Lapse function and spacetime splicing

In summary, the time vector is orthogonal to the space-like Cauchy surfaces if and only if the vector gradient is orthogonal to the level sets of the function.
  • #1
Matterwave
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Hi, I have a question regarding the foliation of space-time by slices of constant "time".

I know that such a foliation is possible given a globally hyperbolic manifold, and one can define a "time function" t, the level sets of which are 3-D Cauchy surfaces which foliate the spacetime. My question is, if we define a "time vector" to this foliation by the requirement that [itex]t^\mu\nabla_\mu t=1[/itex], why is this "time vector" not (in general) orthogonal to the space-like Cauchy surfaces? The lapse function and shift vector measure the amount by which this time vector fails to be orthogonal to the Cauchy surfaces, but it seems to me that due to that requirement above, the vector should always be orthogonal shouldn't it?

My logic is such: since t is a scalar function, the covariant derivative of it is nothing other than its (one form) gradient, and the (vector) gradient of a function is always orthogonal to the level sets of the function is it not?
 
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  • #2
Wait a moment, I think I figured it out. The vector gradient is not, in general, [itex]t^\mu[/itex] right, it's [itex]g^{\mu\nu}\nabla_\nu t\neq t^\mu[/itex] as defined above. So it seems that The definition of the time vector via [itex]t^\mu\nabla_\mu t=1[/itex] is requiring me to rotate away from the normal such that the vector pierces only 1 level set of the gradient.

Why is this desirable or necessary? Is it impossible to define my "time vector" as some normalized version of [itex]g^{\mu\nu}\nabla_\nu t[/itex] so that at least it's still orthogonal?
 
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  • #3
Maybe it's simply worded poorly. I find a lot of the older physicists' presentation of GR concepts very tedious and hard to follow.

Any metric [itex]g_{\mu\nu} dx^\mu dx^\nu[/itex] can always be written

[tex]g_{\mu\nu} dx^\mu dx^\nu = g_{tt} dt^2 + 2 g_{ti} dt dx^i + g_{ij} dx^i dx^j,[/tex]
so that is how the lapse and shift functions turn up. The rest is a matter of finding a time function to give you the coordinate t.
 
  • #4
So, basically I can just think of them as a method to break down the metric from a 4x4 matrix into a block of 1, 1x3, 3x1, and 3x3 matrix?

I also see Wald make statements like "[itex]h_{ab}[/itex] is the induced spatial metric given by: [tex]h_{ab}=g_{ab}+n_a n_b[/tex]", and I can't make heads or tails of this equation because h should be a 3x3 matrix while g should be 4x4. What does this mean? This issue seems related.
 
  • #5
[tex]n^a n^b h_{ab}=?[/tex]
 
  • #6
I don't think I follow, sorry...I just don't see I can have a 3-D object on the left and a 4-D object on the right.

Maybe if we just took a really simple example, a trivial example, of a Minkowski space-time which is foliated by the global coordinate time t.

In this case g=diag(-1,1,1,1), n=(1,0,0,0), and h=diag(1,1,1) right? Or is h=diag(0,1,1,1) so I should interpret h as a 4-D object but with one of the rows and columns all 0?
 
  • #7
[tex]n^a n^b h_{ab}=?[/tex]
Now suppose [itex]v^a[/itex] is orthogonal to [itex]n^a[/itex].
[tex]v^a v^b h_{ab}=?[/tex]

What are the results of these calculations?
 
  • #8
George Jones said:
[tex]n^a n^b h_{ab}=?[/tex]

[tex]n^a n^b h_{ab}=0[/tex]
George Jones said:
Now suppose [itex]v^a[/itex] is orthogonal to [itex]n^a[/itex].

[tex]v^a v^b h_{ab}=?[/tex]

[tex]v^a v^b h_{ab}=v^a v^b g_{ab}[/tex]

Now choose an othonormal that has [itex]n^a[/itex] as one element. What does the matrix for [itex]h[/itex] with respect to this basis look like?
 
  • #9
"reading" this thread makes me wish I learned math. :(
 
  • #10
Can you tell me if h is a 3x3 or a 4x4 matrix first? I think I'm pretty much just stuck on that. I don't even know what it is, is I guess what I'm saying, so if you ask me to calculate anything with it, I'm pretty lost.
 
  • #11
Right now, think 4x4.
 
