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I was trying to read about 3+1 decomposition of spacetime from section 12.2 of Padmanabhan's book Gravitation Foundations and Frontiers. However, other sources can also provide the context for my question.

Once the coordinate system $(t,y^\alpha)$ has been adopted on the spacetime from the foliation, $x^a=x^a(t,y^\alpha)$, then we can write (also the book uses the convention $a=0,1,2,3$; $\alpha=1,2,3$ or latin indices represent spacetime and greek indices only space), \begin{align} dx^a&=\frac{\partial x^a}{\partial t}dt+\frac{\partial x^a}{\partial y^\alpha}dy^\alpha\nonumber\\ &=t^adt+e^a_\alpha dy^\alpha\nonumber\\ &=\left(Nn^a+N^\alpha e^a_\alpha\right)dt+e^a_\alpha dy^\alpha\nonumber\\ &=\left(Ndt\right)n^a+\left(N^\alpha dt+dy^\alpha\right)e^a_\alpha \end{align} Where we have used the fact that the tangent to the curves parametrized by $t$ is $t^a=\partial x^a/\partial t=Nn^a+N^\alpha e^a_\alpha$; and $N$ is called the lapse function and $N^\alpha$ is called the shift vector. $e^a_\alpha=\partial x^a/\partial y^\alpha$ are the tangent to the hypersurface called tetrads.

The line element (squared) now becomes, \begin{align} ds^2&=g_{mn}dx^mdx^n\nonumber\\ &=g_{mn}\left[\left(Ndt\right)n^m+\left(N^\alpha dt+dx^\alpha\right)e^m_\alpha\right]\left[\left(Ndt\right)n^n+\left(N^\beta dt+dx^\beta\right)e^n_\beta\right]\nonumber\\ &=-N^2dt^2+h_{\alpha\beta}\left(dx^\alpha+N^\alpha dt\right)\left(dx^\beta+N^\beta dt\right), \end{align} here, \begin{align} h_{\alpha\beta}=g_{mn}e^m_\alpha e^n_{\beta}=g_{\alpha\beta}. \end{align} The metric can be read out from the above line element, \begin{align} g_{00}=-N^2+N_\gamma N^\gamma,\quad g_{0\alpha}=N_\alpha,\quad g_{\alpha\beta}=h_{\alpha\beta} \end{align} In matrix form, \begin{align} g_{mn}=\begin{pmatrix} -N^2+N_\gamma N^\gamma & N_\alpha\\ N_\alpha & h_{\alpha\beta} \end{pmatrix} \end{align}

My question is how to calculate the inverse of this metric?

I have tried to do that but not succeeded except for the component $g^{00}$ and I am not sure whether that derivation is correct. So let me describe the process in the following.

Now, as $\partial_a t=\delta^t_a=\delta^0_a$ in the coordinate system $(t,y^\alpha)$. Thus, \begin{align} g^{00}&=g^{ab}\partial_a t\partial_b t\nonumber\\ &=\frac{1}{N^2}g^{ab}n_an_b\nonumber\\ &=-N^{-2}. \end{align} Where I have used the fact that the normal vectors are defined as $n_a=-N\partial_a t$ and the normalization for spacelike hypersurfaces is such that $n^an_a=-1$.

In the book, the components for the inverse metric are given to be, \begin{align} g^{00}=-N^{-2},\quad g^{0\alpha}=N^{-2}N^{\alpha},\quad g^{\alpha\gamma}=h^{\alpha\gamma}-N^{-2}N^\alpha N^\gamma \end{align}

Therefore the answer I am looking for is the step by step derivation of the inverse metric given the components of the metric and also one should verify whether my calculation for $g^{00}$ is correct. Thank you.

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Let me do it for once and for all. Though the question has been answered by spiridon, I would like to give a formal derivation as spiridon's answer involves guess work. We have a situation where we need to calculate the inverse of a partitioned matrix. So let us first derive a general formula for the inverse of partitioned matrices and then we shall apply it to the metric.

