化简方程

• 
  改写二阶为一阶：
  令$\displaystyle k=\partial_t \phi+\frac{g^{0i}}{g^{00}}\partial_i \phi,\xi_i=\partial_i \phi,$那么$\displaystyle \partial_t \xi_i=\partial_i\partial_t \phi=\partial_i(k-\frac{g^{0i}}{g^{00}}\partial_i \phi)$
  记$T=2N\varphi^{cd}(g^{ij}\Phi_{ica}\Phi_{jdb}-\Pi_{ca}\Pi_{bd}-\varphi^{ef}\Gamma_{ace}\Gamma_{bdf})-\frac{1}{2}Nt^ct^d\Pi_{cd}\Pi_{ab}-Nt^c\Pi_{ci}g^{ij}\Phi_{jab}-2N\nabla_{(a}H_{b)}$
  令$\Gamma_b=H_b$就是广义调和,令$\Phi_{iab}=\partial_i\varphi_{ab},\ \ -N\Pi_{ab}=\partial_t \varphi_{ab}-N^i\Phi_{iab}$，$t=(t^0,t^1,t^2,t^3)=(\frac{1}{N},\frac{-N^1}{N},\frac{-N^2}{N},\frac{-N^3}{N})$
  现在的变量有$\{\varphi_{ab},\Pi_{ab},\Phi_{iab},\phi,k,\xi_i\}$
  $$\begin{aligned} \partial_t\varphi_{ab}-N^k\partial_k\varphi_{ab}&=-N\Pi_{ab}\\ \partial_t\Pi_{ab}-N^k\partial_k\Pi_{ab}+Ng^{ki}\partial_k\Phi_{iab}&=T+(\partial_a\phi )(\partial_b \phi)+\varphi_{ab}V\\ \partial_t\Phi_{iab}-N^k\partial_k\Phi_{iab}&=\frac{1}{2}Nt^ct^d\Phi_{icd}\Pi_{ab}+Ng^{jk}t^c\Phi_{ijk}\Phi_{kab}\\ \partial_t \phi-N^i\partial_i \phi&=k\\ \partial_t k -N^{i}\partial_i k-g^{ij}N^2\partial_i \xi_j&=Ng^{ij}[N(H_j+g^{kl}\Gamma_{ikl})-\partial_jN]+\frac{\partial V}{\partial \phi}\\ \partial_t \xi_i-N^j\partial_j \xi_i-\partial_i k&=\partial_i(N^j)\xi_j \end{aligned}$$
  边界条件：我们先选择度量：
  $$\begin{aligned} ds^2&=-N^2dt^2+\gamma(dr+\beta dt)^2\\ &=(-N^2+\beta^2\gamma)dt^2+\gamma dr^2+2\beta\gamma drdt \end{aligned}$$
  $$\begin{aligned} \phi(0,r)&=\phi_0e^{-(r-r_0)^2/{\Delta^2}}\\ N(0,r)&=\gamma(0,r)^{-2}\\ \beta(0,r)&=0\\ \gamma'(0,r)|_{r=0}&=0\\ \gamma(0,r)|_{r \to \infty}&=1\\ \partial_r\varphi_{ab}&=0\\ \partial_r\Pi_{ab}&=0 \end{aligned}$$
  这个边值条件和施瓦希差多远？
  $\left( \begin{array}{cccccccccccc} -N^1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -N^1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -N^1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -N^1 & 0 & 0 & g^{11} N & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -N^1 & 0 & 0 & g^{11} N & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -N^1 & 0 & 0 & g^{11} N & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -N^1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -N^1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -N^1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -N^1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -N^1 & -g^{11} N^2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & -N^1 \\ \end{array} \right)$
