I What's the underlying frame of the Einstein's Field Equation?

Pyter
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In what coordinate system is set up and solved the Einstein's Field Equation and how does it relate to our continuum?
Hello all,

I have a question on a pivotal concept of GR that I've never managed to fully grasp.

In what coordinate system is the Einstein's Field Equation set up and solved?

I've always assumed it's an Euclidean 4D space, whose metric is irrelevant because we are dealing with scalar functions, solutions of the ten independent differential equations.
But if it's so, how does it map to our 4D curved continuum? I've always seen our curved spacetime only mapped to a "local" flat space in the infinitesimal region around a given point, but never "globally".

What if a want to solve the EFE in an extended spacetime region, i.e. our solar system for several days?

From a procedural standpoint, should I have to take a "snapshot" of the masses in my curved spacetime to express the stress-energy tensor as a function of my curvilinear coordinates, map it to a 4D flat space, solve the EFE to obtain the metric tensor g as a function of Euclidean coordinates, then map it back to my curved spacetime?

What boundary conditions should I use i.e. for velocities, expressed in which coordinate system?

And what would be the use for such a g? To compute the motion equations of the masses in my curved spacetime?
 
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GR is generally covariant, i.e., the Einstein(-Hilbert) field equations look the same in all coordinates (reference frames). If it comes to the solution of these non-linear (and thus complicated) equations of motion, of course a clever choice of coordinates is crucial. All the exact solutions (like the Schwarzschild, Kerr, and FLRW solutions) of course result from having many symmetries and the choice of coordinates is according to these symmetries.
 
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Pyter said:
Summary:: In what coordinate system is set up and solved the Einstein's Field Equation and how does it relate to our continuum?

In what coordinate system is the Einstein's Field Equation set up and solved?
You can use any coordinate system. It is valid in all of them.

Pyter said:
Summary:: In what coordinate system is set up and solved the Einstein's Field Equation and how does it relate to our continuum?

What if a want to solve the EFE in an extended spacetime region, i.e. our solar system for several days?
For something like that you are probably going to want to use the ADM formalism:

https://en.m.wikipedia.org/wiki/ADM_formalism

This is the usual approach for solving the EFE in terms of boundary conditions and so forth.
 
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In addition you need such a (1+3)-formalism to formulate the corresponding initial-value problems and solve them numerically. For the application in relativistic transport/hydro simulations to predict gravitational-wave signals, see

L. Rezzolla and O. Zanotti, Relativistic hydrodynamics,
Oxford University Press, Oxford (2013),
https://dx.doi.org/10.1093/acprof:oso/9780198528906.001.0001
 
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Pyter said:
I've always assumed it's an Euclidean 4D space
Spacetime is Lorentzian, not Euclidean.

Pyter said:
whose metric is irrelevant because we are dealing with scalar functions, solutions of the ten independent differential equations.
Those ten solutions are the metric (each of the ten functions is one of the ten independent components of the metric, and the ten independent differential equations are differential equations involving various combinations of those functions), so the metric is not irrelevant.
 
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So any frame is good. But since I observe the stress-energy tensor in my curvilinear frame, am I forced to use that frame? Or should I come up with any arbitrary map that simplifies the EFE, solve it in that frame, then map it back to my lab frame?

And once I have the metric tensor, I solve the equations of motion of my masses by imposing that they follow geodesics? The gravity is not thus considered a force, but it only warps the spacetime through the metric?
 
PeterDonis said:
Spacetime is Lorentzian, not Euclidean.Those ten solutions are the metric (each of the ten functions is one of the ten independent components of the metric, and the ten independent differential equations are differential equations involving various combinations of those functions), so the metric is not irrelevant.
I meant that I don't need the metric of the 4D flat (Euclidean, Minkowski or whatnot) space I assumed the EFE had to be solved into, since I'm not using vectors. I know that the solution is the metric of my warped continuum.
 
Pyter said:
I observe the stress-energy tensor in my curvilinear frame
You make actual measurements of tensor components using particular measuring devices, which can be used to define a particular frame, but you can transform those measured components into any frame you like for the purpose of making calculations, including the calculations required to solve the EFE.

Pyter said:
once I have the metric tensor, I solve the equations of motion of my masses by imposing that they follow geodesics?
Assuming that there are no forces present, yes. But that won't always be the case.

Pyter said:
gravity is not thus considered a force, but it only warps the spacetime through the metric?
In GR, yes, that is correct.
 
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Pyter said:
I meant that I don't need the metric of the 4D flat (Euclidean, Minkowski or whatnot) space I assumed the EFE had to be solved into
You assumed wrong. You solve the EFE for your 4D spacetime; the solution gives you the metric of that spacetime. That metric won't be flat unless your stress-energy tensor vanishes everywhere, and even then flat Minkowski spacetime is not the only possible vacuum solution.

Pyter said:
I'm not using vectors.
How are you not using vectors? Tensors are vectors--more precisely, what you are calling "vectors" are just 1st-rank contravariant tensors, but tensors of any rank form a vector space.
 
