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  Killing spinor equation

+ 3 like - 0 dislike

The Supersymmetry transformation is:

$$\delta \psi_\mu^i=(\partial_\mu +1/4 \gamma^{ab}\omega_{\mu ab})\epsilon^i -1/8\sqrt{2}\kappa \gamma^{ab}F_{ab}\epsilon^{ij} \gamma_\mu \epsilon_j$$ For the time part, we substitute $\mu$ by $t$ so equation (2) is:

$$ \delta \psi_{tA} = \partial_t \epsilon _A + \frac{1}{4} \gamma^{ab} \omega_{tab} \epsilon_A - \frac{1}{8} \sqrt{2} \kappa F_{ab} \gamma^{ab} \gamma_t \epsilon_{AB} \epsilon^B =0$$

Automatically since the spin connections I found to be:

$\omega^{0i} = e^U \partial_iU e^0$ and $\omega ^{ij} = -dx^i\partial_jU+dx^j\partial_iU$

Implies that the ${a, b}$ indices in equation (2) should either be ${0,i}$ or ${i, j}$ $$ \delta \psi_{tA} = \partial_t \epsilon _A + \frac{1}{4} \gamma^{0i} \omega_{t0i} \epsilon_A - \frac{1}{8} \sqrt{2} \kappa F_{0i} \gamma^{0i} \gamma_t \epsilon_{AB} \epsilon^B + \frac{1}{4} \gamma^{ij} \omega_{tij} \epsilon_A - \frac{1}{8} \sqrt{2} \kappa F_{ij} \gamma^{ij} \gamma_t \epsilon_{AB} \epsilon^B =0$$

From here, I am having difficulty in proceeding. Some papers like http://arxiv.org/abs/hep-th/0608139 see (4.3.27) separate the $\gamma^{0i}$ into $\gamma^{0}\gamma^{i}$.

This is the first time I encounter such problem, I have no idea how to proceed from here in order to find 2 conditions for ERN BH.

This post imported from StackExchange Physics at 2014-12-03 15:54 (UTC), posted by SE-user beyondtheory

asked Dec 3, 2014 in Theoretical Physics by beyondtheory (15 points) [ revision history ]
edited May 30, 2015 by Dilaton
This is a (admittedly very high level) homework question. Please check the homework policy. Specifically, why are you not sure how to proceed? What conceptual issues are you having?

This post imported from StackExchange Physics at 2014-12-03 15:54 (UTC), posted by SE-user Sean
@Sean if I continued from where I stopped, this would be: $$\partial_t \epsilon_A + \frac{1}{4} \gamma^{0i} e^U \partial_iUe^0 \epsilon_A + \frac{1}{4} \gamma^{ij}(-dx^i \partial_jU+dx^j \partial_iU )\epsilon_A -\frac{1}{8} \sqrt{2} \kappa (-\partial_i A_t) \gamma^{0i} \gamma_t \epsilon_{AB} \epsilon^B = 0$$ But the final answer written in the book (actually this is not a homework but an exercise I was following) is: $$\delta \psi_{tA} = \partial_t \epsilon _A +1/2 e^{2U} \partial_i U\gamma^i \gamma^0 \epsilon_A -1/4 \sqrt{2}\kappa e^u \partial_i A_t \gamma^i \epsilon_{AB} \epsilon^b =0$$

This post imported from StackExchange Physics at 2014-12-03 15:54 (UTC), posted by SE-user beyondtheory

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