What is second-order 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...


1 Answer 1


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 Levi-Civita connection, i.e. the Christoffel symbols, the covariant derivative will be:

$$\nabla_i \nabla_j f=\nabla_i \partial_jf=\partial_i \partial_j f-\partial_k f~\Gamma^{k}_{ij}$$

  • $\begingroup$ Thank you very much, elfmotat. I have a further question: isn't the Levi-Civita connection built into e definition of the covariant derivative? Is an extra assumption required? -- I may have got something confused. Thanks again! $\endgroup$
    – Harrold
    Commented Feb 21, 2013 at 22:27
  • $\begingroup$ In GR the connection is assumed to be torsion-free, so the Levi-Civita 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. $\endgroup$
    – Jold
    Commented Feb 21, 2013 at 22:36

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