Moncrief Invariants and Hamiltonian Reduction

The Regge–Wheeler and Zerilli equations describe the physical part of a much larger metric perturbation. Hamiltonian reduction explains precisely which part survives: for every radiative spherical harmonic, the linearized constraints and their gauge transformations remove all but one canonical pair in each parity sector. We construct those pairs, check their brackets, and obtain their normalization from the gravitational action. The calculation is local to the vacuum Schwarzschild exterior; boundary charges and the nonradiative low multipoles require separate treatment.

Required background. PDEs and Green Functions for Gravitation supplies the evolution and boundary-value structure of the reduced equations.

Helpful background. Black-Hole Geometry and Mechanics fixes the Schwarzschild foliation; Formulations of General Relativity develops the ADM constraints; Presymplectic Potentials and Currents supplies the canonical and flux language; Black-Hole Perturbation Theory supplies gauge-invariant perturbation theory.

Canonical data on a Schwarzschild slice

Write $m=GM$, $f=1-2m/r$, and use Schwarzschild time slices with $\bar\gamma_{ij}dx^i dx^j=f^{-1}dr^2+r^2d\Omega^2$ and lapse $\bar N=\sqrt f$. A prime means $\partial_r$, a dot means $\partial_t$, and $x=r_*$ satisfies $dr=f\,dx$. We work with smooth perturbations and gauge parameters of compact radial support in $2m\lt r\lt\infty$. This permits integration by parts without choosing a horizon or infinity phase space.

The canonical momentum includes the gravitational coupling. With $K_{ij}=-(\dot\gamma_{ij}-D_i\beta_j-D_j\beta_i)/(2N)$, it is the weight-one density

$$\pi^{ij}=\frac{\sqrt\gamma}{16\pi G} \left(K\gamma^{ij}-K^{ij}\right).$$

Here $D_i$ is the spatial covariant derivative, $K=\gamma^{ij}K_{ij}$, and $\pi=\gamma_{ij}\pi^{ij}$. The ADM action and its constraints are

$$S=\int dt\,d^3x\, \left(\pi^{ij}\dot\gamma_{ij}-N\mathcal H-\beta^i\mathcal H_i\right),$$ $$\mathcal H=\frac{16\pi G}{\sqrt\gamma} \left(\pi^{ij}\pi_{ij}-\frac12\pi^2\right) -\frac{\sqrt\gamma}{16\pi G}\,{}^{(3)}R, \qquad \mathcal H_i=-2\gamma_{ij}D_k\pi^{jk}.$$

Boundary terms are suppressed only under the compact-support assumption. These formulas follow from the ADM Legendre transform in Wald 1984, Appendix E.2, pp. 463–467, eqs. (E.2.29)–(E.2.34): Wald uses the opposite sign for $K_{ij}$ and suppresses the overall $1/(16\pi G)$. Translating both gives the momentum written above.

The static background has $\bar\pi^{ij}=0$. In the quadratic perturbation action, lapse and shift perturbations multiply the linearized constraints; they have no conjugate momenta of their own. For a radial canonical pair, $\{q(r),p(s)\}=\delta(r-s)$. A smeared constraint acts by $\delta_\xi F=\{F,C[\xi]\}$. Reduction means both imposing the constraint and identifying points on its gauge orbit, not simply deleting a multiplier.

Use one real, unit-normalized scalar spherical harmonic $Y$, so $\int Y^2d\Omega=1$. Suppress its mode label and set

$$L=\ell(\ell+1),\qquad \mu=L-2,\qquad n=\frac L2, \qquad \lambda=\frac\mu2,\qquad \Lambda=\mu+\frac{6m}{r},\qquad \ell\geq2.$$

On the unit sphere, $Y_A=D_A Y$ and $Y_{AB}=D_A D_B Y+(L/2)\Omega_{AB}Y$ are the even vector and trace-free tensor harmonics. Their odd partners are $X_A=-\epsilon_A{}^B D_B Y$ and $X_{AB}=D_{(A}X_{B)}$. Their norms are

$$\int Y_A Y^A d\Omega=\int X_A X^A d\Omega=L, \qquad \int Y_{AB}Y^{AB}d\Omega=\int X_{AB}X^{AB}d\Omega=\frac{L\mu}{2}.$$

