Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 311 / 3

The four independent Killing vector fields are the stationary field ${\partial }_t$ and the three generators of spatial rotations on the two-spheres. In the Schwarzschild interior, $f(r)<0$, soEvery rotational Killing field is tangent to the positive-definite round-sphere metric. All four fields are tangent to a surface of constant $r$, whose induced metricis positive definite. Consequently every nonzero linear combination of the Killing fields is spacelike wherever it does not vanish. A pure rotational Killing field can vanish on its rotation axis, but it is nowhere timelike.

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Inside the horizon, $g^{ab}{\nabla }_{a}r{\nabla }_{b}r=f<0$, so $\nabla r$ is timelike. The chosen black-hole time orientation declares $-\nabla r$ future directed; therefore every future-directed timelike vector $V$ has $V(r)<0$. Thus $r$ decreases strictly along every future-directed timelike curve.

Write $F=2M/r-1=-f>0$. Along such a curve,It cannot remain at any $r>0$ indefinitely, because $r$ is a time function and the displayed bound gives finite remaining proper time. From a starting radius $r_p<2M$,Putting $r=2M{\sin }^2\chi$ evaluates the last integral as $\pi M$. Hence every such curve reaches the $r=0$ curvature singularity with

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The two future null directions are proportional to $k_{+}=(1,1)$ and $k_{-}=(1,-1)$. The null energy condition requiresEquivalently,

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In two dimensions every future causal vector is a nonnegative linear combination of $k_{+}$ and $k_{-}$. The dominant energy condition is therefore equivalent to nonnegativity of $T(k_{+},k_{+})$, $T(k_{-},k_{-})$, and the mixed pairingThus the necessary and sufficient conditions are

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A future unit timelike vector is proportional to $(1,v)$ with $|v|<1$. The measured energy density has the sign ofUnder the assumption $c>a$, the weak energy condition first requires $a=q(0)\ge 0$, and hence $c>0$. The convex quadratic has its minimum at $v=-b/c$. Nonnegativity for every $|v|<1$ is then equivalent toIndeed this inequality implies $|b|/c\le \sqrt{a/c}<1$, so the minimum lies inside the allowed interval. Thereforeis necessary and sufficient when $c>a$.

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Let ${\Sigma }_t$ be a Cauchy hypersurface with induced metric $h_{ij}$, lapse $N$, shift $N^i$, and future unit normal $n^a$. For the normalization of the action in the question, the canonical momentum density isThe equal-time canonical commutation relations areandWith the conventional extra factor $1/2$ in the action, $\Pi$ loses the factor two.

For complex classical solutions, the Klein-Gordon inner product isThe integrand is a conserved current because both fields obey the Klein-Gordon equation. Applying the divergence theorem between two Cauchy hypersurfaces shows that the value is independent of the foliation, provided there is no boundary flux.

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A strictly stationary spacetime has an everywhere timelike Killing field $K$. Choose a complete orthonormal set of positive-frequency solutions satisfyingwith positive Klein-Gordon inner product. Expandwith the sum replaced by an integral for continuous labels. Equivalently, the coefficients are projections using the Klein-Gordon product: $a_j=(u_j,\hat{\Phi }{)}_{KG}$ and $a_j^{†}=-(u_j^{\ast },\hat{\Phi }{)}_{KG}$.

The stationary vacuum is uniquely selected, up to degeneracies and unitary changes of positive-frequency basis, byfor every $j$. Acting with the $a_j^{†}$ constructs the bosonic Fock space, the symmetric direct sum of all particle-number sectors.

In a nonstationary spacetime no preferred timelike Killing flow exists, so there is no canonical split into positive and negative frequencies. Different splits mix creation and annihilation operators by Bogoliubov transformation and lead to the vacuum ambiguity in a nonstationary spacetime.

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