The Connection Formulation of General Relativity

26 年 4 月 12 日 星期日
2494 字
13 分钟

系列文章:LQG Notes

It has long been known that general relativity can be formulated by applying a constraint on the BF theory. Plebanski successfully obtained the Riemannian action:

S[B,ω,λ]=1κM[( ⁣BIJ+1γBIJ)FIJ(ω)+λIJKLBIJBKL].S[B, \omega, \lambda] = \frac{1}{\kappa}\int_\mathcal{M}\left[\left(^\star\!B^{IJ} + \frac{1}{\gamma}B^{IJ}\right)\wedge F_{IJ}(\omega) + \lambda_{IJKL} B^{IJ}\wedge B^{KL}\right].

This can be thought of as a Spin(4)\mathrm{Spin}(4) BF theory and a constraint imposed by the lagrangian multiplier λ\lambda. It depends on a so(4)\mathfrak{so}(4) connection ω\omega and a Lie-algebra valued 2-form BB.

The symmetry of BB and \wedge imposes a constraint on λ\lambda as well. Since switching between the indices of BIJB^{IJ} is antisymmetric and that the operation \wedge is symmetric, we have the following relation:

λIJKL=λJIKL=λIJLK=λKLIJ.\lambda_{IJKL} = -\lambda_{JIKL} = -\lambda_{IJLK} = \lambda_{KLIJ}.

This restrains the dimension of λ\lambda down to 21. However, if λIJKLϵIJKL\lambda_{IJKL}\propto \epsilon_{IJKL} the lagrangian multiplier will give ϵIJKLBIJBKL\epsilon_{IJKL}B^{IJ}\wedge B^{KL}, which implies a degenerate metric. Therefore ϵIJKLλIJKL=0\epsilon^{IJKL}\lambda_{IJKL} = 0, and that dimλ=20\mathrm{dim}\lambda = 20.

To actually calculate the constraint given by the lagrangian multiplier, we introduce

λIJKL=ΛIJKL14!ΛMNOPϵMNOPϵIJKL.\lambda_{IJKL} = \Lambda_{IJKL} - \frac{1}{4!}\Lambda_{MNOP}\epsilon^{MNOP}\epsilon_{IJKL}.

We make this substitution because if we assume Λ\Lambda to be the unconstrained lagrangian multiplier and

Λ[IJKL]=cϵIJKL\Lambda_{[IJKL]} = c\cdot\epsilon_{IJKL}

we can multiply both sides with ϵ\epsilon and have

Λ[IJKL]ϵIJKL=cϵ2    c=14!Λ[MNOP]ϵMNOP.\Lambda_{[IJKL]}\epsilon^{IJKL} = c\cdot\epsilon^2\implies c = \frac{1}{4!}\Lambda_{[MNOP]}\epsilon^{MNOP}.

This gives the derivative of the action with respect to Λ\Lambda:

δSδΛIJKL=BIJBKL14!ϵIJKLϵMNOPBMNBOP=0.\frac{\delta S}{\delta \Lambda_{IJKL}} = B^{IJ}\wedge B^{KL} - \frac{1}{4!}\epsilon^{IJKL}\epsilon_{MNOP}B^{MN}\wedge B^{OP} = 0.

which implies

ϵμνρσBμνIJBρσKL=14!ϵIJKLϵMNOPBμνMNBρσOPϵμνρσ.\epsilon^{\mu\nu\rho\sigma} B^{IJ}_{\mu\nu} B^{KL}_{\rho\sigma} = \frac{1}{4!} \epsilon^{IJKL} \epsilon_{MNOP} B^{MN}_{\mu\nu} B^{OP}_{\rho\sigma} \epsilon^{\mu\nu\rho\sigma}.

This equation constraints the degrees of freedom of BB, limiting its dimension down to 16=4×416 = 4\times 4. So if we define:

e:=14!ϵMNOPBμνMNBρσOPϵμνρσe := \frac{1}{4!} \epsilon_{MNOP} B^{MN}_{\mu\nu} B^{OP}_{\rho\sigma} \epsilon^{\mu\nu\rho\sigma}

It behaves exactly as the local tetrad. The solution of BB would be

B=±(ee)and±ee.B = \pm ^\star(e\wedge e) \quad\text{and}\quad \pm e\wedge e.

Here ee is the new variable, and the term λBB\lambda B\wedge B in the action is simply the constraint. So if we put B=±(ee)B = \pm^\star(e\wedge e) back into the action SS, we obtain

S[e,ω]=1κMTr[((ee)+1γee)F(ω)].S[e, \omega] = \frac{1}{\kappa}\int_\mathcal{M} \operatorname{Tr}\left[\left(^\star(e\wedge e) + \frac{1}{\gamma}e\wedge e\right) \wedge F(\omega)\right].

