It has been several years since I looked at General relativity, and I am trying to brush up on it because it was always interesting and I am in need of it for my research.
Specifically, I am looking at linearized gravity in "Spacetime and Geometry" by Sean Carrol, page 275.
My issue is, I just don't remember tensor manipulation very well, so I apologize if this is a trivial question. After finding the linearized Christoffel symbols, he calculates the Riemann tensor to be:
$$R_{\mu\nu\rho\sigma}=\frac{1}{2}\left( \partial_\rho\partial_\nu h_{\mu\sigma} + \partial_\sigma \partial_{\mu} h_{\nu\rho} - \partial_{\sigma}\partial_\nu h_{\mu\rho} - \partial_{\rho}\partial_{\mu} h_{\nu\sigma} \right)$$
Which I was able to get after I realized the linearized metric tensor $h$ is symmetric and I can flip the indices ($h_{\mu\nu} = h_{\nu\mu}$). The part I am having trouble with is the next part. He says "The Ricci tensor comes from contracting over $\mu$ and $\rho$, giving:"
$$R_{\mu\nu} = \frac{1}{2}\left( \partial_{\sigma}\partial_\nu h^{\sigma}\;_{\mu} + \partial_{\sigma}\partial_{\mu}h^{\sigma}\;_{\nu}-\partial_{\mu}\partial_{\nu}h - \square h_{\mu\nu} \right)$$
where $\square = \partial^{\mu}\partial_{\mu}$ and $h=\eta^{\mu\nu}h_{\mu\nu}=h^{\mu}\;_{\mu}$. When I attempt to do the above calculation to get the Ricci tensor, I set $\rho = \mu$ and get:
$$R_{\mu\nu\sigma}=\frac{1}{2}\left( \partial_{\mu}\partial_{\nu} h_{\mu\sigma} + \partial_{\sigma} \partial_{\mu} h_{\nu\mu} - \partial_{\sigma}\partial_{\nu} h_{\mu\mu} - \partial_{\mu}\partial_{\mu} h_{\nu\sigma} \right).$$
I just can't figure out how to get this in the same form as the book. I don't think I can raise indices on partial derivatives since they don't transform as tensors, but maybe I am wrong in that. Would anyone be able to walk me through how the book got this form of the Ricci? Again, I apologize if this is trivial but I can only remember the broad strokes of GR tensor mathematics.