Indices in these integrals are raised with $\Omega^{AB}$, not the physical sphere metric. The resulting constants will be

$$\mathcal N_\ell=\frac{L}{16\pi G},\qquad C_\ell=\frac{\mathcal N_\ell\mu}{4} =\frac{L\mu}{64\pi G}.$$

Using real harmonics avoids treating a complex mode and its conjugate as independent real degrees of freedom. A complex harmonic expansion is equivalent once its reality condition is imposed.

Odd parity: exposing the constraint pair

The odd spatial amplitudes and shift are $\delta\gamma_{rA}=h_rX_A$, $\delta\gamma_{AB}=h_2X_{AB}$, and $\delta g_{tA}=A X_A$. Define

$$k=h_r-\frac12h_2'+\frac{h_2}{r},\qquad \chi=A'-\dot h_r-\frac{2A}{r},\qquad b=A-\frac12\dot h_2.$$

The Einstein–Hilbert reduction developed on the Regge–Wheeler Equation page gives the radial Lagrangian density

$$\mathscr L_{\mathrm{odd}} =\frac{\mathcal N_\ell}{2} \left[\chi^2+\frac{\mu}{r^2f}b^2-\frac{\mu f}{r^2}k^2\right].$$

It is important to keep $h_2$ until after the Legendre transform. Fixing $h_2=0$ at the outset would conceal the gauge pair that we want to identify. Differentiating with respect to the two velocities gives

$$p_1=-\mathcal N_\ell\chi,\qquad p_2=-\frac{\mathcal N_\ell\mu}{2r^2f}\,b.$$

These signs can also be checked directly from ADM: the odd extrinsic curvature has $\delta K_{rA}=\chi X_A/(2\sqrt f)$ and $\delta K_{AB}=bX_{AB}/\sqrt f$, with zero trace. Contracting its momentum with $\delta\dot\gamma_{ij}$, including both symmetric $rA$ entries, reproduces $p_1\dot h_r+p_2\dot h_2$.

A canonical change of variables

Set

$$p=p_1,\qquad q_g=-\frac{h_2}{2},\qquad C_g=p'+\frac{2p}{r}-2p_2.$$

The symplectic potential provides a short, decisive test:

$$\int dr\,(p_1\delta h_r+p_2\delta h_2) =\int dr\,(p\delta k+C_g\delta q_g)-[p\delta q_g].$$

Brackets therefore take the canonical form $\{k(r),p(s)\}=\{q_g(r),C_g(s)\}=\delta(r-s)$, with all cross brackets zero. For example, $C_g[\xi]=\int dr\,\xi C_g$ generates

$$\delta_\xi h_r=-\xi'+\frac{2\xi}{r},\qquad \delta_\xi h_2=-2\xi,\qquad \delta_\xi q_g=\xi,\qquad \delta_\xi k=\delta_\xi p=0.$$

The derivative in $C_g$ generates the derivative in the coordinate transformation. In particular, $\{C_g[\xi],C_g[\eta]\}=0$: the constraint is first class and the proposed physical pair commutes with it.

The Legendre transform, with its spatial endpoint term displayed, is

$$\mathscr H_{\mathrm{odd}} =\frac{p^2}{2\mathcal N_\ell} +\frac{2r^2f}{\mathcal N_\ell\mu}p_2^2 +\frac{\mathcal N_\ell\mu f}{2r^2}k^2 -A C_g+\partial_r(Ap).$$

Varying $A$ imposes $C_g=0$. Hence $p_2=(p'+2p/r)/2$; quotienting $q_g$ leaves

$$\mathscr H_{\mathrm{odd,red}} =\frac{p^2}{2\mathcal N_\ell} +\frac{r^2f}{2\mathcal N_\ell\mu} \left(p'+\frac{2p}{r}\right)^2 +\frac{\mathcal N_\ell\mu f}{2r^2}k^2.$$

