Contracted Bianchi Identity – [PROOF]

The contracted Bianchi Identity in GTR can be written as:
\nabla^\mu G_{\mu \nu}=0
where G_{\mu \nu} is the Einstein Tensor such that,
G_{\mu \nu}=R_{\mu \nu}- \frac{1}{2}g_{\mu \nu}R
and R_{\mu \nu} is the Ricci Tensor and R is the Ricci Scalar.

In this post, I will be demonstrating a simple way to prove the contracted Bianchi Identity.

We start with the usual Bianchi Identity in GTR:

\nabla_{\mu}R^{\quad \enspace \delta}_{\nu \lambda \rho}+\nabla_{\nu}R^{\quad \enspace \delta}_{\lambda \mu \rho}+\nabla_{\lambda}R^{\quad \enspace \delta}_{\mu \nu \rho} = 0  \qquad (\mathrm{eq. 1})

where R_{\nu \lambda \rho}^{\quad \enspace \delta} is the Riemann Tensor,
and \nabla_{\mu} is the covariant derivative.

We see that there are 5 indices in the above tensors. So we go ahead and contract alternate indices twice, to get a single index tensor.

Remember the following properties of the Riemann Tensor:
R_{\mu \nu \lambda}^{\quad \enspace \rho}=-R_{\nu \mu \lambda}^{\quad \enspace \rho}
R_{\mu \nu \lambda}^{\quad \enspace \rho}=-R_{\mu \nu \enspace \lambda}^{\enspace \enspace \rho}
R_{\mu \nu \lambda}^{\quad \enspace \rho}=R_{\lambda \enspace \mu  \nu}^{\enspace \rho}

Now, swap (interchange) the following in the first term of \mathrm{eq1} :
\delta with \rho ,
\nu with \lambda ,
and \enspace ^\delta_\rho with \lambda \nu
\implies \nabla_{\mu}R^{\delta}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R^{\quad \enspace \delta}_{\lambda \mu \rho}+\nabla_{\lambda}R^{\quad \enspace \delta}_{\mu \nu \rho} = 0

[Remember Riemann Tensor is pair symmetric]

Perform a similar swapping of indices, as in the last step, on the second and third term too, to get:
\implies \nabla_{\mu}R^{\delta}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R^{\delta}_{\enspace \rho \mu \lambda}+\nabla_{\lambda}R^{\delta}_{\enspace \rho \nu \mu} = 0

Operate g^{\mu}_{\enspace \delta}=\delta^\mu_\delta on the above equation,

[Note that g is covariantly constant and can be taken inside the covariant derivative]

g^{\mu}_{\enspace \delta}\left(\nabla_{\mu}R^{\delta}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R^{\delta}_{\enspace \rho \mu \lambda}+\nabla_{\lambda}R^{\delta}_{\enspace \rho \nu \mu}\right) = 0

\implies \nabla_{\mu}R^{\mu}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R^{\mu}_{\enspace \rho \mu \lambda}+\nabla_{\lambda}R^{\mu}_{\enspace \rho \nu \mu} = 0

\implies \nabla_{\mu}R^{\mu}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R_{\rho \lambda}-\nabla_{\lambda}R^{\mu}_{\enspace \rho \mu \nu} = 0

\implies \nabla_{\mu}R^{\mu}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R_{\rho \lambda}-\nabla_{\lambda}R_{\rho \nu} = 0

Now, operate g^{\nu \rho} on the above,

\implies g^{\nu \rho}\left(\nabla_{\mu}R^{\mu}_{\enspace \rho \lambda \nu}+\nabla_{\nu}R_{\rho \lambda}-\nabla_{\lambda}R_{\rho \nu}\right) = 0

\implies \nabla_{\mu}R^{\mu}_{\enspace \lambda}+\nabla_{\nu}R_{\enspace \lambda}^{\nu}-\nabla_{\lambda}R = 0

Since, \nabla_{\mu}R^{\mu}_{\enspace \lambda}=\nabla^{\mu}R_{\mu \lambda}

\implies \nabla^{\mu}R_{\mu \lambda}+\nabla^{\nu}R_{\nu \lambda}-\nabla_{\lambda}R = 0

Since, \mu and \nu are dummy indices, therefore, we can write

\implies 2\nabla^{\mu}R_{\mu \lambda}-\nabla_{\lambda}R = 0

\implies 2\nabla^{\mu}R_{\mu \lambda}-g_{\mu \lambda}\nabla^{\mu}R = 0

\implies \nabla^{\mu} \left( R_{\mu \lambda}-\frac{g_{\mu \lambda}}{2}R \right)=0

\implies \boxed{\nabla^{\mu}G_{\mu \lambda}=0}

Hence Proved.


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