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

[Pyter]

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.

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.

 

[PeterDonis]

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.

(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".

 

[PeroK]

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.

 

[Pyter]

You don't need to know the curvature to set up coordinates. Coordinates are just arbitrary numbers labeling points in spacetime.

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.

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.

 

[Nugatory]

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.

 

[Pyter]

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?

 

[DrGreg]

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.

 

[Nugatory]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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.

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.

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.

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.

 

[cianfa72]

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.

 

[Pyter]

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.

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.

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.

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?

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.

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.

You need some chart.

Yes, but at least one.

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".

 

[PeterDonis]

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.

 

[PeterDonis]

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.

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.

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.

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.

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.

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.

 

[PeterDonis]

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.

 

[Pyter]

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?

 

[Pyter]

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.

 

[Nugatory]

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.

 

[Pyter]

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.

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.

@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.

 

[cianfa72]

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.

 

[PeterDonis]

Am I on the right track?

No.

I think I've spent on them quite some time already :).

It doesn't seem like it from what you're posting.

I've found those charts in those textbooks.

What 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.

Latitude and longitude on a sphere. These are coordinates just on the sphere, and do not either account for or require the sphere to be embedded in a higher dimensional space.

The same is true of every chart used in GR.

So far I've looked into the most famous one, the Schwartzschild solution

It doesn't seem like it, since none of the descriptions you have given of how you think the EFE is solved bear any resemblance at all to how the EFE is actually solved to obtain the Schwarzschild solution.

that didn't answer my questions on the general case.

If your understanding of that one special case is wrong, as it seems to me to be, of course you're not going to learn anything about the general case from it.

 

[Ibix]

I wrote an Insight on solving the EFEs for a fairly simple non-vacuum case, here, which may be of interest. It's a fairly bull-at-a-gate approach, but there are ways to make it more elegant discussed at the end.

 

[Pyter]

No.

Closer at least?

What textbooks?

For instance, in Intro to Differential Geometry and General Relativity - S. Waner, page 10, I've found:

Latitude and longitude on a sphere. These are coordinates just on the sphere, and do not either account for or require the sphere to be embedded in a higher dimensional space.

The same is true of every chart used in GR.

See above. I count n+1 dimensions for the ambient space, n dimensions for the sphere.

If your understanding of that one special case is wrong, as it seems to me to be, of course you're not going to learn anything about the general case from it.

That's why I'm on this forum asking you GR megaexperts for help :)

 

[PeterDonis]

A couple of my Insights articles are also relevant to this thread's topic:

https://www.physicsforums.com/insig...-in-a-static-spherically-symmetric-spacetime/

https://www.physicsforums.com/insights/short-proof-birkhoffs-theorem/

 

[PeterDonis]

Closer at least?

No.

For instance, in Intro to Differential Geometry and General Relativity - S. Waner, page 10, I've found:

Hm. Is this the only GR textbook you've looked at? Its approach is completely different from that of any GR textbook I've read. Part of the problem may be that it's written by a mathematician, not a physicist; but it's still surprising to me because a mathematician should understand even better than a physicist the theory of the intrinsic curvature of manifolds, without regard to the usage or even the existence of an embedding in a higher dimensional space.

In short, I do not think this is a good reference.

That's why I'm on this forum asking you GR megaexperts for help :)

But you're not listening to the help we're giving you. You keep making the same wrong assertions even after you've been told they're wrong multiple times. If you want help, you have to listen to it.

For some specific examples, you can look at the Insights articles that @Ibix and I linked to. Also here are some examples from GR textbooks and lecture notes:

Sean Carroll's lecture notes on GR [1], Chapter 7, discusses the Schwarzschild solution and how it can be obtained. (Carroll also has a textbook which contains all the material in the lecture notes plus more, but the notes are nice because they're available for free online.)

Carroll's notes, Chapter 8, discusses the FRW solutions used in cosmology and how they can be obtained.

Misner, Thorne, & Wheeler, Chapter 23, discusses how to construct solutions describing static, spherically symmetric stars.

MTW, Chapter 27, discusses how to construct idealized cosmological models of the universe.

MTW, Chapters 31 and 32, discuss black hole solutions and various coordinate charts used for them.

Wald, Chapter 5, discusses cosmological models and solutions.

Wald, Chapter 6, discusses the Schwarzschild solution and its derivation.