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  • #10
PeterDonis said:
You assumed wrong
I'm aware of that now, thank you all.
PeterDonis said:
How are you not using vectors?
In the sense that to solve the differential scalar equations I don't need them. I'll use the ds² later when I set up the geodesics equations with the solved g.
 
  • #11
Pyter said:
Or should I come up with any arbitrary map that simplifies the EFE, solve it in that frame, then map it back to my lab frame?
That is usually the best approach. If there is some symmetry, then using coordinates with that same symmetry is often helpful
 
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  • #12
PeterDonis said:
That metric won't be flat unless your stress-energy tensor vanishes everywhere, and even then flat Minkowski spacetime is not the only possible vacuum solution.
That's interesting, maybe topic for another thread. I haven't delved into it so far, but I figured that the homogeneous EFE only admitted the identically flat space solution or the planar wave solution, for analogy with the homogeneous Maxwell equations.
 
  • #13
Pyter said:
I figured that the homogeneous EFE only admitted the identically flat space solution
As you have seen, that is not correct. Other important vacuum solutions besides flat Minkowski spacetime include Schwarzschild spacetime and Kerr spacetime.

Pyter said:
for analogy with the homogeneous Maxwell equations.
The key difference between them is that Maxwell's equations are linear, while the EFE is nonlinear.
 
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  • #14
PeterDonis said:
Other important vacuum solutions besides flat Minkowski spacetime include Schwarzschild spacetime and Kerr spacetime.
And of course most of the typical gravitational wave spacetimes.
 
  • #15
Dale said:
And of course most of the typical gravitational wave spacetimes.
Yes. The OP in post #12 appeared to be including those by mentioning "planar wave solutions".
 
  • #16
PeterDonis said:
Other important vacuum solutions [...] include Schwarzschild spacetime and Kerr spacetime
I thought that those implied the presence of a mass. Weren't they found in relation with stars/black holes, even though they hold true in the empty space outside the fluid?
 
  • #17
Pyter said:
I thought that those implied the presence of a mass.
No, they don't. They are vacuum solutions. Most physicists believe that these solutions will only describe a portion of spacetime, not all of spacetime, but that's a judgment about physical reasonableness, not a mathematical requirement of the solution. See further comments below.

A general property of solutions of the EFE that you might not be taking into account here is that any solution of the EFE describes an open region of spacetime. If all of spacetime is an open region, then a solution can describe all of spacetime. But any solution can also describe an open region of a spacetime that also contains other regions described by different solutions, where the solutions are matched at the boundaries of the regions according to certain constraints (called "junction conditions" in the literature).

So no solution by itself can ever "imply" the presence of particular other solutions. There are always multiple ways to construct global spacetimes from one or more EFE solutions, either by having a spacetime that is an open region which can be covered by a single solution, or by matching together multiple open regions in which there are different solutions.

Pyter said:
Weren't they found in relation with stars/black holes, even though they hold true in the empty space outside the fluid?
No. Those solutions were originally discovered by simply doing math to solve the EFE under certain assumptions.

Schwarzschild, who discovered the vacuum solution that now bears his name, also discovered a solution for the interior of a spherically symmetric massive body, and it was not difficult to show that those two solutions can be fit together at the boundary of the massive body, so the vacuum solution describes spacetime outside the body. However, he did that only after discovering the two solutions separately.

We actually do not have any known construction of the interior of a rotating body that matches at the boundary to Kerr spacetime. The only thing we know for sure that Kerr spacetime describes is a rotating black hole.
 
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  • #18
Though it's not usually explicitly stated in textbooks, it's good to remember, what to do to define observable quantities within a gauge theory. A gauge theory is a theory, where you use more "computational degrees of freedom" than in principle necessary, and that's why the description has some redundancy.

E.g., in classical electrodynamics you use the four-vector potential to describe the observable electromagnetic field (with electric and magnetic components). The equations of motion for these observable fields are Maxwell's equations, and you introduce the four-potential to fulfill the two homogeneous ones identically. So you have to bother only about the inhomogeneous Maxwell equations in terms of the four-potential, but these equations do not uniquely determine the four-potential, because the four-potential is defined only modulo a gauge transformation. So to solve the inhomogeneous equations you can fix the gauge (in this case by imposing one gauge-fixing constraint like choosing the Coulomb or Lorenz gauge). That shows that the four-potentials do not have a clear physical meaning as observables but you have to calculate the electromagnetic field, which is observable, and these do not depend on the choice of gauge to solve for the four-potential. That's clear, because the original Maxwell equations for the observable fields have a unique solution (given initial and appropriate boundary conditions to solve them).

Now also GR is a gauge theory. The gauge freedom is the general covariance, i.e., the invariance of all observable quantities under arbitrary diffeomorphisms between coordinates.

Now in physical term's what's observable are usually local observables, e.g., the cosmic background radiation being detected by the Planck satellite. You can define observables most easily by measuring local scalar, vector, or tensor quantities (like the temperature of the background radiation as a function of direction as the Planck satellite most successfully did). To that end you introduce a local inertial reference frame in terms of choosing an appropriate tetrade (in this case fixed at the free-falling satellite). Then you can define your observables in a way that is independent of the choice of coordinates.

