# Calculate lapse function from the metric

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.
• How is $\nabla_af$ calculated? Does one treat $f$ as a scalar? – Ryan Unger Jan 15 '16 at 16:18
• Yes, $f$ is a scalar — it's a real-valued function on the spacetime manifold $\mathcal{M}$. – Michael Seifert Jan 15 '16 at 17:44