Ngô Quốc Anh

December 7, 2012

Derivation of the Einstein constraint equations: The Hamitonian constraint

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 16:41

This entry is devoted to a derivation of the Einstein constraint equations. In fact, this is an expanded version of the previous entry. For the sake of simplicity, here we only derive the Hamiltonian constant equation. The momentum constaint equation will be considered in the coming entry.

Before we start, let us recall the following form of the Einstein equation with the cosmological constant $\Lambda$, that is $\displaystyle \text{Ric}_{\overline g}-\frac{1}{2}\overline g \text{Scal}_{\overline g}+\Lambda\overline g = \kappa T$

where $\kappa$ is a constant. The above equation is understood over the Lorentzian manifold $(V,\overline g)$ of the dimension $n+1$. We shall use $\langle\cdot, \cdot\rangle$ to denote $\overline g(\cdot, \cdot)$.

Let us take $(M,g)$ a submanifold of $V$ of the dimension $n$. It is well-known that the Levi-Civita connection $\nabla$ verifies the following decomposition $\displaystyle\overline\nabla_X Y=\nabla_X Y + \mathrm{I\!I}(X,Y)$

for any smooth vector fields $X$ and $Y$ tangent to $M$ and $\mathrm{I\!I}$ is the second fundamental form.

We now mention the so-called Gauss equation. Before we do that, let us recall the Riemann curvature tensor $R$ given by $\displaystyle R(X,Y)Z = \nabla_X\nabla_Y Z-\nabla_Y\nabla_X Z - \nabla_{[X,Y]}Z.$

Furthermore, there holds $\displaystyle\nabla_X Y = \top(\overline\nabla_X Y),\quad \mathrm{I\!I}(X,Y) = \bot(\overline\nabla_X Y).$

Then the Gauss equation is given by $\displaystyle\left\langle {\overline R (X,Y)Z,W} \right\rangle = \left\langle {R\left( {X,Y)Z} \right),W} \right\rangle + \left\langle {\mathrm{I\!I}(X,Z)} , {\mathrm{I\!I}(Y,W)} \right\rangle - \left\langle {\mathrm{I\!I}(Y,Z)} , {\mathrm{I\!I}(X,W)} \right\rangle.$

We now let $e_1,...,e_n$ be a local orthonormal frame field for $M$. Using the Gauss equation, we arrive at $\displaystyle\left\langle {\overline R \left( {{e_i},{e_j}} \right){e_i},{e_j}} \right\rangle = \left\langle {R({e_i},{e_j}){e_i},{e_j}} \right\rangle + \left\langle {\mathrm{I\!I}({e_i},{e_i})} ,{\mathrm{I\!I}({e_j},{e_j})} \right\rangle - \left\langle {\mathrm{I\!I}({e_i},{e_j})}, {\mathrm{I\!I}({e_i},{e_j})} \right\rangle,$

which yields $\displaystyle\begin{array} {lcl}\displaystyle\sum\limits_{i,j = 1}^n {\left\langle {\overline R \left( {{e_i},{e_j}} \right){e_i},{e_j}} \right\rangle } &=&\displaystyle \sum\limits_{i,j = 1}^n {\left\langle {R({e_i},{e_j}){e_i},{e_j}} \right\rangle } \hfill \\&&\displaystyle + \sum\limits_{i,j = 1}^n {\left\langle {\mathrm{I\!I}({e_i},{e_i})}, {\mathrm{I\!I}({e_j},{e_j})} \right\rangle } -\sum\limits_{i,j = 1}^n {\left\langle {\mathrm{I\!I}({e_i},{e_j}),\mathrm{I\!I}({e_i},{e_j})} \right\rangle }.\end{array}$

By definition of the scalar curvature, we have $\displaystyle \text{Scal}_g = \sum\limits_{i,j = 1}^n {\left\langle {R({e_i},{e_j}){e_i},{e_j}} \right\rangle }.$

Besides, by the definition of the (scalar-valued) second fundamental form $K(\cdot,\cdot)$ with respect to the unit normal vector $e_0$, i.e., $\displaystyle K(X,Y)=\langle\mathrm{I\!I}(X,Y),e_0\rangle.$

The respective mean curvature $H$ which is nothing but the trace of the above second fundamental form $K$ is $\displaystyle H=\text{trace}_gK =\sum\limits_{i= 1}^nK(e_i,e_i).$

In the case of a hypersurface, there holds $\displaystyle \mathrm{I\!I}(X,Y)=\langle e_0,e_0 \rangle K(X,Y)e_0.$

