Covariant and partial derivative of metric determinant

[the_doors]

Homework Statement

is this statement is true : $\nabla_\mu \nabla_\nu \sqrt{g} \phi = \partial_\mu \sqrt{g} \partial_\nu \phi$

Homework Equations

The Attempt at a Solution

well we know $\nabla_\mu \sqrt{g} =0$ so it moves back : $\nabla_\mu \sqrt{g} \nabla_\nu \phi =\sqrt{g} \nabla_\mu \nabla_\nu \phi$ . but because metric determinant is not transfers like scalar we can not write $\partial_\mu$ instead of$\nabla_\mu$.

 

[mark726]

This statement is not necessarily true. While it is true that $\nabla_\mu \sqrt{g} = 0$, this does not mean that $\nabla_\mu \nabla_\nu \sqrt{g} = 0$. The derivative operator $\nabla_\mu$ acts on both the metric determinant and the scalar field, so we cannot simply move it outside of the derivative.

Additionally, the statement is missing a factor of $\sqrt{g}$ on the right-hand side, which would make it equal to the left-hand side. So, the correct statement would be: $\nabla_\mu \nabla_\nu \sqrt{g} \phi = \sqrt{g} \partial_\mu \partial_\nu \phi$.