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 }.

See also:

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