- #1

quantum123

- 306

- 1

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter quantum123
- Start date

- #1

quantum123

- 306

- 1

- #2

coalquay404

- 217

- 1

It looks like this:

[tex]R_{ab} - \frac{1}{2}g_{ab}R = 8\pi T_{ab}.[/tex]

You're free to use Cartesian coordinates if you wish, but you may run into some difficulty eventually.

On a more serious note, plenty of work has been done to cast Einstein's equations in various PDE forms. Obviously, most of this has been centered on exploiting conformal equivalence classes of three-metrics so that one reaches something along the lines of the Lichnerowicz-York equation. More specifically, if we perform the usual 3+1 decomposition of a spacetime so that [itex]\mathcal{M}\simeq\Sigma\times I[/itex] (where [itex]\Sigma[/itex] is a space-like three-manifold and I is some interval in [itex]\mathbb{R}[/itex]) then the question of solving Einstein's equations essentially reduces to that of trying to find solutions to the following quasilinear elliptic equation for [itex]\phi[/itex]:

[tex]8\Delta\phi - R\phi + K_{ij}^{TT}K^{ij}_{TT}\phi^{-7} - \frac{2}{3}K^2\phi^5 = 0[/tex]

Here [itex]R[/itex] is the Ricci scalar on [itex]\Sigma[/itex] and [itex]K_{ij}^{TT}[/itex] are the transverse-traceless components of the extrinsic curvature. There are

There are also some rather nice attempts at putting the system in first-order hyperbolic form, but I don't know a great deal about those.

Last edited:

- #3

quantum123

- 306

- 1

I know is has just 4 independent variables (x,y,z,t) and 10 (16-6 due to symmetric matrix) dependent variables right?

- #4

coalquay404

- 217

- 1

But how many degree and how many order is this EFE PDE?

Well, let's take a look at it:

[tex]8\Delta\phi - R\phi + K_{ij}^{TT}K^{ij}_{TT}\phi^{-7} - \frac{2}{3}K^2\phi^5 = 0[/tex]

This is a quasilinear elliptic PDE. What we want to solve it for is the conformal factor [itex]\phi[/itex]. Thus, it's a PDE which, if we're very clever, we can solve for a scalar function [itex]\phi[/itex]. So, in your language, it's a second-order PDE (it's a second-order PDE because of the Laplacian [itex]\Delta=\nabla_k\nabla^k[/itex] which is a second-order differential operator that acts on [itex]\phi[/itex]).

I know is has just 4 independent variables (x,y,z,t) and 10 (16-6 due to symmetric matrix) dependent variables right?

It might be helpful to state explicitly that the Einstein field equations, when written in the form

[tex]G_{ab} = R_{ab} - \frac{1}{2}g_{ab}R = 0[/tex]

are a system of PDEs which one wishes to solve for the metric tensor [itex]g_{ab}[/itex]. The Einstein field equations involve the Ricci tensor, which is itself a contraction of the Riemann tensor (the usual convention is that [itex]R_{ab} = R^{c}_{\phantom{c}acb}[/itex]). However, the Riemann tensor involves second-derivatives of the metric, so one concludes that the EFEs are actually a second-order system of nonlinear, coupled PDEs. You can indeed, as you say, use the symmetries of the Riemann tensor to recognise that there are not in fact ten separate PDEs contained within the EFEs. But if you actually want to

Last edited:

- #5

- 10,180

- 1,343

I think what the topic of disucsson is what's known in the literature as "the intial value formulation of General relativity".

One starts by specifying a 3-d spatial metric (this is the metric induced on some particular spatial hypersurface by the full 4-d metric for the Lorentz interval) on some particular 3-d hypersurface which is a "surface of constant time". One also needs to specify something equivalent to the "time derivative" of the metric on this hypersurface. Defining this is a bit tricky, but it's discussed in the textbooks.

So the dynamical variables in this theory are the metric coefficients - more particularly, the spatial metric coefficients.

Given these initial conditions, one can calculate how the metric coefficients change as a function of time (which is a global coordinate).