From metric momentum to the master coordinate

The Cunningham–Price–Moncrief field is proportional to the old momentum:

$$\Psi_{\mathrm{odd}}=-\frac{2rp}{\mathcal N_\ell\mu},\qquad \Pi_{\mathrm{odd}}=\frac{\mathcal N_\ell\mu}{2r}k.$$

Indeed, $p\delta k=\Pi_{\mathrm{odd}}\delta\Psi_{\mathrm{odd}} +\delta(pk)$, so this exchange is canonical. The exact variation changes the action by a time-endpoint term, not the symplectic form. In spacetime notation this is the usual curl invariant

$$\Psi_{\mathrm{odd}}=\frac{2r}{\mu}\epsilon^{ab} \left(\nabla_a\widetilde h_b-\frac{2r_a}{r}\widetilde h_b\right), \qquad \epsilon^{tr}=-1,$$

where $\widetilde h_a=h_a-\partial_a h_2/2+r_a h_2/r$ and $r_a=\nabla_a r$. Substitution in the reduced Hamiltonian gives

$$\begin{aligned} \mathscr H_{\mathrm{odd,red}}={}& \frac{f}{2C_\ell}\Pi_{\mathrm{odd}}^2 +\frac{C_\ell}{2} \left[f(\Psi_{\mathrm{odd}}')^2 +\left(\frac L{r^2}-\frac{6m}{r^3}\right)\Psi_{\mathrm{odd}}^2\right]\\ &+\partial_r\!\left(\frac{C_\ell f}{2r}\Psi_{\mathrm{odd}}^2\right). \end{aligned}$$

Now change the radial measure and normalize the pair:

$$Q_{\mathrm{odd}}=\sqrt{C_\ell}\Psi_{\mathrm{odd}},\qquad P_{\mathrm{odd}}=\frac{f}{\sqrt{C_\ell}}\Pi_{\mathrm{odd}} =\frac{2\sqrt{C_\ell}f}{r}k.$$

Then $\{Q_{\mathrm{odd}}(x),P_{\mathrm{odd}}(y)\}=\delta(x-y)$. Notice that $P_{\mathrm{odd}}=2\sqrt{C_\ell}\Psi_{\mathrm{RW}}$, where $\Psi_{\mathrm{RW}}=fk/r$. Hamilton’s equation $\dot Q_{\mathrm{odd}}=P_{\mathrm{odd}}$ therefore gives $\dot\Psi_{\mathrm{odd}}=2\Psi_{\mathrm{RW}}$. No division by frequency was used; stationary solutions have not been lost.

Even parity: isolating all three constraints

Write the even spatial perturbation as

$$\delta\gamma_{rr}=f^{-1}HY,\qquad \delta\gamma_{rA}=jY_A,\qquad \delta\gamma_{AB}=r^2(K\Omega_{AB}Y+GY_{AB}).$$

There are four configuration amplitudes. The lapse perturbation and two even shift amplitudes will impose three first-class constraints. First remove the two spatial gauge directions from the metric combinations:

$$\varepsilon=j-\frac{r^2}{2}G',\qquad k=K+\frac L2G-\frac{2f}{r}\varepsilon,\qquad h=H-2f\varepsilon'-f'\varepsilon.$$

Here $k$ and $h$ are the values of $K$ and $H$ in spatial Regge–Wheeler gauge, not extra metric amplitudes. Their invariance follows directly under a spatial diffeomorphism with covariant components $\xi_r=aY$, $\xi_A=bY_A$:

$$\begin{aligned} \delta H&=2fa'+f'a,&\quad \delta j&=a+b'-2b/r,\\ \delta K&=2fa/r-Lb/r^2,& \delta G&=2b/r^2. \end{aligned}$$

This display uses $\delta\gamma=\mathcal L_\xi\bar\gamma$; replacing $\xi$ by $-\xi$ gives the opposite passive-coordinate convention with the same invariants. Since the background extrinsic curvature vanishes, a linearized normal displacement does not change the spatial metric; its constraint will instead generate a momentum direction.

