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?
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?
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.
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.
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?
[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?
[tex]n^a n^b h_{ab}=0[/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?
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.
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?
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?
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]
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.
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.
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?
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...
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.