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Why is allowed decompose the spacetime metric into a spatial part + temporal part like this for example

$$ds^2 ~=~ (-N^2 + N_aN^a)dt^2 + 2N_adtdx^a + q_{ab}dx^adx^b$$

($N$ is called lapse, $N_a$ is the shift vector and $q_{ab}$ is the spatial part of the metric.)

in order to arrive at a Hamiltonian formulation of GR? How is a breaking of Lorentz invariance avoided by doing this ?

Sorry it this is a dumb question; maybe I should just read on to get it but I`m curious about this now ... :-)

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NB: really we should be discussing "diffeomorphism invariance" when working with general relativity. In the special case of Minkowski flat spacetime, the diffeomorphism invariance amounts to Lorentz invariance. (I'm a mathematician: I'm pathological about these terminological issues!) –  Alex Nelson Jun 24 '12 at 16:16

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Well, perhaps one should consider reading The Hamiltonian formulation of General Relativity: myths and reality for further mathematical details. But I would like to remind to you with most constrained Hamiltonian systems, the Poisson bracket of the constraint generates gauge transformations.

For General Relativity, foliating spacetime $\mathcal{M}$ as $\mathbb{R}\times\Sigma$ ends up producing diffeomorphism constraints $\mathcal{H}^{i}\approx0$ and a Hamiltonian constraint $\mathcal{H}\approx 0$. Note I denote weak equalities as $\approx$.

This is first considered in Peter G. Bergmann and Arthur Komar's "The coordinate group symmetries of general relativity" Inter. J. The. Phys. 5 no 1 (1972) pp 15-28.

Since you asked, I'll give you a few exercises to consider!

Exercise 1: Lie Derivative of the Metric

The Lie derivative of the metric along a vector $\xi^{a}$ is $$ \mathcal{L}_{\xi}g_{ab}=g_{ac}\partial_{b}\xi^{c}+g_{bc}\partial_{a}\xi^{c}+\xi^{c}\partial_{c}g_{ab} $$ Show that this may be rewritten as $$ \mathcal{L}_{\xi}g_{ab}=\nabla_{a}\xi_{b}+\nabla_{b}\xi_{a} $$ where $\nabla$ is the standard covariant derivative.

Exercise 2: Constraints generate diffeomorphisms

Recall that the Hamiltonian and momentum constraints are $$\mathcal{H} = \frac{16\pi G}{\sqrt{q}}\left(\pi_{ij}\pi^{ij}-\frac{1}{2}\pi^{2}\right)-\frac{\sqrt{q}}{16\pi G}{}^{(3)}\!R,\quad\mathcal{H}^{i} = -2D_{j}\pi^{ij}$$ and $\pi^{ij}=\displaystyle\frac{1}{16\pi G}\sqrt{q}(K^{ij}-q^{ij}K)$ with $K_{ij}=\displaystyle\frac{1}{2N}(\partial_{t}q_{ij}-D_{i}N_{j}-D_{j}N_{i})$. Let $$H[\widehat{\xi}] = \int d^{3}x\left[\hat{\xi}^{\bot}\mathcal{H}+\widehat{\xi}^{i}\mathcal{H}_{i} \right]$$ Show that $\mathcal{H}[\widehat{\xi}]$ generates (spacetime) diffeomorphisms of $q_{ij}$, that is, $$\left\{H[\widehat{\xi}],q_{ij}\right\}=(\mathcal{L}_{\xi}g)_{ij}$$ where $\mathcal{L}_{\xi}$ is the full spacetime Lie derivative and the spacetime vector field $\xi^{\mu}$ is given by $$\widehat{\xi}^{\bot}=N\xi^{0}, \quad \widehat{\xi}^{i}=\xi^{i}+N^{i}\xi^{0}$$ The parameters $\{\widehat{\xi}^{\bot},\widehat{\xi}^{i}\}$ are known as "surface deformation" parameters.