I do see from my textbook that one winds up with results similar to coalquay404's, but I'm not quite sure where [itex]\Phi[/itex] enters the picture or what it represents.

- #6

Chris Hillman

Science Advisor

- 2,353

- 9

It depends upon what coordinate chart and what spacetime model you are using, and what kind of solution you are looking for (what you plug in on the right hand side for the stress-energy tensor).

To keep things simple, I'll assume you are asking about the vacuum EFE [itex]G^{\hat{a}\hat{b}} = 0[/tex], where the hats suggest that I plan to write out the components, not in a coordinate basis, but with respect to a suitable frame field.

Next, you can write down some general coordinate chart for an arbitrary Lorentzian spacetime, and then, writing down a suitable coframe and computing the Einstein tensor, you will obtain an explicit version of the vacuum EFE such that, in principle, the solution space consists of all local vacuum solutions in gtr. This will look fairly awful, although you can organize the terms so that it takes only two pages or so to write out in all its glorious detail (allowing for algebraic symmetries); see for example Frankel,

A much more common procedure would be to make additional assumptions, such as symmetry assumptions, which make greatly reduce the complexity of the EFE. For example, if you assume that your solution is static spherically symmetric, and write down a simple chart consistent with those assumptions, typically the Schwarzschild chart, the EFE reduces to a simple system of ODEs, which is easily solved, yielding the familiar Schwarzschild vacuum.

Here is a somewhat more typical example: as it happens, I solved the vacuum EFE a few hours ago (not for the first time!), while reading a new eprint (see http://www.arxiv.org/abs/gr-qc/0701161) in which Gowdy and Edmonds introduce a new type of Gowdy vacuum. Following Gowdy's classic papers from 1971-4, they start with the chart (actually I'll make some minor changes in notation)

[tex]ds^2 = -2 \, e^{2 \, a(u,v)} \, du \, dv + b(u,v) \, \left( e^{2 \, h(u,v)} \, dx^2 + e^{-2 \, h(u,v)} \, dy^2 \right) [/tex]

[tex]-1 < u,v < 1, \; -\infty < x,y < \infty [/tex]

For those who like NP tetrads, one can take as an alternative definition of the Gowdy model the tetrad

[tex]\vec{\ell} = \partial_u, \; \; \vec{n} = e^{-2a} \, \partial_v, \; \; \vec{m} = \frac{1}{\sqrt{2 \, b}} \, \left( e^{-h} \, \partial_x + i \, e^{h} \, \partial_y \right)[/tex]

(plus mbar).

Then, computing either the Einstein tensor or (in the Newman-Penrose formalism) the Ricci spinor components, the vacuum EFE reduces to

[tex]b_{uv} = 0[/tex]

[tex]h_{uv} = \frac{-1}{2 b} \, \left( b_u \, h_v + b_v \, h_u \right)[/tex]

[tex]a_u = \frac{b_{uu}}{2 b_u} - \frac{b_u}{4 b} + \frac{b \, h_u^2}{b_u}[/tex]

[tex]a_v = \frac{b_{vv}}{2 b_v} - \frac{b_v}{4 b} + \frac{b \, h_v^2}{b_v}[/tex]

where the subscripts indicate partial derivatives, a very common notion in the literature when PDEs are mentioned. (Yes, for those studying the original eprint, there is a missing + in Equation (4) of the current version as of this morning.)

Here, the first equation is just the usual one-dimensional wave equation in double null coordinates [itex]u,v[/itex]. Once one chooses a solution of the wave equation and plugs into the second equation, one can solve for h. Finally, a is then completely determined by quadrature.

It turns out that the pattern of a pair of master equations (second order coupled nonlinear PDEs), plus two first order equations giving a third function in terms of the first two, is fairly typical for spacetimes with two independent Killing vectors, except that here the master equations also partially uncouple.

(Computing the Weyl spinor components, the Gowdy vacuums turn out to be Petrov I in general.)

