Metric compatibility and covariant derivative

[George Keeling]

TL;DR Summary

Sean Carroll says that if we have metric compatibility then we may lower the index on a vector in a covariant derivative. Why?

Sean Carroll says that if we have metric compatibility then we may lower the index on a vector in a covariant derivative. As far as I know, metric compatibility means $\nabla_\rho g_{\mu\nu}=\nabla_\rho g^{\mu\nu}=0$, so in that case $\nabla_\lambda p^\mu=\nabla_\lambda p_\mu$. I can't see why one follows from the other. I can do this with the Leibnitz rule and then using metric compatibility$$
\nabla_\lambda p^\mu=\nabla_\lambda\left(g^{\mu\nu}p_\nu\right)=p_\nu\nabla_\lambda g^{\mu\nu}+g^{\mu\nu}\nabla_\lambda p_\nu=0+\nabla_\lambda p^\mu
$$which goes nowhere.
Also I get the same result much slower if I expand $\nabla_\lambda\left(g^{\mu\nu}p_\nu\right)$ using the usual rules for the covariant the covariant derivative and Christoffel symbols.

Can anybody tell me why metric compatibility $\Rightarrow\nabla_\lambda p^\mu=\nabla_\lambda p_\mu$?

It is vital for proving something about conservation of momentum.

 

[martinbn]

Can anybody tell me why metric compatibility $\Rightarrow\nabla_\lambda p^\mu=\nabla_\lambda p_\mu$?

That is not what he means, and this of course isn't true. The two tensors are of different type, they cannot be equal. What he means is that to lower the index in $\nabla_\lambda p^\mu$ you can do it inside the derivative i.e. $g_{\mu\nu}(\nabla_\lambda p^\nu)=\nabla_\lambda (g_{\mu\nu}p^\nu)$.

 

[Orodruin]

That is not what he means, and this of course isn't true. The two tensors are of different type, they cannot be equal. What he means is that to lower the index in $\nabla_\lambda p^\mu$ you can do it inside the derivative i.e. $g_{\mu\nu}(\nabla_\lambda p^\nu)=\nabla_\lambda (g_{\mu\nu}p^\nu)$.

Just to add that this is easy to verify as
$$
\nabla_\lambda(A_{\mu\nu} B^\nu) = B^\nu \nabla_\lambda A_{\mu\nu} + A_{\mu\nu} \nabla_\lambda B^\nu
$$
in general. Letting $A = g$ and $B = p$ and using $\nabla_\lambda g_{\mu\nu} = 0$ directly gives you the result.

 

[George Keeling]

The two tensors are of different type, they cannot be equal.

How embarrassing! But I should have told the whole truth: What Carroll actually said was that metric compatibility means that$$
p^\lambda\nabla_\lambda p^\mu=0\Rightarrow p^\lambda\nabla_\lambda p_\mu=0
$$so perhaps I am forgiven for my indexing faux pas. Now in baby steps:$$
p^\lambda\nabla_\lambda p^\mu=0
$$$$
\Rightarrow g^{\mu\nu}p^\lambda\nabla_\lambda p_\nu=0
$$$$
\Rightarrow g_{\rho\mu}g^{\mu\nu}p^\lambda\nabla_\lambda p_\nu=0
$$$$
\Rightarrow\delta_\rho^\nu p^\lambda\nabla_\lambda p_\nu=0
$$$$
\Rightarrow p^\lambda\nabla_\lambda p_\rho=0
$$I hope I haven't made another terrible blunder and perhaps metric compatibility was not important? The first step could use it or more elegantly use the rule for covariant derivatives that they '3. Commute with contractions'

Here's the way I understand #3.
$g^{\lambda \nu} (\nabla_\mu T_{\nu \lambda \rho}) = \nabla_\mu (g^{\lambda \nu} T_{\nu \lambda \rho})$

or you could just say that $\nabla_\lambda p^\mu$ is a tensor so you can contract with the metric anyway.

 

[Pencilvester]

What Carroll actually said was that metric compatibility means that
$$p^\lambda\nabla_\lambda p^\mu=0\Rightarrow p^\lambda\nabla_\lambda p_\mu=0$$

Carroll should have the caveat that this only applies to vector fields whose integral curves are geodesics, (i.e. if the vector is the velocity vector of a geodesic). So in general you will still have $$p^\lambda\nabla_\lambda p^\mu \neq p^\lambda\nabla_\lambda p_\mu$$

 

[Orodruin]

Carroll should have the caveat that this only applies to vector fields whose integral curves are geodesics, (i.e. if the vector is the velocity vector of a geodesic). So in general you will still have $$p^\lambda\nabla_\lambda p^\mu \neq p^\lambda\nabla_\lambda p_\mu$$

There is no caveat to the given statement. The given statement was a particular implication.

 

[Pencilvester]

There is no caveat to the given statement.

You’re absolutely right, I did not phrase my thought correctly. I should have also included what I was specifically replying to:

so perhaps I am forgiven for my indexing faux pas.

Not that anyone here would hold a grudge, but the statement made in the OP that X = Y still does not generally apply even when the derivative is taken in the direction of the vector.