How is the secondorder covariant derivative of a scalar computed?
The covariant derivative of a scalar is just its gradient because scalars don't depend on your basis vectors:
$$\nabla_j f=\partial_jf$$
Now it's a dual vector, so the next covariant derivative will depend on the connection. Assuming the LeviCivita connection, i.e. the Christoffel symbols, the covariant derivative will be:
$$\nabla_i \nabla_j f=\nabla_i \partial_jf=\partial_i \partial_j f\partial_k f~\Gamma^{k}_{ij}$$
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Harrold
Updated on August 01, 2022Comments

Harrold 10 months
What is secondorder covariant derivative $$\nabla_i\nabla_jf(r)$$ in terms of $r,\theta, g(r)$ and partial derivative, given that the metric takes the form $$ds^2=dr^2+g(r)d\theta^2$$ and $f$ is a scalar function of $r$?
For Cartesians, I know that the covariant derivatives reduce to partial derivatives. However, since this is in polar coordinates...

Harrold over 10 yearsThank you very much, elfmotat. I have a further question: isn't the LeviCivita connection built into e definition of the covariant derivative? Is an extra assumption required?  I may have got something confused. Thanks again!

Jold over 10 yearsIn GR the connection is assumed to be torsionfree, so the LeviCivita connection is all that is discussed in most introductory texbooks. Other connections are possible though, and the covariant derivative depends on the type of connection. If you're just learning GR then you don't really need to worry about any of that though.