In the new eprint, Gowdy and Edmonds suggest starting with the following solution of the wave equation:

[tex]b(u,v) = u^2-v^2[/tex]

(In Gowdy's previous papers, he started with some other simple solutions of the wave equation.) Then a "three parameter solution" (some of this alleged freedom actually involves choice of coordinate chart) of the remaining equations is given by separation of variables:

[tex]h(u,v) = \frac{1}{\sqrt{ (A \, u^2 + B) \, (A \, v^2 + B) }} [/tex]

[tex]a(u,v) = \log \left( \frac{\sqrt{uv}}{ (u+v)^{1/4} \, (u-v)^{1/4} } \right) + \frac{A^2 \, (u+v)^2 \, (u-v)^2}{8 \, (A \, u^2 + B)^2 \, (A \, v^2 + B)^2} + C[/tex]

So this is a simple explicit example of a Gowdy-Edmonds vacuum.

This only takes a few seconds (casesplit followed by pdsolve in GRTensorII), so we've come a long way since Einstein gloomily predicted in 1915 that his new gravitational field equations were so complicated that they would probably never be solved in closed form. This prognostication was almost immediately disproven by Karl Schwarzschild; to be fair, Einstein no doubt meant that he doubted that the

Gowdy and Edmonds point out that their new family of solutions model a linear polarized gravitational wave propagating on a Kasner vacuum background. Some Gowdy vacuums also turn out to admit interpretations as the interaction zone of a colliding plane wave (CPW) model, although I haven't yet had a chance to check whether the necessary condition for this to happen is satisfied here.

One last point: the approach I have taken here illustrates the most elementary thing one might mean by "EFE written out in terms of partial derivatives". The initial value formulation provides a very important and somewhat more sophisticated approach. One reason why casting the EFE into "hyperbolic" form is so important is that this allows us to quote some general results in order to conclude that the initial value formulation of general relativity is (in a technical sense) well-posed.

Last edited:

- #7

coalquay404

- 217

- 1

I think what the topic of disucsson is what's known in the literature as "the intial value formulation of General relativity".

One starts by specifying a 3-d spatial metric (this is the metric induced on some particular spatial hypersurface by the full 4-d metric for the Lorentz interval) on some particular 3-d hypersurface which is a "surface of constant time". One also needs to specify something equivalent to the "time derivative" of the metric on this hypersurface. Defining this is a bit tricky, but it's discussed in the textbooks.

So the dynamical variables in this theory are the metric coefficients - more particularly, the spatial metric coefficients.

Given these initial conditions, one can calculate how the metric coefficients change as a function of time (which is a global coordinate).

I do see from my textbook that one winds up with results similar to coalquay404's, but I'm not quite sure where [itex]\Phi[/itex] enters the picture or what it represents.

This is reasonably correct apart from the claim that "one starts by specifying a 3-d spatial metric." The

Look, in the 3+1 approach to GR, one has four constraint equations. Three of these are encapsulated in what is called the momentum constriaint:

[tex] \mathcal{H}^i = -2\nabla_j\pi^{ij} = 0[/tex]

where [itex]\pi^{ij}[/itex] is the momentum conjugate to the three-metric, and [itex]\nabla[/itex] is a [itex]g_{ij}[/itex]-compatible covariant derivative. The momentum constraint is well understood and really presents no difficulty when trying to, for example, study numerical evolutions of the field equations. The remaining constraint, known as the Hamiltonian constraint, is given by

[tex]\mathcal{H} = \frac{1}{\sqrt{g}}(\pi_{ij}\pi^{ij} - \frac{1}{2}\pi^2) - gR = 0[/tex].

