Weyl type N solutions with null electromagnetic fields in the Einstein–Maxwell p-form theory

Abstract

We consider d-dimensional solutions to the electrovacuum Einstein–Maxwell equations with the Weyl tensor of type N and a null Maxwell $(p+1)$-form field. We prove that such spacetimes are necessarily aligned, i.e. the Weyl tensor of the corresponding spacetime and the electromagnetic field share the same aligned null direction (AND). Moreover, this AND is geodetic, shear-free, non-expanding and non-twisting and hence Einstein–Maxwell equations imply that Weyl type N spacetimes with a null Maxwell $(p+1)$-form field belong to the Kundt class. Moreover, these Kundt spacetimes are necessarily ${ CSI}$ and the $(p+1)$-form is ${ VSI}$. Finally, a general coordinate form of solutions and a reduction of the field equations are discussed.

Appendix A: Curvature invariants in Kundt spacetimes of Weyl and traceless Ricci type N

Let us study scalar curvature invariants in the Kundt subclass of Weyl type N spacetimes with the Ricci tensor of the form

$$\begin{aligned} R_{ab} = \tilde{\Lambda }g_{ab} + \eta k_a k_b, \end{aligned}$$

(A.1)

where $\tilde{\Lambda }$ is a constant, $\eta$ is a scalar and $\varvec{k}$ is a null vector. In general, a Kundt vector in a Ricci type II spacetime is necessarily a multiple WAND (see e.g. Proposition 2 of [18]). Thus, in our case, it must coincide with the common AND $\varvec{\ell }$ of the Weyl and the Ricci tensor. Hence, the spacetime is of Riemann type II (i.e. the Riemann tensor is of type II) with the Kundt AND $\varvec{\ell }$. In addition to this, all boost weight zero null frame components of the Riemann tensor are constant (depending on $\tilde{\Lambda }$).

If $\tilde{\Lambda }= 0$, the Riemann tensor is of type N and it is aligned with the Kundt vector $\varvec{\ell }$. Thus, according to Theorem 1 of [15] on characterization of VSI spacetimes, the corresponding spacetime is VSI.

If $\tilde{\Lambda }$ is non-vanishing, then the spacetime is at least ${ CSI}_0$, i.e. all curvature invariants constructed solely from the Riemann tensor (without incorporating its covariant derivatives) are constant. Indeed, any full contraction of a tensor $\varvec{T}$ is completely determined by its boost weight zero part $\varvec{T}_{(0)}$. Since in our case, the boost weight zero part $\varvec{R}_{(0)}$ of the Riemann tensor $\varvec{R}$ possesses only constant null frame components, also any full contraction of a tensor given by a series of tensor products of the Riemann tensor with itself or with the metric (which consists only of its boost weight zero part) is necessarily constant. See also Sect. 2.3 of [11].

In order to study higher-order invariants constructed using also covariant derivatives of $\varvec{R}$, we make use of the balanced scalar approach introduced in [14], see also [22]. First, let us define k-balancedness of a tensor.

Definition A.1

We say that a tensor $\varvec{T}$ is k-balanced if there exists a null vector $\varvec{\ell }$ such that $bo _{\varvec{\ell }}(\varvec{T}) < -k$ and for any of its null frame components $\eta$ with boost weight $b<-k$, the derivative $D^{-b-k} \eta$ is zero.

Now, let us prove the following result on k-balancedness of covariant derivatives $\nabla ^{(n)} \varvec{R}$ of the Riemann tensor.

Proposition A.2

In a Weyl type N Kundt spacetime with the Ricci tensor of the form (A.1), an arbitrary covariant derivative of the Riemann tensor is 1-balanced and thus it is VSI.

Proof

Let us stress out again that such a spacetime is necessarily aligned [10], and hence without loss of generality one can assume that $\varvec{k}= \varvec{\ell }$. Also, using the notation of the GHP formalism [8], the boost weight $(-2)$ component of the Ricci tensor reads $\omega ^{\prime }= \eta$.

Consider the Ricci decomposition of the Riemann tensor [7],

$$\begin{aligned} \varvec{R} = \varvec{G} + \varvec{E}+\varvec{C}, \end{aligned}$$

(A.2)

where $\varvec{G}$ is the so-called scalar part ($G_{abcd} \propto R g_{a[c}g_{d]b}$ ), $\varvec{E}$ is the semi-traceless part ($E_{abcd} \propto g_{a[c}S_{d]b}-g_{b[c}S_{d]a}$, $S_{ab}$ being the traceless part of the Ricci tensor) and the Weyl tensor $\varvec{C}$ is the fully traceless part of the Riemann tensor.

Since the Ricci scalar R is constant ($R \propto \tilde{\Lambda }$), we have $\nabla \varvec{G} = \varvec{0}$ and hence

$$\begin{aligned} \nabla \varvec{R} = \nabla \varvec{C} + \nabla \varvec{E}. \end{aligned}$$

(A.3)

For Weyl type N Kundt spacetimes with the Ricci tensor of the form (A.1), Eqs. (28)–(30) in [23] are still valid together with $D\omega '=0$ and thus the following generalizations of Lemmas 4.2 and 4.3 in [23] are also valid.³

Lemma A.3

In a Weyl type N Kundt spacetime with the Ricci tensor of the form (A.1), for a 1-balanced scalar $\eta$, scalars $L_{11} \eta$, $\tau _i \eta$, $L_{1i}\eta$, $\kappa '_i \eta$, $\rho '_{ij} \eta$, $\mathop {M}\limits ^{i}_{j1} \eta$, $\mathop {M}\limits ^{i}_{kl} \eta$ and $D \eta ,\ \delta _i \eta ,\ \bigtriangleup \eta$ are also 1-balanced scalars.

Lemma A.4

In a Weyl type N Kundt spacetime with the Ricci tensor of the form (A.1), a covariant derivative of a 1-balanced tensor is again a 1-balanced tensor.

Since the Weyl tensor and the traceless part of the Ricci tensor, $\eta \ell _a\ell _b$, as well as $\varvec{E}$ are 1-balanced, their arbitrary derivative is also 1-balanced and so is any covariant derivative of the Riemann tensor, i.e. $\nabla ^{(k)} \varvec{R}$ is 1-balanced for any $k \in \mathbb {N}$. $\square$

Thus, although the Riemann tensor is not VSI, its first covariant derivative is. In particular, this means that the only non-trivial scalar curvature invariants are precisely those constructed from the Riemann tensor itself. However, we already know that all these invariants are necessarily constant. Therefore, we immediately obtain the following result.

Corollary A.5

A Weyl type N Kundt spacetime with the Ricci tensor of the form (A.1) is ${ CSI}$.