- #1
JustMeDK
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Let [itex]\gamma^{\rho} \in M_{4}(\mathbb{R})[/itex] be the Majorana representation of the Dirac algebra (in spacetime signature [itex]\eta_{00} = -1[/itex]), and consider the Majorana Lagrangian [tex]\mathcal{L} = \mathrm{i} \theta^{\mathrm{T}} \gamma^{0} (\gamma^{\rho} \partial_{\rho} - m) \theta,[/tex] where [itex]\theta[/itex] is a Grassmann-valued four-spinor. The associated gravitational energy-density, the 00-component of the Belinfante energy-momentum tensor, I calculate to be [tex]\Theta^{00} = \frac{\mathrm{i}}{2} [ \theta^{\mathrm{T}} (\partial^{0} \theta) - (\partial^{0} \theta)^{\mathrm{T}} \theta].[/tex] Inserting into it the stationary plane, wave solution [itex]\theta = \mathrm{exp}(\gamma^{0}Et)\eta[/itex], where [itex]\eta[/itex] is some spacetime-independent, Grassmann-valued four-spinor, yields [itex]\Theta^{00} = \mathrm{i} E \eta^{\mathrm{T}} \gamma^{0} \eta[/itex]. Due to [itex](xy)^{*} \equiv y^{*}x^{*}[/itex] for Grassmann-valued quantities, this expression for [itex]\Theta^{00}[/itex] is complex self-conjugate (and nonvanishing), as it should be, but it is not real-valued.
In comparison, for the Dirac Lagrangian, [tex]\mathcal{L}_{D} = -\mathrm{i} \psi^{\dagger} \gamma^{0} (\gamma^{\rho} \partial_{\rho} - m) \psi,[/tex] also in spacetime signature [itex]\eta_{00} = -1[/itex], a similar calculation of the gravitational energy-density yields for a plane wave solution [itex]\psi = \mathrm{exp}(-\mathrm{i}Et)\psi_{0}[/itex] the real-valued expression [itex]\Theta^{00} = E \psi_{0}^{\dagger} \psi_{0}[/itex]. The exact sign of [itex] -\mathrm{i} \psi^{\dagger} \gamma^{0} \gamma^{0} \partial_{0} \psi = +\mathrm{i} \psi^{\dagger} \partial_{0} \psi [/itex] in [itex]\mathcal{L}_{D}[/itex] is essential for this energy-density to be positive-definite.
And now to my question: Is it nonsensical to analogously contemplate what the sign of the time derivative should be in the case of the Majorana Lagrangian? And if not, what is it?
In comparison, for the Dirac Lagrangian, [tex]\mathcal{L}_{D} = -\mathrm{i} \psi^{\dagger} \gamma^{0} (\gamma^{\rho} \partial_{\rho} - m) \psi,[/tex] also in spacetime signature [itex]\eta_{00} = -1[/itex], a similar calculation of the gravitational energy-density yields for a plane wave solution [itex]\psi = \mathrm{exp}(-\mathrm{i}Et)\psi_{0}[/itex] the real-valued expression [itex]\Theta^{00} = E \psi_{0}^{\dagger} \psi_{0}[/itex]. The exact sign of [itex] -\mathrm{i} \psi^{\dagger} \gamma^{0} \gamma^{0} \partial_{0} \psi = +\mathrm{i} \psi^{\dagger} \partial_{0} \psi [/itex] in [itex]\mathcal{L}_{D}[/itex] is essential for this energy-density to be positive-definite.
And now to my question: Is it nonsensical to analogously contemplate what the sign of the time derivative should be in the case of the Majorana Lagrangian? And if not, what is it?
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