How is the second-order covariant derivative of a scalar computed?

homework-and-exercises general-relativity differential-geometry metric-tensor differentiation

7,438

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:

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

7,438

Harrold

Updated on August 01, 2022

Comments

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 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 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.