An invertible coordinate map

Choose four new configuration variables:

$$\begin{aligned} q_1&=r\Lambda k+2rf(h-rk'),\\ q_2&=q_1'+L(h-rk'),\qquad q_3=G,\qquad q_4=j. \end{aligned}$$

They are not merely names for unknown canonical variables. The inverse is explicit:

$$k=\frac{q_1}{r\Lambda}-\frac{2f}{L\Lambda}(q_2-q_1'), \qquad h=rk'+\frac{q_2-q_1'}{L},$$ $$G=q_3,\qquad j=q_4,\qquad K=k-nG+\frac{2f}{r}\varepsilon,\qquad H=h+2f\varepsilon'+f'\varepsilon.$$

All denominators are nonzero in the stated exterior radiative sector. In particular, no radial integration or frequency inversion is needed. The physical coordinate is related to the Zerilli–Moncrief field by

$$\Psi_{\mathrm{even}} =\frac{2r}{L}\left[k+\frac{2f}{\Lambda}(h-rk')\right], \qquad q_1=n\Lambda\Psi_{\mathrm{even}}.$$

For comparison directly in metric amplitudes,

$$\Psi_{\mathrm{even}} =\frac{2r}{L}\left[K+\frac{2f}{\Lambda}(H-rK')\right] +rG-\frac{4f}{\Lambda}j.$$

The harmonic dictionary matters here. If an angular metric is instead expanded using $D_AD_BY$, its trace coefficient is $\bar K=K+nG$. The $K,G,h_1^*$ relations and the second expression for $H_2$ in Yang, Miao, and Chen 2014, Appendix B, eq. (B1), p. 104050-17, PDF are consistent with this traceful coefficient, whereas their Appendix A prints a trace-free tensor harmonic. We use the explicit shift $\bar K=K+nG$ when translating those relations. Their first printed expression for $H_2$ contains an additional $q_1$ term that does not agree with their second expression after substitution; it is not used here. A useful check is the pure angular gauge perturbation $G=2\beta$, $j=r^2\beta'$, $K=-L\beta$, $H=0$: the displayed invariant vanishes identically. Importing the traceful formula with $\bar K$ replaced by the site’s $K$ would fail this test.

The cotangent map and canonical brackets

Let $p_K,p_H,p_G,p_j$ be the momenta paired with $K,H,G,j$ in the measure $dr$. The trace shift must also act on momenta: in the $(\bar K,G)$ description they are $p_{\bar K}=p_K$ and $p_G^{\mathrm{trace}}=p_G-np_K$. Define

$$W=p_K-(rp_H)',\qquad b_0=\frac{f}{n\Lambda}.$$

The momenta conjugate to the four $q_i$ are

$$\begin{aligned} \Pi_1&=\frac{W}{r\Lambda}-(b_0W)'+\frac{p_H'}{2n},\\ \Pi_2&=-b_0W+\frac{p_H}{2n},\\ \Pi_3&=p_G-np_K+(rfW)'+[(r-3m)p_H]',\\ \Pi_4&=p_j+\frac{2f}{r}W+\frac{2(r-3m)}{r^2}p_H. \end{aligned}$$

These formulas are obtained by substituting the inverse coordinate map into the old symplectic potential and integrating by parts. More explicitly,

$$\int dr\,(p_K\delta K+p_H\delta H+p_G\delta G+p_j\delta j) =\int dr\,\sum_{i=1}^4\Pi_i\delta q_i+[\mathcal B_\Theta],$$ $$\begin{aligned} \mathcal B_\Theta={}&rp_H\delta\bar K +\left(b_0W-\frac{p_H}{2n}\right)\delta q_1\\ &-\left[rfW+(r-3m)p_H\right]\delta q_3. \end{aligned}$$

Thus the interior brackets are $\{q_i(r),\Pi_j(s)\}=\delta_{ij}\delta(r-s)$, and all coordinate–coordinate and momentum–momentum brackets vanish. An invariant coordinate alone would not establish this result: its conjugate momentum and the endpoint term are essential.

