Einstein Group

1. Jun 29, 2007

Klaus_Hoffmann

if we consider the Group structure of $$[ \pi _{ij} , g_{ab}]= \delta _{ia,jb}$$

of the Einstein equation for the momenta and metric, could we obtain its Casimir Operator, conserved quantities and so on?

2. Jun 30, 2007

shoehorn

It may be a misunderstanding on my part, by the expression you've given above makes no sense. Care to give us some context?

3. Jul 3, 2007

Klaus_Hoffmann

Yes of course , if you have the Einstein Lagrangian $$\sqrt{-g}R$$

then the Einstein Group (as my definition) would be the group that leaves the Lagrangian invariant so if you change the metric in the from $$Fg_{ab} \rightarrow h_{ab}$$ where h and g are the transformed metric under the group G so the Lagrangian remains the same but a surface term, so using NOether's theorem the infinitesimal generators of the Group will be the (physical) conserved quantities.

4. Jul 3, 2007

olgranpappy

I shoehorn may be refering to the $$\left[\;\;,\;\;\right]$$ symbol. What do you mean by that?

5. Jul 3, 2007

pervect

Staff Emeritus
The [ , ] symbol is most likely a commutator (see for instance Wald, GR, pg 18). Given two smooth vector fields v and w, a commutator defines a new vector field:

[v,w](f) = v(w(f)) - w(v(f))

Of course in this case v and w are rank 2 tensor fields, not vector fields. I believe it's sensible to talk about the commutator of a rank 2 tensor field, but I'd have to look up the details.

I would furthermore assume that the $\pi$ are cannonical momenta in the Hamiltonian formulation of GR.

Unfortunately I don't know the answer to the original poster's questions offhand - I was hoping that someone else would tackle it.

6. Jul 3, 2007

shoehorn

I'm mystified. The OP seems to be defining an algebra using a three-momentum ($\pi^{ij}$) and a four-metric ($g_{ab}$). This is obviously fruitless since, without further restrictive qualification, such an algebra can't even be defined.

What's even more confusing is that he's mixing in the Einstein-Hilbert Lagrangian with manifestly three-dimensional quantities such as the three-momentum.

7. Jul 3, 2007

pervect

Staff Emeritus

Maybe he meant the metric induced on 3-d space, but I usually see that written as $\Sigma_{ij}$.

8. Jul 3, 2007

shoehorn

Care to give me a link to where the three-metric is denoted by a capital sigma? In the literate it is almost always either g or lower case gamma. Regardless, even if he was talking about the three-metric, his expression still doesn't make any sense since any conceivable algebra would have to respect the symmetry of the three-metric and its momentum. If you look at the expression the OP gave, the rhs isn't symmetric in (i,j) or (a,b).

What's more, he should have some form of distribution on the rhs also.

9. Jul 3, 2007

pervect

Staff Emeritus
I thought Wald did, but I see now that I was confused, $\Sigma_t$ is a spacelike slice at time t, not a metric.

<snip some silly stuff I wrote earlier>

So I have to agree it's not really clear what the original question is about, but it seems to me that we can be polite about asking the OP what he had in mind.

BTW, I'm not really confident that [] represents commutators, maybe they're Poisson brackets.

Last edited: Jul 4, 2007
10. Jul 4, 2007

masudr

Surely

$$\pi_{ab} = \frac{\partial \mathcal{L}}{\partial g_{ab}}$$

which is not three-momenta, but instead the canonical momenta, given that $g_{ab}$ is the field variable.

11. Jul 4, 2007

shoehorn

When I say "three-momenta" I'm taking it as given that I'm referring to "field momentum canonically conjugate to the three-metric on a spatial slice".

12. Jul 4, 2007

masudr

Ah, my mistake. My ignorance has been publically displayed.

13. Jul 4, 2007

shoehorn

Let me clarify it even further for you. When I say "three-momentum" I use it in exactly the same sense that almost every paper I've ever seen on canonical general relativity uses.

Also, your definition of the canonical momentum

$$\pi_{ab}= \frac{\partial\mathcal{L}}{\partial g_{ab}}$$

is incorrect for several reasons.

14. Jul 4, 2007

Klaus_Hoffmann

in GR or its quantum version physicist asume that metric can be written by means of a foliation.

$$g_{ab}=dt^{2}+g_{ij}dx^{i}dx^{j}$$ hence i and j runs only over space variables x,y,z

15. Jul 4, 2007

pervect

Staff Emeritus
I've seen $$\frac{\pi^{ij}}{16 \pi} = \pi_{\mathrm{true}}^{ij} = \frac{\delta L}{\delta g_{ij}}$$

for instance, MTW pg 521

so you're very close, but apparently you won't be getting any cigar from shoehorn :-). This assumes $g_{ij}$ is a three metric, in the sense that's already been discussed. The difference betweeen $\partial$ and $\delta$ is the difference between the partial derivative of a function and a functional derivative (the action is a functional, i.e. it's a map from a function to a scalar).

I hope this is both more helpful than shoehorn's terse reply and also close to being correct (the textbook quote is definitely quoted correctly) unfortunately I don't really work much with the ADM formalism of which this is a part.

If you have Goldstein, Classical Mechanics, you might also look at the section on pg 420 about the symmetry groups of mechanical systems - rereading this makes it seem pretty likely to me that the [ ] are indeed Poisson brackets, though I can't quite justify the specific relation in the original question in my mind.

Last edited: Jul 4, 2007
16. Jul 4, 2007

shoehorn

Please don't take my abrupt comments as being rude; I certainly don't intend to be so and if I've come across that way, my apologies.

You claim that MTW say that the momentum is defined by:

$$\pi^{ij} = \frac{\delta L}{\delta g_{ij}}$$

You're correct when you claim that MTW say this is how the momentum is defined. However, the important point is that MTW are wrong! This is not as heretical as it may sound. Consider, for example, an ordinary classical system with generalised coordinates $q^i[/tex], generalised velocities [itex]\dot{q}^i$, and a Lagrangian $L$. In this trivial case we don't define the momentum conjugate to $q^i$ as

$$\pi_i = \frac{\partial L}{\partial q^i}$$

This is the analogue of what you've written for the momentum in general relativity. The correct way to define the momentum is of course

$$\pi_i = \frac{\partial L}{\partial\dot{q}^i}$$

This is a trivial observation. The correct way to define the three-momentum in general relativity is then

$$\pi^{ij} = \frac{\delta \mathcal{L}}{\delta\dot{g}_{ij}}$$

where the Lagrangian density (and note that the correct thing to use is the Lagrangian density, not the Lagrangian itself due to the possible presence of surface terms) is understood to have been expressed in terms of the first and second fundamental forms of a hypersurface in the spacetime.

If you work through the calculations you'll find that my correct way of defining the three-momentum gives

$$\pi^{ij} = \sqrt{g}(g^{ij}K - K^{ij})$$

On the other hand, if you use MTW's incorrect definition, you'll get something a whole lot more complicated (you'll get the euler-lagrange equations actually, as you could have noticed had you thought a little bit about the equation in MTW before transcribing it).

Last edited: Jul 4, 2007
17. Jul 4, 2007

masudr

General cigar shortage, it appears (:

18. Jul 4, 2007

pervect

Staff Emeritus
LOL, yep, but educational.