[1] https://arxiv.org/abs/gr-qc/9712019

 

[Pyter]

The Schwartschild solution in the textbook I've just mentioned is built to find g** of the form:

(So I was right, the SET is also function of g**).
After setting some more constraints, T** becomes:

Then, after imposing that the equations of motions:

are satisfied, we have the additional constraint:

Then it goes on to solve the differential equations for the single components, finding:

Just looking at this, you'd think that the r in the equations is the actual radius of the star measured in our continuum, while it's a linear (local) coordinate, and that enough is a source of confusion to me.
Also the variable t is a linear/local coordinate, which doesn't match the actual time measured by an external observer. You compute the latter through the metric.

 

[Pyter]

Besides, I really don't know how I could infer the general case from that special case.

a mathematician should understand even better than a physicist the theory of the intrinsic curvature of manifolds, without regard to the usage or even the existence of an embedding in a higher dimensional space.

Intuitively I thought that the very concept of "curvature" implied the embedding on a higher space.
If you only have a 1D space, you wouldn't have "curved" or "straight" lines, you'd only have "lines", because the curvature only appears in a 2D space or higher. Same thing for the 2D, and higher dimensions, surfaces.
After all, the curvature of a surface is the rate of change of its normal vector, and that is "outside" the surface, in a higher dimension space.

 

[Pyter]

I wrote an Insight on solving the EFEs for a fairly simple non-vacuum case, here, which may be of interest. It's a fairly bull-at-a-gate approach, but there are ways to make it more elegant discussed at the end.

@Ibix what do you think of my understanding so far of the general approach?

https://www.physicsforums.com/threa...einsteins-field-equation.1010389/post-6579533

Care to comment on the various points?

 

[PeterDonis]

The Schwartschild solution in the textbook I've just mentioned is built to find g** of the form:

Yes, that's one of the possible ways of writing the metric for a spherically symmetric spacetime. Note that it has two unknown functions in it, $\Lambda$ and $\Phi$.

(So I was right, the SET is also function of g**).

That SET is the general form of the SET for a perfect fluid. In order to see why the metric appears in it, write it this way:

$$
T^{ab} = \rho u^a u^b + p \left( g^{ab} + u^a u^b \right)
$$

The tensor multiplying $p$ is the projection tensor into spacelike hypersurfaces orthogonal to the fluid. In other words, the metric appears in it in order to reflect the physical fact that, in the rest frame of the fluid, the pressure is purely spatial. (Note that this expression is valid in any coordinate chart.)

Just looking at this, you'd think that the r in the equations is the actual radius of the star measured in our continuum

No, you wouldn't. Look at the metric. The increment of radial distance is not $dr$, it's $e^\Lambda dr$. The coordinate $r$ is, again looking at the metric, the "areal radius", i.e., the area of a 2-sphere with radial coordinate $r$ is $4 \pi r^2$.

while it's a linear (local) coordinate

No, it isn't. It's a global coordinate on the spacetime.

and that enough is a source of confusion to me.

Your confusion appears to be the result of an incorrect understanding of what you are reading. See above.

Also the variable t is a linear/local coordinate

No, it isn't. It's a global coordinate on the spacetime. The chart you are looking at is a global chart, not a local one.

which doesn't match the actual time measured by an external observer.

Yes, it does, for the usual meaning of "external observer". The $t$ coordinate matches proper time for an observer at infinity. It does not match proper time for an observer at finite $r$; the increment of proper time for such an observer is $e^\Phi dt$, not $dt$. But such an observer is a "local" observer, not an "external" observer.

You compute the latter through the metric.

You compute the proper time for an observer at finite $r$ using the metric, yes, as I just did above. (Note, though, that that expression is only valid for an observer who "hovers" at the same $r$, $\theta$, $\phi$ forever.)

 

[PeterDonis]

I really don't know how I could infer the general case from that special case.

If you look at the Insights articles or the textbook references that have been given, you will see specific solutions being derived using the same general method.

Intuitively I thought that the very concept of "curvature" implied the embedding on a higher space.

Then you thought wrong. It doesn't. As you have already been told repeatedly in this thread.

If you only have a 1D space, you wouldn't have "curved" or "straight" lines, you'd only have "lines", because the curvature only appears in a 2D space or higher.

That is correct, but it does not generalize the way you claim.

Same thing for the 2D, and higher dimensions, surfaces.

Wrong. For any manifold of 2 or more dimensions, there is a well-defined intrinsic curvature that does not require the usage or even the existence of an embedding in any higher dimensional space.

the curvature of a surface is the rate of change of its normal vector

That is extrinsic curvature, not intrinsic curvature (intrinsic curvature is sometimes also called Gaussian curvature in the literature). The spacetime curvature referred to in GR is intrinsic curvature. If all of your knowledge of curvature up to now has been of extrinsic curvature, no wonder you're having problems. You need to study the definition and examples of intrinsic curvature so that you understand it. It is a different concept from extrinsic curvature.