5
$\begingroup$

The identity,

$$ -\gamma^b{\mathcal{R}}_{ab} = {\mathcal{R}}_{ab}\gamma^b = \frac{1}{2}\gamma^b R_{ab}$$ is presented in the answer to the question Dirac Equation in General Relativity.

How does one prove the identity?

$\endgroup$
5
$\begingroup$

The curvature two-form is defined by $$\newcommand{\Rcal}{\mathcal{R}} \Rcal_{ab} \Psi = [D_a, D_b] \Psi.$$ Here $D_a$ is the covariant derivative without gauge terms. $\Rcal_{ab}$ takes values in the Dirac representation of the Lorentz Lie algebra. Thus, really the relation is $$\mathcal R_{ab \alpha\beta} \Psi_\beta = [D_a, D_b] \Psi_\alpha$$ where $\alpha, \beta$ are Dirac spinor indices. Compare this with the more familiar Riemann tensor $$R_{ab}{}^\mu{}_\nu x^\nu = [D_a, D_b] x^\mu. $$ Since the Riemann tensor is antisymmetric in $\mu,\nu$ we can consider it too to be a 2-form taking values in a representation of the Lorentz Lie algebra.. Of course, this representation is the 4-vector representation.

The Dirac representation of the Lie algebra is realized by $$\epsilon_{\mu\nu} \mapsto \frac{1}{4} \epsilon_{\mu\nu}\gamma^\mu\gamma^\nu$$ that is under an infinitesimal Lorentz transformation by $\epsilon_{\mu\nu}$, $$\Psi \mapsto \Psi + \frac{1}{4} \epsilon_{\mu\nu}\gamma^\mu\gamma^\nu \Psi.$$ This means that $$\Rcal_{ab} = \frac{1}{4}R_{abst}\gamma^s \gamma^t. $$

Now let us define the Ricci tensor by $$R_{ab} = R_a{}^\mu{}_{\mu b} = R_{astb} g^{st}.$$ Then from the fundamental anticommutation relation of the gamma matrices, we can write \begin{align} R_{ab} \gamma^b & = \frac{1}{2} R_{astb} (\gamma^s \gamma^t + \gamma^t \gamma^s) \gamma^b \\ & = \frac{1}{2} R_{astb}\gamma^s\gamma^t \gamma^b - \frac{1}{2} (R_{abst} + R_{atbs}) \gamma^t\gamma^s \gamma^b \tag{1}.\end{align} Here I have used the symmetry of the Riemann tensor, $R_{astb} + R_{abst} + R_{atbs} = 0$. Note that the first term is precisely $2\Rcal_{as}\gamma^s$. Now $R_{atbs} = -R_{atsb}$ and so a relabeling of contracted indices in the last term shows that this term also contributes $2\Rcal_{as}\gamma^s$. For the middle term, use the anticommutation relation to find $$\gamma^t \gamma^s \gamma^b = \gamma^b \gamma^t \gamma^s - 2g^{tb}\gamma^s + 2g^{sb}\gamma^t \tag{2}.$$ Hence, \begin{align}R_{abst}\gamma^t\gamma^s \gamma^b & = -R_{abts} \gamma^b \gamma^t \gamma^s - 2R_a{}^\mu{}_{s \mu} \gamma^s + 2R_a{}^\mu{}_{\mu t}\gamma^t \\ & = -\Rcal_{as}\gamma^s + 4R_{as}\gamma^s. \end{align}

We now have that (1) is $$R_{ab}\gamma^b = 6 \Rcal_{ab}\gamma^b - 2R_{ab}\gamma^b$$ so clearly $$\frac{1}{2}R_{ab}\gamma^b = \Rcal_{ab}\gamma^b \tag{3}$$ which is one of the identities in your question. The other should follow from using the anticommutation relations and (3).

$\endgroup$
  • $\begingroup$ Why so $\Rcal_{ab} = \frac{1}{4}R_{abst}\gamma^s \gamma^t$ from $\Psi \mapsto \Psi + \frac{1}{4} \epsilon_{\mu\nu}\gamma^\mu\gamma^\nu \Psi$ ? $\endgroup$ – user48875 Jun 19 '14 at 0:08
  • $\begingroup$ And is there this calculations in the literature? What literature is? $\endgroup$ – user48875 Jun 19 '14 at 0:11
  • $\begingroup$ Because in the Dirac representation, the element $\epsilon_{\mu\nu}$ of the Lorentz Lie algebra is represented by $\frac{1}{4}\epsilon_{\mu\nu}\gamma^\mu\gamma^\nu$. This is a fairly standard result, if nothing else it should be in Penrose and Rindler. I don't know if the other calculations are anywhere in the literature. $\endgroup$ – Robin Ekman Jun 19 '14 at 0:16
  • $\begingroup$ My first question may reformulate so: "How does the curvature from element of the Lorentz Lie algebra arise?" I don't know this fairly standart result, unfortunately. Whence does it follow? $\endgroup$ – user48875 Jun 21 '14 at 9:47
  • $\begingroup$ The Dirac representation is a direct sum of a left-handed and a right-handed spin $1/2$ fermion. Write the $\gamma^\mu$ matrices in the Weyl representation and check that for example $\gamma^1\gamma^2$ generates rotations around the $3$-axis, and $\gamma^0 \gamma^1$ boosts along the $1$-axis (by comparing to how they are represented in for right- and left-handed particles). $\endgroup$ – Robin Ekman Jun 21 '14 at 10:14

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.