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

  • Thread starter pellman
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  • #1
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Main Question or Discussion Point

Why the factor of 1/2 in the Einstein-Hilbert action?

The Einstein field equation is

[tex]\frac{1}{\kappa}\left(R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}\right)=T_{\mu\nu}[/tex]

where [itex]\kappa=\frac{c^4}{8\pi G}[/itex]

This can be derived from extremizing the action

[tex]S=\int{(\mathcal{L}_{G}+\mathcal{L}_{matter})d^4x}[/tex]

with respect to variation in the metric. Typically the gravitational Lagrangian is given as

[tex]\mathcal{L}_{G}=\frac{c^4}{16\pi G}R\sqrt{-g}=\frac{1}{2\kappa}R\sqrt{-g}[/tex]

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

[tex]\delta\mathcal{L}_{matter}=-\frac{1}{2}T_{\mu\nu}\delta g^{\mu\nu}[/tex]

Any idea why the 1/2s? Why not just [itex]\mathcal{L}_{G}=\frac{1}{\kappa}R\sqrt{-g}[/itex] and [itex]\delta\mathcal{L}_{matter}=-T_{\mu\nu}\delta g^{\mu\nu}[/itex] ?
 
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Answers and Replies

  • #2
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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.
 
  • #3
666
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Yes! I had forgotten about the Hamiltonian.
 

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