# Two questions about the ADM formalism

1. Mar 6, 2008

### kdv

In that formalism, one writes

$$g_{\alpha \beta} dx^\alpha dx^\beta = (-N^2 + N_i N^i) dt^2 + 2 N_j dt dx^j + h_{ij} dx^i dx^j$$

First, a simple question: the lapse function N here is unrelated to the shift vector $N_^i [/tex], right? I mean, clearly [itex] N^2 \neq N_i N^i [/tex] but is there any other relation? I am assuming no. How are the Latin indices raised and lowered? Is it true that [itex] N_i = h_{ij} N^j$ ?

Now, I have read that a conformal transformation

$$g_{\alpha \beta} \rightarrow \Omega^2 g_{\alpha \beta}$$

corresponds to the following transformations of the laspe function and shift vectors:

$$N \rightarrow \Omega N, ~~~~N^i \rightarrow N^i ~~~~ h_{ij} \rightarrow \Omega^2 h_{ij}$$

This would make sense if $N_i = h_{ij} N^j$ so that we would have $$N_i \rightarrow \Omega^2 N_i$$ but I am not sure if I am correct about this way of lowering the indices on N.

Thanks

2. Mar 6, 2008

### George Jones

Staff Emeritus
At each point, $N$ and $N^i$ are the components of a particular timelike 4-vector with respect to a splitting of spacetime into time and space, where time is 4-orthogonal to space, but where the spactial basis vectors are not necessarily orthonormal (with respect to each other), so they have to satisfy

$$-N^2 + h_{ij} N^i N^j < 0.$$

Yes.

This is used, for example, in the above, and to go back and forth between the metric in the form that you wrote and an equivalent version of the metric, i.e.,

$$(-N^2 + N_i N^i) dt^2 + 2 N_j dt dx^j + h_{ij} dx^i dx^j = -N^2 dt^2 + h_{ij} \left(dx^i + N^i dt \right) \left(dx^j + N^j dt \right).$$

Hmm, new funky quote notation.

Last edited: Mar 6, 2008
3. Mar 6, 2008

### kdv

Ah, ok! This is all very clear now. Thank you!

I am trying to find a low level introduction to the ADM formalism but haven't succeeded so far. Any recommendation?

Thanks a lot

4. Mar 6, 2008

### George Jones

Staff Emeritus
Yikes, now it's not clear to me! The N's are the components of a particular 4-vector, but now I'm not so sure that this 4-vector has to be timelike.

Have to catch my bus; will check tonight or tomorrow.

5. Mar 7, 2008

### Old Smuggler

This 4 vector field is the timelike orthonormal unit vector field of the spatial hypersurfaces defined by holding the time coordinate constant. Note that the coordinate vector field d/dt is not necessarily
orthogonal to the hypersurfaces.

6. Mar 7, 2008

### George Jones

Staff Emeritus
I don't think so.

I'm in the middle of composing a long post about this stuff. After I put this post up, let me know if I screwed things up.

Lunch break.

7. Mar 7, 2008

### kdv

Thanks, I am looking forward to reading it.

Is it possible to continue the discussion of general coordinate transformations vs Lorentz transformations in the other thread?

Thanks George

8. Mar 7, 2008

### George Jones

Staff Emeritus
Maybe A Relativist's Toolkit: The Mathematics of Black-Hole Mechanics by Eric Poisson, http://books.google.com/books?id=v-...s_brr=0&source=gbs_summary_s&cad=0#PPA129,M1".

There is a scalar field $t$ that is a global "time" function in a region V of spacetime, and $\left\{x^\alpha\right\}$ is a coordinate system that covers V, and, in general, $t$ is not necessarily $x^0$. V could be all of spacetime. The idea is to set up a global foliation of spacetime into space and time throughout V, and to set up a new coordinate system $\left(t, y^a \right)$ (a = 1, 2,3 ) in V that reflects this foliation.

Look at Figure 4.2. Each (roughly) horizontal line $\Sigma_t$ is a 3-dimensional hypersurface that is space at a fixed value of $t$. On each (roughly) vertical line, the $y^a$ are held fixed, and only $t$ varies. $t$ labels which $\Sigma$ and the $y^a$ label where on $\Sigma$.

Now look at Figure 4.3. $n$ is a timelike unit vector orthogonal to $\Sigma_t$ that has components $n^\alpha$ with respect to the coordinate system $\left\{x^\alpha\right\}$, and $T$ ($t^\alpha$ on the diagram, but I don't want to confuse it with the scalar $t$) is a 4-vector tangent to one of the lines of constant $y^a$. If this lines of constant $y^a$ is parameterized by $t$, then $T = d/dt$. Where this line intersects $\Sigma_t$, a basis $\left\{e_a\right\}$ of spacelike vectors is set up with these spacelike vectors to the $y^a$ coordinate curves.