BTW: Historically this was the greatest obstacle for Einstein to find the final and very successful form of his GR. He had already the right idea concerning general covariance and also the field equations but then thought that this is not the right thing yet, because the solutions are not unique. It took him some more years to get back to these right equations and understand the physical meaning of the theory. This has been figured out in great detail in

J. Renn (Ed.), The Genesis of General Relativity, 4 vols., Springer (2007)
 
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  • #19
vanhees71 said:
To that end you introduce a local inertial reference frame in terms of choosing an appropriate tetrade (in this case fixed at the free-falling satellite). Then you can define your observables in a way that is independent of the choice of coordinates
How so? The position and time of the observable mass/energies still depend on the IFR's choice.
 
  • #20
PeterDonis said:
Those solutions were originally discovered by simply doing math to solve the EFE under certain assumptions.

I'd hesitate to describe the derivation of the Kerr metric as simple. :oldeyes:
 
  • #21
Pyter said:
How so? The position and time of the observable mass/energies still depend on the IFR's choice.
Of course the components of four-vectors (the energy is a temporal component of the four-momentum of the particle) depend on the basis. A vector does not depend on the basis (momentum is a four-vector and thus basis independent). Mass is a scalar and thus basis-independent anyway.
 
  • #22
ergospherical said:
I'd hesitate to describe the derivation of the Kerr metric as simple. :oldeyes:
Well, it turned out to be pretty simple once Roy Kerr did it. :wink:
 
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  • #23
Just when I thought I had got it, this topic eluded me again.
I stated earlier that initially you take a snapshot of the stress-energy tensor in curvilinear coordinates to solve the EFE.
But it occurred to me that you can't do that because you don't know the curvature yet, not until you solve the EFE for ##G^{ \mu \nu }##.
As written by @vanhees71, you initially assign the spacetime coordinates of the masses/flows in your extended region assuming you're in an in IFR (flat space).
But then your representation of the SET is inaccurate, because you can't account for aberrations in the positions (and times) due to the space curvature, like gravitational lens effects.
So your first solution of the EFE can't be the correct one.
And once you get a first solution for ##G^{\mu \nu}##, do you repeat the process with your new SET coordinates until the solution converges?
Is this what the ADM formalism somehow does?

(As an aside, once you have a solution for ##G^{\mu \nu}##, can you univocally determine the mapping of your curved spacetime to Euclidean spacetime, in order to be able to use curvilinear coordinates?)
 
  • #24
It depends on what you want to do. The point is that it's not so simple to find solutions of these highly nonlinear equations of motion. If you have only continuum mechanical sources (usual simple models are "dust matter" or "perfect fluids"; already viscous fluids are not so simple anymore) if you have found a solution of the EFEs you also have a solution for the fluid motion, because the Bianchi identity ensures that ##\nabla_{\mu} T^{\mu \nu}=0##, and these are the equations of motion for the matter. If you have in addition electromagnetic fields you have to solve for the Maxwell equations simultaneously.

If you look in textbooks with exact, analytic solutions like the Schwarzschild, Nordstrom-Reissner, Kerr, Friedmann-Lemaitre-Robertson-Walker metrics, the idea is just to make an ansatz for the metric components under assumptions of symmetries, choosing corresponding coordinates and solve for the EFEs with a correspondingly parametrized energy-momentum tensor (assuming also an equation of state or the fluid).

For numerical solutions you need to set up an initial-boundary-value problem with some foliation of spacetime. For that see

L. Rezzolla and O. Zanotti, Relativistic hydrodynamics,
Oxford University Press, Oxford (2013),
https://dx.doi.org/10.1093/acprof:oso/9780198528906.001.0001
 
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  • #25
Pyter said:
But it occurred to me that you can't do that because you don't know the curvature yet,
I believe that the usual approach for arbitrary spacetimes (e.g. when trying to predict gravitational wave signatures for LIGO) is to write down an approximation to the metric (e.g., model a pair of black holes as some sort of linear superposition of Kerr metrics) then use numerical methods to refine that to an actual solution on some spacelike slice, then evolve it. Large computers needed.
 
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  • #26
vanhees71 said:
It depends on what you want to do
Let's suppose I want to solve the EFE for the whole solar system, from the Earth's (or a geostationary satellite's) vantage point. There is no fluid, only (approximately) point-like masses whose 4D coordinates I observe through light beams or radio waves.
 
  • #27
That would be "dust" ;-)).
 
  • #28
@vanhees71 ok.
Another question came to my mind from what you just wrote about the motion of the SET constituents.
Given that to get a "snapshot" of the SET in your desired spacetime region you have basically to track the planets' positions during the time interval delimiting your region, say from ##t_{start}## to ##t_{finish}##, the net effect is that you already have determined, through observation, the equations of motions of the "dust".
So what would be the purpose of solving the EFE? Just to verify that, with the found ##G^{\mu \nu}##, the EOM are geodesics (assuming only gravitational forces act on the masses)?
 
