Take the 2-minute tour ×
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.

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 $\delta A^{\nu}=\Gamma^{\mu}_{\alpha\beta}A^{\mu} dx^{\beta}$

share|improve this question

1 Answer 1

$dx^a=\frac {\partial x^a}{\partial x^b}dx^b,$ $d^2x^a=d(\frac {\partial x^a}{x^b}dx^b)=\frac {\partial x^a}{\partial x^b}d^2x^b+\frac {\partial^2 x^a}{\partial x^b \partial x^c}dx^b dx^c,$ where the $\frac {\partial x^a}{\partial x^b}d^2x^b=d^2x^a$ then $d^2x^a=d(\frac {\partial x^a}{x^b}dx^b)=d^2x^a+\frac {\partial^2 x^a}{\partial x^b \partial x^c}dx^b dx^c=d^2x^a+\Gamma^a_{bc}dx^bdx^c,$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.