  • #12
Ok, then does that mean that if I choose n=(1,0,0,0), then h is a block matrix with 0's for the 0th row and column, and a 3x3 submatrix? If I don't choose such nice coordinates, doesn't that mean my h will not have this nice structure?
 
  • #13
I think I get where you're going. The defining aspects of the induced metric is [itex]n^a n^b h_{ab}=0[/itex] and [itex]v^a v^b h_{ab}=v^a v^b g_{ab}[/itex] right, and then because this is so, if I choose coordinates in which the normal vectors are (1,0,0,0) then I can simply reduce h into a nice 3x3 matrix with extra padded 0's. If I don't choose to do this, h is simply some 4x4 symmetric matrix with 4 additional constraints. Is that so?
 
  • #14
Some more help on this would be nice, thanks.
 
  • #15
Yes, note more precisely that [itex]n^ah_{ab}=0[/itex] (and itex]v^av^bh_{ab}=v^av^bg_{ab}[/itex]). So this is really what you see as the induced metric: it gives the correct inner product for vectors tangent to the hypersurfaces (and it's not a metric on the 4D space, since there it is degenerate).

On the lapse and shift thing: note that there is no reason for [itex]\nabla_\mu t[/itex] to be parallel to the normal (or orthogonal to the tangent space to the hypersurfaces). So just call the component tangent to the hypersurface the shift vector, and the component of the normal the shift vector
[tex]\nabla_\mu t = N n_\mu + N_\mu.[/tex]
 
  • #16
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 [itex]t[/itex] with gradient [itex]dt[/itex]. You want to find the induced metric on the level sets of [itex]dt[/itex]. The level sets of [itex]dt[/itex] are generated by a triplet of linearly-independent vector fields X, Y, Z such that

[tex]dt(X) = dt(Y) = dt(Z) = 0.[/tex]
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 [itex]t[/itex] is a global time function, and X, Y, Z are everywhere perpendicular to [itex]dt[/itex].

Then the induced metric [itex]h[/itex] can be given by (where X and Y are some vectors within the level surface)

[tex]\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}[/tex]
So perhaps this is what George means by "think of [itex]h[/itex] as 4x4". Note that I'm assuming the vectors X and Y are already tangent to the level surfaces of [itex]dt[/itex]. One can imagine instead a 4x4 metric on general vectors that includes some extra terms to project those vectors onto the level surfaces of [itex]dt[/itex]. 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 [itex]\nabla_\mu t[/itex] 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.
 
  • #17
Ben Niehoff said:
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 [itex]t[/itex] with gradient [itex]dt[/itex]. You want to find the induced metric on the level sets of [itex]dt[/itex]. The level sets of [itex]dt[/itex] are generated by a triplet of linearly-independent vector fields X, Y, Z such that

[tex]dt(X) = dt(Y) = dt(Z) = 0.[/tex]
While you can find 3 independent vectors satisfying this, they simply do not span the space tangent to your hypersurface.

Since [itex]h_{\mu\nu}n^\mu[/itex] is 0, this [itex]h[/itex] really behaves as a 3x3 metric on the hypersurfaces.
 
  • #18
eendavid said:
Yes, note more precisely that [itex]n^ah_{ab}=0[/itex] (and itex]v^av^bh_{ab}=v^av^bg_{ab}[/itex]). So this is really what you see as the induced metric: it gives the correct inner product for vectors tangent to the hypersurfaces (and it's not a metric on the 4D space, since there it is degenerate).

On the lapse and shift thing: note that there is no reason for [itex]\nabla_\mu t[/itex] to be parallel to the normal (or orthogonal to the tangent space to the hypersurfaces). So just call the component tangent to the hypersurface the shift vector, and the component of the normal the shift vector
[tex]\nabla_\mu t = N n_\mu + N_\mu.[/tex]

Isn't [itex](\nabla_\mu t) dx^\mu=(\partial_\mu t) dx^\mu= dt[/itex] so that this is simply the gradient and is therefore normal to the level sets of the function t?
 
  • #19
eendavid said:
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 [itex]dt[/itex]. The subspace of the tangent space orthogonal to [itex]dt[/itex] is 3-dimensional. Therefore it is a simple fact of linear algebra that these 3 linearly-independent vectors span this space...
 
