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?


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 Jan 15 '16 at 16:18
  • $\begingroup$ Yes, $f$ is a scalar — it's a real-valued function on the spacetime manifold $\mathcal{M}$. $\endgroup$ – Michael Seifert Jan 15 '16 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 Jan 22 '16 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. $\endgroup$ – Michael Seifert Jan 22 '16 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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