A $1+n$ dimensional semi-Riemannian metric is called "regularly sliced" if it can be written as,

$ds^2 = -N^2 (\theta^0)^2 + g_{ij}\theta ^i \theta ^j$

where $N$ is called the ``Lapse Function" and the $\theta's$ are defined as follows,

$\theta ^0 = dt$

$\theta ^i = dx^i + \beta ^i dt$

where $\beta ^i$ is called the ``Shift vector"

I would like to know how this particular form of the metric is motivated. Is there something canonical or natural about this?

I am aware of Yvonne-Costakis' proof that being regularly sliced is an iff condition for the manifold to be globally hyperbolic and their proofs which use such metrics to establish conditions about the future completeness of non-spacelike geodesics depending on the behaviour of the Lapse Function.

For this metric the Christoffel symbols are apparently of the form,

$\Gamma ^i _{00} = Ng^{ij}\partial _j N $

$\Gamma ^0 _{ij} = \frac{1}{2N^2} (\partial _0 g_{ij} - g_{hj}\partial _i \beta ^h -g_{ih} \partial _j \beta ^h)$

$\Gamma ^0 _{io} = \frac{\partial_i N}{N}$

$\Gamma ^0 _{00} = \frac{\partial _0 N}{N}$

I would like to know if there is any slick way of deriving the above equations.

I had asked a similar question here.

  • 1
    $\begingroup$ The easiest way I know to calculate Christoffel symbols in general is by varying the energy functional $I=\frac{1}{2}\int g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}d\tau$ to derive the geodesic equations and get the Christoffel symbols by comparison. Is this not slick enough? $\endgroup$
    – j.c.
    Sep 21, 2010 at 15:25

2 Answers 2


This form of a semi-Riemannian metric is beloved of both numerical and mathematical relativists, but especially the former. The starting point is usually a globally hyperbolic spacetime $(M,h)$. Then $M$ can be foliated by surfaces $\Sigma_t$ of constant $t$, where $t$ is a global time coordinate i.e. a $C^1$ function $t:M\to\mathbb{R}$ such that the 1-form $dt$ is non-vanishing and everywhere timelike: $h^{-1}(dt,dt)<0$. Each $\Sigma_t$ is a Riemannian 3-manifold with metric $g$ inherited from $(M,h)$. The most convenient way to write this metric is $g=h+n\otimes n$, where $n$ is the normal 1-form of $\Sigma_t$ with unit length. The lapse and shift are not canonical or unique. They are defined with respect to a choice of future pointing timelike vector field $\vec{t}$, conventionally normalized by $dt(\vec{t})=1$. Then the lapse is $N=-n(\vec{t})$ and the shift 1-form is defined by $\beta(\vec{v})=g(\vec{t},\vec{v})$ for all tangent vectors $\vec{v}$. Note that $\beta(\vec{n})=0$. Following the integral curves of $\vec{t}$ allows one to construct a diffeomorphism relating the different $\Sigma_t$. Then one can interpret the lapse as measuring the proper time between different slices $\Sigma_t$, and $\beta$ as measuring the relative velocity of observers that follow the normal $\vec{n}$ and those that follow $\vec{t}$.

The Einstein equations can be decomposed with respect to a regular slicing, yielding evolution and constraint equations. Different gauge choices – that is, different choices of $\vec{t}$ - are used by relativists to produce different decompositions that have favourable mathematical properties, usually relating to the particular flavour of hyperbolicity of the evolution equations.

So lapse and shift certainly both carry meaning, but neither is canonically defined inasmuch as they depend upon the choice of $\vec{t}$. The regular sliced form of the metric is indeed natural, in that it arises as described above once one has identified a global time coordinate on the spacetime.

Added: This attempts to answer the first question(s). I don't think that there is a better response to the second question - a slick way of calculating the Christoffel symbols - than that given by jc. The alternatives are also standard: a direct coordinate calculation or calculation of the connection 1-forms, from which one can read off the Christoffel symbols.

Edit: (In response to Anirbit's queries.) First, about the normal 1-form $n$. This is a 1-form field on $M$ that is fixed and smooth once the (smooth) foliation by $\Sigma_t$'s is specified. So there is no choice at this level. The same holds for the form of the 3-metric $g$ on $\Sigma_t$: this is just the spacetime 4-metric $h$ restricted to tangent vectors to $\Sigma_t$: $g=h+n\otimes n$ is just the most convenient way to wite this metric.