  • #20
Ben Niehoff said:
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 [itex]t[/itex] with gradient [itex]dt[/itex]. You want to find the induced metric on the level sets of [itex]dt[/itex]. The level sets of [itex]dt[/itex] are generated by a triplet of linearly-independent vector fields X, Y, Z such that

[tex]dt(X) = dt(Y) = dt(Z) = 0.[/tex]
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 [itex]t[/itex] is a global time function, and X, Y, Z are everywhere perpendicular to [itex]dt[/itex].

Then the induced metric [itex]h[/itex] can be given by (where X and Y are some vectors within the level surface)

[tex]\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}[/tex]
So perhaps this is what George means by "think of [itex]h[/itex] as 4x4". Note that I'm assuming the vectors X and Y are already tangent to the level surfaces of [itex]dt[/itex]. One can imagine instead a 4x4 metric on general vectors that includes some extra terms to project those vectors onto the level surfaces of [itex]dt[/itex]. 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 [itex]\nabla_\mu t[/itex] 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 [itex]h_{ij} x^i y^j=g_{ij} x^i y^j[/itex] 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 [itex]h_{ab}=g_{ab} + n_a n_b[/itex] 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.
 
  • #21
Perhaps the confusion comes from the fact that the spacelike slices do not have to be level surfaces of [itex]dt[/itex]. All they have to be is spacelike; i.e., all you really need is a foliation of your manifold by spacelike surfaces.
 
  • #22
I think Wald specifically says that the space-like splices are slices of constant t. That's what he used in his construction.
 
  • #23
OK, I am sorry for the confusion. You are obviously right that [itex]dt[/itex] is orthogonal to the hypersurfaces. The lapse and shift vectors turn up from decomposing [itex](\frac{\partial}{\partial t})^a = N n^a + N^a[/itex] (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 [itex]dt[/itex] is orthogonal to [itex]dx^i+N^idt[/itex]. 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.
 
  • #24
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?
 
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  • #25
That's right.
 
  • #26
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.
 
  • #27
Ben Niehoff said:
I also see that I haven't used the covariant derivative [itex]\nabla_\mu t[/itex] 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 [itex]\nabla_\mu t[/itex] because you haven't introduced a congruence of curves that intersect the hypersurfaces, with each curve parametrized by [itex]t[/itex]. I think that the condition [itex]t^\mu \nabla_\mu t = 1[/itex] says that integral curves of the vector field [itex]t^\mu[/itex] have [itex]t[/itex] as curve parameter.
 
  • #28
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:

[tex]x_{new} ^i=x^i-N^idt[/tex]

Which seems to suggest that these "normal struts" (normal vectors) do not start and end at the same coordinates.
 
  • #29
Matterwave said:
The set of coordinates on the initial hypersurface has some origin somewhere.

A coordinate system doesn't have to have an origin.
Matterwave said:
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.
Matterwave said:
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 [itex]t[/itex] and spatial coordinates that are constant along the integral curves of [itex]t^\mu[/itex].
Matterwave said:
but MTW makes a statement that the shift vectors tell you where these "normal struts" end up on the above hypersurface via the equation:

[tex]x_{new} ^i=x^i-N^idt[/tex]

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 [itex]t^\mu[/itex].
 
  • #30
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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1. What is a lapse function?

A lapse function is a mathematical concept used in general relativity to describe the rate at which time passes in a given reference frame. It is a scalar function that determines the relationship between the coordinate time and the proper time of an object moving in spacetime.

2. How is a lapse function related to spacetime?

The lapse function is an essential component of the metric tensor, which describes the geometry of spacetime. It is used to calculate the spacetime interval between two events, which is a fundamental concept in general relativity.

3. What is spacetime splicing?

Spacetime splicing is a theoretical concept in general relativity that suggests the possibility of joining two different spacetimes together at a specific point. This idea has been explored in various thought experiments and has implications for the study of black holes and the nature of the universe.

4. How does spacetime splicing affect our understanding of the universe?

Spacetime splicing challenges our traditional understanding of the universe as a continuous and unchanging entity. It suggests that there may be regions of spacetime that are disconnected or joined in unexpected ways, which could have significant implications for our understanding of cosmology and the nature of space and time.

5. Is there any evidence for spacetime splicing?

Currently, there is no direct evidence for spacetime splicing. However, some theories, such as loop quantum gravity, suggest that spacetime may be discrete rather than continuous, which could potentially support the concept of spacetime splicing. Further research and experiments are needed to explore this idea further.

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