2
$\begingroup$

On page 101 of Freedman and Van Proeyen's book on Supergravity they find the propagator of the gravitino, however I'm not sure how to work through the steps in (5.30), and hints or answers would be great help.

It begins with the equation of motion

$$\gamma^{\mu\nu\rho}\partial_\nu\Psi_\rho=J^\mu\tag{5.18}$$

and an ansatz solution of the form,

$$\Psi_\mu(x)=-\int d^DyS_{\mu\nu}(x-y)J^\nu(y).\tag{5.26}$$

Subbing this into the equation of motion gives,

$$i\gamma^{\mu\sigma\rho}p_\sigma S_{\rho\nu}(p)=i{\delta}{^\mu_\nu}-ip_\nu\Omega^\mu(p) \tag{5.28}$$

in momentum space where $\Omega^\mu$ is a pure gauge term and contains all the dependence on $p_\mu$ only. They then go on to produce an ansatz for this equation of the form,

$$\require{cancel}iS_{\rho\nu}(p)=A(p^2)\eta_\rho\nu\cancel{p}+B(p^2)\gamma_\rho\cancel{p}\gamma_\nu\tag{5.29}$$

which they then sub into the LHS of $(5.28)$ to obtain the following,

$$\begin{align} i\gamma^{\mu\sigma\rho}p_\sigma S_{\rho\nu}(p)&=A\gamma^{\mu\sigma}_\nu\cancel{p}p_\sigma+(D-2)B\gamma^{\mu\sigma}\cancel{p}\gamma_\nu p_\sigma\\ &=A(p^\mu\gamma^\sigma_\nu-p^\sigma\gamma^\mu_\nu)p_\sigma+(D-2)B(-p^\mu\gamma^\sigma+p^\sigma\gamma^\mu)\gamma_\nu p_\sigma+...\\ &=[A-(D-2)B](p^\mu\gamma^\sigma_\nu-p^\sigma\gamma^\mu_\nu)p_\sigma+(D-2)Bp^2\delta^\mu_\nu+... \end{align}\tag{5.30}$$

where the ... represents terms that are proportional to the vector $p_\nu$. I'm just unsure of how to go from the first to second and then to the third line of the above aligned equations.

$\endgroup$
  • 1
    $\begingroup$ have you tried gamma matrix identities introduced thus far? $\endgroup$ – Kosm Jun 3 at 15:23
  • $\begingroup$ I've used the one they highlight in that chapter but it's the slash that I'm struggling to manipulate into the correct form. $\endgroup$ – huntercallum Jun 3 at 15:46
  • 2
    $\begingroup$ Undo the slashes as $\gamma_\kappa p^\kappa$. Do the simpler version of their (3.21), with just one γ hitting on the right. The B term is even simpler: symmetrize and antisymmetric the two γ s involved. They really expect you to master the technique if you are to trust them. $\endgroup$ – Cosmas Zachos Jun 3 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.