Canonical Analysis

We perform the 3+1 decomposition. We use labels a,b=1,2,3a,b = 1,2,3 as spacial indices on Σt\Sigma_t. Introduce

ΠabIJ=BabIJ+1γBabIJ    pure spacial; lives on ΣtHaIJ=B0aIJ+1γB0aIJ    space-time.\begin{aligned} \Pi_{ab}^{IJ} &= ^\star B_{ab}^{IJ} + \frac{1}{\gamma}B_{ab}^{IJ}\implies \text{pure spacial; lives on }\Sigma_t\\ H_a^{IJ} &= ^\star B_{0a}^{IJ} + \frac{1}{\gamma}B_{0a}^{IJ}\implies \text{space-time}. \end{aligned}

Also, we decompose the curvature form FμνF_{\mu\nu}:

Fab=[aωb]+[ωa,ωb]F0a=0ωaaω0+[ω0,ωa]=ω˙a(aω0[ωa,ω0])=:ω˙aDaω0.\begin{aligned} F_{ab} &= \partial_{[a}\omega_{b]} + [\omega_a, \omega_b]\\ F_{0a} &= \partial_0\omega_a - \partial_a\omega_0 + [\omega_0, \omega_a] = \dot{\omega}_a - (\partial_a\omega_0 - [\omega_a, \omega_0]) =: \dot{\omega}_a - \mathcal{D}_a\omega_0. \end{aligned}

The action is originally:

S[Π,H,ω,λ]=1κM[(BIJ+1γBIJ)FIJ+λIJKLBIJBKL]S[\Pi, H, \omega, \lambda] = \frac{1}{\kappa}\int_\mathcal{M}\left[\left(^\star B^{IJ} + \frac{1}{\gamma}B^{IJ}\right)\wedge F_{IJ} + \lambda_{IJKL} B^{IJ}\wedge B^{KL}\right]

Substituting the 3+1 decomposed variables:

S=1κM[(ΠabIJFIJ,0a)+(HaIJFIJ,ab)+λIJKLB0aIJBabKL]S = \frac{1}{\kappa}\int_\mathcal{M} \left[(\Pi_{ab}^{IJ}\wedge F_{IJ,0a}) + (H_{a}^{IJ}\wedge F_{IJ,ab}) + \lambda_{IJKL}B_{0a}^{IJ}\wedge B_{ab}^{KL}\right]

After some algebra, this leads to:

S=1κM(ΠabIJω˙IJ,cΠabIJDaω0,IJ+HaIJFIJ,ab+λIJKL{γ2γ21(γHaIJHaIJ)γγ21(γΠbcKLΠbcKL)}).\begin{aligned} S &= \frac{1}{\kappa}\int_\mathcal{M}\left(\Pi_{ab}^{IJ}\wedge \dot{\omega}_{IJ,c} - \Pi_{ab}^{IJ}\wedge \mathcal{D}_a \omega_{0,IJ} + H_a^{IJ}\wedge F_{IJ,ab} \right. \\ &+ \left.\lambda_{IJKL}\left\{\frac{\gamma^2}{\gamma^2-1}(\gamma^\star H_{a}^{IJ} - H_{a}^{IJ}) \wedge \frac{\gamma}{\gamma^2-1}(\gamma\Pi_{bc}^{KL} - ^\star\Pi_{bc}^{KL})\right\}\right). \end{aligned}

By integrating by parts, the second term gives ω0DωΠ\omega_0 \mathcal{D}_\omega\wedge \Pi, and if we redefine λ=γ2(γ21)2λ\lambda' = \frac{\gamma^2}{(\gamma^2-1)^2}\lambda, the action is:

S=dtΣtTr[Πω˙+ω0DωΠ+HF+λ(γHH)(γΠΠ)].S = \int dt \int_{\Sigma_t} \operatorname{Tr} \left[ \Pi \wedge \dot{\omega} + \omega_0 \mathcal{D}_\omega \wedge \Pi + H \wedge F + \lambda' (\gamma ^\star H - H) \wedge (\gamma ^\star \Pi - \Pi) \right].

The canonical pair is (ΠabIJ,ωcKL)(\Pi^{IJ}_{ab}, \omega^{KL}_c). If we introduce

P(γ,±)abi:=14ϵ jkiΠabjk±12γΠab0i±Aci:=12ϵ mniωcmn±γωc0iH(γ,±)ai:=14ϵ jkiHajk±12γHa0i\begin{aligned} \mathbb{P}(\gamma, \pm)^i_{ab} &:= \frac{1}{4} \epsilon^i_{\ jk} \Pi^{jk}_{ab} \pm \frac{1}{2\gamma} \Pi^{0i}_{ab} \\ {}^{\pm}\mathbb{A}_c^i &:= \frac{1}{2} \epsilon^i_{\ mn} \omega^{mn}_c \pm \gamma \omega^{0i}_c \\ \mathbb{H}(\gamma, \pm)^i_a &:= \frac{1}{4} \epsilon^i_{\ jk} H^{jk}_a \pm \frac{1}{2\gamma} H^{0i}_a \end{aligned}

where A\mathbb{A} is a function of ω\omega, its time derivative corresponds to

0±Aci=12ϵ mni0ωcmn±γ0ωc0i=12ϵ mniω˙cmn±γω˙c0i=:±A˙ci.\partial_0 {}^{\pm}\mathbb{A}_c^i = \frac{1}{2} \epsilon^i_{\ mn} \partial_0 \omega^{mn}_c \pm \gamma \partial_0 \omega^{0i}_c = \frac{1}{2} \epsilon^i_{\ mn} \dot{\omega}^{mn}_c \pm \gamma \dot{\omega}^{0i}_c =: {}^{\pm}\dot{\mathbb{A}}_c^i.

Also notice that

P(γ,+)abi=14ϵijkΠabjk+12γΠab,i0,\mathbb{P}(\gamma, +)^i_{ab} = \frac{1}{4} \epsilon_{ijk} \Pi^{jk}_{ab} + \frac{1}{2\gamma} \Pi^{0}_{ab, i}, P(γ,)abi=14ϵijkΠabjk12γΠab,i0.\mathbb{P}(\gamma, -)^i_{ab} = \frac{1}{4} \epsilon_{ijk} \Pi^{jk}_{ab} - \frac{1}{2\gamma} \Pi^{0}_{ab, i}.

