, because in the second fraction there is that 1 in the numerator that I don't know how to collect with respect to ##dx^{\mu}##.

I suppose to express ##1## as ##\frac{dx_{\mu}dx^{\mu}}{\sqrt{dx^2}}## but I don't know if is it correct because I leave inside the ##K_{\mu \nu}## this term ##\frac{dx_{\mu}}{\sqrt{dx^2}}##