Let two non-singular $n\times n$ matrices $A$ and $B$ be partitioned as follows, \begin{align} A=\begin{pmatrix} A_{11} & A_{12}\\ A_{21} & A_{22} \end{pmatrix},\quad B=\begin{pmatrix} B_{11} & B_{12}\\ B_{21} & B_{22} \end{pmatrix}. \end{align} Let $A_{11}$ and $B_{11}$ be $k\times k$ matrices with $k<n$. We shall also assume, \begin{align} \det (A_{11})\neq0;\quad\det (A_{22})\neq0. \end{align} Now, if $B=A^{-1}$, then we shall find the component matrices of $B$ in terms of the component matrices of $A$. We have, \begin{align} \begin{pmatrix} A_{11} & A_{12}\\ A_{21} & A_{22} \end{pmatrix}\begin{pmatrix} B_{11} & B_{12}\\ B_{21} & B_{22} \end{pmatrix}=\begin{pmatrix} I_{k\times k} & O_{k\times n-k}\\ O_{n-k\times k} & I_{n-k\times n-k} \end{pmatrix} \end{align} This matrix relation reduces to, \begin{align} A_{11}B_{11}+A_{12}B_{21}&=I_{k\times k}\qquad &&(1)\\ A_{11}B_{12}+A_{12}B_{22}&=O_{k\times n-k}\qquad &&(2)\\ A_{21}B_{11}+A_{22}B_{21}&=O_{n-k\times k}\qquad &&(3)\\ A_{21}B_{12}+A_{22}B_{22}&=I_{n-k\times n-k}\qquad &&(4) \end{align} From (2) and (3) we have, \begin{align} B_{12}=-A_{11}^{-1}A_{12}B_{22}\\ B_{21}=-A_{22}^{-1}A_{21}B_{11} \end{align} Substituting these into (1) and (4), we get, \begin{align} \left(A_{11}-A_{12}A^{-1}_{22}A_{21}\right)B_{11}&=I_{k\times k}\\ \left(A_{22}-A_{21}A^{-1}_{11}A_{12}\right)B_{22}&=I_{n-k\times n-k} \end{align} Hence, \begin{align} B_{11}&=\left(A_{11}-A_{12}A^{-1}_{22}A_{21}\right)^{-1}\\ B_{22}&=\left(A_{22}-A_{21}A^{-1}_{11}A_{12}\right)^{-1} \end{align} Now substituting these into (2) and (3), we get, \begin{align} B_{12}&=-A_{11}^{-1}A_{12}B_{22}=-A_{11}^{-1}A_{12}\left(A_{22}-A_{21}A^{-1}_{11}A_{12}\right)^{-1}\\ B_{21}&=-A_{22}^{-1}A_{21}B_{11}=-A_{22}^{-1}A_{21}\left(A_{11}-A_{12}A^{-1}_{22}A_{21}\right)^{-1} \end{align} Therefore, \begin{align} B=A^{-1}=\begin{pmatrix} \left(A_{11}-A_{12}A^{-1}_{22}A_{21}\right)^{-1} & -A_{11}^{-1}A_{12}\left(A_{22}-A_{21}A^{-1}_{11}A_{12}\right)^{-1}\\ -A_{22}^{-1}A_{21}\left(A_{11}-A_{12}A^{-1}_{22}A_{21}\right)^{-1} &\left(A_{22}-A_{21}A^{-1}_{11}A_{12}\right)^{-1} \end{pmatrix} \end{align} For our purpose it would be convenient to expand, $\left(A_{22}-A_{21}A^{-1}_{11}A_{12}\right)^{-1}$ in terms of Woodbury matrix identity. First, let us derive the identity. Note that, \begin{align} U+UCVM^{-1}U=UC\left(C^{-1}+VM^{-1}U\right)=\left(M+UCV\right)M^{-1}U \end{align} This implies, \begin{align} \left(M+UCV\right)^{-1}UC=M^{-1}U\left(C^{-1}+VM^{-1}U\right)^{-1}, \end{align} given all the required inverses exist! Then, \begin{align} M^{-1}&=\left(M+UCV\right)^{-1}\left(M+UCV\right)M^{-1}\nonumber\\ &=\left(M+UCV\right)^{-1}\left(I+UCVM^{-1}\right)\nonumber\\ &=\left(M+UCV\right)^{-1}+\left(M+UCV\right)^{-1}UCVM^{-1}\nonumber\\ &=\left(M+UCV\right)^{-1}+M^{-1}U\left(C^{-1}+VM^{-1}U\right)^{-1}VM^{-1} \end{align} Thus, \begin{align} \left(M+UCV\right)^{-1}=M^{-1}-M^{-1}U\left(C^{-1}+VM^{-1}U\right)^{-1}VM^{-1} \end{align} The above identity is called the Woodbury matrix identity. Now, identifying $M=A_{22}$, $U=-A_{21}$, $C=A_{11}^{-1}$ and $V=A_{12}$, we get, \begin{align} \left(A_{22}-A_{21}A_{11}^{-1}A_{12}\right)^{-1}=A_{22}^{-1}+A_{22}^{-1}A_{21}\left(A_{11}-A_{12}A_{22}^{-1}A_{21}\right)^{-1}A_{12}A_{22}^{-1}. \end{align} Therefore, we finally have, \begin{align} A^{-1}= \left( \begin{array}{c|c} \left(A_{11}-A_{12}A^{-1}_{22}A_{21}\right)^{-1} & -A_{11}^{-1}A_{12}\left(A_{22}-A_{21}A^{-1}_{11}A_{12}\right)^{-1}\\ \hline -A_{22}^{-1}A_{21}\left(A_{11}-A_{12}A^{-1}_{22}A_{21}\right)^{-1} &A_{22}^{-1}+A_{22}^{-1}A_{21}\left(A_{11}-A_{12}A_{22}^{-1}A_{21}\right)^{-1}A_{12}A_{22}^{-1} \end{array} \right) \end{align} After deriving this general formula, let us go back to calculating the inverse of the metric. We have, \begin{align} g_{mn}= \left( \begin{array}{c|c} -N^2+N_\gamma N^\gamma & N_\alpha \\ \hline N_{\alpha} & h_{\alpha\beta} \end{array} \right) \end{align} Now, \begin{align} A_{11}=-N^2+N_\gamma N^\gamma, \quad A_{12}=N_\alpha,\quad A_{21}=N_\alpha,\quad A_{22}=h_{\alpha\beta}. \end{align} We also note that, $A_{22}^{-1}=(h_{\alpha\beta})^{-1}=h^{\alpha\beta}$. Then, \begin{align} g^{00}=\left(A_{11}-A_{12}A^{-1}_{22}A_{21}\right)^{-1}=(-N^2+N_\gamma N^\gamma-N_\alpha h^{\alpha\beta}N_\beta)^{-1}=-N^{-2}, \end{align} and \begin{align} g^{\alpha 0}&=-A_{22}^{-1}A_{21}\left(A_{11}-A_{12}A^{-1}_{22}A_{21}\right)^{-1}\nonumber\\ &=-h^{\alpha\beta}N_\beta(-N^2+N_\gamma N^\gamma-N_\alpha h^{\alpha\beta}N_\beta)^{-1}=N^{-2}N^\alpha=g^{0\alpha}, \end{align} and finally, \begin{align} g^{\alpha\beta}&=A_{22}^{-1}+A_{22}^{-1}A_{21}\left(A_{11}-A_{12}A_{22}^{-1}A_{21}\right)^{-1}A_{12}A_{22}^{-1}\\ &=h^{\alpha\beta}+h^{\alpha\gamma}N_\gamma\left(-N^2+N^\sigma N_\sigma-N_\xi h^{\xi\mu}N_\mu\right)^{-1}N_\rho h^{\rho\beta}\nonumber\\ &=h^{\alpha\beta}-N^{-2}N^\alpha N^\beta \end{align} Voila! Enjoy!