Therefore, $\displaystyle\begin{array} {lcl}\displaystyle\sum\limits_{i,j = 1}^n {\left\langle {\mathrm{I\!I}({e_i},{e_i}),\mathrm{I\!I}({e_j},{e_j})} \right\rangle } &=&\displaystyle \sum\limits_{i,j = 1}^n {\left\langle {\left\langle {{e_0},{e_0}} \right\rangle K({e_i},{e_i}){e_0},\left\langle {{e_0},{e_0}} \right\rangle K({e_i},{e_j}){e_0}} \right\rangle } \hfill \\&=&\displaystyle \left\langle {{e_0},{e_0}} \right\rangle \sum\limits_{i,j = 1}^n {K({e_i},{e_i})K({e_j},{e_j})} \hfill \\&=&\displaystyle - \left( {\sum\limits_{i = 1}^n {K({e_i},{e_i})} } \right)\left( {\sum\limits_{j = 1}^n {K({e_j},{e_j})} } \right) \hfill \\&=& - {H^2}. \hfill \\ \end{array}$

For the remaining term, we observe that $\displaystyle\begin{array} {lcl}\displaystyle\sum\limits_{i,j = 1}^n \left\langle {\mathrm{I\!I}({e_i},{e_j}),\mathrm{I\!I}({e_i},{e_j})} \right\rangle &=&\displaystyle \sum\limits_{i,j = 1}^n {\left\langle {\left\langle {{e_0},{e_0}} \right\rangle K({e_i},{e_j}){e_0},\left\langle {{e_0},{e_0}} \right\rangle K({e_i},{e_j}){e_0}} \right\rangle } \hfill \\&=&\displaystyle \left\langle {{e_0},{e_0}} \right\rangle \sum\limits_{i,j = 1}^n {K({e_i},{e_j})K({e_i},{e_j})} \hfill \\&=&\displaystyle - \sum\limits_{i,j = 1}^n {K{{({e_i},{e_j})}^2}} \hfill \\&=&\displaystyle - |K|_g^2.\end{array}$

In other words, we have shown that $\displaystyle\sum\limits_{i,j = 1}^n {\left\langle {\overline R \left( {{e_i},{e_j}} \right){e_i},{e_j}} \right\rangle } = \text{Scal}_g - |K|_g^2 + {H^2}.$

By definition of the Ricci curvature, we have $\displaystyle\text{Ric}_{\overline g}({e_i},{e_i}) = - \left\langle {\overline R \left( {{e_i},{e_0}} \right){e_i},{e_0}} \right\rangle + \sum\limits_{j = 1}^n {\left\langle {\overline R \left( {{e_i},{e_j}} \right){e_i},{e_j}} \right\rangle },$

since $\langle e_0,e_0 \rangle=-1$. Therefore, $\displaystyle\begin{array} {lcl}\displaystyle\sum\limits_{i,j = 1}^n \left\langle {\overline R \left( {{e_i},{e_j}} \right){e_i},{e_j}} \right\rangle &=&\displaystyle\sum\limits_{i = 1}^n {\left( {\text{Ric}_{\overline g}({e_i},{e_i}) + \left\langle {\overline R \left( {{e_i},{e_0}} \right){e_i},{e_0}} \right\rangle } \right)} \hfill \\&=&\displaystyle \text{Ric}_{\overline g}({e_0},{e_0}) + \sum\limits_{i = 1}^n {\text{Ric}_{\overline g}({e_i},{e_i})} \hfill \\&=&\displaystyle 2\text{Ric}_{\overline g}({e_0},{e_0}) + \text{Scal}_{\overline g}. \end{array}$

Since $\displaystyle\text{Ric}_{\overline g} - \frac{1}{2}{\overline g} \text{Scal}_{\overline g} = \kappa T-\Lambda\overline g,$

we obtain $\displaystyle\underbrace{\text{Ric}_{\overline g}({e_0},{e_0}) - \frac{1}{2}\overline g({e_0},{e_0})\text{Scal}_{\overline g}}_{\text{Ric}_{\overline g}({e_0},{e_0}) + \frac{1}{2}\text{Scal}_{\overline g}} = \kappa T({e_0},{e_0})-\Lambda\overline g(e_0,e_0).$

Thus, we have proved that $\displaystyle \text{Scal}_g - |K|_g^2 + (\text{trace}_gK)^2 = 2\kappa \rho + 2\Lambda,$

known as the Hamitonian constraint equation.

Reference: Justin Corvino, Introduction to General Relativity and the Einstein Constraint Equations, Lecture notes.

1. Làm sao mà QA lại viết được công thức toán hay như vậy, có dùng lệnh \$latex … không?

Comment by doanchi — December 7, 2012 @ 17:10

• Có dùng chứ anh, ý anh là sao mà lại bảo em là hay như vậy :(…

Comment by Ngô Quốc Anh — December 7, 2012 @ 17:12

• À, vì anh thấy các công thức đó rất tự nhiên, gần như kiểu mình gõ tex, chứ không có cảm giác gõ tex trong WP nên hỏi thôi mà…

Comment by doanchi — December 10, 2012 @ 12:17

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