$T$ is split into a time parts orthogonal and tangent to $\Sigma_t$,

$$T = N n + N^a e_a,$$

so $N$ and the $N^a$ taken together "make up" a 4-vector, but not one that is hypersurface-orthogonal.

Originally, I thought $T$ had to be timelike, but now I don't think so.

Each of the vectors $e_a$ has components $e^\alpha_a$ with respect to the coordinates $\left\{x^\alpha\right\}$, i.e., $a$ labels which vector and $$\alpha$$ labels which component.

Now, imagine on Figure 4.3 a second spatial hypersurface that is roughly parallel to $\Sigma_t$, and such that has the tip of $Nn$ touching it. In the infinitesimal case, $N^a e_a$ is a shift in space of the $y$ coordinates in a direction away from the hypersurface orthogonal $n$ as one goes from one hypersurface to the next, while $N n$ is proportional to the lapse of proper time for an observer who has $n$ as the tangent to his worldline.

Look at http://books.google.com/books?id=C0...r=0&sig=AQTcR2RTo3cpPISrxmhuBAsXxzs#PPA12,M1"-13 of Carlip's Quatum gravity in 2+1 Dimensions, and note how he naturally derives the form of the metric (his (2.5)) that I posted yesterday,

interval^2 = - time^2 + space^2.

Warning: Carlip's $x$ is Poisson's $y$, and Carlip's $g_{ij}$ is Poisson's $h_{ab}$.

Last edited by a moderator: Apr 23, 2017
9. Mar 7, 2008

### George Jones

Staff Emeritus
I think we're talking past each other. What I meant in post #4 is that together N and the N^i make up the vector T (t^alpha in Poisson's diagram), and that T doesn't have to be timelike, i.e, that the spacelike N^i can dominate the timelike N.

Is this true?

10. Mar 7, 2008

### Old Smuggler

My fault, I misunderstood. However, g(T,T) = N*N g(n,n) + 2 g(n,e_a)N"a + N''a N''b g(e_a,e_b) =
-N*N + N''a N''b h(e_a,e_b) = -N*N + N"a N_a. (Here h is the intrinsic hypersurface metric.)

g(T,T) is less than 0 since -N*N + N''a N_a is less than 0, as it must since t (and x''0)
by hypothesis can be used as a time coordinate in the relevant region of spacetime. That is, T is necessarily timelike.

11. Mar 7, 2008

### George Jones

Staff Emeritus
Sorry, I still don't see it.

I agree that g(T,T) = -N*N + N''a N_a, but I don't see why this negative.

In order for t to be a time coordinate, Isn't it enough that $\partial^\alpha t$ be timelike, since this vector field is hypersurface orhogonal?

Consider the following example.

Let V be all of a 2-dimensional Minkowski spacetime, and let t be the usual time coordinate for a particular inertial frame. The hypersurfaces are now lines of simultaneity for this particular inertial frame. Let the congruence of curves that connect the hypersurfaces be given by (t, 2t + A), where A is any real number.

Here t is a time coordinate, n = (1, 0), and T = (N, N^1) = (1, 2); for every delta t lapse of 1, there is a spatial shift of 2.

Is this a valid 1+1 decomposition?

12. Mar 10, 2008

### Old Smuggler

I don't think that is sufficient. Rather, I believe that the congruence used (when constructing the
coordinate system (t,y^i) from the general coordinate system (x^0,x^i)) is implicitly assumed to be
timelike. In that case, the congruence can be identified with the world lines of a family of observers,
and both t and the proper time of these observers parametrize the congruence.

Besides, the coordinate 3-velocity of an observer moving normal to the hypersurfaces is given
by n^i*c = - (N^i/N)*c. This corresponds to a 3-speed squared relative to the grid of spatial coordinates of (N^i*N_i/N^2)*c^2 = v^2. We see that if g(T,T) = 0, which corresponds to a null
congruence, then v = c. So in this case the coordinates (t,y^i) are null coordinates. Such
coordinates are certainly permitted, but perhaps not very convenient in the 3+1 formalism.
Moreover, if g(T,T) is positive, which corresponds to a spacelike congruence, then v/c exceeds
1. This means that (t,y^i) is not a suitable coordinate system in this case.

I'm afraid that this is the best answer that I am able to give to your question. Maybe someone
else can give a better answer.

No, I believe that it isn't since v = 2*c in this case, so the congruence is spacelike.

Last edited: Mar 10, 2008
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