Problem deriving Einstein tensor form Bianchi identity

[jstrunk]

1. I can't understand one step in the derivation of the Einstein tensor from the Bianchi identity.I have looked in a lot of books and all over the internet and everyone glosses over the same point as if its obvious, but it isn't obvious to me.




2. Below is the entire derivation. It seems to be the usual one that most authors use but I have spelled it out in more detail. The problem is in the third term going from step 9 to step 10. I haven't seen any justification for it. All the identities I have to work with involve Riemann tensors with one or zero upper indexes, none have two upper indexes. I tried to work it out explicitly by expanding the Riemann and Ricci tensors in terms of metrics but that quickly became a morass. I don't think that is how the authors are doing it. I think they are using some
identity or technique that I don't know.




3.
$$\[ \begin{array}{l} R_{\alpha \beta \mu \nu ;\lambda } + R_{\alpha \beta \lambda \mu ;\nu } + R_{\alpha \beta \nu \lambda ;\mu } = 0 \\ \left[ {g^{\alpha \mu } R_{\alpha \beta \mu \nu } } \right]_{;\lambda } + \left[ {g^{\alpha \mu } R_{\alpha \beta \lambda \mu } } \right]_{;\nu } + \left[ {g^{\alpha \mu } R_{\alpha \beta \nu \lambda } } \right]_{;\mu } = 0 \\ \left[ {R_{\beta \mu \nu }^\mu } \right]_{;\lambda } + \left[ {R_{\beta \lambda \mu }^\mu } \right]_{;\nu } + \left[ {R_{\beta \nu \lambda }^\mu } \right]_{;\mu } = 0 \\ R_{\beta \mu \nu ;\lambda }^\mu + R_{\beta \lambda \mu ;\nu }^\mu + R_{\beta \nu \lambda ;\mu }^\mu = 0 \\ R_{\beta \mu \nu ;\lambda }^\mu - R_{\beta \mu \lambda ;\nu }^\mu + R_{\beta \nu \lambda ;\mu }^\mu = 0 \\ g^{\beta \nu } R_{\beta \nu ;\lambda } - g^{\beta \nu } R_{\beta \lambda ;\nu } + g^{\beta \nu } R_{{\rm{ }}\beta \nu \lambda ;\mu }^\mu = 0 \\ \left[ {g^{\beta \nu } R_{\beta \nu } } \right]_{;\lambda } - \left[ {g^{\beta \nu } R_{\beta \lambda } } \right]_{;\nu } + \left[ {g^{\beta \nu } R_{{\rm{ }}\beta \nu \lambda }^\mu } \right]_{;\mu } = 0 \\ \left[ R \right]_{;\lambda } - \left[ {R_{{\rm{ }}\lambda }^\nu } \right]_{;\nu } + \left[ {R_{{\rm{ }}\nu \lambda }^{\mu \nu } } \right]_{;\mu } = 0 \\ R_{;\lambda } - R_{{\rm{ }}\lambda }^\nu _{;\nu } + R_{{\rm{ }}\nu \lambda }^{\mu \nu } _{;\mu } = 0 \\ R_{;\lambda } - R_{{\rm{ }}\lambda }^\nu _{;\nu } - R_{{\rm{ }}\lambda }^\mu _{;\mu } = 0 \\ R_{;\lambda } - R_{{\rm{ }}\lambda }^\mu _{;\mu } - R_{{\rm{ }}\lambda }^\mu _{;\mu } = 0 \\ R_{;\lambda } - 2R_{{\rm{ }}\lambda }^\mu _{;\mu } = 0 \\ 2R_{{\rm{ }}\lambda }^\mu _{;\mu } - R_{;\lambda } = 0 \\ Define{\rm{ }}G^{\alpha \beta } \equiv g^{\alpha \nu } \left[ {R_{{\rm{ }}\nu }^\beta - \frac{1}{2}\delta _{{\rm{ }}\nu }^\beta R} \right] = R^{\alpha \beta } - \frac{1}{2}g^{\alpha \beta } R = G^{\beta \alpha } \\ G_{{\rm{ }};\beta }^{\alpha \beta } = \left[ {2R_{{\rm{ }}\lambda }^\mu - \delta _{{\rm{ }}\lambda }^\mu R} \right]_{;\beta } = 0 \\ \end{array} \]$$

 

[arkajad]

It is hard to count. Can you now write it down separately which particular equality causes you a problem?

Like: "Is R... really equal to R... ? Why?"

