# For what covariant derivative?

1. Jan 23, 2014

### Jhenrique

I will take the differential form of position vector r:

$\vec{r}=r\hat{r}$

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

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?

2. Jan 27, 2014

### IBphysicsy

You need Christoffel symbols to take derivatives on curved surfaces. Therefore they appear whenever you're varying any quantity in a curved space. It should look like this for differentiating covariant terms, $$\nabla_\mu x^\nu = \partial_\mu x^\nu + \Gamma_{\mu \alpha}^\nu x^\alpha$$ and $$\nabla_\mu x_\nu = \partial_\mu x_\nu - \Gamma_{\mu \nu}^\alpha x_\alpha$$ for contravariant terms.