- #1

PhyAmateur

- 105

- 2

I was understanding the theory until he stated with the calculations:

He said that the Supersymmetry transformation (21.49) in his book is:

$$\delta \psi_\mu^i=(\partial_\mu +1/4 \gamma^{ab}\omega_{\mu ab})\epsilon^i -1/8\sqrt{2}\kappa \gamma^{ab}F_{ab}\epsilon^{ij} \gamma_\mu \epsilon_j$$

He then started "It is convenient to work with the chiral projections of the two Majo-rana spinors. We thus use the up/down position of the R-symmetry indices, now denoted by **(A, B=1, 2)**, to specify the chirality. Thus, for the SUSY transformation parameters,

we have"

$$\gamma_* \epsilon^A= P_L \epsilon^A = \epsilon^A$$ and $$\gamma_* \epsilon_A= - P_R \epsilon_AA = - \epsilon_A$$ where $$\gamma_*$$ is the usual $$\gamma_5$$ and $$P_L, P_R$$ are projection operators to define chiral parts.

So back to the first equation that I wrote here: He substituted it with:

$$ \delta \psi_{tA} = \partial_t \epsilon _A +1/2 e^{2U} \partial_i U\gamma^i \gamma^0 \epsilon_A -1/4 \sqrt{2}\kappa e^u \partial_i A_t \gamma^i \epsilon_{AB} \epsilon^b =0$$

It is specifically the substitutions that I could not follow, where did the gamma matrices in the last equation come from?

Concerning the spin conection I found them to be: $$\omega^{0i} = e^U \partial_iU e^0$$ and $$\omega ^{ij} = -dx^i\partial_jU+dx^j\partial_iU$$ and they each have 2 indices while in the supersymmetry transformation equation the omega has 3 indices. I know I am missing something and I hope you can help me understand the substitution better.