The factor of 1/2 in the Einstein-Hilbert action

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SUMMARY

The factor of 1/2 in the Einstein-Hilbert action is essential for maintaining consistency with established conventions in theoretical physics. The gravitational Lagrangian is expressed as \(\mathcal{L}_{G}=\frac{c^4}{16\pi G}R\sqrt{-g}=\frac{1}{2\kappa}R\sqrt{-g}\), where \(\kappa=\frac{c^4}{8\pi G}\). This scaling ensures that the variation in the matter Lagrangian, \(\delta\mathcal{L}_{matter}=-\frac{1}{2}T_{\mu\nu}\delta g^{\mu\nu}\), aligns with the overall structure of the theory. Dropping the factor of 1/2 would disrupt the relationship with the Hamiltonian and the kinetic energy terms, leading to inconsistencies.

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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} ?
 
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The factor is arbitrary and doesn't affect the theory but note that any factor present in the Lagrangian also shows in the Hamiltonian. Dropping that factor of 1/2 would ultimately force you to drop the factor of 1/2 in the kinetic energy. So, to keep every thing consistent with long held conventions, that 1/2 factor is kept.
 
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Yes! I had forgotten about the Hamiltonian.
 

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