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?
 A: 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.
