Ngô Quốc Anh

January 1, 2013

The harmonic slicing

Filed under: Uncategorized — Ngô Quốc Anh @ 17:46

As we have already seen in this post, the Einstein equations are essentially hyperbolic if one assumes that the harmonic condition holds, i.e. $\square_g x^\alpha=0$. This type of condition was first introduced by De Donder in 1921 and have played an important role in theoretical developments, notably in the Choquet-Bruhat work of the well-posedness of the Cauchy problem for $3+1$ Einstein equations.

The harmonic slicing is defined by requiring that the harmonic condition holds only for the coordinate $x^0=t$, i.e. $\square_g t=0$, leaving freedom to choose any coordinate $(x^\alpha)$, $\alpha>0$ in each hypersurface $M$.

Using the formula for the d’Alembertian operator, which is the Laplace-Beltrami operator in the Minkowski space, we obtain

$\displaystyle\frac{1}{{\sqrt { - \det g} }}\frac{\partial }{{\partial {x^\mu }}} \Big(\sqrt { - \det g} {g^{\mu \nu }}\underbrace {\frac{{\partial t}}{{\partial {x^\nu }}}}_{\delta _\nu ^0}\Big) = 0,$

that is to say

$\displaystyle\frac{\partial }{{\partial {x^\mu }}}(\sqrt { - \det g} {g^{\mu 0}}) = 0.$

Thanks to the identity $\sqrt { - \det g} = N\sqrt {\det \gamma }$, one can see that

$\displaystyle\frac{\partial }{{\partial t}}(N\sqrt {\det \gamma } {g^{00}}) + \frac{\partial }{{\partial {x^i}}}(N\sqrt {\det \gamma } {g^{i0}}) = 0.$

Since

$\displaystyle {g^{00}} = - \frac{1}{{{N^2}}}, \quad {g^{i0}} = \frac{{{\beta ^i}}}{{{N^2}}}$

we find that

$\displaystyle - \frac{\partial }{{\partial t}}\Big(\frac{{\sqrt {\det \gamma } }}{N}\Big) + \frac{\partial }{{\partial {x^i}}}\Big(\frac{{\sqrt {\det \gamma } }}{N}{\beta ^i}\Big) = 0.$

By expanding, we obtain

$\displaystyle\frac{{\partial N}}{{\partial t}} - {\beta ^i}\frac{{\partial N}}{{\partial {x^i}}} - N\left( {\frac{1}{{\sqrt {\det \gamma } }}\frac{\partial }{{\partial t}}(\sqrt {\det \gamma } ) - \frac{1}{{\sqrt {\det \gamma } }}\frac{\partial }{{\partial {x^i}}}(\sqrt {\det \gamma } {\beta ^i})} \right) = 0.$

In view of the divergence for vector field, there holds

$\displaystyle\frac{1}{{\sqrt {\det \gamma } }}\frac{\partial }{{\partial {x^i}}}(\sqrt {\det \gamma } {\beta ^i}) = \text{div}_\gamma \beta = {\nabla _\gamma } \cdot \vec\beta .$

Thanks to the evolution equation of the spatial metric, i.e.

$\displaystyle\frac{\partial }{{\partial t}}{\gamma _{ij}} = - 2N{K_{ij}} + {\mathcal L_{\vec \beta} }{\gamma _{ij}},$

we find that

$\displaystyle {\gamma ^{ij}}\left( {\frac{{\partial {\gamma _{ij}}}}{{\partial t}} - {\mathcal L_{\vec \beta} }{\gamma _{ij}}} \right) = - 2N\underbrace {{\gamma ^{ij}}{K_{ij}}}_K.$

Since $\mathcal L_{\vec \beta}\gamma_{ij}=D_i\beta_j + D_j\beta_i$, we obtain

$\displaystyle \text{div}_\gamma \vec\beta - NK = \frac{1}{2}{\gamma ^{ij}}\frac{{\partial {\gamma _{ij}}}}{{\partial t}},$

equivalently,

$\displaystyle \text{div}_\gamma\vec\beta - NK = \frac{1}{{\sqrt {\det \gamma } }}\frac{\partial }{{\partial t}}(\sqrt {\det \gamma } ).$

By collecting all above formulas, we find that

$\displaystyle\frac{{\partial N}}{{\partial t}} - {\beta ^i}\frac{{\partial N}}{{\partial {x^i}}} + K{N^2} = 0.$

Using the formula for the Lie derivative for scalar functions, we get that

$\displaystyle\left( {\frac{\partial }{{\partial t}} - {\mathcal L_{\vec \beta} }} \right)N = - K{N^2}.$

Hence, we have got an evolution equation for the lapse function $N$. Thanks to

$\displaystyle {K_{ij}} = - \frac{1}{{2N}}\left( {\frac{\partial }{{\partial t}} - {\mathcal L_{\vec\beta} }} \right){\gamma_{ij}},$

the above equation is nothing but

$\displaystyle \frac{1}{N}\left( {\frac{\partial }{{\partial t}} - {L_\beta }} \right)N = \frac{1}{2}{\gamma ^{ij}}\left( {\frac{\partial }{{\partial t}} - {\mathcal L_{\vec\beta} }} \right){\gamma _{ij}} = \frac{1}{{\sqrt {\det \gamma } }}\left( {\frac{\partial }{{\partial t}} - {\mathcal L_{\vec\beta} }} \right)(\sqrt {\det \gamma } ),$

which is

$\displaystyle\left( {\frac{\partial }{{\partial t}} - {\mathcal L_{\vec\beta} }} \right)\log \left( {\frac{N}{{\sqrt {\det \gamma } }}} \right) = 0.$

Thus, for the scalar density $\alpha$ depending on the shift vector $\vec\beta$, the lapse function $N$ is

$\displaystyle N = \alpha \sqrt {\det \gamma }.$