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

[Pyter]

Of course it is solvable, at least numerically.

But if you don't have enough constraints to reduce the degrees of freedom, the solution will be far from unique. In the Schwa. case you use the symmetry of coordinates and the continuity equation to reduce the DOFs, in the "general" case you can't rely on the symmetry: you have more degrees of freedom and the same constraints as in the Schwa. case.
Or are there supplementary constraints I'm not aware of?

 

[PeterDonis]

You mean that in the "general" (neutron stars) scenario, the physical observations (radial and angular positions, estimated (?) masses) are useless in setting up the EFE, let alone solve it?

Yes. More precisely, they don't help in either reducing the number of equations you have to solve or the number of unknown functions you have to solve for. All those physical observations do is give you particular parameters to plug into the solution after you've obtained it. (For example, knowing the mass of some spherically symmetric body does not help you obtain the Schwarzschild solution for the spacetime geometry outside the body; all it does is give you the number $M$ to plug into the solution once you've got it.)

Does it mean that in that case the EFE is not even solvable in theory?

Of course not. See above.

if you don't have enough constraints to reduce the degrees of freedom, the solution will be far from unique.

You are misstating this. See below.

In the Schwa. case you use the symmetry of coordinates and the continuity equation to reduce the DOFs

None of those are physical observations. See my comment above about the Schwarzschild solution for the vacuum region outside a spherically symmetric massive body. You use spherical symmetry plus vacuum to obtain a mathematical expression for the metric. But this expression does not describe just one metric; it describes a family of them, all having the same form, but with different numbers for the constant $M$ that appears in the solution. Your knowledge of the actual mass of the particular body you are interested in tells you what number to plug in for $M$ in the metric.

In other words: even after you have used symmetries, etc., to reduce the number of degrees of freedom, you still don't have a "unique" solution; you will have an expression for the metric that will have some undetermined parameters in it (like $M$ in the Schwarzschild case). If you want a unique solution, you need to fill in values for those parameters. Physical observations of the particular system you are interested in can help you to do that; but they can't help you to get to that point, where you have an expression for the metric that only needs some parameters filled into be unique.

in the "general" case you can't rely on the symmetry: you have more degrees of freedom and the same constraints as in the Schwa. case.

I have no idea what you are talking about here. Perhaps rethinking your understanding in the light of the above will help.

 

[vanhees71]

But if you don't have enough constraints to reduce the degrees of freedom, the solution will be far from unique. In the Schwa. case you use the symmetry of coordinates and the continuity equation to reduce the DOFs, in the "general" case you can't rely on the symmetry: you have more degrees of freedom and the same constraints as in the Schwa. case.
Or are there supplementary constraints I'm not aware of?

As any gauge theory the equations of motion for GR do not lead to unique solutions. That's the same for electromagnetism, where Maxwell's equations do not lead to unique solutions for the four-potential, which is a gauge field and thus contains redundant degrees of freedom. This does, however, not mean that you don't have a unique description of the physical situation. It's just that different solutions, connected by a gauge transformation, describe the same physical situation. In GR the gauge transformations are the (local) diffeomorphisms between different coordinates, but the physical situation doesn't depend on it. That's what's behind the general covariance of GR, and as in any gauge theory you can fix the gauge in any way you like to facilitate your calculation.

 

[PeterDonis]

different solutions, connected by a gauge transformation, describe the same physical situation.

Note that while this is of course true in GR for arbitrary diffeomorphisms, it is also the case that solving the field equation in a given case can give you a family of solutions even after the gauge freedom is taken into account. For example, as I said in a previous post, solving the EFE for the case of spherical symmetry and vacuum gives the Schwarzschild solution, but "the" Schwarzschild solution is really an infinite family of solutions, one for each possible value of $M$. Those solutions are physically different; you can't transform a Schwarzschild solution with $M_1$ into a Schwarzschild solution with $M_2 \neq M_1$ by a gauge transformation. But they are both solutions of the EFE under the same set of assumptions.

 

[vanhees71]

Of course not, because there you have physically different situations, i.e., vacuum Schwarzschild solutions for "matter"/"black holes" of different "mass", $M$. If there is vacuum in the region $r>R$, then there can be anything in $r<R$, as long as it's spherically symmetric. In the extreme singular case, you have a (non-rotating) black hole with a "point mass" $M$ at rest in the origin of the symmetry center.