Identifying the scalar and momentum constraints

The meaning of $q_2$ can be checked without guessing a canonical transformation. The linearized spatial scalar curvature, evaluated on the invariant amplitudes, is

$$\begin{aligned} \frac{\delta{}^{(3)}R}{Y}={}& \frac{2f}{r}h'-2fk''-\left(\frac{6f}{r}+f'\right)k'\\ &+\frac{L+2}{r^2}h+\frac{\mu}{r^2}k =\frac{q_2}{r^2}. \end{aligned}$$

To derive it, insert the diagonal invariant perturbation into $\delta{}^{(3)}R=D_iD_j\delta\gamma^{ij}-D^2(\bar\gamma^{ij}\delta\gamma_{ij}) -\bar R^{ij}\delta\gamma_{ij}$, where the raised perturbation means $\delta\gamma^{ij}=\bar\gamma^{ik}\bar\gamma^{jl}\delta\gamma_{kl}$, not the variation of the inverse metric. The background has $\bar R_{rr}=-2m/(fr^3)$, $\bar R_{AB}=(m/r)\Omega_{AB}$, and zero scalar curvature. Expanding $q_1'+L(h-rk')$ reproduces the expression above.

Define the lapse and shift amplitudes by $\delta N=-\sqrt f H_0Y/2$, $\delta g_{tr}=H_1Y$, and $\delta g_{tA}=j_tY_A$. The multiplier part of the Hamiltonian is

$$\mathscr H_{\mathrm{mult}} =\frac{H_0}{32\pi G}q_2+H_1\Pi_4 +\frac{j_t}{r^2}\left[2\Pi_3-(r^2\Pi_4)'\right].$$

The lapse coefficient follows immediately by multiplying the linearized ADM scalar constraint by $\delta N$ and integrating $Y^2$ over the sphere. For an independent check of the shifts, contracting the old momenta with the spatial gauge variations above gives the two generators

$$\begin{aligned} C_a&=-(2fp_H)'+f'p_H+\frac{2f}{r}p_K+p_j=\Pi_4,\\ C_b&=\frac{2p_G-Lp_K}{r^2}-p_j'-\frac{2p_j}{r} =\frac{2\Pi_3-(r^2\Pi_4)'}{r^2}. \end{aligned}$$

Consequently the three constraints are precisely

$$q_2=0,\qquad \Pi_3=0,\qquad \Pi_4=0.$$

Their mutual brackets vanish, as do their brackets with $q_1$ and $\Pi_1$. The three gauge pairs can be displayed without an ellipsis:

$$(-\Pi_2,q_2),\qquad(q_3,\Pi_3),\qquad(q_4,\Pi_4).$$

In the first pair the constraint is a momentum after exchanging coordinate and momentum with the indicated minus sign. Imposing the constraints and quotienting these three directions leaves exactly $(q_1,\Pi_1)$. The full source Hamiltonian and its three constraint equations appear in Yang, Miao, and Chen 2014, § II.E, eqs. (2.32)–(2.36), pp. 104050-6–104050-7, PDF.

The reduced Hamiltonians and their normalization

The odd calculation has already fixed $C_\ell$. We now check the even coefficient directly from ADM, rather than borrowing a radiation-flux normalization. Integrating the ADM momentum-quadratic term over the real unit harmonic gives

$$\mathscr H_{\mathrm{kin}} =\frac{16\pi Gf}{r^2} \left[\frac{p_H^2}{2}-p_Hp_K +\frac{2p_G^2}{L\mu}+\frac{r^2p_j^2}{2fL}\right].$$

Choose the representative $\Pi_2=0$ of the normal gauge orbit. On all three constraints the inverse momentum map becomes

$$\begin{aligned} p_H&=2rf\Pi_1,& p_K&=r\Lambda\Pi_1+(2r^2f\Pi_1)',\\ p_G&=nr\Lambda\Pi_1,& p_j&=-2Lf\Pi_1. \end{aligned}$$

Substitution and one integration by parts yield

$$\mathscr H_{\mathrm{kin}} =8\pi G\frac L\mu f\Lambda^2\Pi_1^2 +\partial_r(-32\pi G\,rf^3\Pi_1^2).$$

