- #1
jdstokes
- 523
- 1
The Einstein field equations [itex]\mathsf{G} = \kappa \mathsf{T}[/itex] can be derived by considering stationary metric variations of the Einstein Hilbert action,
[itex]S = \int \mathrm{d}^4x \sqrt{-g} (R/2\kappa + \mathcal{L}_\mathrm{M})[/itex].
[itex]0 = \delta S = \int\mathrm{d}^4 x\left(\frac{1}{2\kappa}\frac{\partial (\sqrt{-g} R)}{\partial g^{\alpha\beta}}+ \frac{\partial (\sqrt{-g}\mathcal{L}_\mathrm{M})}{\partial g^{\alpha\beta}}\right)\delta g^{\alpha\beta}[/itex]
etc.
In conventional field theory, however, we consider variations of the action integrand with respect to both the field [itex]\varphi[/itex] as well as its first derivative [itex]\partial_\mu \varphi[/itex].
Why can we avoid doing this when [itex]\varphi = g^{\alpha\beta}[/itex]?.
[itex]S = \int \mathrm{d}^4x \sqrt{-g} (R/2\kappa + \mathcal{L}_\mathrm{M})[/itex].
[itex]0 = \delta S = \int\mathrm{d}^4 x\left(\frac{1}{2\kappa}\frac{\partial (\sqrt{-g} R)}{\partial g^{\alpha\beta}}+ \frac{\partial (\sqrt{-g}\mathcal{L}_\mathrm{M})}{\partial g^{\alpha\beta}}\right)\delta g^{\alpha\beta}[/itex]
etc.
In conventional field theory, however, we consider variations of the action integrand with respect to both the field [itex]\varphi[/itex] as well as its first derivative [itex]\partial_\mu \varphi[/itex].
Why can we avoid doing this when [itex]\varphi = g^{\alpha\beta}[/itex]?.