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  • #29
For dust you automatically get that the trajectories of the corresponding particles are geodesics. That's because "dust" by definition is just "free streaming", i.e., non-interacting particles:

See the solution of the 2nd problem on the following problem sheet:

https://itp.uni-frankfurt.de/~hees/art-ws16/lsg05.pdf
 
  • #30
Pyter said:
Let's suppose I want to solve the EFE for the whole solar system, from the Earth's (or a geostationary satellite's) vantage point. There is no fluid, only (approximately) point-like masses whose 4D coordinates I observe through light beams or radio waves.
The first approximation of the solar system is to have the Sun as the only gravitational mass and the planets as "test particles". That gets you started and provides a test of the theory in terms of the correction to the precession of Mercury. However, most of the solar system dynamics are approximately unchanged from the Newtonian approximation. We can obtain approximate Schwarzschild coordinates from the distances obtained by previous observation.

To go further must become a very complicated task. I found this paper, for example:

https://arxiv.org/abs/1607.06298

There must be techniques of finding ever closer numerical approximations. I don't know if that addresses your question.

Note that if the solar system had some bizarre geometry that could hardly be guessed at, then the task of reconciling observational data with a theorectical model would be much harder. But, we know it's approximately Schwarzschild geometry and that gives us something to start from.
 
  • #31
vanhees71 said:
For dust you automatically get that the trajectories of the corresponding particles are geodesics. That's because "dust" by definition is just "free streaming", i.e., non-interacting particles:
That doesn't model the general case, because the planets are also sources of gravity.
PeroK said:
The first approximation of the solar system is to have the Sun as the only gravitational mass and the planets as "test particles"
You mean that you only include the Sun in the SET, compute the curvature, then find the equations of motions of planets imposing they're geodesics? Yes it surely is an approximation, but how would you refine it further to include the planets as sources of curvature? The paper you linked seems to contain a methodology for that.

The idea I've got so far is that a "trivial" (or even "feasible") solution of the EFE is only possible with a SET containing a single source of gravity. In that case, you place it at the origin of the coordinate system and compute the curvature of the space around it as if the other lesser masses were zero.

That also means that it's hard to solve the EFE in any spacetime region where only comparable masses are present.
 
  • #32
Pyter said:
I stated earlier that initially you take a snapshot of the stress-energy tensor in curvilinear coordinates to solve the EFE.
But it occurred to me that you can't do that because you don't know the curvature yet, not until you solve the EFE for ##G^{ \mu \nu }##.
You don't need to know the curvature to set up coordinates. Coordinates are just arbitrary numbers labeling points in spacetime. There is no requirement that they have to somehow "match up" with anything else.

In spacetimes with particular symmetries, it is usually helpful to choose coordinates that match those symmetries, but that still doesn't require you to know the curvature, just the symmetries.

Pyter said:
(As an aside, once you have a solution for ##G^{\mu \nu}##, can you univocally determine the mapping of your curved spacetime to Euclidean spacetime, in order to be able to use curvilinear coordinates?)
Coordinates are not a "mapping of curved spacetime to Euclidean spacetime".
 
  • #33
Pyter said:
The idea I've got so far is that a "trivial" (or even "feasible") solution of the EFE is only possible with a SET containing a single source of gravity. In that case, you place it at the origin of the coordinate system and compute the curvature of the space around it as if the other lesser masses were zero.
That's patently not the case. Binary star systems are studied, for example.
 
  • #34
PeterDonis said:
You don't need to know the curvature to set up coordinates. Coordinates are just arbitrary numbers labeling points in spacetime.
PeterDonis said:
Coordinates are not a "mapping of curved spacetime to Euclidean spacetime"
Bear with me @PeterDonis, I'm sure you'll correct me if I'm wrong.
What I know about curvilinear coordinates system, i.e. to make a concrete example locating points on the surface of a 3D sphere (that is a 2D curved surface immersed in a 3D space), is that you map them to Cartesian coordinates. For the spherical surface you could use latitude and longitude, or the projection from one of the poles to a Cartesian 2D plane, and so on. In both cases you have an invertible map that biunivocally connects a point on the curved N-dimensional manifold to a point in the N-dimensional Cartesian space.
In that sense I've stated that to use curvilinear coordinates, you need a "map" to/from an Euclidean spacetime (maybe a "Cartesian space" would have been clearer). You use a point in the Cartesian space to identify a point on the manifold, through the (inverted) map.
What you call "arbitrary numbers" depend nevertheless from the choice of the coordinate system, i.e. where I place the origin, how I measure the distance, if I use straight "rulers" (radio/light waves) or not.
PeroK said:
That's patently not the case. Binary star systems are studied, for example.
Perhaps, but for three or more bodies I bet things get more difficult. OTOH, the three-body problem is notoriously hard to solve even in Newtonian physics.
 
