
#19
Feb2312, 04:06 PM

Sci Advisor
P: 1,562

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 3dimensional. Therefore it is a simple fact of linear algebra that these 3 linearlyindependent vectors span this space... 



#20
Feb2312, 04:08 PM

P: 2,043





#21
Feb2312, 04:11 PM

Sci Advisor
P: 1,562

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
Feb2312, 04:13 PM

P: 2,043

I think Wald specifically says that the spacelike splices are slices of constant t. That's what he used in his construction.




#23
Feb2312, 04:48 PM

P: 17

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
Feb2312, 04:49 PM

P: 2,043

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? 



#25
Feb2312, 04:53 PM

P: 17

That's right.




#26
Feb2312, 07:25 PM

P: 2,043

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
Feb2312, 10:23 PM

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P: 6,037





#28
Feb2312, 11:27 PM

P: 2,043

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^iN^idt[/tex] Which seems to suggest that these "normal struts" (normal vectors) do not start and end at the same coordinates. 



#29
Feb2412, 12:05 AM

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#30
Feb2412, 12:07 AM

P: 2,043

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 comoving observer then right? A comoving 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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