These simplify the canonical trace term Tr[Πω˙]\operatorname{Tr}[\Pi\wedge\dot{\omega}]:

P++A˙+PA˙=(Π+Π0)(ω˙+ω˙0)+(ΠΠ0)(ω˙ω˙0)=2Πω˙+2Π0ω˙0=12ϵijkΠjk12ϵmniω˙mn+1γΠi0γω˙i0=12(δjmδknδjnδkm)Πjkω˙mn+Πi0ω˙i0=12Πijω˙ij+Πi0ω˙i0=Tr[Πω˙].\begin{aligned} \mathbb{P}^{+} \wedge {}^{+}\dot{\mathbb{A}} + \mathbb{P}^{-} \wedge {}^{-}\dot{\mathbb{A}} &= (\Pi + \Pi^0) \wedge (\dot{\omega} + \dot{\omega}^0) + (\Pi - \Pi^0) \wedge (\dot{\omega} - \dot{\omega}^0) \\ &= 2 \Pi \wedge \dot{\omega} + 2 \Pi^0 \wedge \dot{\omega}^0 \\ &= \frac{1}{2} \epsilon_{ijk} \Pi^{jk} \wedge \frac{1}{2} \epsilon^i_{\, mn} \dot{\omega}^{mn} + \frac{1}{\gamma} \Pi^{0}_i \wedge \gamma \dot{\omega}^{i0} \\ &= \frac{1}{2} (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}) \Pi^{jk} \wedge \dot{\omega}^{mn} + \Pi^{0}_i \wedge \dot{\omega}^{i0} \\ &= \frac{1}{2} \Pi_{ij} \wedge \dot{\omega}^{ij} + \Pi^0_{i} \wedge \dot{\omega}^{i0} = \operatorname{Tr} [ \Pi \wedge \dot{\omega} ]. \end{aligned}

The term ω0DωΠ\omega_0 \mathcal{D}_\omega \wedge \Pi acts as a Lagrangian constraint, with ω0\omega_0 serving as the multiplier. We can express the connection variables as:

+A+A2=12ϵ mniωcmnand(+AA)j=2γωc0j.\frac{{}^{+}\mathbb{A} + {}^{-}\mathbb{A}}{2} = \frac{1}{2} \epsilon^i_{\ mn} \omega^{mn}_c \quad \text{and} \quad ({}^{+}\mathbb{A} - {}^{-}\mathbb{A})_j = 2\gamma \omega^{0j}_c.

Consequently, we can decompose the constraint term:

ω0DωΠ=+A0(D+AP+)i+A0(DAP)i.\begin{aligned} \omega_0 \mathcal{D}_\omega \wedge \Pi &= {}^{+}\mathbb{A}_0 (\mathcal{D}_{{}^{+}\mathbb{A}} \mathbb{P}^{+})_i + {}^{-}\mathbb{A}_0 (\mathcal{D}_{{}^{-}\mathbb{A}} \mathbb{P}^{-})_i. \end{aligned}

Expanding the first term, we find:

+A0(dP+++A×P+)=+A0[dP++12ϵijk(ϵ mnjωmn+γω0j)P+k]=+A0[dP++12ϵijkϵ mnjωmnP+k+γϵijkω0jP+k]=+A0D(12ϵω)P+++A0ϵijk+AA2P+kηiBi+NiGi.\begin{aligned} {}^{+}\mathbb{A}_0 \left( d \mathbb{P}^{+} + {}^{+}\mathbb{A} \times \mathbb{P}^{+} \right) &= {}^{+}\mathbb{A}_0 \left[ d \mathbb{P}^{+} + \frac{1}{2}\epsilon_{ijk} \left( \epsilon^j_{\ mn} \omega^{mn} + \gamma \omega^{0j} \right) \mathbb{P}^{+k}\right] \\ &= {}^{+}\mathbb{A}_0 \left[ d \mathbb{P}^{+} + \frac{1}{2} \epsilon_{ijk} \epsilon^j_{\ mn} \omega^{mn} \mathbb{P}^{+k} + \gamma \epsilon_{ijk} \omega^{0j} \mathbb{P}^{+k} \right] \\ &= {}^{+}\mathbb{A}_0 \, \mathcal{D}_{\left(\frac{1}{2}\epsilon \omega\right)} \mathbb{P}^{+} + {}^{+}\mathbb{A}_0 \, \epsilon_{ijk} \frac{{}^{+}\mathbb{A} - {}^{-}\mathbb{A}}{2} \mathbb{P}^{+k} \\ &\sim \eta_i \mathbb{B}^i + N_i \mathbb{G}^i. \end{aligned}

Here we introduce new constraints Bi\mathbb{B}^i and Gi\mathbb{G}^i. Hence, we can rewrite the action as:

S[P(γ,±),±A,λ]=dtΣt{P++A˙+PA˙+NiGi+ηiBi+A0DAP+HF+λ(γHH)(γΠΠ)}.\begin{aligned} S[\mathbb{P}(\gamma, \pm), {}^{\pm}\mathbb{A}, \lambda] &= \int dt \int_{\Sigma_t} \Big\{ \mathbb{P}^{+} \wedge {}^{+}\dot{\mathbb{A}} + \mathbb{P}^{-} \wedge {}^{-}\dot{\mathbb{A}} + N_i \mathbb{G}^i + \eta_i \mathbb{B}^i \\ &+ {}^{-}\mathbb{A}_0 \, \mathcal{D}_{{}^{-}\mathbb{A}} \mathbb{P}^{-} + H \wedge F + \lambda' (\gamma ^\star H - H) \wedge (\gamma ^\star \Pi - \Pi) \Big\}. \end{aligned}

