Type D vacuum solutions: a new intrinsic approach

Abstract

We present a new approach to the intrinsic properties of the type D vacuum solutions based on the invariant symmetries that these spacetimes admit. By using tensorial formalism and without explicitly integrating the field equations, we offer a new proof that the upper bound of covariant derivatives of the Riemann tensor required for a Cartan–Karlhede classification is two. Moreover we show that, except for the Ehlers–Kundt’s C-metrics, the Riemann derivatives depend on the first order ones, and for the C-metrics they depend on the first order derivatives and on a second order constant invariant. In our analysis the existence of an invariant complex Killing vector plays a central role. It also allows us to easily obtain and to geometrically interpret several known relations. We apply to the vacuum case the intrinsic classification of the type D spacetimes based on the first order differential properties of the 2 + 2 Weyl principal structure, and we show that only six classes are compatible. We define several natural and suitable subclasses and present an operational algorithm to detect them.

Appendices

Appendix 1: Proof of Lemmas 1, 3 and 4

1.1 Proof of Lemma 1

Suppose that $(Z,Z)=0$. If we differentiate this condition and take into account (19) we obtain:

$$\begin{aligned} 2 w^{\frac{2}{3}} Z = \left[ \bar{w}^{\frac{1}{3}} (Z, \bar{Z}) - 2 m\right] \varPi (Z). \end{aligned}$$

(29)

From here we have $Z \wedge \varPi (Z) = 0$, that is, $Z$ is a null vector which lies on the time-like principal plane: $Z = \varPi (Z)$. Then (12) implies $Z \wedge \bar{Z} =0$ and consequently $(Z, \bar{Z})=0$, and (29) is equivalent to:

$$\begin{aligned} Z = \varPi (Z), \qquad m + w^{\frac{2}{3}} = 0. \end{aligned}$$

(30)

Now if we differentiate the first constraint in (30) (or equivalently $i(Z)\mathcal{U} = i(Z)\bar{\mathcal{U}}$), and take into account (19) and $Z \wedge \bar{Z} =0$, we arrive at $m = w^{\frac{2}{3}}$ which is not compatible with the second constraint in (30).

1.2 Proof of Lemma 3

Condition $v(Z,Z)=0$ equivalently states $(Z,Z) + \varPi (Z,Z) =0$. If we differentiate this scalar condition and we take into account the expression of the covariant derivatives (19) of $Z$ and (10) of $\mathcal{U}$, we obtain the following expression for $m$:

$$\begin{aligned} m = - \frac{1}{2} \bar{w}^{\frac{1}{3}} (Z, \bar{Z}) - \nu , \quad \nu \equiv w^{\frac{2}{3}} + \frac{1}{2} w^{\frac{1}{3}} (Z,Z). \end{aligned}$$

(31)

Then Eq. (20) becomes:

$$\begin{aligned} \nu h(\bar{Z}) + \bar{\nu } h(Z) = 0. \end{aligned}$$

(32)

On the other hand, if we differentiate (31) and make use of (10) and (19) we obtain a new scalar condition:

$$\begin{aligned} (Z, \bar{Z}) + 2m \bar{w}^{-\frac{1}{3}} + 2 \bar{m} w^{-\frac{1}{3}} = 0. \end{aligned}$$

(33)

Finally, if we differentiate this equation and we take into account (10) and (19) we arrive at:

$$\begin{aligned} \nu v(\bar{Z}) + \bar{\nu } v(Z) = 0. \end{aligned}$$

(34)

Constraints (32) and (34) imply $\nu \bar{Z} + \bar{\nu } Z = 0$. Consequently $Z \wedge \bar{Z}= 0$ or $\nu =0$. This last condition and (31) and (33) lead to $(Z, \bar{Z})=0$. This condition, hypothesis $(Z,Z) + \varPi (Z,Z) =0$, and identity (12) imply $\varPi (Z) = Z$, which also implies, with (12), $Z \wedge \bar{Z} = 0$.

Note that (31) gives the scalar $m$ in terms of 0th- and 1st-order Riemann derivatives for the type D vacuum solutions satisfying $v(Z,Z)=0$. Thus, we could state for them a specific theorem similar to Theorems 1 and 2. Nevertheless we prefer use this lemma 3 and consider this case as included in theorem 1.