  • #35
Pyter said:
Let's suppose I want to solve the EFE for the whole solar system, from the Earth's (or a geostationary satellite's) vantage point. There is no fluid, only (approximately) point-like masses whose 4D coordinates I observe through light beams or radio waves.
If by “solve the EFE for the whole solar system” you mean “write down the metric in closed form, including the gravitational effects of the planets” you can’t. We can’t even find an exact closed-form solution to Newton's equation for the three-body problem and you're talking about a ten-body EFE problem. But the next sentence suggests that you don't really want to solve an intractably hairy problem of multiple coupled nonlinear differential equations - you want to get to a sufficiently accurate solution using coordinates in which the Earth is at rest.

The general recipe is the same whether you use GR or classical gravitation: First, find coordinates in which the planetary motion can be easily solved to the level of accuracy you need. A heliocentric frame in which the sun is at rest and the planetary masses are negligible may be good enough (and is the practical choice using GR - that's how we deal with the nomalous precession of Mercury). Second, find a coordinate transformation from your original coordinates to the coordinates you want, and apply it. Note that for the solar system this is already a solved problem - you can look online to see which planets will appear where in the night sky at any given time.
 
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  • #36
My main problems boil down to these:

1) how can you build the stress-energy tensor if you don't have a complete Riemannian manifold yet?
2) If you have a complete Riemannian manifold, you don't need to solve the EFE because you can compute the metric tensor directly from the manifold chart(s).

I'll try to explain it in other words.
(In my previous posts, "map" should be read as "charts", which is the proper terminology).

In relativity, our continuum is modeled after a 4D Riemannian manifold (with Minkowskian metric). The latter is - simplifying - a N-dimensional surface, immersed in a N+1 dimension space, equipped with one or more invertible charts which map a point on the surface to a linear N-space (not necessarily Cartesian, they could be polar or spherical or cylindrical coordinates, but they are linear) and back.

In SR, there are no gravity sources and the motion is inertial, so the chart coincides with the identity chart. We know that ##g_{\mu \nu} = \eta_{\mu\nu}## everywhere.

So far so good.

In GR, instead, the charts depend on the gravity sources (or, equivalently, the accelerated/rotatory motion of the observer).

To determine the position of a point in the manifold our continuum is modeled with, you need its charts, just like when you want to determine the position of a point on a 2D spherical surface with fixed radius, you may use the linear coordinates ##\theta, \phi## and the sine and cosine charts mapping the linear coordinates to the 3D coordinates of the surface points.

To build the SET, you need to specify the position and times of the masses (or of a single extended mass, like a star) in the 4D manifold, and in order to do so you need its charts.
But those charts depends on the manifold's curvature and you don't know it because you haven't solved the EFE yet.
But you can't solve the EFE until you build the SET.

And if you somehow find the right charts to build the SET, why should you need to solve EFE, since you can express the metric tensor analytically by taking the partial derivatives of the chart functions?

And even if you solve the EFE for the ##g_{\mu\nu}## (as a function of the linear coordinates), is it mathematically guaranteed that you can univocally find a chart expressing that metric tensor?

Can you see my conundrum?
 
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  • #37
Pyter said:
In relativity, our continuum is modeled after a 4D Riemannian manifold (with Minkowskian metric). The latter is - simplifying - a N-dimensional surface, immersed in a N+1 dimension space, equipped with one or more invertible charts which map a point on the surface to a linear N-space (not necessarily Cartesian, they could be polar or spherical or cylindrical coordinates, but they are linear) and back.
I think the words I have highlighted in bold might be where you are going wrong. The point about differential geometry is that you don't immerse (or embed) your 4D manifold into any 5D (or larger) space. I suspect you are thinking that you need to find the embedding map that maps 5D coordinates into 4D coordinates. You don't. You just have 4D coordinates and you need to find the manifold's metric expressed in terms of those coordinates. That's it, there is nothing more to do.
 
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  • #38
Pyter said:
1) how can you build the stress-energy tensor if you don't have a complete Riemannian manifold yet?
Often we know the distribution of matter and energy so we know the stress-energy tensor. The derivation of the Schwarzschild metric is an example: the SE tensor is zero everywhere. A more complex example with a known but non-zero SE tensor would be the Schwarzschild interior solution. We solve the EFEto find the metric that is consistent with the known SE tensor and the symmetries and boundary conditions of our problem.
In relativity, our continuum is modeled after a 4D pseudo-Riemannian manifold (with locally Minkowskian metric). The latter is - simplifying - a N-dimensional surface, immersed in a N+1 dimension space
Bold text above has been added are added for the sake of precision, ignore if you agree. More important: that bit about embedding in an N+1 dimensional space is incorrect. GR has no notion of embedding four-dimensional spacetime in five-dimensional space; everything is described in terms of points and curves in the four-dimensional manifold. Knowing the metric allows us to calculate the geometrical relationships between these points and curves without introducing a higher-dimensional space.

