User: diener Date: 2012/04/03 04:52 PM

Modified: /numerical_relativity/ numerical_relativity.tex

Log: Add some slides on finite differencing.

File Changes:

Directory: /numerical_relativity/ =================================

File [modified]: numerical_relativity.tex Delta lines: +89 -27 =================================================================== --- numerical_relativity/numerical_relativity.tex 2012-04-03 18:30:44 UTC (rev 28) +++ numerical_relativity/numerical_relativity.tex 2012-04-03 21:52:37 UTC (rev 29) @@ -15,7 +15,7 @@ % \item 3+1 spacetime decomposition. \item The BSSN formulation. \item Puncture data. -% \item Finite Differencing. + \item Finite Differencing. % \item The method of lines. \end{itemize} } @@ -31,33 +31,33 @@ \begin{align} {\cal H} & \equiv R + K^2 - K_{ij} K^{ij} = 0, \ {\cal M}^i & \equiv D_j (K^{ij} - \gamma^{ij} K) = 0. -\end{align} -These equations were used a lot in the early years of numerical relativity. +\end{align}\pause +These equations were used a lot in the early years of numerical relativity.\pause

However, it was later discovered that the ADM evolution equations are only -weakly hyperbolic. +weakly hyperbolic.\pause

In the mid to late 90's a new formulation was introduced that proved to be much more robust and stable: The BSSN formulation. }

\frame{\frametitle{The BSSN formulation (continued)} -\setlength{\vs}{-1.5ex} +\setlength{\vs}{-0.5ex} Introduce a conformal rescaling of the three metric\vspace{\vs} \begin{equation} \gamma_{ij} = \psi^4 \tg_{ij}.\vspace{\vs} -\end{equation} -We choose $\psi = \gamma^{1/12}$ such that the determinant of $\tg_{ij}$ is 1\vspace{\vs} +\end{equation}\pause +We choose $\psi = \gamma^{1/12}$ such that the determinant of $\tg_{ij}$ is 1%\vspace{\vs} \begin{equation} \mathrm{det}(\tg_{ij}) = \mathrm{det}(\psi^{-4}\gamma_{ij}) = \mathrm{det}(\gamma^{-1/3}\gamma_{ij}) = -\gamma^{-1} \mathrm{det}(\gamma_{ij}) = 1. \vspace{\vs} -\end{equation} +\gamma^{-1} \mathrm{det}(\gamma_{ij}) = 1.\vspace{\vs} +\end{equation}\pause In addition we introduce a trace decomposition of the extrinsic curvature.\vspace{\vs} \begin{align} K & = \gamma^{ij} K_{ij}, \ A_{ij} & = K_{ij} - \frac{1}{3} \gamma_{ij} K.\vspace{\vs} -\end{align} +\end{align}\pause We then promote the following variables to evolution variables\vspace{\vs} \begin{align} \phi &= \ln \psi = \frac{1}{12} \ln \gamma, @@ -75,16 +75,17 @@ \begin{equation} \tG^i = \tg^{jk} \tG^i{}_{jk} = - \partial_j \tg^{ij}, \end{equation} -to evolved variables as well. +to evolved variables as well.\pause + The final set of evolution variables are $\phi$, $K$, $\tg_{ij}$, $\tA_{ij}$ -and $\tG^i$. +and $\tG^i$.\pause

The BSSN evolution equations can be derived from the ADM equations. As an example take the equation for $\phi$. \begin{equation} \dt \phi = \partial_0\phi = \partial_0 \left (\frac{1}{12}\ln\gamma\right) = \frac{1}{12}\frac{1}{\gamma}\partial_0\gamma. -\end{equation} +\end{equation}\pause Using the expression for the derivative of the determinant of the metric in terms of the derivatives of the metric ($\partial_0\gamma = \gamma \gamma^{ij} \partial_0\gamma_{ij}$) we find @@ -114,13 +115,11 @@ & - 2 \tilde{A}^{ij} \partial_j\alpha + 2 \alpha ( \tilde{\Gamma}^i{}_{jk} \tilde{A}^{jk} + 6 \tilde{A}^{ij} \partial_j \phi - \frac{2}{3} \tg^{ij} \partial_j K). -\end{align}%\vspace{\vs} +\end{align} }

\frame{\frametitle{The BSSN formulation (continued)} -%\raisebox{-1.2\vs}[0pt][0pt]{ Here $R_{ij} = \tilde{R}_{ij} + R^{\phi}_{ij}$, where\vspace{1.0\vs} -%}\vspace{\vs} \begin{align} R^{\phi}_{ij} = {} & - 2 \tilde{D}_i \tilde{D}_j \phi - 2 \tilde{\gamma}_{ij} @@ -142,18 +141,18 @@ \tilde{{\cal G}} & \equiv \tg-1 = 0, \ \tilde{{\cal A}} & \equiv \tg^{ij}\tA_{ij} = 0, \ \tilde{{\cal L}}^{i} & \equiv \tG^{i}+\partial_j\tg^{ij} = 0. -\end{align} +\end{align}\pause The constraints $\tilde{{\cal G}}$ and $\tilde{{\cal A}}$ are enforced actively -at each timestep. +at each timestep.\pause

The other constraints ($\tilde{{\cal H}}$, $\tilde{{\cal M}}^i$ and -$\tilde{{\cal L}}^{i}$) are not enforced. +$\tilde{{\cal L}}^{i}$) are not enforced.\pause

