I have a technical question about the lapse function:

Assume I have some given (Lorentzian) metric $g$. I have seen the following definition of the lapse function

$\alpha^{-2}=-g(\nabla f, \nabla f)$

where $\nabla f$ is non-zero everywhere timeline vector field. Now, this definition is very abstract to me, in the sense, what exactly is $\nabla f$? How do I compute it? I understand the physical intuition behind the lapse function as a function that "measures" the distance between the space like hyper surfaces (in 3+1 decomposition).

To sum up the question: assume I'm given a metric, how do I explicitly calculate the lapse function?


1 Answer 1


The lapse function is not defined by the metric alone, but instead depends on both the metric $g_{ab}$ and its slicing into timelike hypersurfaces. One way to "slice" a spacetime $\mathcal{M}$ into timelike hypersurfaces is to define a timelike coordinate $f$, which is just a function $f: \mathcal{M} \to \mathbb{R}$ such that $\nabla_a f$ is a timelike vector field everywhere in $\mathcal{M}$. The "slices" of the spacetime are then those hypersurfaces for which $f = \text{const.}$, and the lapse function for this foliation is given by the formula you have.

However, a different time coordinate $g$ would lead to a different foliation of $\mathcal{M}$ and a different lapse function. As a trivial example, suppose that $g$ is just a function of $f$, so that the hypersurfaces are in fact the same, and $\nabla_a g = g' \nabla_a f$. Then the lapse function $\alpha_g$ corresponding to the foliation given by $g$ will be related to $\alpha_f$ by $$ \alpha_g = \frac{1}{g'} \alpha_f. $$

The fact that the foliation is arbitrary ends up manifesting itself if you look at the Lagrangian in terms of this 3+1 decomposition. The Lagrangian $\mathcal{L} = \sqrt{-g} R$ can be written in terms of the lapse, shift, 3-surface metric, and extrinsic curvature of the the 3-surfaces. However, if you do this, you find that the time derivative of the lapse (and the shift) vanish identically. This suggests that these functions are arbitrary; they can be thought of as unphysical "gauge" degrees of freedom.

  • $\begingroup$ How is $\nabla_af$ calculated? Does one treat $f$ as a scalar? $\endgroup$
    – Ryan Unger
    Commented Jan 15, 2016 at 16:18
  • $\begingroup$ Yes, $f$ is a scalar — it's a real-valued function on the spacetime manifold $\mathcal{M}$. $\endgroup$ Commented Jan 15, 2016 at 17:44
  • $\begingroup$ @MichaelSeifert a follow up question: is there a relation between the lapse function and asymptotic flatness of a spacetime? I read that for an asymptotically flat spacetime the proper time coincides with coordinate time, i.e. in limit the lapse function should go to 1. However, it also goes to 1 for an asymptotically non-flat spacetime, e.g. Kerr-NUT. Could you maybe give an insight if there is a relation? $\endgroup$
    – user46446
    Commented Jan 22, 2016 at 9:10
  • $\begingroup$ @user46446: That's probably worth asking as a new question. I would point out that the lapse function is completely arbitrary in any spacetime, as it depends on the slicing you use. More interesting questions would be: "Under what conditions does there exist a smooth global slicing whose associated lapse goes to 1 asymptotically? And if the spacetime has symmetries such as rotational or time-translation invariance, can we always choose the slicing to have these symmetries as well?" But I don't know the answers to these questions. $\endgroup$ Commented Jan 22, 2016 at 15:23

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