There is an analogy with navigating around on the curved two-dimensional surface of the earth. As long as we’re only moving in two dimensions we work with latitude and longitude and the metric relationships (ship captains call these relationships “spherical trigonometry” but they’re a specific example of the general concept of the metric at work) between these coordinates. The cartesian x,y,z coordinates of the three-dimensional space around the Earth are completely unhelpful - we don’t need them, we don’t calculate them, we don’t think about them.
In SR… we know that ##g_{\mu \nu} = \eta_{\mu\nu}## everywhere.
Another small point, you can ignore if you agree: that statement is only true if we choose to use Minkowski coordinates. Use, for example, Rindler or spherical coordinates and the components of ##g## will take on different values. But it’s the same flat spacetime and the same (trivial) solution of the EFE.
But those charts depends on the manifold's curvature and you don't know it because you haven't solved the EFE yet.
But you can't solve the EFE until you build the SET.
You do have your ansatz, and you can match it to your boundary conditions - and as I said above you often do know the SE tensor.

You are right that there is no general recipe for finding the metric for an arbitrary configuration of stress-energy. As with most problems involving complex differential equations, we need an inspired guess as to the form of the solution before we can make progress. It’s worth looking at the derivations of some of the exact solutions of the EFE to see what techniques are useful.
 
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  • #39
Pyter said:
What I know about curvilinear coordinates system, i.e. to make a concrete example locating points on the surface of a 3D sphere (that is a 2D curved surface immersed in a 3D space), is that you map them to Cartesian coordinates.
In some cases you can, but there is no requirement that you must always do so or even that you must always be able to do so. Since your arguments appear to hinge on the assumption that you must always do this, your arguments are based on a false premise.
 
  • #40
Pyter said:
1) how can you build the stress-energy tensor if you don't have a complete Riemannian manifold yet?
The EFE is a differential equation. You can solve a differential equation in a local neighborhood of some point without having to know the full manifold. In fact, in many cases, including famous solutions of the EFE such as Schwarzschild, it took decades from the initial discovery of the solution before the full maximal analytic extension of the manifold was known.

Pyter said:
If you have a complete Riemannian manifold, you don't need to solve the EFE because you can compute the metric tensor directly from the manifold chart(s).
That's true, but you also have no guarantee that when you compute the Einstein tensor of this known metric and use the EFE to see what it implies about the stress-energy tensor, that you will get anything that is physically reasonable. That's why physicists work the other way: they start with some physically reasonable stress-energy tensor and solve the EFE to see what metric it implies.

Pyter said:
To build the SET, you need to specify the position and times of the masses (or of a single extended mass, like a star) in the 4D manifold, and in order to do so you need its charts.
You need some chart. There are an infinite number of possible charts on any manifold. If you know that the manifold has some symmetries, you can pick charts that match up with those symmetries (for example, the standard Schwarzschild-type chart for spherical symmetry), but that still leaves quite a bit of freedom of choice of chart.

Pyter said:
But those charts depends on the manifold's curvature
No, they don't. The metric in a given chart depends on the manifold's geometry, but knowing a chart is not the same as knowing the full metric. You might be able to write down an expression for the metric that has fewer than ten unknown functions in it (ten is the most general number possible since the metric is a symmetric 2nd-rank tensor), and/or with functions of fewer than all four coordinates.

For example, if you write the most general metric for a spherically symmetric spacetime in the standard Schwarzschild chart, your metric will have two unknown functions in it, functions of only the radial and time coordinates, which you then have to determine by solving the EFE, which in turn means you have to make some assumption about the stress-energy tensor. You can do all that without knowing the full metric.
 
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  • #41
PeterDonis said:
You need some chart. There are an infinite number of possible charts on any manifold. If you know that the manifold has some symmetries, you can pick charts that match up with those symmetries (for example, the standard Schwarzschild-type chart for spherical symmetry), but that still leaves quite a bit of freedom of choice of chart.
So, for example, I believe you can solve EFE for the orbits of solar system's planets starting from Schwarzschild spacetime in Schwarzschild chart with origin (coordinate radius ##r=0##) in the Sun.
 
  • #42
DrGreg said:
I think the words I have highlighted in bold might be where you are going wrong. The point about differential geometry is that you don't immerse (or embed) your 4D manifold into any 5D (or larger) space. I suspect you are thinking that you need to find the embedding map that maps 5D coordinates into 4D coordinates. You don't. You just have 4D coordinates and you need to find the manifold's metric expressed in terms of those coordinates. That's it, there is nothing more to do.
So far I've only seen examples of charts mapping from a N+1-dimension Euclidean space to a N-dimension space. Anyway you're right, the main purpose of the metric tensor is to get rid of the N+1-dimension (so-called ambient) space and to work only with N coordinates.

My point is that to map your manifold to the linear space you either need the charts (which give you a superior knowledge over the metric tensor, since you can obtain its components by taking the partial derivatives with respect to the linear coordinates and doing scalar products of the resulting vectors), or the metric tensor. Before solving the EFE you don't have any of them, and to solve the EFE you need to build a SET, function of the linear coordinates, for which you need either the charts or the metric tensor.
Nugatory said:
A more complex example with a known but non-zero SE tensor would be the Schwarzschild interior solution. We solve the EFEto find the metric that is consistent with the known SE tensor
This only shifts the problem from a discrete number of points with finite mass to an infinite number of points with infinitesimal mass (the interior of the fluid). You need to specify their linear coordinates in the SET anyway.