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  • $\begingroup$ This is lengthy but very rigorous otherwise. I was having a bit hard time understanding why the guesses taken by @spiridon_the_sun_rotator should be correct, but I understand this direct computation better. Thanks! $\endgroup$ – Faber Bosch Sep 3 '20 at 6:05
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Well, maybe there is a more clear way to do this, without some guessing. I would start from the definition of an inverse matrix: $$ g^{\mu \alpha} g_{\alpha \nu} = \delta_{\nu}^{\mu} $$ Or more concretely: $$ \begin{pmatrix} -N^2+N_\gamma N^\gamma & N_\alpha\\ N_\alpha & h_{\alpha\beta} \end{pmatrix} \begin{pmatrix} g^{00} & g^{0 \alpha}\\ g^{0 \alpha} & g^{\alpha \beta} \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $$ Written in components: $$ \begin{align} (-N^2+N_\gamma N^\gamma) g^{00} + N_\alpha g^{0 \alpha} = 1 \\ (-N^2+N_\gamma N^\gamma) g^{0\alpha} + N_\beta g^{\beta \alpha} = 0 \\ N_\alpha g^{0\beta} + h_{\alpha \gamma} g^{\gamma \beta } = \delta_\alpha^{\beta} \end{align} $$ Now, using the symmetry of $g_{\mu \nu}, h_{\mu \nu}$ under exchange of $\mu \leftrightarrow \nu$, one may see, that there are $ D(D+1) / 2$ linear equations on the same number of unknows, which can be in principle solved.

