Recently I have started to study the classical theory of gravity. In Landau, Classical Theory of Field, paragraph 84 ("Distances and time intervals") , it is written

We also state that the determinanats $g$ and $\gamma$, formed respectively from the quantities $g_{ik}$ and $\gamma_{\alpha\beta}$ are related to one another by $$-g=g_{00}\gamma $$ I don't understand how this relation between the determinants of the metric tensors can be obtained. Could someone explain, or make some hint, or give a direction?

In this formulas $g_{ik}$ is the metric tensor of the four-dimensional space-time and $\gamma_{\alpha\beta}$ is the corresponding three-dimensional metric tensor of the space. These tensors are related to one another by the following formulas $$\gamma_{\alpha\beta}=(-g_{\alpha\beta}+\frac{g_{0\alpha}g_{0\beta}}{g_{00}})$$ $$\gamma^{\alpha\beta}=-g^{\alpha\beta}$$

Thanks a lot.


3 Answers 3


In my opinion it is better to work in an explicit covariant form. In my answer I will use two different definitions, the Greek indexes always run from $0$ to $3$ and Latin indexes from $1$ to $3$ and the metric $g_{\mu\nu}$ has signature $(-1,1,1,1)$.

To translate the expressions to a explicit covariant form we define some timelike vector field $v^\mu$. We can define an adapted coordinate system such that $v^\mu = \delta^\mu{}_0$ and, therefore, $$g_{00} = g_{\mu\nu}v^\mu{}v^\nu.$$ However, it is better (in my opinion) to not work in such coordinate system, in this case we always use expression like the right hand side of the above, i.e., $v^2 \equiv g_{\mu\nu}v^\mu{}v^\nu$.

The projected metric from your equation can be expressed as (where I reworked the signs to conform to my definitions) $$\gamma_{ij} = g_{ij} - \frac{g_{0i}g_{0j}}{g_{00}} = g_{ij} - \frac{v^\mu v^\nu g_{\mu i}g_{\nu j}}{v^2} = g_{ij} + \frac{v_i v_j}{\sqrt{-v^2}\sqrt{-v^2}} = g_{ij} + n_i n_j,$$ where we defined, naturally, $v_\mu = g_{\mu\nu}v^\nu$ and the normalized version of $v^\mu$, i.e., $n^\mu \equiv v^\mu/\sqrt{-v^2}$. Note that here we introduced the rest of the coordinate basis given by the vector fields $e_i{}^\mu$ such that $g_{ij} = e_i{}^\mu{}e_j{}^\nu{}g_{\mu\nu}$, $v_i = v_\mu{}e_i{}^\mu$, etc. Then, to avoid the introduction of coordinates we define the projector $$\gamma_{\mu\nu} = g_{\mu\nu} + n_\mu{}n_\nu,$$ which reduces to the spatial metric in the adapted coordinate system and in general works as a projector $\gamma_{\mu\nu}n^\nu = 0$.

One can introduce the determinant in a covariant format using the space of totally antisymmetric tensors of type $(4,0)$, e.g. $\epsilon^{\mu\nu\alpha\beta}$ which is antisymmetric in any two adjacent indexes. We define $\epsilon^{\mu\nu\alpha\beta}$ using the expression $$\epsilon^{\mu\nu\alpha\beta}\epsilon_{\mu\nu\alpha\beta} = -4!,$$ where the indexes were lowered using the metric $g_{\mu\nu}$, this defines $\epsilon^{\mu\nu\alpha\beta}$ up to a signal since the space of totally antisymmetry tensors is one dimensional. Using the definition of determinant in terms of the Levi-Civita symbol it is easy to show (you can take a look at Appendix B of Wald 1984) that $$v^\mu{}e_1{}^\nu{}e_2{}^\alpha{}e_3{}^\beta\epsilon_{\mu\nu\alpha\beta} = \epsilon_{0123} = \sqrt{-g},$$ where $g$ is the determinant of $g_{\mu\nu}$ calculated in the coordinate basis formed by $(v^\mu,e_i{}^\mu)$.

