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.


PhD researcher at Friedrich-Schiller University Jena, Germany. I'm a physicist specializing in theoretical, computational and experimental condensed matter physics. I like to develop Physics related apps and softwares from time to time. Can code in most of the popular languages. Like to share my knowledge in Physics and applications using this Blog and a YouTube channel.
[wpedon id="7041" align="center"]

Leave a Reply

Your email address will not be published. Required fields are marked *