The Lagrangian multipliers N,η,λN, \eta, \lambda' impose the following constraints:

{Bi:=D+A+A2P(γ,+)i0Gi:=12γϵijk(+AA)jP(γ,+)k0(γHH)(γΠΠ)=0    ϵμνρσBμνIJBρσKLeϵIJKL=0.\begin{cases} \mathbb{B}^i := \mathcal{D}_{\frac{{}^{+}\mathbb{A} + {}^{-}\mathbb{A}}{2}} \wedge \mathbb{P}(\gamma, +)^i \approx 0 \\ \mathbb{G}^i := \frac{1}{2\gamma} \epsilon_{ijk} ({}^{+}\mathbb{A} - {}^{-}\mathbb{A})_j \wedge \mathbb{P}(\gamma, +)_k \approx 0 \\ (\gamma^\star H - H)\wedge(\gamma^\star \Pi - \Pi) = 0 \implies \epsilon_{\mu\nu\rho\sigma} B^{IJ}_{\mu\nu} B^{KL}_{\rho\sigma} - e\epsilon^{IJKL} = 0. \end{cases}

The index pair [IJ,KL][IJ, KL] must be decomposed. We separate it into three cases:

{IJKL=0i0jIJKL=ijklIJKL=0ijk.\begin{cases} IJKL = 0i0j\\ IJKL = ijkl\\ IJKL = 0ijk. \end{cases}

For case 1 (IJKL=0i0jIJKL = 0i0j):

B0iB0j=0    CH0iCΠ0j+(ij)=0    (γH0iH0i)(γΠ0jΠ0j)+(ij)=0    (γ2ϵ jkiHjkH0i)(γ2ϵ kljΠklΠ0j)+(ij)=0.\begin{gathered} B^{0i} \wedge B^{0j} = 0 \\ \implies C^{0i}_H \wedge C^{0j}_\Pi + (i \leftrightarrow j) = 0 \\ \implies (\gamma ^\star H^{0i} - H^{0i}) \wedge (\gamma ^\star \Pi^{0j} - \Pi^{0j}) + (i \leftrightarrow j) = 0 \\ \implies \left(\frac{\gamma}{2}\epsilon^i_{\ jk}H^{jk} - H^{0i}\right) \wedge \left(\frac{\gamma}{2}\epsilon^j_{\ kl}\Pi^{kl} - \Pi^{0j}\right) + (i \leftrightarrow j) = 0. \end{gathered}

For case 2 (IJKL=ijklIJKL = ijkl):

BijBkl=0    CHijCΠkl+(ijkl)=0    (γHijHij)(γΠklΠkl)+(ijkl)=0    ϵ   ij(p(γ2ϵ  mijH0mHij)ϵ   klq)(γ2ϵ  nklΠ0nΠkl)=0.\begin{gathered} B^{ij} \wedge B^{kl} = 0 \\ \implies C^{ij}_H \wedge C^{kl}_\Pi + (ij \leftrightarrow kl) = 0 \\ \implies (\gamma ^\star H^{ij} - H^{ij}) \wedge (\gamma ^\star \Pi^{kl} - \Pi^{kl}) + (ij \leftrightarrow kl) = 0 \\ \implies \epsilon^{(p}_{\ \ \ ij}\left(\frac{\gamma}{2}\epsilon^{ij}_{\ \ m} H^{0m} - H^{ij}\right) \wedge \epsilon^{q)}_{\ \ \ kl}\left(\frac{\gamma}{2}\epsilon^{kl}_{\ \ n} \Pi^{0n} - \Pi^{kl}\right) = 0. \end{gathered}

For case 3 (IJKL=0ijkIJKL = 0ijk):

B0iBjk=νϵ0ijk    CH0iCΠjk+(0ijk)=νϵ0ijk    (γH0iH0i)(γΠjkΠjk)+(0ijk)=νϵ0ijk    (γ2ϵ mniHmnH0i)(γ2ϵ  ljkΠ0lΠjk)+(γ2ϵ mniΠmnΠ0i)(γ2ϵ  ljkH0lHjk)νϵ0ijk=0.\begin{gathered} B^{0i} \wedge B^{jk} = \nu \epsilon^{0ijk} \\ \implies C^{0i}_H \wedge C^{jk}_\Pi + (0i \leftrightarrow jk) = \nu \epsilon^{0ijk} \\ \implies (\gamma ^\star H^{0i} - H^{0i}) \wedge (\gamma ^\star \Pi^{jk} - \Pi^{jk}) + (0i \leftrightarrow jk) = \nu \epsilon^{0ijk} \\ \begin{aligned} \implies &\left(\frac{\gamma}{2}\epsilon^i_{\ mn} H^{mn} - H^{0i}\right) \wedge \left(\frac{\gamma}{2}\epsilon^{jk}_{\ \ l} \Pi^{0l} - \Pi^{jk}\right) +\\ &\left(\frac{\gamma}{2}\epsilon^i_{\ mn}\Pi^{mn} - \Pi^{0i}\right) \wedge \left(\frac{\gamma}{2}\epsilon^{jk}_{\ \ l}H^{0l} - H^{jk}\right) - \nu \epsilon^{0ijk}= 0. \end{aligned} \end{gathered}