It's always good to consider the electromagnetic analogy. In the case of electromagnetism, when you assume a spherically symmetric vacuum solution of the Maxwell equations (in Minkowski space) you get a Coulomb field (assuming that there's vacuum outside some sphere of radius $R$). The only relevant parameter is the electric charge in the "interiour", $r<R$, which can take any value and can be due to completely different charge distributions. The singular case is that of a point charge sitting at rest in the symmetry center.

 

[PeterDonis]

[Edit: Removed statement based on misreading of quoted post.]

In the extreme singular case, you have a (non-rotating) black hole with a "point mass" $M$ at rest in the origin of the symmetry center.

This is not a correct description of a black hole. The "symmetry center" is not a point in space, it's a moment of time (it's a spacelike curve, not a timelike curve), and there is no "point mass" there, it's vacuum all the way down to the singularity (which is itself not part of the manifold).

It's always good to consider the electromagnetic analogy.

Not in cases where it is misleading, as in the case of a "point charge", which does not have an analogue of a "point mass" in GR (although, ironically, it does in Newtonian mechanics).

 

[Orodruin]

Sure there can, it just can't be static.

I believe he is intending R to be arbitrary, not the Schwarzschild radius.

 

[PeterDonis]

I believe he is intending R to be arbitrary, not the Schwarzschild radius.

Ah, yes, I see I misread "can" for "can't" in his post. I'll edit my previous post to correct.

 

[Dale]

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.

Not quite. There is indeed a smooth invertible map from events in the manifold to points in R4. But the points in R4 are not Cartesian. The “distance” between them is given by the metric, not the Pythagorean theorem.

But if you don't have enough constraints to reduce the degrees of freedom, the solution will be far from unique.

That isn’t really a problem. Many different mathematical solutions to many different differential equations can describe the same physics.

 

[vanhees71]

[Edit: Removed statement based on misreading of quoted post.]This is not a correct description of a black hole. The "symmetry center" is not a point in space, it's a moment of time (it's a spacelike curve, not a timelike curve), and there is no "point mass" there, it's vacuum all the way down to the singularity (which is itself not part of the manifold).Not in cases where it is misleading, as in the case of a "point charge", which does not have an analogue of a "point mass" in GR (although, ironically, it does in Newtonian mechanics).

You are right. I forgot that in standard Schwarzschild coordinates $r$ becomes the time-like coordinate for $r<r_{\text{S}}$.

 

[Pyter]

you still don't have a "unique" solution; you will have an expression for the metric that will have some undetermined parameters in it (like in the Schwarzschild case). If you want a unique solution, you need to fill in values for those parameters. Physical observations of the particular system you are interested in can help you to do that;

That's what I was going to ask next: you have to plug in the observable quantities into the metric to make it "usable". Not only M but also r, I believe, since it's a parameter too: If you wanted to compute the Schwa. metric induced by the Sun at the Earth's orbit, you can't simply replace r with Earth's observed distance from the Sun. Do you find this value by solving other differential equations, this time involving the observations?

 

[Pyter]

Not quite. There is indeed a smooth invertible map from events in the manifold to points in R4. But the points in R4 are not Cartesian. The “distance” between them is given by the metric, not the Pythagorean theorem.

According to the manifold's definition I had posted, which I repeat here for convenience:

all the "local coordinates" $\in E_1$, in the sense that they're rectlinear.
Example: latitude and longitude for a $S_2$ sphere. If you want to organize them as Cartesian coordinates (i.e. consider the $(\phi, \theta)$ plane, or not), it's up to you. Then you could define the "distance" between two points on the manifold as the differences between their latitudes and longitudes. In that case you only use the metric to map the points in the manifold to their (latitude, longitude).
If you instead need the length of an arbitrary path connecting two points on the manifold and lying entirely on the manifold, you will integrate the $ds^2$, which as you say is a function of the metric, along this path.

 

[Dale]

Then you could define the "distance" between two points on the manifold as the differences between their latitudes and longitudes.

Since this distance has no physical meaning I would not call that space “Cartesian”.

 

[martinbn]

According to the manifold's definition I had posted, which I repeat here for convenience:

View attachment 295032
all the "local coordinates" $\in E_1$, in the sense that they're rectlinear.

No, they are in $E_1$ in the sense that they have values in the real numbers. It has nothing to do with rectlinear, which doesn't make sense in genereal.

Example: latitude and longitude for a $S_2$ sphere. If you want to organize them as Cartesian coordinates (i.e. consider the $(\phi, \theta)$ plane, or not), it's up to you. Then you could define the "distance" between two points on the manifold as the differences between their latitudes and longitudes.

