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

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


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Updated on August 01, 2022


  • Harrold
    Harrold 10 months

    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...

  • Harrold
    Harrold over 10 years
    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!
  • Jold
    Jold over 10 years
    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.