Recall that we've defined

H(γ,±)ai:=14ϵ jkiHajk±12γHa0iP(γ,±)abi:=14ϵ jkiΠabjk±12γΠab0i.\begin{aligned} \mathbb{H}(\gamma, \pm)^i_a &:= \frac{1}{4} \epsilon^i_{\ jk} H^{jk}_a \pm \frac{1}{2\gamma} H^{0i}_a \\ \mathbb{P}(\gamma, \pm)^i_{ab} &:= \frac{1}{4} \epsilon^i_{\ jk} \Pi^{jk}_{ab} \pm \frac{1}{2\gamma} \Pi^{0i}_{ab}. \end{aligned}

Therefore we can rewrite the constraints as:

Bi=D(+A+A)/2P(γ,+)i0Gi=12γϵijk(+AA)jP(γ,+)k0Iij=H(γ,)(iP(γ,)j)0IIij=H(1γ,)(iP(1γ,)j)0IIIij=H(γ,)iP(γ,+)j+H(1γ,)iP(γ,)j13δijTr[]=0\begin{aligned} &\mathbb{B}^i = \mathcal{D}_{({}^{+}\mathbb{A} + {}^{-}\mathbb{A})/2} \wedge \mathbb{P}(\gamma, +)^i \approx 0 \\ &\mathbb{G}^i = \frac{1}{2\gamma} \epsilon^{ijk} ({}^{+}\mathbb{A} - {}^{-}\mathbb{A})_j \wedge \mathbb{P}(\gamma, +)_k \approx 0 \\ &\mathbb{I}^{ij} = \mathbb{H}(\gamma, -)^{(i} \wedge \mathbb{P}(\gamma, -)^{j)} \approx 0 \\ &\mathbb{II}^{ij} = \mathbb{H}\left(\frac{1}{\gamma}, -\right)^{(i} \wedge \mathbb{P}\left(\frac{1}{\gamma}, -\right)^{j)} \approx 0 \\ &\mathbb{III}^{ij} = \mathbb{H}(\gamma, -)^{i} \wedge \mathbb{P}(\gamma, +)^{j} + \mathbb{H}\left(\frac{1}{\gamma}, -\right)^{i} \wedge \mathbb{P}(\gamma, -)^{j} - \frac{1}{3} \delta^{ij} \operatorname{Tr}[\dots] = 0 \end{aligned}

The last three constraints are derived from the original lagrangian multiplier λ\lambda, therefore include 20 constraints in total and are simplicity constraints. In the 3+1 setting, by choosing the time gauge to be nI=(1,0,0,0)n^I = (1, 0, 0, 0), eμIe^I_\mu is

{e00=N(lapse)ea0=0(pure spacial, no time-like component)e0i=Ni(shift)eai=standard 3-d tetrad\begin{cases} e^0_0 = N & (\text{lapse}) \\ e^0_a = 0 & (\text{pure spacial, no time-like component}) \\ e^i_0 = N^i & (\text{shift}) \\ e^i_a = \text{standard 3-d tetrad} \end{cases}

(Indices: I,J0,1,2,3I, J \in 0, 1, 2, 3 local; μ,ν0,1,2,3\mu, \nu \in 0, 1, 2, 3 global; a,b1,2,3a,b \in 1, 2, 3 global spacial; i,j1,2,3i,j \in 1, 2, 3 local spacial). We consider the solution B=ee    BμνIJ=eμIeνJeμJeνIB = e \wedge e \implies B^{IJ}_{\mu\nu} = e^I_\mu e^J_\nu - e^J_\mu e^I_\nu.

Globally spacial components:

Bab0i=ea0ebieb0eai=0    Bab0i=12ϵ   jk0iBabjk=12ϵ   jk0i(eajebkebjeak)=ϵ  jkieajebkBabjk=eajebkeakebj=2ea[jebk]    Babjk=12ϵ   0ijkBab0i=0.\begin{aligned} B^{0i}_{ab} &= e^0_a e^i_b - e^0_b e^i_a = 0 \implies ^\star B^{0i}_{ab} = \frac{1}{2} \epsilon_{\ \ \ jk}^{0i} B^{jk}_{ab} = \frac{1}{2} \epsilon^{0i}_{\ \ \ jk} (e^j_a e^k_b - e^j_b e^k_a) = \epsilon^{i}_{\ \ jk} e^j_a e^k_b \\ B^{jk}_{ab} &= e^j_a e^k_b - e^k_a e^j_b = 2 e^{[j}_a e^{k]}_b \implies ^\star B^{jk}_{ab} = \frac{1}{2} \epsilon^{jk}_{\ \ \ 0i} B^{0i}_{ab} = 0. \end{aligned}

Globally timelike components:

B0a0i=e00eaiea0e0i=Neai    B0a0i=12ϵ   jk0iB0ajk=12ϵ  jk0i(e0jeake0keaj)=ϵ  jkiNjeakB0ajk=e0jeake0keaj=2N[jeak]    B0ajk=12ϵ   0ijkB0a0i=12ϵ   0ijkNeai=Nϵ   0ijke  ai.\begin{aligned} B^{0i}_{0a} &= e^0_0 e^i_a - e^0_a e^i_0 = N e^i_a \implies ^\star B^{0i}_{0a} = \frac{1}{2} \epsilon_{\ \ \ jk}^{0i} B^{jk}_{0a} = \frac{1}{2} \epsilon^{0i}_{\ \ jk} (e^j_0 e^k_a - e^k_0 e^j_a) = \epsilon^{i}_{\ \ jk} N^j e^k_a \\ B^{jk}_{0a} &= e^j_0 e^k_a - e^k_0 e^j_a = 2 N^{[j} e^{k]}_a \implies ^\star B^{jk}_{0a} = \frac{1}{2} \epsilon^{jk}_{\ \ \ 0i} B^{0i}_{0a} = \frac{1}{2} \epsilon^{jk}_{\ \ \ 0i} N e^i_a = N \epsilon^{jk}_{\ \ \ 0i} e^{i}_{\ \ a}. \end{aligned}