Doing these directly seems a tedious task, so there can be an educated guess. Supposing we knew that $g^{00}$ is $-N^2$, in general ansatz could be $\alpha N^2 + \beta N_\alpha N^{\alpha}$, then the first equation is immediately resolved by setting : $$ g^{0 \alpha} = N^{-2} N^{\alpha} $$ Then one may look on the second line. Here is also natural to assume, that $g^{\mu \nu} = h^{\mu \nu} + b^{\mu \nu}$, where $b^{\mu \nu}$ is also symmetric. This substitution gives: $$ -N^{\alpha} - N^{-2} N_\gamma N^\gamma N^{\alpha} + N^{\alpha} + N_\beta b^{\beta \alpha} = 0 $$ Here also, one may see, that the $b^{\mu \nu} = -N^{-2} N^{\beta} N^{\alpha}$ does the job.

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  • $\begingroup$ Wow! This is much cleaner approach. Thank you very much. $\endgroup$ – Faber Bosch Aug 16 '20 at 10:51
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This answer slightly extends the one of spiridon, and rephrases parts of OP's setup in slightly different language.

The inverse metric $g^{-1}$, being a tensor, is coordinate independent. Thus, one way to determine the components of the inverse metric in a particular coordinate system, is to derive it from a coordinate independent representation. To wit, if the inverse metric in a basis $\{{\bf e}_a\}$ is given by $$ g^{-1} = g^{ab}\, {\bf e}_a \otimes {\bf e}_b, $$ then its components are given by the action of $g^{-1}$ on the dual basis $\{{\bf e}^a\}$: $$ g^{ab} = g^{-1}({\bf e}^a,{\bf e}^b). $$ The 3+1 decomposition of spacetime is realised by the level surfaces (really hypersurfaces) of a scalar field $f$. A unit normal is $n^a = - N g^{ab} \nabla_b f$. From the unit normal $n^a$ one can construct projectors parallel ($P_\parallel$) and orthogonal ($P_\perp$) to it. Their components are given by the expressions $$ P_\parallel{}^{a}{}_{b} \equiv - n^a n_b, \qquad P_\perp{}^{a}{}_{b} \equiv \delta^a_b - P_\parallel{}^{a}{}_{b} = \delta^a_b - n^a n_b. $$ With these projectors one can determine the components of the metric $g_{ab}$ in terms of the hypersurface foliation: $$ g_{ab} = h_{ab} - n_a n_b \equiv P_\perp{}^{c}{}_{a} P_\perp{}^{d}{}_{b}g_{cd} - n_a n_b. $$ The tensor field $h_{ab}$ is the induced metric on the hypersurfaces, as every contraction of it with the unit normal vanishes. Similarly one can verify that the components of the inverse metric satisfy $$ g^{ab} = h^{ab} - n^a n^b \equiv P_\perp{}^{a}{}_{c} P_\perp{}^{b}{}_{d}g^{cd} - n^a n^b.\tag{1}\label{inverse} $$ On a given hypersurface $f=t$, one introduces a set of one-parameter coordinates $y^\alpha$ that vary smoothly as a function of $t$. This generates a set of vector fields $e_\alpha{}^a \equiv \partial x^a/\partial y^\alpha$ tangential to the hypersurface, that serve as an embedding map from the hypersurface to spacetime. In particular, the induced metric can be expressed in terms of these new coordinates via the relation $h_{\alpha\beta}=h_{ab}e_\alpha{}^a e_\beta{}^b$. In this coordinate system, the time vector $t^a$ is generally not orthogonal to the hypersurface, but can be decomposed into orthogonal $N$ and tangential $N^\alpha$ parts: $$ t^a = Nn^a + N^\alpha e_\alpha{}^a.\tag{2}\label{decomposition} $$ Note that $\nabla_a f = -N^{-1}n_a$ is dual to the time vector $t^a$. Substitution of \eqref{decomposition} into \eqref{inverse} then yields $$ g^{\mu\nu} = - N^{-2} t^\mu t^\nu + N^{-2} t^\mu N^\alpha e_\alpha{}^\nu + N^{-2} t^\nu N^\alpha e_\alpha{}^\mu + \left(h^{\alpha\beta} - N^{-2} N^\alpha N^\beta \right) e_\alpha{}^a e_\beta{}^b. $$ The components of the inverse metric in the given coordinate system can then be found by contraction: \begin{align} g^{00} &= g^{ab}\nabla_a f \nabla_b f = - N^{-2},\\ g^{0\alpha} &= g^{ab} \nabla_a f\; e_b{}^\alpha = N^{-2} N^\alpha= g^{a0},\\ g^{\alpha\beta} &= g^{ab}e_a{}^\alpha e_b{}^\beta = h_{\alpha\beta} - N^{-2} N^\alpha N^\beta. \end{align}

References:

  • E. Poisson (2007), A Relativist’s Toolkit - chapters 3, 4
  • E. Gourgoulhon (2012), 3+1 Formalism and Bases of Numerical Relativity - chapters 2, 3
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