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.


As @octonion says, I think the confusion here comes from the fact that Carol re-labels $\sigma, \nu$ back to $\mu, \nu$. First I'll do the computation as Caroll defines it (contracting $\mu$ and $\rho$ in $R_{\mu\nu\rho\sigma}$): $$ R_{\nu \sigma} := R^{\rho}_{\ \; \nu\rho\sigma} = g^{\rho\mu}R_{\mu\nu\rho\sigma} $$ Using the definition: $$ R_{\nu \sigma} = g^{\rho\mu} \tfrac{1}{2} \left( \partial_{\rho} \partial_{\nu} h_{\mu\sigma} + \partial_{\sigma} \partial_{\mu} h_{\nu\rho} - \partial_{\sigma} \partial_{\nu} h_{\mu\rho} -\partial_{\rho} \partial_{\mu} h_{\nu\sigma} \right) $$ Multiply through the $g^{\rho\mu}$ in through the bracket $$ R_{\nu \sigma} = \tfrac{1}{2} \left( \partial_{\rho} \partial_{\nu} h^{\rho}_{\; \sigma} + \partial_{\sigma} \partial^{\rho} h_{\nu\rho} - \partial_{\sigma} \partial_{\nu} h^{\rho}_{\ \; \rho} - \partial_{\rho} \partial^{\rho} h_{\nu\sigma} \right) $$ Cleaning this up using Carrol's definitions gives $$ R_{\nu \sigma} = \tfrac{1}{2} \left( \partial_{\rho} \partial_{\nu} h^{\rho}_{\; \sigma} + \partial^{\rho} \partial_{\sigma} h_{\nu\rho} - \partial_{\nu} \partial_{\sigma} h - \Box h_{\nu\sigma} \right) $$ This is exactly the equation you gave for $R_{\mu\nu}$ (except you have to switch $\nu \to \mu$ and $\sigma \to \nu$).

Side note: every metric tensor is symmetric (not just the linearized one).

  • $\begingroup$ Thank you for your help! $\endgroup$ – user41178 Jul 23 '19 at 11:48

It is the same form as in the book, you just have some differences in notation. Since you decided to contract $\mu,\rho$ the free indices you are left over with are $\sigma, \nu$. Carrol chose to rename $\sigma,\nu$ back to $\mu,\nu$. This doesn't change anything as long as you do it consistently, and you can see he made this choice by looking at the left hand side. You have a $\mu$ which shouldn't be there (since this is the Ricci tensor not the Riemann tensor, and $\mu$ was contracted), but otherwise the left hand side tells you what the two free indices are called.

The rest of the notation differences are just that $h$ without any indices is defined as the trace $h_{\mu\mu}$ (usually there are distinctions between upper and lower indices, by the way, so watch out. But here we are dealing with linearized gravity so they don't matter). And the box is just the contracted second derivative $\partial_\mu \partial_\mu$

  • $\begingroup$ Thanks! Makes sense! $\endgroup$ – user41178 Jul 23 '19 at 11:48

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