- #1
pellman
- 684
- 5
torsion --> covariant deriv of det(g) non-zero?
I am missing something here. This paper makes the case (on page 5) that for non-vanishing torsion, the usual invariant volume element [tex]\sqrt{-g}d^4x[/tex] is not appropriate because the covariant derivative of sqrt(-g) is non-zero. This perplexes me. The determinant of g is simply a function of the individual components of g, and their covariant derivatives vanish by assumption. So how can the derivative of the determinant be non-zero?
Here is a summary of the math. I have checked all this and my results match those of the linked paper (though I use a different notation). So maybe I am wrong in expecting the covariant derivative of g to vanish?
(1) If we assume a metric connection with [tex]\nabla_\gamma g_{\alpha\beta}=0[/tex], then it follows that the connection coefficients take the form
[tex]{\Gamma^\gamma}_{\alpha\beta}=\frac{1}{2}g^{\gamma\lambda}\left(\partial_\alpha g_{\lambda\beta} +\partial_\beta g_{\alpha\lambda}-\partial_\lambda g_{ \beta\alpha }\right)+\frac{1}{2} \left( {{T_\alpha}^\gamma}_\beta + {{T_\beta }^\gamma}_ \alpha + {T^\gamma}_{\alpha\beta}\right)[/tex]
where [tex]{T^\gamma}_{\alpha\beta}\equiv 2{\Gamma^\gamma}_{[\alpha\beta]}[/tex] is the torsion tensor.
(2) The square root of the determinant of the metric is a weight 1 scalar density. As such its covariant derivative is [tex]\nabla_\gamma \sqrt{-g}=\partial_\gamma\sqrt{-g}-{\Gamma^\rho}_{\rho\gamma}\sqrt{-g}[/tex]
(3) The result of this has a non-zero term dependent on the torsion:
[tex]\partial_\gamma\sqrt{-g}-{\Gamma^\rho}_{\rho\gamma}\sqrt{-g}=-{T^\rho}_{\rho\gamma}\sqrt{-g}\neq 0[/tex]
I am missing something here. This paper makes the case (on page 5) that for non-vanishing torsion, the usual invariant volume element [tex]\sqrt{-g}d^4x[/tex] is not appropriate because the covariant derivative of sqrt(-g) is non-zero. This perplexes me. The determinant of g is simply a function of the individual components of g, and their covariant derivatives vanish by assumption. So how can the derivative of the determinant be non-zero?
Here is a summary of the math. I have checked all this and my results match those of the linked paper (though I use a different notation). So maybe I am wrong in expecting the covariant derivative of g to vanish?
(1) If we assume a metric connection with [tex]\nabla_\gamma g_{\alpha\beta}=0[/tex], then it follows that the connection coefficients take the form
[tex]{\Gamma^\gamma}_{\alpha\beta}=\frac{1}{2}g^{\gamma\lambda}\left(\partial_\alpha g_{\lambda\beta} +\partial_\beta g_{\alpha\lambda}-\partial_\lambda g_{ \beta\alpha }\right)+\frac{1}{2} \left( {{T_\alpha}^\gamma}_\beta + {{T_\beta }^\gamma}_ \alpha + {T^\gamma}_{\alpha\beta}\right)[/tex]
where [tex]{T^\gamma}_{\alpha\beta}\equiv 2{\Gamma^\gamma}_{[\alpha\beta]}[/tex] is the torsion tensor.
(2) The square root of the determinant of the metric is a weight 1 scalar density. As such its covariant derivative is [tex]\nabla_\gamma \sqrt{-g}=\partial_\gamma\sqrt{-g}-{\Gamma^\rho}_{\rho\gamma}\sqrt{-g}[/tex]
(3) The result of this has a non-zero term dependent on the torsion:
[tex]\partial_\gamma\sqrt{-g}-{\Gamma^\rho}_{\rho\gamma}\sqrt{-g}=-{T^\rho}_{\rho\gamma}\sqrt{-g}\neq 0[/tex]