 

[jstrunk]

This is the point I am having trouble with
$$\[ R_{{\rm{ }}\nu \lambda }^{\mu \nu } _{;\mu } = R_{{\rm{ }}\lambda }^\mu _{;\mu } \]$$

 

[arkajad]

First observation: in your notation it is hard to follow the order of indices. Your definition of the Ricci tensor seems to be

$$R_{\beta\nu}=R^\mu_{\beta\mu\nu}$$

I guess that means

a) $$R_{\beta\nu}={R^\mu}_{\beta\mu\nu}$$

but it could also mean

b) $$R_{\beta\nu}={{R_\beta\,}^\mu}_{\mu\nu}$$

Probably your confusion is the result of this lack. Could you write your definition of the Ricci tensor paying attention to the order? And maybe recheck your calculations paying attention to the order of indices?

 

[ebank]

Hello

The Riemann Tensor is anti symmetric on the first 2 indices. So switch them and get the minus sign. then the index nu is in the first and third slot, on the top and bottom, as required for the Ricci contraction. So apply the definition of the Ricci tensor and you are done. Hope this helps.

:)

 

[ebank]

$$\[ \begin{array}{l} R_{\alpha \beta \mu \nu ;\lambda } + R_{\alpha \beta \lambda \mu ;\nu } + R_{\alpha \beta \nu \lambda ;\mu } = 0 \\ \left[ {g^{\alpha \mu } R_{\alpha \beta \mu \nu } } \right]_{;\lambda } + \left[ {g^{\alpha \mu } R_{\alpha \beta \lambda \mu } } \right]_{;\nu } + \left[ {g^{\alpha \mu } R_{\alpha \beta \nu \lambda } } \right]_{;\mu } = 0 \\ \left[ {R_{\beta \mu \nu }^\mu } \right]_{;\lambda } + \left[ {R_{\beta \lambda \mu }^\mu } \right]_{;\nu } + \left[ {R_{\beta \nu \lambda }^\mu } \right]_{;\mu } = 0 \\ R_{\beta \mu \nu ;\lambda }^\mu + R_{\beta \lambda \mu ;\nu }^\mu + R_{\beta \nu \lambda ;\mu }^\mu = 0 \\ R_{\beta \mu \nu ;\lambda }^\mu - R_{\beta \mu \lambda ;\nu }^\mu + R_{\beta \nu \lambda ;\mu }^\mu = 0 \\ g^{\beta \nu } R_{\beta \nu ;\lambda } - g^{\beta \nu } R_{\beta \lambda ;\nu } + g^{\beta \nu } R_{{\rm{ }}\beta \nu \lambda ;\mu }^\mu = 0 \\ \left[ {g^{\beta \nu } R_{\beta \nu } } \right]_{;\lambda } - \left[ {g^{\beta \nu } R_{\beta \lambda } } \right]_{;\nu } + \left[ {g^{\beta \nu } R_{{\rm{ }}\beta \nu \lambda }^\mu } \right]_{;\mu } = 0 \\ \left[ R \right]_{;\lambda } - \left[ {R_{{\rm{ }}\lambda }^\nu } \right]_{;\nu } + \left[ {R_{{\rm{ }}\nu \lambda }^{\mu \nu } } \right]_{;\mu } = 0 \\ R_{;\lambda } - R_{{\rm{ }}\lambda }^\nu _{;\nu } + R_{{\rm{ }}\nu \lambda }^{\mu \nu } _{;\mu } = 0 \\ R_{;\lambda } - R_{{\rm{ }}\lambda }^\nu _{;\nu } - R_{{\rm{ }}\nu \lambda }^{\nu \mu } _{;\mu } = 0 Antisymetry -of -Riemann -in -index -1-and-2\\ R_{;\lambda } - R_{{\rm{ }}\lambda }^\nu _{;\nu } - R_{{\rm{ }}\lambda }^\mu _{;\mu } = 0 \\ R_{;\lambda } - R_{{\rm{ }}\lambda }^\mu _{;\mu } - R_{{\rm{ }}\lambda }^\mu _{;\mu } = 0 \\ R_{;\lambda } - 2R_{{\rm{ }}\lambda }^\mu _{;\mu } = 0 \\ 2R_{{\rm{ }}\lambda }^\mu _{;\mu } - R_{;\lambda } = 0 \\ 2R_{{\rm{ }}\lambda }^\mu _{;\mu } - g^\mu \ _\lambda R_{;\mu } = 0 change-derivitave-with-\lambda-to-\mu \\ ( 2R_{{\rm{ }}\lambda }^\mu - g^\mu \ _\lambda R ) _{;\mu } = 0 \\ Raising an index...\\ ( 2R ^{\mu\lambda } - g^{\mu \lambda} R ) _{;\mu } = 0 \\ \\ \end{array} \]$$

 

[ebank]

$$\[ \begin{array}{l}Define{\rm{ }}G^{\alpha \beta } = R^{\alpha \beta } - \frac{1}{2}g^{\alpha \beta } R = G^{\beta \alpha } so that G_{{\rm{ }};\beta }^{\alpha \beta } = 0 \\ \end{array} \]$$