Purely kinetic initial data for Schwarzschild

I am at MSRI at the moment attending a program on mathematical relativity, and today heard something interesting from Mu-Tao Wang during his discussion of quasi-local mass.

To motivate: consider Einstein’s (vacuum) equations in general relativity. The initial data formulation requires specifying a manifold $\Sigma$, a Riemannian metric $g$, and a symmetric two tensor $k$. Remember that we are interested in solving for a space-time $(M,\bar{g})$ satisfying $\mathrm{Ric}(\bar{g}) = 0$; the induced metric $g$ can be thought of as the initial data and the tensor $k$ (which will turn out to be the second fundamental form of the embedding of $\Sigma$ into $M$) the initial first time derivative of the data (Einstein’s equation is second order, and so we specify both the data and its first time derivative on the initial slice).

The nature of Einstein’s equations is that the initial data needs to satisfy a set of constraint equations, for it to be compatible with any solution. In any dimension we have the Hamiltonian constraint (assuming vacuum, so no stress-energy tensor)

$\displaystyle R(g) - |k|^2_g + \mathrm{tr}_g k = 0$

and the momentum constraint

$\displaystyle \mathrm{div}_g k - \mathrm{d}~\mathrm{tr}_g k = 0$

A particular class of initial data is those called time-symmetric, which are those with $k \equiv 0$. The physical interpretation is clear in view of the above formal considerations: $k$ corresponds to the instantaneous “velocity” of the space-time as a whole, and its vanishing represents that the spacetime is instantaneously “not moving”. To draw an analogy with classical mechanics, this is the situation where velocities are zero and all the energy of the system is contained in the potentials.

One may ask the question whether we can write down initial data for Einstein’s equation that represents, again using the classical mechanics analogy, a system that is instantaneously completely relaxed, and so all its energy are placed in the kinetic part. A reasonable interpretation for “completely relaxed” would be that the initial data has zero spatial curvature, like the standard slice in Minkowski space; all the energy will live inside the “kinetic term” $k$.

As it turns out: we can in fact do this. To exhibit a solution, let us work in spherical symmetry to reduce the constraint equations to an ordinary differential equation. Since the spatial metric is flat, we can choose spherical coordinates so that

$\displaystyle g = \mathrm{d}r^2 + r^2 \mathrm{d}\omega^2$

and by spherical symmetry (assuming spatial dimension $n \geq 3$) we have that the second fundamental form must decompose as

$\displaystyle k = A(r) \mathrm{d} r^2 + r^2 B(r) \mathrm{d}\omega^2$

which implies that

$\displaystyle |k|^2_g = A^2 + (n-1) B^2 \qquad \mathrm{tr}_g k = A + (n-1) B$

so that the Hamiltonian constraint yields

$\displaystyle B \equiv 0 \qquad \text{or} \qquad 2A + (n-2)B = 0$

(note that if $A = B = 0$ then we reduce down to Minkowski initial data). Next we consider the momentum constraint. A direct computation (using Christoffel symbols or otherwise) of the divergence term yields

$\displaystyle (\mathrm{div} K)(\partial_r) = \partial_r A + \frac{n-1}{r} (A - B)$

Now, in either of the cases we can reduce the momentum constraint to a one variable ODE:

$B \equiv 0 \implies A \equiv 0 \implies \text{Minkowski}$

and more interestingly

$\displaystyle 2A + (n-2)B = 0 \implies \partial_r A + \frac{n-1}{r} (1+ \frac{2}{n-2}) A = \partial_r A - (n-1)\frac{2}{n-2} \partial_r A$

which we can simplify to get

$\displaystyle \partial_r A + \frac{n}{2r} A = 0$

Using integrating factors we solve to get

$\displaystyle A = c r^{-n/2} \qquad B = - \frac{2c}{n-2} r^{-n/2}$

Now where does this slice sit in Schwarzschild? If we solve, in Boyer-Lindquist coordinates for $t = f(r)$ such that the induced metric is flat, we easily come to the conclusion that, in the case $n = 3$ (I may add the computation for general $n$ later) we have $f'(r) = \pm \sqrt{2m/r}(1 - 2m/r)^{-1}$. This slice goes from space-like infinity inward, and cuts through the (future or past, depending on sign) event horizon as $r \to 2m$.

Interestingly, we note that the decay of the second fundamental form $k$ is in fact critical, at least in the context of the positive mass theorem (the momentum integral formally diverges!), so at least on that level this doesn’t give a contradiction to the PMT.

For more discussions one can see this paper.