Finally, $\epsilon^{\mu\nu\alpha\beta}\epsilon_{\mu\nu\alpha\beta}$ is expressed as follows, $$\begin{align} \epsilon_{\mu\nu\alpha\beta}\epsilon_{\lambda\sigma\phi\psi}g^{\mu\lambda}g^{\nu\sigma}g^{\alpha\phi}g^{\beta\psi} &= \epsilon_{\mu\nu\alpha\beta}\epsilon_{\lambda\sigma\phi\psi}(\gamma^{\mu\lambda}-n^{\mu}n^{\lambda})(\gamma^{\nu\sigma}-n^{\nu}n^{\sigma})(\gamma^{\alpha\phi}-n^{\alpha}n^{\phi})(\gamma^{\beta\psi}-n^{\beta}n^{\psi}), \\ &= -4n^{\mu}n^{\lambda}\epsilon_{\mu\nu\alpha\beta}\epsilon_{\lambda\sigma\phi\psi}\gamma^{\nu\sigma}\gamma^{\alpha\phi}\gamma^{\beta\psi}, \end{align}$$ where we used that any contraction of two $n^\mu$ with $\epsilon_{\mu\nu\alpha\beta}$ is null. Now, using that $\epsilon^{\mu\nu\alpha\beta}\epsilon_{\mu\nu\alpha\beta} = -4!$, we obtain $$n^{\mu}n^{\lambda}\epsilon_{\mu\nu\alpha\beta}\epsilon_{\lambda\sigma\phi\psi}\gamma^{\nu\sigma}\gamma^{\alpha\phi}\gamma^{\beta\psi} = -v^{-2}\epsilon_{0ijk}\epsilon_{0lmn}\gamma^{il}\gamma^{jm}\gamma^{kn} = 3!,$$ using again the formula of the determinant through the antisymmetric symbol we obtain $$v^2\gamma = -(\epsilon_{0123})^2 = g \qquad \Rightarrow\qquad -g = -g_{00}\gamma,$$ where the difference of signs comes from the different signature definition.

In terms of volume forms this result is equivalent to $$\tilde{\epsilon} = -4\tilde{n}\wedge{}^3\tilde{\epsilon},$$ where $\tilde{\epsilon}$ is just the four volume form with its components given by $\epsilon_{\mu\nu\alpha\beta}$ and ${}^3\tilde{\epsilon}$ is the three dimensional induced volume form with its components given by $n^\mu\epsilon_{\mu\nu\alpha\beta}$.

  • $\begingroup$ About the last equation: shouldn't it be $4! \tilde{\epsilon} = -3! \tilde{n}\wedge {}^3\tilde{\epsilon}$ by the definition of the wedge product? $\endgroup$
    – auxsvr
    Jun 30, 2014 at 9:18
  1. Consider the $4\times 4$ matrix $g_{\mu\nu}$ with zeroth row $g_{0\nu}$.

  2. Now for $i=1,2,3$, add to the $i$'th row the zeroth row times $-g_{i0}/g_{00}$.

  3. This produces the following matrix $$\begin{bmatrix} g_{00} & g_{01} & g_{02}& g_{03} \\ 0 & -\gamma_{11} & -\gamma_{12}& -\gamma_{13} \\ 0 & -\gamma_{21} & -\gamma_{22}& -\gamma_{23} \\ 0 & -\gamma_{31} & -\gamma_{32}& -\gamma_{33} \end{bmatrix}.$$

  4. Such row manipulations do not change the determinant. So it is still $g=\det(g_{\mu\nu})$.

  5. On the other hand the determinant can be expanded in the zeroth column to yield $g_{00}\times \det(-\gamma_{ij})$.

  6. Hence we obtain the result $g=-g_{00}\det(\gamma_{ij})$.


You have to be extremely careful about what conventions you are using for defining your 4-metric and your 3-metric and your time vector.

In particular, if you are using coordinates where your metric has off-diagonal components, be aware about what the value for your unit normal is, and I would stronlgy advice choosing your slicing condition as one of your four coordinates, so that you're choosing surfaces of $\tau = $constant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.