8
$\begingroup$

Dirac equation for the massless fermions in curved spase time is $γ^ae^μ_aD_μΨ=0$, where $e^μ_a$ are the tetrads. I have to show that Dirac spinors obey the following equation: $$(−D_μD^μ+\frac{1}{4}R)Ψ=0\qquad(1)$$

where $R$ is the Ricci scalar.

I already know that $[D_\mu,D_\nu]A^\rho={{R_{\mu\nu}}^\rho}_\sigma A^\sigma$, but a key point is to know what $[D_\mu,D_\nu]\Psi$ is.

($D_μΨ=∂_μΨ+A^{ab}_μΣ_{ab}$ is the covariant derivative of the spinor field and $Σ_{ab}$ the Lorentz generators involving gamma matrices).

|cite|improve this question
$\endgroup$
  • 5
    $\begingroup$ The right way to solve it is to act with $D_\mu$ from the left on your Dirac GR equation again. Once you do so, you must realize that the new covariant derivative also acts on the tetrads which means that it effectively differentiates the metric and ultimately produces the Ricci scalar term. However, you must realize that the adjective "covariant" means that it has two new connection terms, one from the curved spacetime metric and one from the electromagnetic potential. The latter isn't needed at all for your desired derivation - the gauge field is just kept everywhere as a part of $D_\mu$. $\endgroup$ – Luboš Motl Jan 16 '13 at 10:09
  • $\begingroup$ Related is physics.stackexchange.com/questions/51269/… Are these really distinct? If not, might consider asking mods to merge them. $\endgroup$ – twistor59 Jan 16 '13 at 10:36
  • $\begingroup$ Thanks for the comment, but $D_\mu e^\nu_a$ is not zero? $\endgroup$ – Gauge Jan 16 '13 at 10:45
  • 1
    $\begingroup$ Aren't the tetrads covariantly constant provided you include both the Christoffel symbols and the spin connection? $\endgroup$ – twistor59 Jan 16 '13 at 11:03
  • 1
    $\begingroup$ I would say the cov derivative of the tetrad is something like $ \partial_{\mu}e_{\nu}^a-\Gamma^{\sigma}_{\mu\nu}e_{\sigma}^a+\omega^{ab}_{\mu}e_{\nu b}$ were $\omega^{ab}_{\mu}$ is the spin connection. (Greek is spacetime and latin is tetrad label, also I'm keeping any EM gauge stuff out of here). (PS I'm not claiming to have solved your problem!) $\endgroup$ – twistor59 Jan 16 '13 at 11:37
12
$\begingroup$

Denoting by $\gamma^a$ the Minkowski space gamma matrices with respect to the Lorentz tetrad $\{e^a\}$, and covariant derivative $D_a$, then the gammas are covariantly constant.

Start with the massless Dirac equation $$ \gamma^{b}D_{b}\Psi = 0$$

Act again with the Dirac operator $$\gamma^{a}D_{a}\gamma^{b}D_{b}\Psi=0 $$ So, since $D$ annihilates $\gamma$ $$\gamma^{a}\gamma^{b}D_{a}D_{b}\Psi = 0 $$ so $$\frac{1}{2}\{\gamma^{a},\gamma^{b}\}D_{a}D_{b}\Psi + \frac{1}{2}\gamma^{a}\gamma^{b}[D_a,D_b]\Psi = 0 \ \ (1) $$ But $$\{\gamma^{a},\gamma^{b}\}=2\eta^{ab} $$ and $$ [D_a,D_b]\Psi = {\mathcal{R}_{ab}}\Psi $$ Where ${\mathcal{R}}_{ab}$ is the spin-curvature (antisymmetric in a and b). ${\mathcal{R}}_{ab}$ satisfies the identity $$ -\gamma^b{\mathcal{R}}_{ab} = {\mathcal{R}}_{ab}\gamma^b = \frac{1}{2}\gamma^b R_{ab}$$ where $R_{ab}$ is the Ricci tensor (in the Lorentz tetrad). so (1) becomes $$ [D^aD_a+\frac{1}{4}\gamma^a\gamma^bR_{ab}]\Psi = 0 $$ i.e. $$ [D^aD_a-\frac{1}{4}R]\Psi = 0 $$

|cite|improve this answer
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.