It is the Hamiltonian constraint which is the very essence of general relativity: it encapsulates the statement that one is free to slice a four-dimensional solution to Einstein's equations in (almost) whatever way one likes -- this is general covariance by another name. (The Hamiltonian constraint is also responsible for almost every single difficulty involved in attempts to quantize general relativity, but that's a different story. )

The point is that, in practice, you do not begin by specifying a three-metric as initial data. You

There are various ways of trying to attack the IV problem. You can approach it from the (thick) sandwich point of view where you specify two three-metrics, so that they bound a four-dimensional region of spacetime. When Wheeler proposed this idea he thought it would be only a matter of time before somebody figured out how to make it work. Unfortunately, forty years later, we still know nothing about the (thick) sandwich approach. We

An alternative approach is called the thin sandwich formulation. Here one specifies a three-metric [itex]g_{ij}[/itex] and another tensor [itex]\dot{g}_{ij}[/itex] on a spatial surface and evolves them forward. (It's crucial to note that despite our suggestive notation, [itex]\dot{g}_{ij}[/itex] is

The solution to these problems was recognised years ago by Jim York (I think he may have been at UNC at the time, with later refinement coming after he moved to Cornell), and is called the

- #8

- 10,180

- 1,343

On some further reading, it looks like [tex]\phi^2 = \Omega[/tex], where the conformal transformation is [tex]g'_{ab} = \Omega^2 g_{ab}[/tex], so your equation for [itex]\phi[/itex] (which also appears in Wald) is the equation for the square root of the conformal factor.

Given that the the conformal factor satisfies the above diffeq, I guess it's supposed to be "easy" to solve for the time evolution of [itex]\dot{g}_{ij}[/itex], though I'm not quite sure I understand all the details. If I'm interpreting the equations correctly, the requirement is that [itex]K_{ab} = \dot{g}_{ab}[/itex] be traceles and also satisfy [itex]\nabla^a K_{ab} = 0[/itex]. Wald writes a D rather than an [itex]\nabla[/itex], so I might be misinterpreting something.

- #9

coalquay404

- 217

- 1

OK, I see that by totally neglecting the constraint equations I glossed over some points which turn out to be crucial.

On some further reading, it looks like [tex]\phi^2 = \Omega[/tex], where the conformal transformation is [tex]g'_{ab} = \Omega^2 g_{ab}[/tex], so your equation for [itex]\phi[/itex] (which also appears in Wald) is the equation for the square root of the conformal factor.

That's essentially correct. Unfortunately, Wald defines a conformally transformed metric as

[tex]\overline{g}_{ij} = \Omega^2g_{ij}.[/tex]

However, when you plough through the calculations (and they're

pervect said:Given that the the conformal factor satisfies the above diffeq, I guess it's supposed to be "easy" to solve for the time evolution of [itex]\dot{g}_{ij}[/itex], though I'm not quite sure I understand all the details. If I'm interpreting the equations correctly, the requirement is that [itex]K_{ab} = \dot{g}_{ab}[/itex] be traceles and also satisfy [itex]\nabla^a K_{ab} = 0[/itex]. Wald writes a D rather than an [itex]\nabla[/itex], so I might be misinterpreting something.

Essentially this is correct. Note, however, that the extrinsic curvature is given more correctly in terms of [itex]\dot{g}_{ij}[/itex] as

[tex]K_{ij} = -\frac{1}{2N}(\dot{g}_{ij} - 2\nabla_{(i}N_{j)})[/tex]

where [itex]N[/itex] is the lapse and [itex]N^i[/itex] is the shift. (I think Wald actually uses a different sign to this but the expression here is the one which most frequently occurs in the literature.) The expression you gave is equivalent to imposing the gauge choice [itex](N,N^i)=(1,0)[/itex], which some people call a stable "geodesic slicing".

The other point to note is that since we're dealing with elliptic PDEs, none of this is "easy".

BTW, a nice, easy introduction to this sort of thing is http://relativity.livingreviews.org/Articles/lrr-2000-5/index.html [Broken].

Last edited by a moderator:

- #10

wandering.the.cosmos

- 22

- 0

Wald writes a D rather than an [itex]\nabla[/itex], so I might be misinterpreting something.

Note, however, that the extrinsic curvature is given more correctly in terms of [itex]\dot{g}_{ij}[/itex] as

[tex]K_{ij} = -\frac{1}{2N}(\dot{g}_{ij} - 2\nabla_{(i}N_{j)})[/tex]

where [itex]N[/itex] is the lapse and [itex]N^i[/itex] is the shift.

