Can I simply reverse the indices in a contraction?

Tags:
1. Jun 4, 2015

unscientific

Suppose I have something like
$$\left( \nabla_\mu \nabla_\beta - \nabla_\beta \nabla_\mu \right) V^\mu = R_{\nu \beta} V^\nu$$
Can since all the terms involving $\mu$ on the left and $\nu$ on the right are contractions, can I simply do:
$$\left( \nabla^\mu \nabla_\beta - \nabla_\beta \nabla^\mu \right) V_{\mu} = R^\nu_{\beta} V_{\nu}$$?

2. Jun 4, 2015

Orodruin

Staff Emeritus
As long as the connection is metric compatible.

3. Jun 4, 2015

unscientific

But my concern is that in the first term on the LHS, the gradient operator is acting on the gradient operator via $\nabla_\mu (\nabla_\beta) V^\mu$, and not on the vector $V^\mu$. I thought for contractions, these have to be acting on one another?

4. Jun 4, 2015

Orodruin

Staff Emeritus
No, this is not how to interpret the LHS. The operators are each acting on $V^\mu$, one after the other.

5. Jun 4, 2015

unscientific

Ok, now I'm confused. I thought the chain rule applies:
$$\nabla^\mu \left( \nabla_\mu A_\nu \right) = \nabla^\mu \left( \nabla_\mu\right) A_\nu + \nabla_\mu \nabla^\mu A_\nu$$

6. Jun 4, 2015

unscientific

If I have something like $F_{\mu \nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu$, what will $\nabla^\mu F_{\mu \nu}$ look like?

7. Jun 5, 2015

Orodruin

Staff Emeritus
What do you mean by $\nabla_\mu(\nabla^\mu)$? This is an operator applied to an operator. The only way to make sense of this is to apply the right operator to something first and then apply the left to the result.

8. Jun 5, 2015

haushofer

You should write out this covariant derivative first in terms of F, and then in that expression write out F in terms of A. If your connection is symmetric, F can be expressed in terms of partial derivatives only.