You could do that if you had a chart that covers the whole manifold otherwise you cannot.

In that case you only use the metric to map the points in the manifold to their (latitude, longitude).

That is not what a metric is!

If you instead need the length of an arbitrary path connecting two points on the manifold and lying entirely on the manifold, you will integrate the $ds^2$, which as you say is a function of the metric, along this path.

This is also not true, even if you ignore the imprecisions you want $ds$.

 

[Orodruin]

That's what I was going to ask next: you have to plug in the observable quantities into the metric to make it "usable". Not only M but also r, I believe, since it's a parameter too: If you wanted to compute the Schwa. metric induced by the Sun at the Earth's orbit, you can't simply replace r with Earth's observed distance from the Sun. Do you find this value by solving other differential equations, this time involving the observations?

No, r is not a parameter of the solution. It is one of the coordinates. For the Sun-Earth system, it is to very good approximation given by the distance from the Sun.

 

[Pyter]

Since this distance has no physical meaning I would not call that space “Cartesian”.

Maybe "Euclidean" would be more appropriate. Since we have n coordinates, each $\in E_1$, we have an $E_n$ space. Every point in this space is identified by a n-ple of values.

No, they are in in the sense that they have values in the real numbers. It has nothing to do with rectlinear, which doesn't make sense in genereal.

"Rectilinear" in the sense that you could consider each one a coordinate axis in an $E_n$ space.

That is not what a metric is!

How else would you map the points of the manifold to the "local coordinates" (so-called even if they may cover the whole manifold) space $E_n$, if you don't have the charts' explicit functions?

No, r is not a parameter of the solution. It is one of the coordinates. For the Sun-Earth system, it is to very good approximation given by the distance from the Sun.

Right, r and t are the "local coordinates" (according to the definition above). As it was stated in a previous post, they exactly match the physical quantities for the observer at infinity only. If you plug in the observed distance from the Sun for r you do, as you said, an approximation.

 

[Orodruin]

If you plug in the observed distance from the Sun for r you do, as you said, an approximation.

In the Sun-Earth system that approximation is much more precise than the variations in the distance due to the orbital characteristics. Accurate enough for the difference not to matter.

Furthermore, when the difference becomes relevant, the entire concept of distance needs a big caveat.

 

[Pyter]

Furthermore, when the difference becomes relevant, the entire concept of distance needs a big caveat.

Exactly, for instance in situations where the curvature gradient is very steep, like in proximity of a neutron star.

EDIT: but in such a situation I could still "observe" a distance from the star, maybe with telemetry, imagining that the star's surface reflects a laser beam back.

 

[Dale]

Maybe "Euclidean" would be more appropriate. Since we have n coordinates, each ∈E1, we have an En space. Every point in this space is identified by a n-ple of values.

Yes, every event in an open region of spacetime is identified by a 4-tuple of values. So I would simply say that: every event in the spacetime manifold is identified by a point in $\mathbb{R}^4$. I wouldn't characterize $\mathbb{R}^4$ further. Anything else, Cartesian or Euclidean, seems to add a flat geometry that is not physically meaningful in the coordinate space.

Does it need to have a name for some reason? Is $\mathbb{R}^4$ not sufficient for some reason?

 

[PeterDonis]

That's what I was going to ask next: you have to plug in the observable quantities into the metric to make it "usable". Not only M but also r, I believe, since it's a parameter too

Strictly speaking, $r$ is a coordinate, not a parameter.

If you wanted to compute the Schwa. metric induced by the Sun at the Earth's orbit, you can't simply replace r with Earth's observed distance from the Sun.

Correct, you have to compute $r$ from the observed distance. (Although in the case of the solar system the correction is very small.)

Do you find this value by solving other differential equations, this time involving the observations?

No. You find it by knowing the relationship between radial distance and the $r$ coordinate in Schwarzschild spacetime, which is given by the metric that you've already solved for.

 

[PeterDonis]

Maybe "Euclidean" would be more appropriate. Since we have n coordinates, each $\in E_1$, we have an $E_n$ space. Every point in this space is identified by a n-ple of values.

No, "Euclidean" is not appropriate, at least not if you want to use standard terminology. The standard terminology for "the space of 4-tuples of real numbers" with no other structure is what @Dale said, $\mathbb{R}^4$. "Euclidean n-space" means you have a particular metric. (The coordinates are not individually in "Euclidean 1-space" either; that would imply a particular metric on the real numbers, which is not valid in the general case.)