Now we calculate the corresponding Π\Pi & HH:

Πab0i=Bab0i+1γBab0i=ϵ  jkieajebkΠabjk=Babjk+1γBabjk=2γea[jebk]Ha0i=B0a0i+1γB0a0i=ϵ  jkiNjeak+NγeaiHajk=B0ajk+1γB0ajk=Nϵ  ijkeai+2γN[jeak].\begin{aligned} \Pi^{0i}_{ab} &= {}^\star B^{0i}_{ab} + \frac{1}{\gamma} B^{0i}_{ab} = \epsilon^{i}_{\ \ jk} e^{j}_{a} e^{k}_{b} \\ \Pi^{jk}_{ab} &= {}^\star B^{jk}_{ab} + \frac{1}{\gamma} B^{jk}_{ab} = \frac{2}{\gamma} e^{[j}_{a} e^{k]}_{b} \\ H^{0i}_{a} &= {}^\star B^{0i}_{0a} + \frac{1}{\gamma} B^{0i}_{0a} = \epsilon^{i}_{\ \ jk} N^{j} e^{k}_{a} + \frac{N}{\gamma} e^{i}_{a} \\ H^{jk}_{a} &= {}^\star B^{jk}_{0a} + \frac{1}{\gamma} B^{jk}_{0a} = N \epsilon_{\ \ i}^{jk} e^{i}_{a} + \frac{2}{\gamma} N^{[j} e^{k]}_{a}. \end{aligned}

Now we can show that

P(γ,±)abi=14ϵ jkiΠabjk±12γΠab0i=14ϵ jki2γe[ajeb]k±12γϵ jkieajebk={P(γ,)abi=0P(γ,+)abi=1γϵ jkieajebkP(1γ,±)abi=14ϵ jkiΠabjk±γ2Πab0i=12γϵ jkieajebk±γ2ϵ jkieajebk=1±γ22γP(γ,+)abiH(γ,)ai=14ϵ jkiHajk12γHa0i=14ϵ jki(Nϵ   ljkeal+2γN[jeak])12γ(ϵ  jkiNjeak+Nγeai)=12Nδlieal+12γϵ jkiNjeak12γϵ  jkiNjeakN2γ2eai=Nγ212γ2eaiH(1γ,)ai=14ϵ jkiHajkγ2Ha0i=12Nδlieal+12γϵ jkiNjeakγ2(ϵ  jkiNjeak+Nγeai)=1γ22γϵ  jkiNbebjeak=1γ22NbP(γ,+)bai.\begin{aligned} \mathbb{P}(\gamma, \pm)^i_{ab} &= \frac{1}{4} \epsilon^i_{\ jk} \Pi^{jk}_{ab} \pm \frac{1}{2\gamma} \Pi^{0i}_{ab} = \frac{1}{4} \epsilon^i_{\ jk} \frac{2}{\gamma} e^j_{[a} e^k_{b]} \pm \frac{1}{2\gamma} \epsilon^i_{\ jk} e^j_a e^k_b = \begin{cases} \mathbb{P}(\gamma, -)^i_{ab} = 0 \\ \mathbb{P}(\gamma, +)^i_{ab} = \frac{1}{\gamma} \epsilon^i_{\ jk} e^j_a e^k_b \end{cases} \\ \mathbb{P}\left(\frac{1}{\gamma}, \pm\right)^i_{ab} &= \frac{1}{4} \epsilon^i_{\ jk} \Pi^{jk}_{ab} \pm \frac{\gamma}{2} \Pi^{0i}_{ab} = \frac{1}{2\gamma} \epsilon^i_{\ jk} e^j_a e^k_b \pm \frac{\gamma}{2} \epsilon^i_{\ jk} e^j_a e^k_b = \frac{1\pm\gamma^2}{2\gamma}\mathbb{P}(\gamma, +)^i_{ab} \\ \mathbb{H}(\gamma, -)^i_a &= \frac{1}{4} \epsilon^i_{\ jk} H^{jk}_a - \frac{1}{2\gamma} H^{0i}_a = \frac{1}{4} \epsilon^i_{\ jk} \left(N \epsilon^{jk}_{\ \ \ l} e_{a}^l + \frac{2}{\gamma} N^{[j} e^{k]}_a\right) - \frac{1}{2\gamma}\left(\epsilon^{i}_{\ \ jk} N^j e_{a}^k + \frac{N}{\gamma} e^i_a\right) \\ &= \frac{1}{2} N \delta^i_l e^l_a + \frac{1}{2\gamma} \epsilon^i_{\ jk} N^j e^k_a - \frac{1}{2\gamma} \epsilon^{i}_{\ \ jk} N^j e_{a}^k - \frac{N}{2\gamma^2}e^i_a = N \frac{\gamma^2 - 1}{2\gamma^2} e^i_a \\ \mathbb{H}\left(\frac{1}{\gamma}, -\right)^i_a &= \frac{1}{4} \epsilon^i_{\ jk} H^{jk}_a - \frac{\gamma}{2} H^{0i}_a = \frac{1}{2} N \delta^i_l e^l_a + \frac{1}{2\gamma} \epsilon^i_{\ jk} N^j e^k_a - \frac{\gamma}{2}\left(\epsilon^{i}_{\ \ jk} N^j e_{a}^k + \frac{N}{\gamma} e^i_a\right) \\ &= \frac{1-\gamma^2}{2\gamma} \epsilon^{i}_{\ \ jk} N^b e^j_b e^k_a = \frac{1 - \gamma^2}{2} N^b \mathbb{P}(\gamma, +)^i_{ba}. \end{aligned}

