# Killing spinor equation

+ 3 like - 0 dislike
1033 views

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
edited May 30, 2015
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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsO$\varnothing$erflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.