Nugatory said:
that bit about embedding in an N+1 dimensional space is incorrect. GR has no notion of embedding four-dimensional spacetime in five-dimensional space; everything is described in terms of points and curves in the four-dimensional manifold. Knowing the metric allows us to calculate the geometrical relationships between these points and curves without introducing a higher-dimensional space.

Nugatory said:
As long as we’re only moving in two dimensions we work with latitude and longitude and the metric relationships (ship captains call these relationships “spherical trigonometry” but they’re a specific example of the general concept of the metric at work) between these coordinates. The cartesian x,y,z coordinates of the three-dimensional space around the Earth are completely unhelpful - we don’t need them, we don’t calculate them, we don’t think about them.
If that's true, from the procedural standpoint, how would you find the latitude and longitude of two points on the Earth's surface without leaving it, using only curved rulers or light/radio waves running along its surface?
PeterDonis said:
In some cases you can, but there is no requirement that you must always do so or even that you must always be able to do so.
If you want to express the SET as function of the linear coordinates, you need either a metric or - even better - the charts.
PeterDonis said:
The EFE is a differential equation. You can solve a differential equation in a local neighborhood of some point without having to know the full manifold.
Since you cite a local neighborhood, I guess you mean to solve the EFE under the initial assumption that the identity chart applies? This would conflict with the requirement that the solution must yield a non-flat space, rendering the identity chart invalid.
PeterDonis said:
You need some chart.
Yes, but at least one.
PeterDonis said:
No, they don't. The metric in a given chart depends on the manifold's geometry, but knowing a chart is not the same as knowing the full metric.
The charts also contain info on the manifold's curvature and more complete, since you can derive the metric tensor from the charts. Or at least you can derive the metric in the part of the manifold covered by that particular chart, if that's what you meant by "knowing the full metric".
 
  • #43
cianfa72 said:
So, for example, I believe you can solve EFE for the orbits of solar system's planets starting from Schwarzschild spacetime in Schwarzschild chart with origin (coordinate radius ##r=0##) in the Sun.
This isn't solving the EFE. You already have the solution: Schwarzschild spacetime (more precisely, Schwarzschild spacetime for the vacuum region outside the sun). All you're doing by solving for planetary orbits is solving the geodesic equation for that known solution of the EFE.

Note, btw, that our Solar System measurements are accurate enough that we can spot inaccuracies in the scheme just described, due to the fact that the planets are not test particles but are significant sources of gravity on their own. The approximation scheme used to address this works by ignoring non-linearities and solving for the motion of each planet by simply adding the contributions to the spacetime geometry from the Sun and all the other planets. But none of this is solving the EFE; it's just solving the geodesic equation.
 
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  • #44
Pyter said:
So far I've only seen examples of charts mapping from a N+1-dimension Euclidean space to a N-dimension space.
Then you need to spend some time with some GR textbooks, working through actual examples from GR, not just using your intuition developed from other fields.

Pyter said:
to map your manifold to the linear space
Which, as you have already been told, you never need to do to solve the EFE. You simply do not understand how the EFE is actually solved. You need to spend some time with some GR textbooks, learning how the EFE is actually solved by working through actual examples.

Pyter said:
You need to specify their linear coordinates in the SET anyway.
No, you don't. You can specify the SET in any coordinate chart you like.

Pyter said:
If you want to express the SET as function of the linear coordinates, you need either a metric or - even better - the charts.
You need a chart, but, as I have already said, the metric expressed in this chart can and will have unknown functions of the coordinates in it. Solving the EFE means solving for these unknown functions.

You keep making assertions that are simply wrong, I suspect because you have simply not spent any time actually looking at a GR textbook and seeing how actual solutions to the EFE work in GR. And you are continuing to make the same wrong assertions even though you have been corrected multiple times. There is no point in continuing to go around in circles.

Pyter said:
Since you cite a local neighborhood, I guess you mean to solve the EFE under the initial assumption that the identity chart applies?
No. A GR textbook will explain what I meant. This is a very basic concept.

Pyter said:
The charts also contain info on the manifold's curvature
No, they don't. The metric contains information on curvature, once all unknown functions in the metric have been solved for. But you don't need to have done all that to have a chart. Once again, you continue to make these wrong assertions even though you have been corrected multiple times already. There is no point in continuing to go around in circles.
 
  • #45
Pyter said:
how would you find the latitude and longitude of two points on the Earth's surface without leaving it, using only curved rulers or light/radio waves running along its surface?
You wouldn't. Latitude and longitude are global coordinates, not local ones. You can't find them by purely local measurements.

