pellman
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Why the factor of 1/2 in the Einstein-Hilbert action?
The Einstein field equation is
\frac{1}{\kappa}\left(R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}\right)=T_{\mu\nu}
where \kappa=\frac{c^4}{8\pi G}
This can be derived from extremizing the action
S=\int{(\mathcal{L}_{G}+\mathcal{L}_{matter})d^4x}
with respect to variation in the metric. Typically the gravitational Lagrangian is given as
\mathcal{L}_{G}=\frac{c^4}{16\pi G}R\sqrt{-g}=\frac{1}{2\kappa}R\sqrt{-g}
I have seen this 1/2kappa scaling of the Einstein-Hilbert action frequently in the literature. The matter Lagrangian must then be scaled in such a way that
\delta\mathcal{L}_{matter}=-\frac{1}{2}T_{\mu\nu}\delta g^{\mu\nu}
Any idea why the 1/2s? Why not just \mathcal{L}_{G}=\frac{1}{\kappa}R\sqrt{-g} and \delta\mathcal{L}_{matter}=-T_{\mu\nu}\delta g^{\mu\nu} ?
The Einstein field equation is
\frac{1}{\kappa}\left(R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}\right)=T_{\mu\nu}
where \kappa=\frac{c^4}{8\pi G}
This can be derived from extremizing the action
S=\int{(\mathcal{L}_{G}+\mathcal{L}_{matter})d^4x}
with respect to variation in the metric. Typically the gravitational Lagrangian is given as
\mathcal{L}_{G}=\frac{c^4}{16\pi G}R\sqrt{-g}=\frac{1}{2\kappa}R\sqrt{-g}
I have seen this 1/2kappa scaling of the Einstein-Hilbert action frequently in the literature. The matter Lagrangian must then be scaled in such a way that
\delta\mathcal{L}_{matter}=-\frac{1}{2}T_{\mu\nu}\delta g^{\mu\nu}
Any idea why the 1/2s? Why not just \mathcal{L}_{G}=\frac{1}{\kappa}R\sqrt{-g} and \delta\mathcal{L}_{matter}=-T_{\mu\nu}\delta g^{\mu\nu} ?
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