A key point here is that each $\Sigma_t$ is a 3 dimensional Riemannian manifold, and so the metric $g$ has signature $(+,+,+)$ at each point. That is, at any point the metric $g$ has three positive eigenvalues - where the eigenvectors are tangent to $\Sigma$. This is natural when we take the 3-dim perspective. When we think of $\Sigma_t$ as being embedded in $M$, and allow $g$ to act on any tangent vector of $M$, we'd say that the signature is $(0,+,+,+)$. The zero eigenvalue corresponds to $n$.

Now we step down one more dimension to consider a spatial 2-surface $S$ embedded in $\Sigma_t$. A typical example is a 2-sphere in $\mathbb{R}^3$ considered as a slice $t=$constant of Minkowski spacetime. The normal $m$ is a spacelike vector orthogonal to $n$. Again, the metric of $S$ is inherited naturally (no choice!) and has the convenient form $g^\prime = h+n\otimes n - m\otimes m$. The minus sign arises to make sure we get the right signature $(0,0,+,+)$ for $g^\prime$: without this signature - and without the minus sign - we would not have a Riemannian metric for $S$. The source of the sign difference is that $n$ is timelike, but $m$ is spacelike.

  • $\begingroup$ Why was this answer was voted down? It addresses the first question of Anirbit's in some detail. $\endgroup$
    – j.c.
    Sep 22, 2010 at 18:40
  • $\begingroup$ I agree with jc. $\endgroup$
    – Deane Yang
    Sep 23, 2010 at 1:36
  • $\begingroup$ I have voted up the answer. (to counter whoever voted it down) Thanks Brien for this beautiful exposition. You motivation of the shift vector was very illuminating. A few points which you made were not very clear to me. Are you choosing an unit normal 1-form $n$ on $\Sigma$ for every time? Or making a choice for $n$ on one space slice is giving you a natural choice on all the space slices since all of them are diffeomorphic to each other? Further when you say, $g = h + n\otimes n$ is this a definition of $g$ given the $h$ and a choice of $n$? And why is this a good choice of $g$ ? $\endgroup$
    – Anirbit
    Oct 2, 2010 at 7:12
  • $\begingroup$ @ Brien I have seen this metric definition of yours being taken to one more level by saying that if say S is a submanifold of the spatial slice Σ with a unit normal m then the metric on S is say g′ such that, g′=h+n⊗n−m⊗m. Here too I would like to what motivates this choice and also this sign flip. $\endgroup$
    – Anirbit
    Oct 2, 2010 at 7:16
  • $\begingroup$ Anirbit - thanks for the upvote. I've responded via an edit. $\endgroup$ Oct 4, 2010 at 22:10

The normal 1-form $n$ is defined by $n(\vec{v})=0$ for all vectors $\vec{v}$ tangent to the surface $\Sigma_t$. This leaves freedom to rescale the length of $n$, and the convenient and conventional choice is to take $n$ to have unit length. In local coordinates, with $\Sigma_t$ given by $t=$ constant, we have (before normalising) $n_a=\nabla_a t$ where $\nabla$ is any derivative operator. Then $\vec{n}$ is the vector field obtained by `raising the index' of $n$: $\vec{n}(f)=n^a\frac{\partial f}{\partial x^a}=h^{ab}n_b\frac{\partial f}{\partial x^a}$.

Your interpretation of the 3-metric $g$ seems reasonable to me.

The embedding is the identity. There is no possible confusion over what metric should be used on $\Sigma_t$ since there is only one metric to play with - the spacetime metric $h$ which $\Sigma_t$ inherits by pull-back. The situation would be different if one started with a 3-dim Riemannian manifold, already equipped with a metric, and then sought to embed this in a 4-dim spacetime. (In fact this is essentially what happens when dealing with the Cauchy problem in General Relativity: in that case, the idea is to evolve the 3-metric $g$ to get the `right' spacetime metric $h$.)

(For some reason I can't add a comment, so I'm replying by way of another answer.)

  • 1
    $\begingroup$ The reason you can't comment is because you haven't yet registered, so the reputation you earned in the other account isn't added to the one you're currently posting with. If you register, you can email a moderator to merge your accounts. $\endgroup$
    – j.c.
    Nov 4, 2010 at 10:51

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