- #1

ergospherical

Gold Member

- 488

- 620

I'm trying to figure out how to do these sorts of calculations but I'm having a lot of trouble figuring out where to start. Take problem

F_{AA' BB'} = \phi_{AB} \bar{\epsilon}_{A'B'} + \bar{\phi}_{A'B'} \epsilon_{AB}

\end{align*}to show that ##\partial^a F_{ab} = 0## and ##\partial_{[a} F_{bc]} = 0## if and only if ##\phi^{AB}## satisfies ##\partial_{A_1' A_1} \phi^{A_1, \dots, A_n} = 0##. I've really not much idea where to start; the spinor equivalent of the first Maxwell equation should be\begin{align*}

\partial^{AA'} F_{AA'BB'} = \phi_{AB} \partial^{AA'} \bar{\epsilon}_{A'B'} + \bar{\epsilon}_{A'B'} \partial^{AA'} \phi_{AB} + \bar{\phi}_{A'B'} \partial^{AA'} \epsilon_{AB} + \epsilon_{AB} \partial^{AA'} \bar{\phi}_{A'B'} = 0

\end{align*}In the text it's mentioned that ##\partial_{AA'} \epsilon_{BC} = 0##, but no proof is given. Maybe as a starter, how can I show that it follows from the definition ##\partial_{\Lambda \Lambda'} \epsilon_{\Sigma \Omega} = \sum_{\mu} {\sigma^{\mu}}_{\Lambda \Lambda'} \dfrac{\partial \epsilon_{\Sigma \Omega}}{\partial x^{\mu}}##? I reckon its simply because ##\epsilon^{\Sigma \Omega} = o^\Sigma \iota^\Omega - \iota^\Sigma o^\Omega## is independent of the spacetime coordinates, with ##\{o, \iota\}## being a fixed basis of ##W##...?

**3**) of Chapter 13 of Wald, i.e. given that a real antisymmetric tensor ##F_{ab}##, corresponding to the spinorial tensor ##F_{AA' BB'}## by the map ##{\sigma^a}_{AA'}##, can be written as\begin{align*}F_{AA' BB'} = \phi_{AB} \bar{\epsilon}_{A'B'} + \bar{\phi}_{A'B'} \epsilon_{AB}

\end{align*}to show that ##\partial^a F_{ab} = 0## and ##\partial_{[a} F_{bc]} = 0## if and only if ##\phi^{AB}## satisfies ##\partial_{A_1' A_1} \phi^{A_1, \dots, A_n} = 0##. I've really not much idea where to start; the spinor equivalent of the first Maxwell equation should be\begin{align*}

\partial^{AA'} F_{AA'BB'} = \phi_{AB} \partial^{AA'} \bar{\epsilon}_{A'B'} + \bar{\epsilon}_{A'B'} \partial^{AA'} \phi_{AB} + \bar{\phi}_{A'B'} \partial^{AA'} \epsilon_{AB} + \epsilon_{AB} \partial^{AA'} \bar{\phi}_{A'B'} = 0

\end{align*}In the text it's mentioned that ##\partial_{AA'} \epsilon_{BC} = 0##, but no proof is given. Maybe as a starter, how can I show that it follows from the definition ##\partial_{\Lambda \Lambda'} \epsilon_{\Sigma \Omega} = \sum_{\mu} {\sigma^{\mu}}_{\Lambda \Lambda'} \dfrac{\partial \epsilon_{\Sigma \Omega}}{\partial x^{\mu}}##? I reckon its simply because ##\epsilon^{\Sigma \Omega} = o^\Sigma \iota^\Omega - \iota^\Sigma o^\Omega## is independent of the spacetime coordinates, with ##\{o, \iota\}## being a fixed basis of ##W##...?

Last edited: