Weyl neutrinos in plane symmetric spacetimes

Abstract

We investigate complex quaternion-valued exterior differential forms over 4-dimensional Lorentzian spacetimes and explore Weyl spinor fields as minimal left ideals within the complex quaternion algebra. The variational derivation of the coupled Einstein–Weyl equations from an action is presented, and the resulting field equations for both first and second order variations are derived and simplified. Exact plane symmetric solutions of the Einstein–Weyl equations are discussed, and two families of exact solutions describing left-moving and right-moving neutrino plane waves are provided. The study highlights the significance of adjusting a quartic self-coupling of the Weyl spinor in the action to ensure the equivalence of the field equations.

Appendix

We start with the identity ${}^*{\bar{e}} = \frac{1}{6} e^{*} \wedge e \wedge e^{*},$ that implies

$$\begin{aligned} \nabla ^*{\bar{e}}= & {} \frac{i}{6}(\nabla e^* \wedge e \wedge e^* - e^* \wedge \nabla e \wedge e^* + e^* \wedge e \wedge \nabla e^*) \\= & {} \frac{i}{6}(T^*\wedge e \wedge e^*- e^* \wedge T \wedge e^* +e^* \wedge e \wedge T^*)\\= & {} \frac{i}{6}{\mathcal {H}}(2T^*\wedge e \wedge e^* - e^*\wedge T \wedge e^*). \end{aligned}$$

Then we use the variational field equations

$$\begin{aligned} 2T= 2 {\mathcal {A}}(h e^{*} \wedge e) - e \wedge {\bar{h}} \wedge e \end{aligned}$$

where $h = -\frac{1}{6} \xi \xi ^{\dagger }$. Then

$$\begin{aligned} \nabla {}^*{\bar{e}}= & {} \frac{i}{12}{\mathcal {H}}\Big ( 2(h^*e \wedge e^* + e^*\wedge eh^* - e^*h\wedge e^*)\wedge e \wedge e^* \\{} & {} \quad -\, e^* \wedge (he^*\wedge e + e\wedge e^*h-eh^*\wedge e) \wedge e^* \Big )\\= & {} \frac{i}{12}{\mathcal {H}}\Big (2h^*\wedge e \wedge e^*\wedge e \wedge e^*+3e^* \wedge eh^*\wedge e \wedge e^* - 4e^*\wedge he^* \wedge e \wedge e^* \Big )\\= & {} \frac{i}{12}{\mathcal {H}}\Big ( 2(h^*e\wedge e^* + e^*\wedge eh^*)\wedge e \wedge e^* + e^* \wedge eh^* \wedge e \wedge e^* - 4e^*\wedge h e^*\wedge e \wedge e^* \Big ). \end{aligned}$$

Now we use the identities

$$\begin{aligned} (h^*e \wedge e^*)^{\dagger }=(e\wedge e^*)^{\dagger }{\bar{h}}=-{\bar{e}}\wedge e^{\dagger }h^* = -e^* \wedge eh^*, \end{aligned}$$

(65)

and

$$\begin{aligned} (e^* \wedge he^*)^{\dagger }=-(he^*)^{\dagger }\wedge {\bar{e}}=-{\bar{e}}h \wedge {\bar{e}}=-e^* \wedge he^*, \end{aligned}$$

(66)

to simplify the above expression:

$$\begin{aligned} \nabla ^*{\bar{e}}=\frac{i}{12}{\mathcal {H}}\Big ( 4{\mathcal {A}}(h^*e\wedge e^*-e^*\wedge eh^*)\wedge e \wedge e^* + e^* \wedge eh^* \wedge e \wedge e^*\Big ). \end{aligned}$$

(67)

The first term inside the big parantheses vanishes identically. The remaining term simplifies:

$$\begin{aligned} \nabla ^*{\bar{e}}= & {} \frac{i}{12}{\mathcal {H}}(e^* \wedge eh^* \wedge e \wedge e^*) \nonumber \\= & {} \frac{i}{12}(e^* \wedge eh^* \wedge e \wedge e^*)\nonumber \\= & {} -\frac{i}{72}(e^* \wedge e\xi )\wedge (\xi ^{\dagger }e \wedge e^*) \nonumber \\= & {} \frac{i}{72}(e^* \wedge e\xi )\wedge (e^* \wedge e\xi )^{\dagger }. \end{aligned}$$

(68)

Now let us write $\xi = \alpha U^1+\beta U^2$ for some complex functions $\alpha$ and $\beta$. After a long straightforward calculation one reaches, on-shell, the equality

$$\begin{aligned} \nabla ^*{\bar{e}}=0. \end{aligned}$$

(69)