# Energy-momentum tensor of Majorana field

The Majorana field action in curved spacetime, in general, is usually written as (see this answer on Physics SE)

$$\mathcal{A}_M = \displaystyle\int_\mathcal{M} d^4x ~ e \left\{\dfrac{1}{2} \bar{\psi}\Gamma^\mu D_\mu\psi - V\left(\psi,\bar{\psi}\right)\right\}$$

where $$\bar{\psi} \equiv \psi^\dagger \mathcal{C}$$ for $$\mathcal{C}$$ being the charge conjugation operator, $$\Gamma^\mu \equiv e^\mu_{~b} \gamma^b$$ is the generalized Dirac-Pauli matrices satisfying $$\left\{\Gamma^\mu, \Gamma^\nu\right\} = 2g^{\mu\nu}$$ and $$D_\mu \equiv \partial_\mu + \dfrac{1}{2} \omega_{\mu ab}\Sigma^{ab}$$ is the generalized covariant derivative; $$\omega_{\mu ab} = e^\nu_{~a} \nabla_\mu e_{\nu b}$$ are the Ricci rotation (or spin) coefficients and $$\Sigma^{ab}=\dfrac{1}{4}\left[\gamma^a,\gamma^b\right]$$ are the generators of the spinor representation of the Lorentz group. Here, $$e$$ is the determinant of the vierbein $$e_\mu^{~a}$$.

Now if one wants to derive the Hilbert energy-momentum tensor of the Majorana spinor (on-shell) from this, what will be that expression? I guess, it would be

$$T_{\mu\nu}^{(M)} = \dfrac{1}{2} \bar{\psi}\Gamma_{(\mu} D_{\nu)}\psi - g_{\mu\nu} \mathcal{L}_M$$

in analogy with the energy-momentum tensor of the Dirac spinor. Will it be so?

My attempt: Following this paper (Spinors, Inflation, and Non-Singular Cyclic Cosmologies, here the energy-momentum tensor of the Dirac spinor has been obtained), one can write the Hilbert energy-momentum tensor as

$$T_{\mu\nu}^{(M)} = \dfrac{e_{\mu a}}{e} \dfrac{\delta \mathcal{A}_M}{\delta e^\nu_{~a}} = \dfrac{e_{\mu a}}{e} \displaystyle\int_\mathcal{M} d^4x ~e \left\{\dfrac{1}{2} \bar{\psi} \dfrac{\delta}{\delta e^\nu_{~a}} \left(\Gamma^\alpha D_\alpha\psi\right) - e_\nu^{~a} \mathcal{L}_M\right\}$$

$$\Longrightarrow T_{\mu\nu}^{(M)} = \dfrac{e_{\mu a}}{e} \displaystyle\int_\mathcal{M} d^4x ~e \left[\dfrac{1}{2} \bar{\psi} \gamma^b \dfrac{\delta}{\delta e^\nu_{~a}} \left\{e^\alpha_{~b} \left(\partial_\alpha + \dfrac{1}{2}\omega_{\alpha cd}\Sigma^{cd}\right)\right\} \psi - e_\nu^{~a} \mathcal{L}_M\right]$$

Now in order to simplify this, one needs the derivative of $$e^\alpha_{~b}$$ and $$\omega_{\alpha cd}$$ w.r.t. $$e^\nu_{~a}$$. But what will be those? I don't see any hope to reach the expression of energy-momentum tensor that I somehow guessed earlier!

• There is some discussion of the derivative of the spin connection in the wikipedia article on the Belinfante-Rosenfeld stress–energy tensor Commented Oct 27, 2023 at 12:33