This is a direct gravitational check of the physical momentum normalization. Together with the spatial-curvature part, the reduced bulk Hamiltonian is

$$\mathscr H_{\mathrm{even,red}} =8\pi G\frac L\mu f\Lambda^2\Pi_1^2 +\frac{\mu f}{32\pi G L\Lambda^2}(q_1')^2 +\frac{\mu^2}{32\pi G r^2\Lambda^3}q_1^2.$$

The last two terms are the constraint-restricted metric terms of Yang, Miao, and Chen 2014, eq. (2.32), p. 104050-6, PDF, with their suppressed action factor restored. Their canonical momentum is $16\pi G\Pi_1$; the coupling convention is explained after Yang, Miao, and Chen 2014, eq. (3.5), p. 104050-8, PDF. The displayed bulk representative is equivalent on compact support; it is not an assertion that the unmodified ADM boundary functional has vanished for arbitrary boundary data.

Set $\Pi_{\mathrm{even}}=n\Lambda\Pi_1$ and $a_0=\Lambda'/\Lambda$. Since $q_1=n\Lambda\Psi_{\mathrm{even}}$, this is a canonical rescaling in $dr$, and the Hamiltonian becomes

$$\mathscr H_{\mathrm{even,red}} =\frac{f}{2C_\ell}\Pi_{\mathrm{even}}^2 +\frac{C_\ell}{2} \left[f(\Psi_{\mathrm{even}}'+a_0\Psi_{\mathrm{even}})^2 +\frac{L\mu}{r^2\Lambda}\Psi_{\mathrm{even}}^2\right].$$

Expanding the square produces a spatial total derivative $\partial_r(C_\ell f a_0\Psi_{\mathrm{even}}^2/2)$ and the potential

$$U_{\mathrm Z}=f a_0^2-(fa_0)'+\frac{L\mu}{r^2\Lambda}, \qquad V_{\mathrm Z}=fU_{\mathrm Z}.$$

Insertion of $\Lambda=\mu+6m/r$ gives

$$\begin{aligned} V_{\mathrm{RW}}&=f\left(\frac L{r^2}-\frac{6m}{r^3}\right),\\ V_{\mathrm Z}&=\frac{2f}{r^3(\lambda r+3m)^2} \left[\lambda^2(\lambda+1)r^3+3\lambda^2mr^2 +9\lambda m^2r+9m^3\right]. \end{aligned}$$

These are exactly the neighboring Regge–Wheeler and Zerilli potentials. Finally define

$$Q_{\mathrm{even}}=\sqrt{C_\ell}\Psi_{\mathrm{even}},\qquad P_{\mathrm{even}}=\frac{fn\Lambda}{\sqrt{C_\ell}}\Pi_1.$$

For either parity, the answer now has the same unit-canonical form:

$$H_{\mathrm{red}}=\frac12\int dx\, \left[P^2+(\partial_xQ)^2+VQ^2\right], \qquad \{Q(x),P(y)\}=\delta(x-y).$$

Hamilton’s equations are $\dot Q=P$ and $\dot P=\partial_x^2Q-VQ$. Eliminating $P$ gives

$$S_{\mathrm{red}}=\frac12\int dt\,dx\, \left[\dot Q^2-(\partial_xQ)^2-VQ^2\right],\qquad (-\partial_t^2+\partial_x^2-V)Q=0.$$

Thus the master equation and its canonical energy come from the same reduction. Both potentials are positive in the open exterior for $\ell\geq2$, so this bulk Hamiltonian is nonnegative on real compactly supported data. That positivity alone is not a proof of decay, horizon regularity, or a complete scattering theory.

What the reduction includes and excludes

The degree count can now be read from explicit canonical pairs:

Parity	Initial spatial pairs	Independent first-class constraints	Remaining physical pair
Odd	2	$C_g$	$(Q_{\mathrm{odd}},P_{\mathrm{odd}})$
Even	4	$q_2,\Pi_3,\Pi_4$	$(Q_{\mathrm{even}},P_{\mathrm{even}})$

Each constraint removes one phase-space direction, and quotienting its gauge orbit removes another. Together the two parities encode the two gravitational degrees of freedom. They are not two independent copies of every complex harmonic coefficient.

