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Take a time 4-dimensional orinted Lorentzian manifold $(M,g)$.

A spacetime is called Strongly Causal at point $p$ if and only if for every neighbourhood $U$ of the point $p$ there exists a neighbourhood $V \subset U$ such that $V$ is causally convex.

Take an arbitrary unit timelike section $u$ of the tangent bundle $TM$: $$g^{ab}u_a u_b = -1$$

A spacetime is called Purely Electric at point $p$ if and only if there exists a neighbourhood around $p$ over which the Weyl tensor $C$ satisfies: $$u_a g^{ab} C_{bc[de} u_{f]}=0$$

where antisymmetrization over the indices is understood.

My question is that:

if a Strongly Causal point is necessarily a Purely Electric one.

If not, would one please give a counter example?

PS: one can equivalently define in 3+1 dimensions Pure Electricity by introduction of the tensor: $$H_{ab} = \frac{1}{2} \epsilon_{acef} {C^{ef}}_{bd} u^c u^d$$

as vanishing of the magnetic part of the Weyl tensor $C_{-}$:

$${(C_{-})^{ab}}_{cd} = 2 \epsilon^{abef} u_{e} u_{[c} H_{d]f} + 2 \epsilon_{cdef} u^e u^{[a} H^{b]f}=0 $$

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These are quite orthogonal conditions. To start, one is a global condition, while the other is a local one.

  1. Every point has a small enough neighborhood that is strongly causal (even globally hyperbolic). Take a small enough neighborhood where the metric is very close to flat, and take the domain of dependence of a sufficiently smaller spatial disk.
  2. A generic Weyl tensor is not purely electric (even in the weaker sense if one allows the $u$ vector to be null) and moreover the subset where this (weaker) condition is violated is open. To see that, picking an auxiliary Riemannian metric at the violating point, the violation of the weaker condition is equivalent to $u^b C_{bc[de} u_{f]} =: \Delta(u) \ne 0$ for all causal $u$ with Riemann norm $\|u\| = 1$. The set of such vectors $u$ is compact, so $0 < \delta \le \|\Delta(u)\|$ for some $\delta$. Hence a sufficiently small perturbation of $C_{bcde}$ (controlled by $\delta$) will not be able to cancel the violation $\Delta(u)$. So the violation will persist in some neighborhood of a given violating point.

So for a counter-example in one direction, it suffices to take small strongly causal neighborhood around a violating point in a generic spacetime. For a counter-example in the other direction, take flat spacetime (which is purely electric in either sense) and do the usual cutting and gluing to violate the strong causality property.

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  • $\begingroup$ It's generally acceptable. But I cannot understand the passage that tries to prove that the non-vanishing subset is open. Specially where you choose an auxiliary Riemannian metric. Shouldn't it be a Lorenzian metric? Since the set of $u^a$ for which the norm is unit, is gonna be quite different. $\endgroup$
    – Bastam Tajik
    Commented Jan 12 at 16:26
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    $\begingroup$ @BastamTajik The condition on the Weyl tensor is homogeneous in $u$, so it is equivalent to state it for all timelike $u$ (or causal $u$ in the weaker version). In turn, it is equivalent to state the weaker condition for only those causal $u$ that have $\|u\|=1$ w.r.t some Riemannian metric, which is now conveniently a compact set. You'll find that the choice of the Riemannian metric can be arbitrary, only the compactness of its unit sphere is used (so that its intersection with the closed light cone is also compact). $\endgroup$
    – Igor Khavkine
    Commented Jan 12 at 16:41

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