Derivation of the `differential Rosenfeld relation' In Ref.1, they mentioned about algebraic Rosenfeld relation
\begin{equation}\boxed{
 \nabla_\mu S^\mu{}_{\nu\lambda} = \theta_{[\lambda\nu]}}
\end{equation}
where
\begin{equation}
 S^{\mu\nu}{}_{\nu\lambda} =  S^\mu{}_{[\nu\lambda]}
\end{equation}
is a spin tensor. This relation can be derived from the symmetric of the metrical tress-energy tensor
$$
T^{\alpha\beta} =  \theta^{\alpha\beta} +  \nabla_\mu \Big( S^{\mu{\alpha\beta}}+S^{{\alpha\beta}\mu}-S^{\beta\mu\alpha}   \Big)
$$ 
But there is also a differential Rosenfeld relation
$$\boxed{\boxed{
\nabla_\alpha \theta^{\alpha\beta} = - R^\beta{}_{\lambda\alpha\mu} S^{\lambda\alpha\mu}}}\;.
$$
The question is: How to obtained it? I have tried using also the algebraic Rosenfeld relation but the terms seem to not recombined properly to the above form. 
References


*

*L.B. Szabados, On canonical pseudotensors, Sparling's form and Noether currents, Class. Quantum Gravity 9 (1992) 2521. The preprint pdf file is available here.

 A: This can be done by the help of the first Bianchi identity:
Since this tensor is divergence-free we have 
\begin{eqnarray}
0=\nabla_\alpha T^{\alpha\beta} &=& \nabla_\alpha \theta^{\alpha\beta} + \nabla_\alpha \nabla_\mu \Big( S^{\mu{\alpha\beta}}+S^{{\alpha\beta}\mu}-S^{\beta\mu\alpha}   \Big)\;,\\
    &=&\nabla_\alpha \theta^{\alpha\beta} + \nabla_\alpha \nabla_\mu B^{\mu{\alpha\beta}}\;,\\
    &=&\nabla_\alpha \theta^{\alpha\beta} + \nabla_{[\alpha} \nabla_{\mu]} B^{\mu{\alpha\beta}}\;,\\
    &=&\nabla_\alpha \theta^{\alpha\beta} + \frac 1 2 R^\mu{}_{\lambda\alpha\mu}B^{\lambda{\alpha\beta}} +\frac 1 2 R^\alpha{}_{\lambda\alpha\mu}B^{\mu\lambda\beta} +\frac 1 2 R^\beta{}_{\lambda\alpha\mu}B^{\mu\alpha\lambda}\;,\\
    &=&\nabla_\alpha \theta^{\alpha\beta} -\overbrace{ \frac 1 2 R_{\lambda\alpha}B^{\lambda{\alpha\beta}}}^0 +\overbrace{ \frac 1 2 R_{\lambda\mu}B^{\mu\lambda\beta}}^0 +\frac 1 2 R^\beta{}_{\lambda\alpha\mu}B^{\mu\alpha\lambda}\;,\\
    &=&\nabla_\alpha \theta^{\alpha\beta} +\frac 1 2 R^\beta{}_{\lambda\alpha\mu}B^{\mu\alpha\lambda}\;,\\
    &=&\nabla_\alpha \theta^{\alpha\beta} +\frac 1 2 R^\beta{}_{\lambda\alpha\mu} \Big(  S^{\mu\alpha\lambda}+S^{\alpha\lambda\mu}-S^{\lambda\mu\alpha}   \Big)\;,\\
    &=&\nabla_\alpha \theta^{\alpha\beta} +\frac 1 2 R^\beta{}_{\lambda\alpha\mu} \Big( ( S^{\mu\alpha\lambda}+S^{\alpha\lambda\mu}+S^{\lambda\mu\alpha})-2S^{\lambda\mu\alpha}   \Big)\;,\\
    &=&\nabla_\alpha \theta^{\alpha\beta} +\frac 1 2 R^\beta{}_{\lambda\alpha\mu}\Big(  -2S^{\lambda\mu\alpha} \Big)\;,\\
\therefore \nabla_\alpha \theta^{\alpha\beta} &=& - R^\beta{}_{\lambda\alpha\mu} S^{\lambda\alpha\mu}\;,
    \end{eqnarray}
where we have use the fact that $R^\beta{}_{\lambda\alpha\mu} \Big(  S^{\mu\alpha\lambda}+S^{\alpha\lambda\mu}+S^{\lambda\mu\alpha}\Big) \sim R^\beta{}_{\lambda\alpha\mu} S^{[\mu\alpha\lambda]}=R^\beta{}_{[\lambda\alpha\mu]}S^{\mu\alpha\lambda}=0$ (by the firstBianchi identity). Noe that $B^{\mu\nu\beta}$ is anti-symmetric in first two indices.