  特征值：$\left\{-N^1_{(10)},-N^1-\sqrt{g^{11}} N,-N^1+\sqrt{g^{11}} N\right\}$
  特征向量：
  $\{0,0,0,0,0,0,0,0,0,1,0,0\},\{0,0,0,0,0,1,0,0,0,0,0,0\},$ $\{0,0,0,0,1,0,0,0,0,0,0,0\},\{0,0,0,1,0,0,0,0,0,0,0,0\},$ $\{0,0,1,0,0,0,0,0,0,0,0,0\},\{0,1,0,0,0,0,0,0,0,0,0,0\},$ $\{1,0,0,0,0,0,0,0,0,0,0,0\},\{0,0,0,0,0,0,0,0,0,0,0,0\},$ $\{0,0,0,0,0,0,0,0,0,0,0,0\},\{0,0,0,0,0,0,0,0,0,0,0,0\},$ $\left\{0,0,0,0,0,0,0,0,0,0,\sqrt{g^{11}} N,1\right\}, \left\{0,0,0,0,0,0,0,0,0,0,-\sqrt{g^{11}} N,1\right\}$
  apparent horizon
  $$\theta=r\partial_t S+(1+r\partial_r S)(-\frac{g_{tr}}{g_{rr}}+\sqrt{\frac{g^2_{tr}}{g^2_{rr}}-\frac{g_{tt}}{g_{rr}}} )=0$$
  在本文简单情况下 $$\theta=-\frac{g_{tr}}{g_{rr}}+\sqrt{\frac{g^2_{tr}}{g^2_{rr}}-\frac{g_{tt}}{g_{rr}}} =0$$

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不把$\phi$考虑进方程组，前9个未知变量
$\left( \begin{array}{ccccccccc} -N^1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -N^1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -N^1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -N^1 & 0 & 0 & g^{11} N & 0 & 0 \\ 0 & 0 & 0 & 0 & -N^1 & 0 & 0 & g^{11} N & 0 \\ 0 & 0 & 0 & 0 & 0 & -N^1 & 0 & 0 & g^{11} N \\ 0 & 0 & 0 & 0 & 0 & 0 & -N^1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -N^1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -N^1 \\ \end{array} \right)$
特征向量是$\{0,0,0,0,0,1,0,0,0\}$,$\{0,0,0,0,1,0,0,0,0\}$,$\{0,0,0,1,0,0,0,0,0\}$,$\{0,0,1,0,0,0,0,0,0\}$,$\{0,1,0,0,0,0,0,0,0\}$,$\{1,0,0,0,0,0,0,0,0\}$,$\{0,0,0,0,0,0,0,0,0\}$,$\{0,0,0,0,0,0,0,0,0\}$,$\{0,0,0,0,0,0,0,0,0\}$
特征值：$\{-1, -1, -1, 1, 1, 1, 0, 0, 0\}$
- 另外一种方案：不把标量场$\phi$放到一阶方程组里。

$$\begin{aligned} \partial_t\varphi_{ab}-N^k\partial_k\varphi_{ab}&=-N\Pi_{ab}\\ \partial_t\Pi_{ab}-N^k\partial_k\Pi_{ab}+Ng^{ki}\partial_k\Phi_{iab}&=T+(\partial_a\phi )(\partial_b \phi)+\varphi_{ab}V\\ \partial_t\Phi_{iab}-N^k\partial_k\Phi_{iab}&=\frac{1}{2}Nt^ct^d\Phi_{icd}\Pi_{ab}+Ng^{jk}t^c\Phi_{ijk}\Phi_{kab}\\ \square \phi&=\frac{\partial V}{\partial \phi} \end{aligned}$$
