广义相对论|PH30101 General relativity代写

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the corresponding null geodesics are
\begin{aligned} &\gamma_{X}: x^{A A^{\prime}} \equiv x_{0}^{A A^{\prime}}+\lambda \bar{X}^{A} X^{A^{\prime}} \ &\gamma_{Y}: x^{A A^{\prime}} \equiv x_{1}^{A A^{\prime}}+\mu \bar{Y}^{A} Y^{A^{\prime}} \end{aligned}
If these intersect at some point, say $x_{2}$, one finds
$$x_{2}^{A A^{\prime}}=x_{0}^{A A^{\prime}}+\lambda \bar{X}^{\mathcal{A}} X^{A^{\prime}}=x_{1}^{A A^{\prime}}+\mu \bar{Y}^{A} Y^{A^{\prime}}$$
where $\lambda \mu \in R$. Hence
$$x_{2}^{A A^{\prime}} \bar{Y}{A} X{A^{\prime}}=x_{0}^{A A^{\prime}} \bar{Y}{A} X{A^{\prime}}=x_{1}^{A A^{\prime}} \bar{Y}{A} X{A^{\prime}}$$

PH30101 COURSE NOTES ：

This enables one to define the spinor field
$$p^{A} \equiv p^{A A^{\prime} B^{\prime}} \pi_{A^{\prime}} \pi_{B^{\prime}}$$
and the patching function
$$f^{A} \equiv p^{A} g\left(p_{B} \omega^{B}, \pi_{B^{\prime}}\right)$$
and the function
$$F\left(x^{a}, \pi_{A^{\prime}}\right) \equiv g\left(i p_{A} x^{A C^{\prime}} \pi_{C^{\prime}}, \pi_{A^{\prime}}\right)$$

广义相对论 General Relativity PHYS97026

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Within this framework (as well as in chapter three) we need to know that $\psi_{A B C D}$ has two scalar invariants:
$$\begin{gathered} I \equiv \psi_{A B C D} \psi^{A B C D} \ J \equiv \psi_{A B}^{C D} \psi_{C D}^{E F} \psi_{E F} A B \end{gathered}$$

Type-II space-times are such that $I^{3}=6 J^{2}$, while in type-III space-times $I=I=$
0 . Moreover, type-D space-times are characterized by the condition
$$\psi_{P Q R\left(A \psi_{B C}\right.}^{P Q} \psi_{D E F)}^{R}=0$$
while in type-N space-times
$$\psi_{(A B}{ }^{E F} \psi_{C D) E F}=0$$

PHYS97026 COURSE NOTES ：

$$\left[\nabla^{a} \nabla^{d}+\nabla^{a} \omega^{d}-\omega^{a} \omega^{d}\right] C_{a b c d}=0$$
We now re-express $\nabla^{a t} \omega^{d}$
$$\nabla^{a} \omega^{d}=\omega^{a} \omega^{d}+\frac{1}{8} T g^{a d}-\frac{1}{2} R^{a d}$$$$\left[\nabla^{a} \nabla^{d}-\frac{1}{2} R^{a d}\right] C_{a b c d}=0$$
This calculation only proves that the vanishing of the Bach tensor, defined as
$$B_{b c} \equiv \nabla^{a} \nabla^{d} C_{a b c d}-\frac{1}{2} R^{a d} C_{a b e d}$$