Notice that P(γ,)=0\mathbb{P}(\gamma, -) = 0 breaks the internal Lorentz gauge. From the above equations, we can parameterize the variables HajkH^{jk}_a and Ha0iH^{0i}_a purely down to P(γ,+)abi\mathbb{P}(\gamma, +)^i_{ab} and variables N,NiN, N^i, which we can see explicitly by solving for them:

H(1γ,)aiH(γ,)ai=(12γγ2)Ha0i=1γ22γHa0i(i)γ2H(γ,)aiH(1γ,)ai=γ214ϵ jkiHajk(ii)\begin{aligned} &\mathbb{H}\left(\frac{1}{\gamma}, -\right)^i_a - \mathbb{H}(\gamma, -)^i_a = \left(\frac{1}{2\gamma} - \frac{\gamma}{2}\right) H^{0i}_a = \frac{1-\gamma^2}{2\gamma} H^{0i}_a \quad &\text{(i)}\\ &\gamma^2 \mathbb{H}(\gamma, -)^i_a - \mathbb{H}\left(\frac{1}{\gamma}, -\right)^i_a = \frac{\gamma^2 - 1}{4} \epsilon^i_{\ jk} H^{jk}_a \quad &\text{(ii)} \end{aligned}

Plugging in our findings for H(γ,)\mathbb{H}(\gamma, -) and H(1/γ,)\mathbb{H}\left(1/\gamma, -\right), we get:

(i)    1γ22NbP(γ,+)bai+N1γ22γ2eai=1γ22γHa0i    Ha0i=Nγeai+γNbP(γ,+)bai(ii)    N1γ22eai1γ22NbP(γ,+)bai=γ214ϵ jkiHajk    12ϵ jkiHajk=Neai+NbP(γ,+)bai\begin{aligned} \text{(i)} &\implies \frac{1 - \gamma^2}{2} N^b \mathbb{P}(\gamma, +)^i_{ba} + N \frac{1 - \gamma^2}{2\gamma^2} e^i_a = \frac{1-\gamma^2}{2\gamma} H^{0i}_a \\ &\implies H^{0i}_a = \frac{N}{\gamma} e^i_a + \gamma N^b \mathbb{P}(\gamma, +)^i_{ba} \\ \text{(ii)} &\implies - N \frac{1 - \gamma^2}{2} e^i_a - \frac{1 - \gamma^2}{2} N^b \mathbb{P}(\gamma, +)^i_{ba} = \frac{\gamma^2 - 1}{4} \epsilon^i_{\ jk} H^{jk}_a \\ &\implies \frac{1}{2} \epsilon^{i}_{\ jk} H^{jk}_a = N e^i_a + N^b \mathbb{P}(\gamma, +)^i_{ba} \end{aligned}

Thus, we have parameterized the Plebanski constraints in terms of P(γ,+)abi\mathbb{P}(\gamma, +)^i_{ab}, a scalar lapse NN, shift vector NaT(Σt)N^a \in T(\Sigma_t), and the 3 parameters in the choice of an internal direction nI=(1,0,0,0)n^I = (1, 0, 0, 0). These sum up to 9+1+3+3=169 + 1 + 3 + 3 = 16 parameters, which is exactly the 16 degrees of freedom in the co-tetrad eμIe^I_\mu.

The constraints are "simple" constraints and they must be conserved in time. By arXiv:0902.3416 [gr-qc], this follows from

Cij=e(jP˙(γ,)i)Ne(j[dei)+12ϵ   lmi)(+A+A)lem]0.\mathbb{C}^{ij} = e^{(j} \dot{\mathbb{P}}(\gamma, -)^{i)} \propto N e^{(j}\wedge \left[\mathrm{d}e^{i)} + \frac{1}{2}\epsilon^{i)}_{\ \ \ lm}\left({}^+\mathbb{A} + {}^-\mathbb{A}\right)^l\wedge e^{m} \right]\approx 0.

Together with Bi0\mathbb{B}^i \approx 0, these 9 constraints are equivalent to

IVai=12(+A+A)aiΓai(e)=0.\mathbb{IV}^i_a = \frac{1}{2} ({}^{+}\mathbb{A} + {}^{-}\mathbb{A})_a^i - \Gamma_a^i(e) = 0.

Where Γai\Gamma_a^i is the spin connection compatible with the triad ee, i.e., the solution to

[aeb]i+ϵ jkiΓ[ajeb]k=0.\partial_{[a} e_{b]}^i + \epsilon^i_{\ jk} \Gamma_{[a}^j e_{b]}^k = 0.

We can also calculate the Poisson bracket:

{IVai(x),P(γ,)bcj(y)}=12ϵabcδijδ(xy).\{ \mathbb{IV}^i_a(x), \mathbb{P}(\gamma, -)^j_{bc}(y) \} = \frac{1}{2} \epsilon_{abc} \delta^{ij} \delta(x-y).

