Solving Spinorial Maxwell's Equations with Wald

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SUMMARY

This discussion focuses on solving Spinorial Maxwell's Equations as presented in Chapter 13 of Wald's text. The primary equation of interest is the relationship between the antisymmetric tensor ##F_{ab}## and the spinorial tensor ##F_{AA' BB'}##, specifically demonstrating that ##\partial^a F_{ab} = 0## and ##\partial_{[a} F_{bc]} = 0## if and only if the tensor ##\phi^{AB}## satisfies the condition ##\partial_{A_1' A_1} \phi^{A_1, \dots, A_n} = 0##. The discussion also highlights the need to prove that ##\partial_{AA'} \epsilon_{BC} = 0##, which follows from the definition of the derivative of the epsilon tensor and the constancy of its components in curved spacetime.

PREREQUISITES
  • Understanding of antisymmetric tensors and their properties
  • Familiarity with spinorial notation and transformations
  • Knowledge of Maxwell's equations in the context of differential geometry
  • Basic concepts of covariant derivatives and connections in curved spacetime
NEXT STEPS
  • Study the derivation of the properties of antisymmetric tensors in Wald's text
  • Learn about the implications of covariant derivatives on spinorial fields
  • Explore the relationship between spinorial equations and classical Maxwell's equations
  • Investigate the unique connection theorem in the context of curved spacetime
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ergospherical
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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 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##...?
 
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Yes, the components are constant, so all the derivatives will be zero. In curved space-time you need the theorem that there is a unique connection with the given properties, one of which is that the ##\epsilon## has zero covariant derivative.
 
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