(Hint: use problem 1 and express the Lie derivative of the spacetime metric in terms of the ADM decomposition.)

Addendum: I'd like to give a few more references on the relation between the diffeomorphism group and the Bergmann-Komar group.

From the Hamiltonian formalism, there are a few references:

  1. C.J. Isham, K.V. Kuchar "Representations of spacetime diffeomorphisms. I. Canonical parametrized field theories". Annals of Physics 164 2 (1985) pp 288–315
  2. C.J. Isham, K.V. Kuchar "Representations of spacetime diffeomorphisms. II. Canonical geometrodynamics" Ann. Phys. 164 2 (1985) pp 316–333

The Lagrangian analysis of the symmetries are presented in:

  1. Josep M Pons, "Generally covariant theories: the Noether obstruction for realizing certain space-time diffeomorphisms in phase space." Classical and Quantum Gravity 20 (2003) 3279-3294; arXiv:gr-qc/0306035
  2. J.M. Pons, D.C. Salisbury, L.C. Shepley, "Gauge transformations in the Lagrangian and Hamiltonian formalisms of generally covariant theories". Phys. Rev. D 55 (1997) pp 658–668; arXiv:gr-qc/9612037
  3. J. Antonio García, J. M. Pons "Lagrangian Noether symmetries as canonical transformations." Int.J.Mod.Phys. A 16 (2001) pp. 3897-3914; arXiv:hep-th/0012094

For more on the hypersurface deformation algebra, it was first really investigated in Hojman, Kuchar, and Teitelboim's "Geometrodynamics Regained" (Annals of Physics 96 1 (1976) pp.88-135).

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Thanks @AlexNelson for this rich answer, since I`ve not yet penetrated deep enough into the Hamiltonian formulation of GR etc it will keep me busy for a while. And it is a nice extension of things only briefly explained in the book I am reading :-) –  Dilaton Jun 24 '12 at 8:42
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@Nemo: you might want to also read Bojowald's book Canonical Gravity and Applications: Cosmology, Black Holes, and Quantum Gravity which is, perhaps, one of the better books on the ADM formalism. Another good book (with a numerical focus) is Baumgarte and Shapiro's Numerical Relativity. –  Alex Nelson Jun 24 '12 at 15:07
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@AlexNelson Quick question: did you omit a $^3R$ term out of the formula for the momentum constraint $\mathcal{H}$? –  twistor59 Apr 13 '13 at 12:05
    
@twistor59: hmm...lemme double check that later today. But for now, I've edited it in. –  Alex Nelson Apr 15 '13 at 16:43
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@twistor59: Yes, you are indeed correct! I set ${}^{(3)}R=0$ to work with flat space without explicitly stating it. (I intended to work with Minkowski spacetime, hoping to relate the Poincare symmetries to the hypersurface deformation algebra --- I assumed that was the intended question of the OP). –  Alex Nelson Jul 8 '13 at 16:38

As I understand it, you're right that splitting the metric into spatial and temporal parts does break the Lorentz symmetry of the metric. When you define a lapse function and shift vector, you start by foliating spacetime into a bunch of spacelike "slices," and these slices can be used to identify a particular reference frame.

However, the key is that you can do this in any manner you want. There's no one "special" choice of the lapse function and shift vector. The trick to showing that the Hamiltonian formulation is Lorentz invariant is making sure that the conclusions you get are equally valid no matter which foliation of spacetime you choose.

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Thanks @DavidZaslavsky, I could have bet that there must be a quite "simple" explanation. Your answer makes a lot of sense to me and in addition it gives me some other keywords to look up :-) –  Dilaton Mar 11 '12 at 9:08
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Thanks, but do keep in mind that I'm not a GR expert, so other people might have something to add here - perhaps you could get a better answer from someone who is more familiar with this stuff. –  David Z Mar 11 '12 at 11:31
    
Jep, additional (and more detailed ?) answers are still welcomed and appreciated of course. –  Dilaton Mar 11 '12 at 11:48

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