1.3 Proof of Lemma 4

From the hypothesis $Z \wedge \bar{Z}\not =0$ we have necessarily $Z \wedge \varPi (Z)\not =0$. Then, condition $Z \wedge \varPi (\bar{Z})=0$ implies that (20) is equivalent to:

$$\begin{aligned} \bar{\mu } Z = \mu \varPi (\bar{Z}), \qquad \bar{\nu } Z = \nu \varPi (\bar{Z}). \end{aligned}$$

(35)

On the other hand, if we differentiate $Z \wedge \varPi (\bar{Z})=0$ and we make use of (10) and (19) we obtain a tensorial equation. Its trace leads to:

$$\begin{aligned} (\bar{m} + \bar{\mu }) Z = (m + \mu ) \varPi (\bar{Z}), \quad (\bar{w}^{\frac{2}{3}} + 2 \bar{\nu }) Z = ( w^{\frac{2}{3}} + 2 \nu ) \varPi (\bar{Z}), \end{aligned}$$

(36)

where in obtaining the second equation we have used the fact that $Z \wedge \varPi (\bar{Z})=0$ implies $\varPi (Z, \bar{Z}) Z = (Z,Z) \varPi (\bar{Z})$ and $\varPi (Z, \bar{Z}) \varPi (\bar{Z}) = (\bar{Z},\bar{Z}) Z$. If we again make use of these relations, from equations (35) (36) we obtain:

$$\begin{aligned} \bar{m} Z&= m \varPi (\bar{Z}), \qquad \bar{w}^{\frac{2}{3}} Z = w^{\frac{2}{3}} \varPi (\bar{Z}),\end{aligned}$$

(37)

$$\begin{aligned} (Z, \bar{Z})V&= 0, \quad \varPi (Z, \bar{Z}) V = 0, \quad V \equiv w^{\frac{1}{3}} Z - \bar{w}^{\frac{1}{3}} \varPi (\bar{Z}). \end{aligned}$$

(38)

Under the hypothesis $Z \wedge \bar{Z}\not =0$, at least one of the scalars $(Z, \bar{Z})$ and $\varPi (Z, \bar{Z})$ does not vanish. Consequently (38) implies $V=0$. This constraint and (37) lead to $\bar{m} = m$, $\bar{w} =w$ and $\varPi (\bar{Z}) = Z$. Moreover, the solution is a strict C-metric because it has real Weyl eigenvalues.

Appendix 2: Proof of Lemma 6

The conditions involved in Lemma 6 can be stated by using the projections $v(\chi )$ and $h(\chi )$ of complex vector $\chi = \frac{1}{2} [\varPhi + \mathrm i \varPsi ]$. In terms of $\{w, \mathcal{U}, Z\}$ these projections take the expression:

$$\begin{aligned} 2 v(\chi ) = w^{\frac{1}{3}}\left[ i(Z)(\mathcal{U}) + i(Z)(\bar{\mathcal{U}})\right] , \qquad 2 h(\chi ) = w^{\frac{1}{3}}\left[ i(Z)(\mathcal{U}) - i(Z)(\bar{\mathcal{U}})\right] . \end{aligned}$$

(39)

Suppose that $v(\varPhi ) = 0$, that is, $v(\chi ) = - v(\bar{\chi })$. If we calculate the covariant derivative of this equation and we take into account expressions (39) and derivatives (10) and (19) we obtain a 2-tensorial equation $E_{\alpha \beta }= 0$. The total projection of this equation on the time-like plane, $v^{\lambda \alpha } v^{\mu \beta } E_{\alpha \beta }= 0$, leads to $v(\varPsi ) \otimes v(\varPsi ) = 0$, and so $v(\varPsi ) = 0$. Consequently point (i) is proven.

Suppose now that $v(\varPsi ) = 0$, that is, $v(\chi ) = v(\bar{\chi })$. If we calculate the covariant derivative of this equation and we take into account expressions (39) and derivatives (10) and (19) we obtain a 2-tensorial equation $F_{\alpha \beta }= 0$. The mixed projection of this equation on the time-like and space-like planes, $v^{\lambda \alpha } h^{\mu \beta } F_{\alpha \beta }= 0$, leads to $v(\varPhi ) \otimes h(\varPsi ) = 0$, and so either $v(\varPhi ) = 0$ or $h(\varPsi ) = 0$. Consequently point (iii) is proven.

The proof of points (ii) and (iv) of Lemma 6 is similar to the proof of points (i) and (iii) by exchanging $v$ for $h$.