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Consider the Ricci tensor given by $R_{\alpha \beta} = \nabla_\alpha \nabla_\beta \phi$ for scalar field $\phi$. I've been trying to use the Ricci and contracted Bianchi identities to show that $$\nabla_\alpha (\nabla_\beta \nabla^\beta \phi) = -2R_{\alpha \beta} \nabla^\beta \phi.$$

[In this case the Ricci identity is $\nabla_\alpha \nabla_\beta V^\gamma - \nabla_\beta \nabla_\alpha V^\gamma = R^\gamma_{\delta \alpha \beta} V^\delta$ and the contracted Bianchi identity is $\nabla^\beta R_{\alpha \beta} = \frac{1}{2}\nabla_\alpha R$.]

My 'attempt':
One can contract the definition with $g^{\alpha \beta}$ to get $R = \nabla_\beta \nabla^\beta \phi$, where $R$ is the Ricci scalar. Applying $\nabla_\alpha$ to both sides and using the contracted Bianchi identity, I got $$2 \nabla^\beta \nabla_\alpha \nabla_\beta \phi = \nabla_\alpha (\nabla_\beta \nabla^\beta \phi ),$$ which is perhaps a non-zero amount of progress. I also played around with the two identities separately but couldn't produce anything that seemed useful. Any help in getting started is greatly appreciated!

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Consider

$$R_{\alpha \beta} = \nabla_\alpha \nabla_\beta \phi\tag{1}$$

Thus, applying $\nabla_\beta$ in the last equality and by Ricci identity and Bianchi identity, we have \begin{align*} \nabla_\beta R_\alpha^\beta&=\nabla_\beta(\nabla_\alpha\nabla_\beta\phi)\\ \frac{1}{2}\nabla_\alpha R&=\nabla_\alpha(\nabla_\beta\nabla^\beta\phi)+R^\beta_{\delta\beta\alpha}\nabla^\delta\phi.\tag{2} \end{align*} By other hand, tracing the equality (1), we obtain $$R=R_\beta^\beta=\nabla_\beta\nabla^\beta\phi.$$ Applying $\nabla_\alpha$ in the last equality, we obtain $$\nabla_\alpha R=\nabla_\alpha\nabla_\beta\nabla^\beta\phi\tag{3}$$

Putting, (3) in (2), we obtain \begin{align*} \frac{1}{2}\nabla_\alpha(\nabla_\beta\nabla^\beta\phi)&=\nabla_\alpha(\nabla_\beta\nabla^\beta\phi)+R_{\delta\alpha}\nabla^\delta\phi\\ -\frac{1}{2}\nabla_\alpha(\nabla_\beta\nabla^\beta\phi)&=R_{\delta\alpha}\nabla^\delta\phi\\ \nabla_\alpha(\nabla_\beta\nabla^\beta\phi)&=-2R_{\alpha\beta}\nabla^\beta\phi \end{align*}

In the last equality, we change $\delta$ by $\beta$ (since is a summand) and the fact that $Ric$ tensor is symmetric.

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