Let us call the spacetime $M$ with a metric $g_{ab}$. There is a unit spacelike vector field $\eta^a$ orthogonal to a hypersurface. So that we can define the so-called gravitational electric and magnetic fields using Weyl tensor:

$E_{ab}=C_{ambn}\eta^m\eta^n$, $B_{ab}=*C_{ambn}\eta^m\eta^n$

where $*C_{ambn}=\epsilon_{ampq}C^{pq}_{\phantom{dd}bn}$ with $\epsilon_{ampq}$ the alternating tensor. This definition is not the standard one, but it does not matter in my question. Plus, I should not call $E_{ab}, B_{ab}$ gravitational electric, magnetic fields because $\eta^a$ is not timelike. But let me use these names to represent the two fields. This is just nomenclature.

The Weyl tensor satisfies the following relation:


I want to show that

$D_{[a}E_{b]c}=0,\quad D_{[a}B_{b]c}=0$

Here, $D_a$ is the covariant derivative compatible with the induced metric $h_{ab}=g_{ab}-\eta_a\eta_b$.

It is easy to prove $D_{[a}E_{b]c}=0$, by simply contracting $\eta^q\eta^c$ with both sides of the equation of Weyl tensor and obtaining:


and then projecting this to the hypersurface to get the result.

But it is not so easy to get the 2nd result: $D_{[a}B_{b]c}=0$. One possible method is to contract $\eta_f\eta^b\epsilon^{cdef}$, then you will get the following complicated expression:


The last bracket will disappear after projected to the hypersurface, but the remaining terms seem to survive. Do you have any idea of how to solve this?


  • $\begingroup$ Isn't $\eta$ a timelike vector field? $\endgroup$ – Muphrid Nov 11 '15 at 5:42
  • $\begingroup$ @Muphrid Sorry, $\eta$ is spacelike. I should not call $E_{ab}, B_{ab}$ gravitational electric, magnetic fields, strickly speaking, but I just want use the names to represent $E_{ab}, B_{ab}$. $\endgroup$ – Drake Marquis Nov 11 '15 at 12:52

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