I will take the differential form of position vector r:(adsbygoogle = window.adsbygoogle || []).push({});

##\vec{r}=r\hat{r}##

##d\vec{r}=dr\hat{r}+rd\hat{r}##

So, now I need find ##d\hat{r}##

##d\hat{r}=\frac{d\hat{r}}{dr}dr+\frac{d\hat{r}}{d\theta}d\theta##

##\frac{d\hat{r}}{dr}=\Gamma ^{r}_{rr}\hat{r}+\Gamma ^{\theta}_{rr}\hat{\theta}=0\hat{r}+0\hat{\theta}=\vec{0}##

##\frac{d\hat{r}}{d\theta}=\Gamma ^{r}_{r\theta}\hat{r}+\Gamma ^{\theta}_{r\theta}\hat{\theta}=0\hat{r}+\frac{1}{r}\hat{\theta}=\frac{1}{r}\hat{\theta}##

So...

##d\hat{r}=\vec{0}dr+\frac{1}{r}\hat{\theta}d\theta=\frac{1}{r}d\theta \hat{\theta}##

Resulting in:

##d\vec{r}=dr\hat{r}+r\frac{1}{r}d\theta \hat{\theta}=dr\hat{r}+d\theta \hat{\theta}##

So, I have 2 question:

1) What the theory of covariant derivative has to do with this? Why I need understand covariante derivative? Where it appears? What expression it simplifies?

To understand the christofell's symbols is necessary because it appears in the process. But I don't see the covariant derivative in process...

2) Why my result is ##d\vec{r}=dr\hat{r}+d\theta \hat{\theta}##? It's wrong! Because ##d\vec{r}=dr\hat{r}+rd\theta \hat{\theta}## (with a factor r in 2nd term). However, I did all computation correctly. Where is the wrong?

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# For what covariant derivative?

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