D is the covariant derivative with respect to the spatial metric [itex]h_{ij}[/itex], whereas [itex]\nabla[/itex] is the covariant derivative with respect to the full spacetime metric. Up to a choice of sign convention, [itex]K_{ij} = (1/2N)(\dot{h}_{ij}-D_i N_j - D_j N_i)[/itex]

Last edited:

- #11

- 10,180

- 1,343

Aha! - that clears up some of the lose ends that were puzzling me, Wald mentioned that the shift & lapse functions weren't considered to be part of the dynamics, now I understand better how they are eliminated by the right gauge choice.

And that clears up the D vs [itex]\nabla[/itex] issue as well.

Thanks, Wandering & coalquay.

And that clears up the D vs [itex]\nabla[/itex] issue as well.

Thanks, Wandering & coalquay.

Last edited:

- #12

wandering.the.cosmos

- 22

- 0

Aha! - that clears up some of the lose ends that were puzzling me, Wald mentioned that the shift & lapse functions weren't considered to be part of the dynamics, now I understand better how they are eliminated by the right gauge choice.

Actually, I have a basic question of my own. Within this ADM formalism, what sort of gauge choices are possible? With and without coupling to matter? For GR, how does one decide whether a particular choice of gauge is legitimate, ADM or not?

Also, I've thought that the lapse function and shift vector are not part of the dynamics simply because there are no time derivatives of them in the lagrangian. I think it's not necessary to gauge them away to see this?

Last edited:

- #13

coalquay404

- 217

- 1

Actually, I have a basic question of my own. Within this ADM formalism, what sort of gauge choices are possible? With and without coupling to matter? For GR, how does one decide whether a particular choice of gauge is legitimate, ADM or not?

There are two separate answers to the allowed sorts of gauge choice.

First, since foliating a spacetime is a purely geometrical exercise, it's pretty easy to convince oneself that essentially any choice of gauge is allowed. In the ADM formalism the most obvious gauge freedom is contained within the pair [itex](N,N^i)[/itex]. Since the lapse is just a function and the shift is a three-vector, this means that you have (at least) four gauge degrees of freedom per space point. The only geometrical restriction on the choice of gauge is that the lapse function must always be positive; apart from this one is free to choose any value of lapse and shift one likes.

Secondly, while you can make any choice of lapse and shift that you want, the real test of whether a gauge choice is "legitimate" is whether or not it makes the equations of motion for the gravitational field easier to understand. For example, recall that the extrinsic curvature is defined by

[tex] K_{ij} = -\frac{1}{2N}(\dot{g}_{ij} - 2\nabla_{(i}N_{j)})[/tex]

Rearranging this immediately gives

[tex]\dot{g}_{ij} = -2NK_{ij} + 2\nabla_{(i}N_{j)}[/tex].

So, on the face of it, you would think that simply choosing [itex]N=1[/itex] and [itex]N^i=(0,0,0)[/itex] would make your life a lot simpler when analysing the dynamics since then the equation of motion for the three-metric reduces to

[tex]\dot{g}_{ij} = -2K_{ij}[/tex].

For a very, very long time, people thought that this choice of gauge was a smart move; after all, getting rid of the lapse and shift from your equations of motion by making this choice of gauge appears to simplify things a great deal, right?

Wrong. It turns out that for many practical applications, the gauge choice [itex](N,N^i)=(1,\vec{0})[/itex] is actually a bad idea since it makes it difficult to interpret the physical evolution. Recently, in other ADM-like formulations (like the BSSN formulation), people have realized that a smart non-zero choice of shift actually makes numerical simulations run for much longer and become much more stable. A good outline of some smart choices of gauge is given in

"Kinematics and Dynamics of General Relativity", James W. York, Jr., in

A guy at the NumRel group at Penn State has a copy of this paper available for download http://gravity.psu.edu/~ksmith/research/research_links.php [Broken]. If you're interested in these sorts of things, it's pretty much required reading.

