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

  1. Sep 26, 2013 #1
    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] ?
     
    Last edited: Sep 26, 2013
  2. jcsd
  3. Sep 26, 2013 #2
    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.
     
  4. Sep 26, 2013 #3
    Yes! I had forgotten about the Hamiltonian.
     
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