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For what covariant derivative?

  1. Jan 23, 2014 #1
    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. jcsd
  3. Jan 27, 2014 #2
    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, [tex]\nabla_\mu x^\nu = \partial_\mu x^\nu + \Gamma_{\mu \alpha}^\nu x^\alpha [/tex] and [tex]\nabla_\mu x_\nu = \partial_\mu x_\nu - \Gamma_{\mu \nu}^\alpha x_\alpha [/tex] for contravariant terms.
     
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