To improve stability and to help maintain $\tilde{{\cal L}}^{i}$ at a low level the following rule is employed in an implementation \begin{itemize} \item Where derivatives of $\tG^{i}$ are needed the evolved $\tG^{i}$ are used - directly. + directly.\pause \item Where $\tG^{i}$ are needed without taking derivatives $\tg^{jk}\tG^{i}{}_{jk}$ are used instead. \end{itemize} @@ -161,7 +160,7 @@

\frame{\frametitle{The BSSN formulation (continued)} In order to evolve a spacetime with the BSSN equations, you have to specify -the gauges $\alpha$ and $\beta^{i}$. +the gauges $\alpha$ and $\beta^{i}$.\pause

Most codes use the ``moving puncture'' gauges \begin{align} @@ -179,12 +178,12 @@ \gamma^{\mathrm{ph}}_{ij} = \psi^4\gamma_{ab}, \mbox{\hspace{2em}} K^{\mathrm{ph}}_{ij} = \psi^{-2} K_{ij}, \end{equation} where $\gamma_{ij}$ is chosen to be the flat metric and $K_{ij}$ is assumed -to be tracefree. \ +to be tracefree.\pause \ The constraint equations become\vspace{\vs} \begin{align} 0 & = \Delta\psi +\frac{1}{8}K^{ij}K_{ij}\psi^{-7} \ 0 & = D_j K^{ij}. -\end{align} +\end{align}\pause The momentum constraint has an analytic solution\vspace{\vs} \begin{align} K^{ij}_{\mathrm{BY}} = {} & \frac{3}{2r^2}\left ( P^i n^j + P^j n^i - @@ -197,21 +196,84 @@

\frame{\frametitle{Puncture data (continued)} \setlength{\vs}{-0.8ex} -For $K_{ij} = 0$ the Hamiltonian constraint has a simple solution for $N$ black holes. +For $K_{ij} = 0$ the Hamiltonian constraint has a simple solution for $N$ +black holes. \begin{equation} \psi = 1+\sum_{i=1}^N\frac{m_i}{2|\vec{r}-\vec{r}_i|}. -\end{equation} +\end{equation}\pause Inspired by this, for $K_{ij}\ne 0$ we make the ansatz \begin{equation} \psi=\frac{1}{\alpha}+u, \mbox{\hspace{2em}} \frac{1}{\alpha} = \sum_{i=1}^N\frac{m_i}{2|\vec{r}-\vec{r}_i|}, -\end{equation} +\end{equation}\pause In which case the Hamiltonian constraint becomes an equation for $u$ \begin{equation} \Delta u+\frac{1}{8}\alpha^7 K^{ij}K_{ij}(1+\alpha u)^{-7} = 0. -\end{equation} +\end{equation}\pause It can be shown that $u$ will be $C^2$ at the location of the `punctures' and that $\psi$ has a unique solution. }

+\frame{\frametitle{Finite Differencing} +With finite differencing we discretize a function by sampling it at a +collection of grid points.\pause + +The grid points are usually (but not necessarily) equally spaced. \pause + +We can then approximate derivatives of a function at a grid point by a +weigthed sum of function values at grid points in the neighbourhood (the +stencil) of the grid point. \pause + +As an example consider a stencil containing the grid point ($f_i$) and it's two +nearest neighbors ($f_{i-1}$ and $f_{i+1}$) with +$\Delta x=x_{i+1}-x_i=x_i-x_{i-1}$. +\begin{equation} + \left. \frac{df}{dx} \right |_{x_i}\approx \frac{1}{\Delta x}\sum_{j=-1}^{j=1} a_j f_{i+j}. +\end{equation} +} +\frame{\frametitle{Finite Differencing (continued)} +The coefficients $a_j$ can be found be expanding $f$ in a Taylor series around +$x_i$ for the grid points in the stencil +\begin{eqnarray} +f(x_{i-1}) & = & f(x_i) - \left. \frac{df}{dx}\right |_{x_i} \Delta x + + \frac{1}{2} \left. \frac{d^2 f}{dx^2}\right |_{x_i} \Delta x^2 + + O((\Delta x)^3) \nonumber \ +f(x_i) & = & f(x_i) \nonumber \ +f(x_{i+1}) & = & f(x_i) + \left. \frac{df}{dx}\right |_{x_i} \Delta x + + \frac{1}{2} \left. \frac{d^2 f}{dx^2}\right |_{x_i} \Delta x^2 + + O((\Delta x)^3) \nonumber +\end{eqnarray} \pause +Requiring that the weighted sum approximates the derivative yields +the following equations for $a_{-1}$, $a_0$ and $a_1$ +\begin{eqnarray} +0 & = & a_{-1}+a_0+a_1 \nonumber \ +1 & = & -a_{-1}+a_1 \nonumber \ +0 & = & a_{-1}+a_1 \nonumber +\end{eqnarray} +with the solution $a_{-1}=-1/2, a_0=0, a_1=1/2$. +} + +\frame{\frametitle{Finite Differencing (continued)} +Thus we find that +\begin{equation} +\left. \frac{df}{dx} \right |_{x_i} = \frac{f_{i+1}-f_{i-1}}{2\Delta x} + + O((\Delta x)^2). +\end{equation}\pause +Similarly we find for the second derivative that +\begin{equation} +\left. \frac{d^2f}{dx^2} \right |_{x_i} = \frac{f_{i+1}-2 f_i+f_{i-1}}{(\Delta x)^2} + + O((\Delta x)^2). +\end{equation}\pause +These finite difference operators are second order accurate.\pause + +Higher order accuracy or higher order derivatives require larger stencils.\pause + +Another way of looking at finite differencing operators is through +interpolating polynomials.\pause + +Either approach gives the same coefficients for the same stencil.\pause + +It is also clear from either approach that the error estimates are only +correct if $f$ is smooth enough. +} \end{document}