Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New features!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

122 submissions , 103 unreviewed
3,497 questions , 1,172 unanswered
4,548 answers , 19,352 comments
1,470 users with positive rep
409 active unimported users
More ...

Lorentz invariance of the 3 + 1 decomposition of spacetime

+ 6 like - 0 dislike
97 views

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 ... :-)

asked Mar 10, 2012 in Theoretical Physics by Dilaton (4,175 points) [ revision history ]
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!)

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Alex Nelson

2 Answers

+ 6 like - 0 dislike

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).

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Alex Nelson
answered Jun 18, 2012 by Alex Nelson (100 points) [ no revision ]
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 :-)

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
@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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Alex Nelson
@AlexNelson Quick question: did you omit a $^3R$ term out of the formula for the momentum constraint $\mathcal{H}$?

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user twistor59
@twistor59: hmm...lemme double check that later today. But for now, I've edited it in.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Alex Nelson
@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).

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Alex Nelson
+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user David Z
answered Mar 11, 2012 by David Z (570 points) [ no revision ]
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 :-)

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user David Z
Jep, additional (and more detailed ?) answers are still welcomed and appreciated of course.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverflo$\varnothing$
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...