# Lapse function and spacetime splicing

by Matterwave
Tags: function, lapse, spacetime, splicing
PF Gold
P: 1,594
 Quote by eendavid While you can find 3 independent vectors satisfying this, they simply do not span the space tangent to your hypersurface.
This is a mysterious statement.

I have 3 linearly independent vectors that are each orthogonal to $dt$. The subspace of the tangent space orthogonal to $dt$ is 3-dimensional. Therefore it is a simple fact of linear algebra that these 3 linearly-independent vectors span this space...
P: 2,953
 Quote by Ben Niehoff I can't figure out what George is getting at, sorry. The induced metric on should only be 3x3. Generically what you're doing is you have some global time function $t$ with gradient $dt$. You want to find the induced metric on the level sets of $dt$. The level sets of $dt$ are generated by a triplet of linearly-independent vector fields X, Y, Z such that $$dt(X) = dt(Y) = dt(Z) = 0.$$ In order that each level set be a surface, this set of vector fields needs to be integrable; that is, the set should be closed under the Lie bracket. This should hold automatically, given that $t$ is a global time function, and X, Y, Z are everywhere perpendicular to $dt$. Then the induced metric $h$ can be given by (where X and Y are some vectors within the level surface) \begin{align}h(X,Y) &= g(X,Y) = g_{tt} dt(X) dt(Y) + g_{ti} \Big( dt(X) dx^i(Y) + dt(Y) dx^i(X) \Big) + g_{ij} dx^i(X) dx^j(Y) \\ &= 0 + 0 + g_{ij} X^i Y^j. \end{align} So perhaps this is what George means by "think of $h$ as 4x4". Note that I'm assuming the vectors X and Y are already tangent to the level surfaces of $dt$. One can imagine instead a 4x4 metric on general vectors that includes some extra terms to project those vectors onto the level surfaces of $dt$. I think that is what George wrote down. But I wouldn't call that the "induced metric", since it acts on a vector space of the wrong dimension. I also see that I haven't used the covariant derivative $\nabla_\mu t$ anywhere; I'm not sure exactly why this is needed. Probably the method I am outlining above is a different route to the same result, rather than the method Wald (?) is using.
I would understand an equation like h(x,y)=g(x,y) for x, y on the hypersurface. And I would also understand an equation like $h_{ij} x^i y^j=g_{ij} x^i y^j$ since i and j explicitly only run from 1 to 3 on both sides of the equation. What I don't seem to get is an equation like $h_{ab}=g_{ab} + n_a n_b$ where now it seems like a and b, if I take them to be indices, would run from 0-3, and now h is a 4x4 matrix.
 Sci Advisor PF Gold P: 1,594 Perhaps the confusion comes from the fact that the spacelike slices do not have to be level surfaces of $dt$. All they have to be is spacelike; i.e., all you really need is a foliation of your manifold by spacelike surfaces.
 Sci Advisor P: 2,953 I think Wald specifically says that the space-like splices are slices of constant t. That's what he used in his construction.
 P: 17 OK, I am sorry for the confusion. You are obviously right that $dt$ is orthogonal to the hypersurfaces. The lapse and shift vectors turn up from decomposing $(\frac{\partial}{\partial t})^a = N n^a + N^a$ (not what I wrote earlier). From this, one can find the usual form for the metric in terms of the lapse and shift vector, and the induced metric on the hypersurfaces (using the identity from Wald). So that's why one writes the projector (acting on the 4D space, but of restricted to the 3D space, acting exactly like the metric) rather then the metric with 3D indices. Obviously both encode the same information. I guess the advantage of this form is that $dt$ is orthogonal to $dx^i+N^idt$. For example, the determinant of g in term of these variables is much easier then the determinant in terms of g_tt, g_ti,g_ij.
 Sci Advisor P: 2,953 Ok I think I figured out the difficulty here. Basically it all has to deal with the fact that the basis vectors "dual" to the basis one forms by the usual requirement that they give the delta function when one acts on the other are not the vectors which are "dual" to the basis one forms as picked out by the metric. Basically, even though the vector gradient is normal to the level sets, the basis vector is not. If the basis vector t were normal to the level sets then my metric would have a form of the first column and row being (1,0,0,0) which it doesn't need to have depending on my choice of coordinates x y and z. EDIT: eendavid, it seems like this is what you were talking about in the first part of your post? Addendum: If my analysis is correct, then it seems that there MUST be a way to choose coordinates such that N=1 and N_a=0 right? We just choose not to do it this way because we don't want to introduce those constraints at this level correct?
 P: 17 That's right.
 Sci Advisor P: 2,953 Sounds good to me. Now I just need to figure out what George was trying to get at with his posts and if my intuitions were correct.
Mentor
P: 6,248
 Quote by Ben Niehoff I also see that I haven't used the covariant derivative $\nabla_\mu t$ anywhere; I'm not sure exactly why this is needed. Probably the method I am outlining above is a different route to the same result, rather than the method Wald (?) is using.
You haven't used $\nabla_\mu t$ because you haven't introduced a congruence of curves that intersect the hypersurfaces, with each curve parametrized by $t$. I think that the condition $t^\mu \nabla_\mu t = 1$ says that integral curves of the vector field $t^\mu$ have $t$ as curve parameter.
 Sci Advisor P: 2,953 I'm a little confused on one last point that I can't seem to figure out myself. The set of coordinates on the initial hypersurface has some origin somewhere. Is this origin carried to the subsequent hypersurface by this vector t or by the normal vector to the hypersurface? Meaning, if I started at the origin and move along this vector t to the next hypersurface, do I stay at the origin, or do I move away from the origin of coordinates? Or is this arbitrary based on how I define my coordinates? I'm confused here because intuitively to me it seems like I would prefer to have my coordinates move along with the normal, but MTW makes a statement that the shift vectors tell you where these "normal struts" end up on the above hypersurface via the equation: $$x_{new} ^i=x^i-N^idt$$ Which seems to suggest that these "normal struts" (normal vectors) do not start and end at the same coordinates.
Mentor
P: 6,248
 Quote by Matterwave The set of coordinates on the initial hypersurface has some origin somewhere.
A coordinate system doesn't have to have an origin.
 Quote by Matterwave Is this origin carried to the subsequent hypersurface by this vector t or by the normal vector to the hypersurface? Meaning, if I started at the origin and move along this vector t to the next hypersurface, do I stay at the origin, or do I move away from the origin of coordinates? Or is this arbitrary based on how I define my coordinates?
It depends on the coordinate system.
 Quote by Matterwave I'm confused here because intuitively to me it seems like I would prefer to have my coordinates move along with the normal
It is more natural to choose a coordinate system that consists of $t$ and spatial coordinates that are constant along the integral curves of $t^\mu$.
 Quote by Matterwave but MTW makes a statement that the shift vectors tell you where these "normal struts" end up on the above hypersurface via the equation: $$x_{new} ^i=x^i-N^idt$$ Which seems to suggest that these "normal struts" (normal vectors) do not start and end at the same coordinates.
Because they don't start and end on the same integral curve of $t^\mu$.
 Sci Advisor P: 2,953 Ok, sounds good. If that is so, then it would be wrong for me to say that the proper time dtau=Ndt is the time on the clock of a co-moving observer then right? A co-moving observer stays stationary in the spatial coordinate system chosen, and so the time Ndt is actually the time experienced by an observer moving orthogonally to the hypersurfaces right? This observer "moves" along the spatial coordinates right.

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