Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 311 / 3

There are no ${\partial }_{t}N$ or ${\partial }_t{N}^i$ terms in the displayed Einstein-Hilbert action, so the lapse function and shift vector have zero canonical momenta. Their equations impose constraint equations in field theory rather than independent propagation; they are Lagrange multipliers in the Hamiltonian formulation of general relativity.

With the question's positive sign for $K_{ij}=({\dot{h}}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$, vary ${\dot{h}}_{ij}$ at fixed $h,N,N^i$. Since $\delta K_{ij}=\delta {\dot{h}}_{ij}/(2N)$,Thus the canonical momentum of the spatial metric is a weight-one tensor density. Write $\pi =h_{ij}{\pi }^{ij}$; in three spatial dimensions $\pi =-2\sqrt{h}K$, soThe action has suppressed its overall $1/(16\pi G)$, and this normalization is retained here. Reversing the sign convention for extrinsic curvature would reverse the momentum relation; the calculation uses exactly the convention supplied in this question.

Perform the Legendre transform in mechanics, using ${\dot{h}}_{ij}=2N{K}_{ij}+D_i{N}_j+D_j{N}_i$. The trace contraction ${\pi }^{ij}{K}_{ij}=\sqrt{h}(K_{ij}{K}^{ij}-K^2)$ givesTo avoid differentiating a tensor density as though it were an ordinary tensor, set $p^{ij}={\pi }^{ij}/\sqrt{h}$. Integration by parts with the Levi-Civita connection of $h$ givesafter dropping the surface term. AlsoTherefore the requested Hamiltonian isVarying the lapse function and shift vector imposes the Hamiltonian constraint $\mathcal{H}=0$ and momentum constraint ${\mathcal{H}}_i=0$. The total canonical generator also contains multipliers for the vanishing lapse and shift canonical momenta. With an asymptotic boundary, surface terms must be restored to define the Arnowitt-Deser-Misner energy.

For a closed universe, the spatial Cauchy hypersurface is compact without boundary. There is consequently no asymptotic surface term in the gravitational Hamiltonian. On solutions, the Hamiltonian constraint and momentum constraint giveThis is the vanishing canonical Hamiltonian of a closed universe: evolution generated by lapse and shift is a gauge transformation, and there is no asymptotic time-translation charge. It does not mean the local stress-energy tensor or Riemann curvature tensor is zero. In an asymptotically flat universe the same bulk constraint equations in field theory hold, but a boundary Arnowitt-Deser-Misner energy can be nonzero.

Work in the Boyer-Lindquist coordinates exterior $r>r_{+}=M+\sqrt{M^2-a^2}$, away from the axes. There $\Delta >0$, $\Sigma >0$, andInverting the $t,\varphi$ block of the Kerr metric givesThus $W^a=-g^{ab}(dt{)}_b=-{\nabla }^{a}t$ has $g(W,W)=g^{tt}<0$, and the smooth nonvanishing timelike vector field $W$ defines a time orientation. Choose it future-directed, matching increasing $t$ near infinity. For every nonzero future causal vector $Z$, the Lorentzian inner product satisfies $g(W,Z)<0$, henceSo $t$ increases strictly along every regular future causal curve in this exterior. Notice that ${\partial }_t$ itself can be spacelike in the Kerr ergoregion; the timelike object used here is $-\nabla t$. Direct substitution also gives $D=\Delta {\sin }^2\theta$.

For a regular causal curve, orient its tangent $Z$ toward the future; reversing its parametrization does not change $d\varphi /dt$. Since $dt(Z)>0$, use $t$ as parameter and set $\Omega =d\varphi /dt$. The causal inequality givesBoth transverse terms are nonnegative in the exterior, so necessarilyBecause $g_{\varphi \varphi }>0$ and the discriminant is $4D>0$, this is equivalent toEquality requires $dr/dt=d\theta /dt=0$ and a null tangent. The limiting vectors are the two local azimuthal null directions; a curve following one at constant radius need not be a null geodesic. For a curve with radial or polar motion these angular bounds are necessary, while the full causal inequality can impose stricter bounds.

In the open exterior Kerr ergoregion, $g_{tt}>0$ andSince $D=g_{t\varphi }^2-g_{tt}{g}_{\varphi \varphi }<g_{t\varphi }^2$,so even the lower permitted angular velocity is positive. Every regular causal trajectory in the ergoregion has $d\varphi /dt>0$: it must co-rotate with the Kerr black hole. At the outer stationary-limit surface $g_{tt}=0$, the lower bound is zero; strict positivity is asserted in the open ergoregion. The sign reverses if the orientation of $a$ is reversed.