Boundary conditions are part of the phase space. The endpoint terms in the symplectic potentials and Hamiltonians have been shown explicitly where used. For general finite intervals or asymptotically flat data, a generator may need a surface term to be differentiable. It is a proper gauge generator only if its admitted charge vanishes throughout the chosen phase space; vanishing on the background alone is insufficient. The present compact-support calculation does not classify those charges or identify arbitrary metric boundary conditions with scalar Dirichlet conditions. Although the formulas hold at every $r\gt2m$, extension through the horizon requires a regular foliation and appropriate function spaces.

Low multipoles are not obtained by taking a singular limit. For $\ell=0,1$, vector or tensor harmonics disappear and factors of $L\mu$ prevent the above inversions. Mass perturbations and odd dipole angular momentum are nonradiative physical data. Even dipole momentum or mass-dipole information depends on the asymptotic frame and the admitted boundary transformations; a choice of center of mass is not an extra local radiative invariant. Low-Multipole Perturbations treats these sectors separately.

The canonical result is compatible with the local metric reconstruction in Even-Parity Metric Perturbations. It explains the physical phase space; it does not by itself solve the initial-value, spectral, or boundary-value problem for a chosen master field.

Exercises

Count the phase-space dimensions

Apply the constraint-and-quotient rule to both rows of the table. Why would subtracting only one dimension per first-class constraint be wrong?

Solution

Odd parity starts with four phase-space dimensions per radius. The constraint surface has one fewer dimension, and quotienting the gauge orbit removes one more: $4-2=2$. Even parity gives $8-2\times3=2$. Each result is one canonical pair, or one configuration degree of freedom. Imposing constraints without the quotient would still count gauge-equivalent data as different physical states.

Change the radial measure canonically

Suppose $\{\Psi(r),\Pi(s)\}=\delta(r-s)$ and $\Theta=\int dr\,\Pi\delta\Psi$. For $dr=f\,dx$ and $Q=\sqrt{C_\ell}\Psi$, find the conjugate $P$ in the measure $dx$ and verify its bracket. Why is $P=\Pi/\sqrt{C_\ell}$ incorrect?

Solution

The symplectic potential becomes

$$\Theta=\int dx\,\frac{f\Pi}{\sqrt{C_\ell}}\delta Q, \qquad P=\frac{f\Pi}{\sqrt{C_\ell}}.$$

Since $f\gt0$, the delta function transforms as

$$\delta(r(x)-r(y))=\frac{\delta(x-y)}{f(x)},\qquad \{Q(x),P(y)\}=\frac{f(y)}{f(x)}\delta(x-y)=\delta(x-y).$$

Omitting $f$ leaves the wrong radial density and the wrong canonical bracket. For the reduced action $\Pi=C_\ell\dot\Psi/f$, the correct transformation also gives $P=\dot Q$.

Test a boundary condition

Vary the unit-canonical Hamiltonian on a finite tortoise interval $a\leq x\leq b$. Identify the boundary term. Does imposing a scalar Dirichlet condition prove that the original gravitational boundary problem has been solved?

Solution

One integration by parts gives

$$\delta H=\int_a^b dx\, \left[P\delta P+(-\partial_x^2Q+VQ)\delta Q\right] +[\partial_xQ\,\delta Q]_a^b.$$

Fixed endpoint values of $Q$ make $\delta Q=0$ there, so this scalar Hamiltonian is differentiable with respect to those variations. The conclusion concerns the reduced scalar problem only. The metric-to-master map contains derivatives, and we have also encountered boundary terms in the canonical transformation. One must translate the gravitational boundary data, permitted gauge transformations, and surface action before claiming equivalence of the two boundary-value problems.

References

• Wald, Robert M. General Relativity. Chicago: University of Chicago Press, 1984. Publisher.
• Yang, Huan, Haixing Miao, and Yanbei Chen. “First-Order Perturbative Hamiltonian Equations of Motion for a Point Particle Orbiting a Schwarzschild Black Hole.” Physical Review D 89 (2014): 104050. DOI. Open PDF.