Covariant derivative of Ricci scalar causing me grief

[ft_c]

Hi all

I'm having trouble understanding what I'm missing here. Basically, if I write the Ricci scalar as the contracted Ricci tensor, then take the covariant derivative, I get something that disagrees with the Bianchi identity:

\begin{align*}
R&=g^{\mu\nu}R_{\mu\nu}\\
\Rightarrow \nabla R&=\nabla(g^{\mu\nu}R_{\mu\nu})\\
&=\nabla(g^{\mu\nu})R_{\mu\nu}+g^{\mu\nu}\nabla(R_{\mu\nu})\\
&=g^{\mu\nu}\nabla(R_{\mu\nu})\\
g_{\mu\nu}\nabla R&=g_{\mu\nu}g^{\mu\nu}\nabla(R_{\mu\nu})\\
&=4\nabla(R_{\mu\nu})\\
\Rightarrow \nabla(R_{\mu\nu}-\tfrac{1}{4}g_{\mu\nu} R)&=0
\end{align*}
whereas the contracted Bianchi identity is
$$\nabla(R_{\mu\nu}-\tfrac{1}{2}g_{\mu\nu}R)=0$$

If anyone could let me know what's going wrong here that would be much appreciated! Thanks very much in advance

 

[Orodruin]

You are multiplying with $g_{\mu\nu}$ when you already have $\mu$ and $\nu$ as dummy indices in your expression and then you are contracting the wrong $\mu$ and $\nu$. This is a very common student mistake.

 

[ft_c]

You are multiplying with $g_{\mu\nu}$ when you already have $\mu$ and $\nu$ as dummy indices in your expression and then you are contracting the wrong $\mu$ and $\nu$. This is a very common student mistake.

Thanks Orodruin! I've actually already tried with different indices already, and I got the same thing, maybe you can take a look?
First write
$$g_{\mu\nu}R=g_{\mu\nu}g^{\rho\sigma}R_{\rho\sigma}$$
Covariant derivative
\begin{align*}
\nabla(g_{\mu\nu}R)&=\nabla(g_{\mu\nu}g^{\rho\sigma}R_{\rho\sigma})\\
g_{\mu\nu}\nabla R&=g_{\mu\nu}g^{\rho\sigma}\nabla R_{\rho\sigma}
\end{align*}

But $$\nabla R_{\rho\sigma} = \tfrac{1}{2}g_{\rho\sigma}\nabla R$$ from the Bianchi identity, so

\begin{align*}
g_{\mu\nu}\nabla R&=g_{\mu\nu}g^{\rho\sigma}(\tfrac{1}{2}g_{\rho\sigma}\nabla R)\\
&=\tfrac{1}{2}g_{\mu\nu}g^{\rho\sigma}g_{\rho\sigma}\nabla R\\
&=2g_{\mu\nu}\nabla R
\end{align*}

Can you see what's going wrong here? Thanks!

 

[TeethWhitener]

I'm just an amateur at this stuff, but maybe you need to index your derivative operator? So the second Bianchi identity ends up looking like:
$$\nabla^{a} (R_{ab} - \frac{1}{2}Rg_{ab}) = 0$$
or
$$\nabla_{a} (R^{ab} - \frac{1}{2}Rg^{ab}) = 0$$

 

[TeethWhitener]

So in post #3 above, the Bianchi identity should read:
$$\nabla^{\rho} R_{\rho\sigma} = \tfrac{1}{2}g_{\rho\sigma}\nabla^{\rho} R$$
but the covariant derivative line will read:
$$g_{\mu\nu}\nabla^{\mu} R=g_{\mu\nu}g^{\rho\sigma}\nabla^{\mu} R_{\rho\sigma}$$
or at least something where the index on $\nabla$ is not $\rho$. So the pattern of indices doesn't allow for the substitution you made in that post. Instead, you'll get:
$$2\nabla^{\mu} R_{\mu\nu} =g_{\mu\nu}g^{\rho\sigma}\nabla^{\mu} R_{\rho\sigma} = \nabla^{\mu}Rg_{\mu\nu}$$
which is just the Bianchi identity again.