EDIT: There is one important caveat here: above, I said that essentially any choice of lapse and shift are allowed. This is true if you add to it the phrase "on the initial Cauchy surface". A common choice for the initial hypersurface is that it be constant mean curvature (CMC), which means that the trace of the extrinsic curvature is constant everywhere on it. If the initial data satisfies the constraints you can then evolve the data in any way you want by choosing any values of the lapse and shift at later time (this is in fact the statement that general relativity has four-dimensional general covariance).

In general, however, subsequent hypersurfaces will not be CMC with such arbitrary choices of gauge. If you want these later hypersurfaces to be CMC also, then the lapse function (note:

wandering.the.cosmos said:Also, I've thought that the lapse function and shift vector are not part of the dynamics simply because there are no time derivatives of them in the lagrangian. I think it's not necessary to gauge them away to see this?

This is correct. Apart from the purely geometrical process of foliating a spacetime (which really implies that the lapse and shift must be gauge), there is another way to see this based on the action. If you write down the 3+1 action (I'm assuming that [itex]\partial\Sigma=\emptyset[/itex] here, otherwise we'd need to include boundary terms):

[tex]S = \int dt\int_\Sigma d^3x \sqrt{g}N(R + K_{ij}K^{ij} - K^2)[/tex]

then no "time" derivatives of either the lapse of shift appear. This, coupled with the Euler-Lagrange equations, immediately tells you that they are gauge variables.

However, there's a third way to see it too. The initially amazing thing about the ADM formalism was that they showed that the Hamiltonian for general relativity vanishes on-shell (by on-shell I mean on that subspace of the phase space for which the constraint equations are satisfied). Suppose that you have a phase space [itex]\Gamma[/itex] for a field theory with phase space coordinates [itex](\phi,\pi)[/itex]. Suppose further that you have some constraints in this theory (for argument's sake, let there be [itex]m[/itex] constraints). The constraints can generally be written as some functions

[tex]\Psi^A(\phi,\pi) = 0,\qquad A=1,\ldots,m.[/tex]

A moment's thought about this will convince you that the constraint equations therefore actually define a

Now, there's a very nice little theorem (see, for example, Dirac's

[tex]F[\phi,\pi;x)|_{\Gamma_1} = 0

\Rightarrow F[\phi,\pi;x) = \int_\Sigma \mu_A\Psi^A[\phi,\pi;x)[/tex].

Now apply this to general relativity: GR is a constrained system so any function that vanishes on-shell in GR (i.e., one which satisfies the Hamiltonian and momentum constraints) can be written as a linear combination of these constraints with Lagrange multipliers. However, crucially, GR is an example of something called a

[tex]\mathcal{H} = \mu\mathcal{H} + \mu_i\mathcal{H}^i[/tex]

where [itex]\mathcal{H}[/itex] and [itex]\mathcal{H}^i[/itex] are the Hamiltonain and momentum constraints, respectively, and the [itex]\mu[/itex]s are Lagrange multipliers. If you go through things to find the equations of motion, you eventually reach the surprising result that these Lagrange multipliers must be the lapse and shift in order for the equations of motion to agree with those obtained from the Lagrangian picture. Thus, working purely from a symplectic geometry perspective, we've shown that the lapse and shift are pure gauge variables.

Last edited by a moderator:

- #14

SoPa

- 1

- 0

[www.hindawi.com/GetPDF.aspx?doi=10.1155/2007/42505 ] . I wanted to know if there are available solutions for that problem or how I can find them using a c++ library.I think I don't need time in this approach,just only space.Finally, this type of EFE has a specific name?Any ideas please?

Share:

- Replies
- 10

- Views
- 405

- Replies
- 186

- Views
- 4K

- Replies
- 1

- Views
- 542

- Replies
- 10

- Views
- 919

- Replies
- 7

- Views
- 627

- Replies
- 57

- Views
- 1K

- Last Post

- Replies
- 5

- Views
- 558

- Last Post

- Replies
- 11

- Views
- 1K

- Replies
- 13

- Views
- 578

- Last Post

- Replies
- 3

- Views
- 450