Which makes them secondary constraints. By setting them strongly to zero, we define Kai=(+AaiAai)/2K_a^i = ({}^{+}\mathbb{A}^i_a - {}^{-}\mathbb{A}^i_a)/2, such that:

+Aai=Γai+γKai=:Aai.{}^{+}\mathbb{A}_a^i = \Gamma_a^i + \gamma K_a^i =: A_a^i.

The action is then reduced to:

S[P(γ,+),+A,N,Ni]=1κdtΣ(P(γ,+)i+A˙i+NiDAP(γ,+)i+ϵ jkiNjP(γ,+)k(γF0i+Fjkϵijk)+Nei(F0iγ+Fjkϵijk)).\begin{aligned} S[\mathbb{P}(\gamma, +), {}^{+}\mathbb{A}, N, N_i] &= \frac{1}{\kappa} \int dt \int_{\Sigma} \Big( \mathbb{P}(\gamma, +)_i \wedge {}^+\dot{\mathbb{A}}^i \\ &+ N_i \mathcal{D}_{A} \wedge \mathbb{P}(\gamma, +)^i + \epsilon^i_{\ jk} N^j \mathbb{P}(\gamma, +)^k \wedge (\gamma F_{0i} + F^{jk}\epsilon_{ijk}) \\ &+ Ne^i \wedge (\frac{F_{0i}}{\gamma} + F^{jk}\epsilon_{ijk} )\Big). \end{aligned}

This is the standard Hamiltonian formulation of general relativity in terms of SU(2) connection variables AaiA_a^i. If we now introduce the densitized triad

Eia=γϵabcP(γ,+)bci,E^a_i = \gamma \epsilon^{abc} \mathbb{P}(\gamma, +)_{bc}^i,

The Poisson bracket between EiaE^a_i and AbjA^j_b is now

{Eia(x),Abj(y)}=κδbaδijδ(xy),{Eia(x),Ejb(y)}={Aai(x),Abj(y)}=0.\{E^a_i(x), A^j_b(y)\} = \kappa \delta^a_b \delta^j_i \delta(x-y), \quad \{E^a_i(x), E^b_j(y)\} = \{A^i_a(x), A^j_b(y)\} = 0.

And the action takes the familiar Ashtekar-Barbero form:

S[E,A,N,Ni,Ni]=1γκdtΣtd3x[EiaA˙aiNbVb(Eia,Aai)NS(Eia,Aai)NiGi(Eia,Aai)]\begin{aligned} S[E, A, N, N^i, N_i] &= \frac{1}{\gamma\kappa} \int dt \int_{\Sigma_t} d^3x \Big[ E^a_i \dot{A}^i_a - N^b V_b(E^a_i, A^i_a) \\ &- N S(E^a_i, A^i_a) - N^i G_i(E^a_i, A^i_a) \Big] \end{aligned}

With the constraints

{Vb=EjaFabj(1+γ2)KbiGiS=EiaEjbdetE(ϵ  kijFabk2(1+γ2)K[aiKb]j)Gi=DaEia\begin{cases} V_b = E^a_j F^j_{ab} - (1+\gamma^2)K^i_b G_i \\ S = \frac{E^a_i E^b_j}{\sqrt{|\det E|}} \left( \epsilon^{ij}_{\ \ k} F^k_{ab} - 2(1+\gamma^2)K^i_{[a} K^j_{b]} \right) \\ G_i = \mathcal{D}_a E^a_i \end{cases}

These three constraints completely dictate both the symmetries and the dynamics of General Relativity. Because it is a fully constrained system, its Hamiltonian is simply a linear combination of these constraints. Physically, they can be interpreted as follows:

1. The Gauss Constraint (Gi0G_i \approx 0)

This is the non-Abelian equivalent of Gauss's Law from Yang-Mills theory (E=0\nabla \cdot \vec{E} = 0). Here, EiaE^a_i acts as the electric field conjugate to the connection AaiA_a^i. It generates local SU(2)SU(2) internal gauge rotations. By setting Gi0G_i \approx 0, we ensure that the physical state of the universe is independent of how we locally orient our internal 3D reference frame (the target space of the triads). Picking a different internal SU(2)SU(2) basis at any point in space doesn't change the physical geometry.

2. The Vector Constraint (Vb0V_b \approx 0)

In the action, this constraint is multiplied by the shift vector NbN^b. It is the generator of spatial diffeomorphisms upon the 3D slice Σ\Sigma. Setting Vb0V_b \approx 0 enforces the principle that coordinates are arbitrary—if you stretch, twist, or remap the spatial coordinate grid on Σ\Sigma, the underlying physical geometry remains exactly the same. The EjaFabjE^a_j F^j_{ab} term essentially comes from the Lie derivative of the connection along the spatial slice, while the (1+γ2)KbiGi- (1+\gamma^2)K^i_b G_i part is a proportional correction term.

3. The Scalar Constraint (S0S \approx 0)

In the action, this is multiplied by the scalar lapse function NN. This constraint generates "time evolution," which in General Relativity corresponds to deforming the spatial slice Σt\Sigma_t forward in the normal direction into the 4D spacetime bulk. Because this must also vanish (S0S \approx 0), it embodies the fact that there is no absolute background time in GR. This is the core dynamical equation of General Relativity.

文章标题:The Connection Formulation of General Relativity

文章作者:Whitney

文章链接:https://phymani.me/posts/diracgr[复制]

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