If you wanted to find the metric of the Earth's surface based on purely local measurements within the surface, here is how you would proceed, by analogy with how the corresponding task would be done in GR:

(1) Set up local coordinates any way you want. The usual method would be to start by labeling the point where you are as ##(0, 0)##, and then marking out two perpendicular straight lines through that point and labeling them as your two different coordinate axes. Strictly speaking, there is no requirement for the coordinate axes to be perpendicular, only linearly independent; but making them perpendicular, at least locally, makes the calculations easier when you want to compute the curvature.

(2) Pick one coordinate axis, call it ##a## (I'll avoid using the standard Cartesian names since we will be finding that the manifold in question is not flat), and mark off equal increments of some convenient unit distance (such as the length of your meter stick) along it from ##(0, 0)##. Then mark out perpendicular straight lines crossing the axis at each of your marked points. At each of the crossing points, these perpendicular straight lines will be tangent to the coordinate grid lines for your other coordinate, call it ##b##.

(3) Extend each of the perpendicular straight lines by purely local means--i.e., just looking at that particular line and keeping it straight by extending it in the same direction the way a surveyor would, without regard to its relationship with any neighboring lines--and see if they converge or diverge. If they do, the manifold is curved, at least in that local region. If they don't, the manifold is flat, at least in that local region.

On Earth, you will find that the lines converge (globally, what you are doing is drawing segments of adjacent great circles that all start off crossing another great circle at right angles, and those will of course converge), indicating that the manifold is curved. By measuring the rate of convergence, you can find the numerical value of the curvature (which, since this is a 2-dimensional manifold, will just be a single number; to put it another way, the Riemann tensor of this manifold has only one component).

In terms of your local coordinate chart, you will find that the perpendicular straight lines you have drawn will not be identical with the coordinate grid lines of ##b##. In terms of the metric written in this coordinate chart, you will find that it looks exactly Euclidean at point ##(0, 0)##, but picks up non-Euclidean terms as you move away from ##(0, 0)##. (Obviously, then, these coordinates are not latitude and longitude, as I said above.)

The method I have described above is analogous to setting up what are called Riemann normal coordinates in GR, centered on a chosen point of spacetime. These coordinates are chosen to "look Minkowskian" at the chosen point, just as the coordinates I described above "look Euclidean" at the chosen point. Most GR textbooks discuss this at a fairly early point in their exposition.
 
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  • #46
Thinking about what was said in this thread, I've found a way out of my impasse.

I considered the RHS/SET in the EFE as the "independent variables" in the equations, and the LHS/Einstein's tensor as the "dependent variables", thinking it was the only side containing ##g_{\mu\nu}##.
So I thought that you needed to fully specify the SET in the spacetime region under observation before solving the equation. But that would be tantamount to completely express (by observation) the equations of motions of the energy/matter/fields between the time boundaries of the region, which would render the solution of the EFE useless.

Now I'm aware that both sides of the EFE contribute to ten differential equations whose unkown are the components of ##g_{\mu\nu}##, and thus also the RHS depends on the linear coordinates through the unknown functions ##g_{\mu\nu}##.

The physical observation of your masses only set the boundary conditions of the EFE, and they contain the unknown ##g_{\mu\nu}## in a parametrized form, for instance in the measurement of your distance from objects of known mass.

Am I on the right track?
 
  • #47
PeterDonis said:
You wouldn't. Latitude and longitude are global coordinates, not local ones. You can't find them by purely local measurements.
You could, but only if you had the charts or the metric tensor on the whole manifold. This way, knowing the length of your measured paths of light from/to the objects, and knowing that they move along a geodesic, you could project these paths on the linear (coordinate) space and find the coordinates of every point.
 
  • #48
Pyter said:
Am I on the right track?
@PeterDonis and I have both suggested that you look at the derivation of some of the known solutions to the EFE. Working through a few of these will tell you more about how these problems are solved than any amount of general principles.
 
  • #49
Pyter said:
So far I've only seen examples of charts mapping from a N+1-dimension Euclidean space to a N-dimension space.
PeterDonis said:
Then you need to spend some time with some GR textbooks, working through actual examples from GR, not just using your intuition developed from other fields.
I think I've spent on them quite some time already :). And I've found those charts in those textbooks. What would be another example of a legit chart on a manifold that doesn't map a N+1 space into a N space? Charts for a 2D spherical surface, for instance.
Nugatory said:
@PeterDonis and I have both suggested that you look at the derivation of some of the known solutions to the EFE. Working through a few of these will tell you more about how these problems are solved than any amount of general principles.
So far I've looked into the most famous one, the Schwartzschild solution, but that didn't answer my questions on the general case.
 
  • #50
PeterDonis said:
This isn't solving the EFE. You already have the solution: Schwarzschild spacetime (more precisely, Schwarzschild spacetime for the vacuum region outside the sun). All you're doing by solving for planetary orbits is solving the geodesic equation for that known solution of the EFE.
Yes definitely, my terms were misleading. So we actually solve the geodesic equation for planet orbit in Schwarzschild coordinates ##(r,\theta,\phi,t)##. From the solution then we can calculate for example the proper time needed to complete an orbit around the Sun.
 
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