There is a problem with Normal derivative in curved space that $dg_{\mu\nu}\ne 0$ but in covariant derivative $Dg_{\mu\nu}=0$ So $Dg_{\mu\nu}A^{\nu}=g_{\mu\nu}DA^{\nu}$ but In this case the Euclidean derivative is broken into two parts, the extrinsic normal component and the intrinsic covariant derivative component. $DA^{\nu}=dA^{\nu}+\delta A^{\nu}$ what i don't know is that how $\delta A^{\nu}=\Gamma^{\mu}_{\alpha\beta}A^{\mu} dx^{\beta}$
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|
$dx^a=\frac {\partial x^a}{\partial x^b}dx^b,$ $d^2x^a=d(\frac {\partial x^a}{x^b}dx^b)=\frac {\partial x^a}{\partial x^b}d^2x^b+\frac {\partial^2 x^a}{\partial x^b \partial x^c}dx^b dx^c,$ where the $\frac {\partial x^a}{\partial x^b}d^2x^b=d^2x^a$ then $d^2x^a=d(\frac {\partial x^a}{x^b}dx^b)=d^2x^a+\frac {\partial^2 x^a}{\partial x^b \partial x^c}dx^b dx^c=d^2x^a+\Gamma^a_{bc}dx^bdx^c,$ |
||||
|
|
