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What is the difference between derivative and covariant derivative?

I realized that $\dfrac {d^2x^a}{ds^2}\ne 0$ while $\dfrac {D^2x^a}{Ds^2}=0$.

Why is it like this?

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up vote 2 down vote accepted

In this case, it may be a bit easier to explain in terms of a manifold in an embedding.

If $\underline P$ is the projection operator onto the manifold, $\nabla$ is the vector derivative of the embedding and $\partial = \underline P(\nabla)$ is the projection of the vector derivative into the manifold, then the covariant derivative $D$ obeys

$$a \cdot D A = \underline P(a \cdot \partial A)$$

So that it lies entirely in the manifold, guaranteed.

In particular, the covariant derivative and projected derivative are related by

$$a \cdot D A = a \cdot \partial A + \underline S(a) \times A$$

where $\underline S(a)$ is the shape tensor. This is why an expression of partial derivatives can be nonzero but the covariant derivative can be zero--the shape tensor term comes into play.

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