给出边界条件：
$$\begin{aligned} \sqrt{2}[k^{(c}\varphi^{d)}_a-\frac{1}{2}k_a\varphi^{cd}]d_{\bot}u_{cd}^{\hat{1}-}&=0\\ d_{\bot}u_{ab}^{\hat 0}&=0\\ d_{\bot}u_{k ab}^{\hat 2}&=0\\ (P^c_aP^d_b-\frac{1}{2}P_{ab}P^{cd})(d_{\bot}u_{cd}^{\hat 1\pm})&=0 \end{aligned}$$

• Lee的笔记：
  $E_k=\Pi_{k}+\Phi_{kt}=\frac{1}{2}(u_k^{1+}+u_k^{1-})+\frac{1}{2}n_k(u_t^{1+}-u_t^{1-})+u^{\hat{2}}_{kt}$
  则 $$n^kE_k=\frac{1}{2}(n^ku_k^{1+}+n^ku_k^{1-}+u_t^{1+}-u_t^{1-})+n^ku_{kt}^{\hat{2}}\tag{35}$$
  关于$\displaystyle n^ku_{kt}^{\hat{2}}$这一项，注意：$n^kP^i_k\Phi_{it}=n^k(\delta^i_k-n^in_k)\Phi_{it}=0$,所以这项必须消失。
  而$\displaystyle \Phi_{ij}=\frac{1}{2}n_i(u_j^{1+}-u^{1-}_j)+u^{\hat{2}}_{ij},B_k=\epsilon^{ij}_k \Phi_{ij}=\frac{1}{2}\epsilon_k^{ij}n_i(u_j^{1+}-u^{1-}_j)+\epsilon_k^{ij}u^{\hat{2}}_{ij}$
  那么$n^kB_k=$
  $$\frac{1}{2}\epsilon_k^{ij}n^kn_i(u_j^{1+}-u^{1-}_j)+n^k\epsilon_k^{ij}u^{\hat{2}}_{ij}=\frac{1}{2}\epsilon^{ijk}n_kn_i(u_j^{1+}-u^{1-}_j)+\epsilon^{ijk} n_ku^{\hat{2}}_{ij}\tag{36}$$
  关于$\frac{1}{2}\epsilon^{ijk}n_kn_i(u_j^{1+}-u^{1-}_j)$这项，注意：$(A\times B)_i=\epsilon_{ijk}A^jB^k=\epsilon^{ijk}A_jB_k$,所以这项必须为0.以下推导（37）式
  $\displaystyle P^k_iE_k=\frac{1}{2}P_i^k(u_k^{1+}+u_k^{1-})+P^k_iu^{\hat{2}}_{kt}=\frac{1}{2}P_i^k(u_k^{1+}+u_k^{1-})+u^{\hat{2}}_{it}$
  $\displaystyle \epsilon^{jk}_in_j B_k= \epsilon^{jk}_in_j \epsilon^{ab}_k u_{ab}^{\hat{2}}+\frac{1}{2}\epsilon^{jk}_in_j \epsilon^{ab}_k n_a(u_b^{1+}-u_b^{1-})$
  $$\begin{aligned} \epsilon^{jk}_in_j B_k\\ &= \epsilon^{jk}_in_j \epsilon^{ab}_k u_{ab}^{\hat{2}}+\frac{1}{2}\epsilon^{jk}_in_j \epsilon^{ab}_k n_a(u_b^{1+}-u_b^{1-})\\ &=\epsilon_{jik}\epsilon^{abk} n^ju^{\hat{2}}_{ab}+\frac{1}{2}\epsilon_{ijk}\epsilon^{abk} n^jn_a(u_b^{1+}-u_b^{1-})\\ &=n^ju^{\hat{2}}_{ab}(\delta^a_j\delta_i^b-\delta^b_j\delta^a_i)+\frac{1}{2}n^jn_a(\delta_i^a\delta_j^b-\delta_i^b\delta_j^a)\\ &=-n^ju_{ij}^{\hat{2}}-\frac{1}{2}P^j_i(u^{1+}_j)+\frac{1}{2}P^j_i(u_j^{1-}) \end{aligned}$$
  可得： $$P^k_iE_k+\epsilon^{jk}_i n_jB_k=P^k_iu^{1-}_k+u^{\hat{2}}_{it}+n^ku_{ik}^{\hat{2}}\tag{37}$$