# D'Alembert operator is commute covariant derivative?

1. Jul 22, 2015

### dhalilsim

For example:

[itex] D_α D_β D^β F_ab= D_β D^β D_α F_ab

is true or not? Are there any books sources?

2. Jul 22, 2015

### JorisL

So you are asking whether $\nabla_\mu\Box F_{ab} = \Box \nabla_\mu F_{ab}$?
Here I wrote $\Box = g^{\mu\nu}\nabla_\mu\nabla_\nu$ following modern notation.

Have you tried finding the commutator? What is $[ \nabla_\mu, \nabla_\nu]$ equal too?

The (free) lecture notes by Carroll might have what you are looking for. At the least they introduce all you need to calculate the commutator. (I know the book that built on those does have the identities you need)

3. Jul 23, 2015

### dhalilsim

Yes my question is whether or not
$∇_μ∇_ν∇^νF_{ab}=∇_ν∇^ν∇_μF_{ab}$ where $F_{ab}$ is electromagnetic field tensor.
Can I simply think,
so d'Alembert operator $∇_ν∇^ν$ is invariant,
then Can I immediately write $[∇_μ,\Box]=0$??

4. Jul 23, 2015

### JorisL

The easiest way to do this is the following $[\nabla_\mu, \Box] = [\nabla_\mu,g^{\alpha\beta}\nabla_\alpha\nabla_\beta] = \ldots$
Now I'll try to guide you through this by asking you some other questions.

$\nabla_\mu g^{\alpha\beta}$ is equal to ... (does metric compatibility ring a bell?)
If you use this you can write the first equation as $g^{\alpha\beta}[\nabla_\mu,\nabla_\alpha\nabla_\beta]$.

Now we need to know how we can write a commutator of the form $[A,BC]$ in terms of $[A,B]$ and $[A,C]$.
The easiest way to do this is expanding the total commutator and looking for the other commutators. It's a standard property that can be found in a lot of places.

Finally look up what $[\nabla_\mu,\nabla_\nu]$ is equal to. This can be found in every text on GR, here is a free link http://preposterousuniverse.com/grnotes/grnotes-three.pdf [Broken]

Last edited by a moderator: May 7, 2017