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Bianchi Identities (Contracted)

Contracting $\lambda$ with $\nu$ in the Bianchi Identities

\begin{displaymath}
R_{\lambda\mu\nu\kappa;\eta}+R_{\lambda\mu\eta\nu;\kappa}+R_{\lambda\mu\kappa\eta;\nu}=0
\end{displaymath} (1)

gives
\begin{displaymath}
R_{\mu\kappa;\eta}-R_{\mu\eta;\kappa}+R^\nu{}_{\mu\kappa\eta;\nu}=0.
\end{displaymath} (2)

Contracting again,
\begin{displaymath}
R_{;\eta}-R^\mu{}_{\eta;\mu}-R^\nu{}_{\eta;\nu}=0,
\end{displaymath} (3)

or
\begin{displaymath}
(R^\mu{}_\eta-{\textstyle{1\over 2}}\delta^\mu{}_\eta R)_{;\mu} = 0,
\end{displaymath} (4)

or
\begin{displaymath}
(R^{\mu\nu}-{\textstyle{1\over 2}}g^{\mu\nu}R)_{;\mu}=0.
\end{displaymath} (5)




© 1996-9 Eric W. Weisstein
1999-05-26