I have question about ADM decomposition of some general scalar-tensor theory of gravity.

Starting with ADM form of the metric:


provided with extrinsic curvature:

$K_{ij}=\frac{1}{2N}(\dot{h_{ij}}-D_iN_j-D_jN_i)$, where $N$ is lapse function and $N^i$'s are shift vector components. Timelike normal vector to the hypersurface is denoted as $n_a$.

I'm considering given action:

$\int d^4x\sqrt{-g}[f(\phi) R-\nabla_\mu\phi\nabla^\mu\phi + U(\phi)]$

and I want to recast this Lagrangian in the 3+1 form which is suitable for discussing Hamiltonian formulation of this theory.

Kinetic term $(X=\nabla_\mu\phi\nabla^\mu\phi)$ decomposes as: $X=-A_*^2+D^i\phi D_i\phi $

where: $A_*=n^\mu\nabla_\mu=\frac {1}{N}(\dot{\phi}-N^iD_i\phi)$ and $D^i$ is 3d covariant derivative associated with metric $h_{ab}$ ($h^a_b\nabla_a=D_b$).

However, i have trouble with the first part of the action:

$\int d^4x\sqrt{-g}f(\phi) R=\int dt\int d^3x \sqrt{-h}N\Big[f(\phi)\big(R^{(3)}+K_{ij}K^{ij}-K^2+2\nabla_{\mu}(n^\mu\nabla_\nu n^\nu-n^\nu\nabla_\nu n^\mu)\big)\Big]$.

First term is left as it is, while term involving derivatives of normal vector $n^\mu$ needs to be integrated by parts (dropping boundary term) - normally in GR this is total divergence and is discarded (this is not the case here):

$\int dt \int d^3x\sqrt{-h}N f(\phi) 2\nabla_{\mu}(n^\mu\nabla_\nu n^\nu-n^\nu\nabla_\nu n^\mu) \\=-2\int dt \int d^3x\sqrt{-h}N(n^\mu\nabla_\nu n^\nu-n^\nu\nabla_\nu n^\mu)\nabla_\mu f(\phi) \\= -2\int dt \int d^3x\sqrt{-h}N(n^\mu\nabla_\nu n^\nu)f_\phi\nabla_\mu\phi+2\int dt \int d^3x\sqrt{-h}N(n^\nu\nabla_\nu n^\mu)f_\phi \nabla_\mu \phi\\=-2\int dt \int d^3x\sqrt{-h}N(n^\mu Kf_\phi\nabla_\mu\phi-n^\nu(\nabla_\nu n^\mu) f_\phi \nabla_\mu \phi)=-2\int dt \int d^3x\sqrt{-h}N (Kf_\phi A_*-(\nabla_\nu n^\mu )n^\nu f_\phi\nabla_\mu \phi)$

where I used identity $K=\nabla_\nu n^\nu$ and $\nabla_\alpha f=f_\phi \nabla_\alpha \phi$ (chain rule).

I have trouble with the last part of the above equation - i have no idea how to simplify this expression and put it into 3+1 form. Here are some references that I'm trying to follow: https://arxiv.org/abs/1101.3403, https://arxiv.org/abs/1708.02951, https://arxiv.org/abs/1812.02667, https://arxiv.org/abs/1512.06820.

  • $\begingroup$ Minor comment to the post (v3): The word Langrangian should be Lagrangian. $\endgroup$ – Qmechanic Mar 7 at 18:17

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