 

[Pyter]

Does it need to have a name for some reason? Is not sufficient for some reason?

I used $E_4$ just because the cited definition of manifold involved n coordinates $\in E_1$, implying now that I think about it that only the metric of the single coordinate is Euclidean. But we agree that the only metric that matters is on the manifold.

You find it by knowing the relationship between radial distance and the coordinate in Schwarzschild spacetime, which is given by the metric that you've already solved for.

By "radial distance" you mean the "observed" distance, measured perhaps through telemetry? Is it connected to r through the geodesic equation?

 

[ergospherical]

By "radial distance" you mean the "observed" distance, measured perhaps through telemetry? Is it connected to r through the geodesic equation?

The Schwarzschild coordinate $r$ of a point $p$ is the area radius function $r = \sqrt{A(p)/4\pi}$, with $A(p)$ the area of the spherical symmetry orbit through $p$. Meanwhile the radial distance comes through $dR =\sqrt{ g_{rr}} dr$.

 

[Pyter]

The Schwarzschild coordinate $r$ of a point $p$ is the area radius function $r = \sqrt{A(p)/4\pi}$, with $A(p)$ the area of the spherical symmetry orbit through $p$. Meanwhile the radial distance comes through $dR =\sqrt{ g_{rr}} dr$.

So the measured distance $R = \int g_{rr}\,dr$? What are the extremes of integration and how do you express r in function of the measured distance?

 

[ergospherical]

So the measured distance $R = \int g_{rr}\,dr$? What are the extremes of integration and how do you express r in function of the measured distance?

what'd you do with my precious square root

 

[PeterDonis]

By "radial distance" you mean the "observed" distance, measured perhaps through telemetry?

Yes.

Is it connected to r through the geodesic equation?

No, it's connected to $r$ through the metric. A radial line of coordinate increment $dr$ covers an actual physical distance of $dr / \sqrt{ 1 - 2M / r }$.

So the measured distance $R = \int g_{rr}\,dr$?

With a square root over $g_{rr}$, as @ergospherical says, yes.

What are the extremes of integration

The extremes of integration are the starting and ending $r$ values. Since $r$ is an "areal radius", a circular orbit at radius $r$ has circumference $2 \pi r$, and that is an actual physical distance, so you could, for example, obtain the radial distance between two circular orbits by measuring both their circumferences $C_1$ and $C_2$ and inverting $C = 2 \pi r$ to obtain $r_1$ and $r_2$.

how do you express r in function of the measured distance?

If you're doing the integral that has been described, you don't, at least not in terms of measured radial distance.

 

[PeterDonis]

I used $E_4$ just because the cited definition of manifold involved n coordinates $\in E_1$,

No, it doesn't. The coordinates are members of $\mathbb{R}$, but that is not the same as $E_1$ because $E_1$ implies a Euclidean metric whereas $\mathbb{R}$ does not.

implying now that I think about it that only the metric of the single coordinate is Euclidean.

Not even that is true. See above. There is no "metric of the single coordinate".

 

[Orodruin]

Yes

No. This will not be the distance measured by the method suggested by the OP (see eg #123) - which was bouncing a light signal and measuring the round trip time - due to Shapiro delay.

Hence my call for caveats in #122.

Ironically, if my early morning brain is not playing jokes on me, that distance would be given by the integral without the square root (but only because $g_{tt} = 1/g_{rr}$) ...

 

[vanhees71]

Yes, every event in an open region of spacetime is identified by a 4-tuple of values. So I would simply say that: every event in the spacetime manifold is identified by a point in $\mathbb{R}^4$. I wouldn't characterize $\mathbb{R}^4$ further. Anything else, Cartesian or Euclidean, seems to add a flat geometry that is not physically meaningful in the coordinate space.

Does it need to have a name for some reason? Is $\mathbb{R}^4$ not sufficient for some reason?

Well, mathematically a differentiable manifold is defined by an equivalence class of complete atlasses of compatible local maps of a Hausdorff topological space to $\mathbb{R}^4$ (as a topological space with the usual topology). That's why you can define derivatives, tangent and cotangent spaces etc. by inheriting these notions from standard tensor analysis in $\mathbb{R}^4$ via the compatible maps of the atlasses to the Hausdorff topological space, making it in this way to a differentiable manifold.

Then you can go further and define more structures on this space by inheriting the corresponding constructions in $\mathbb{R}^4$. For GR you need a connection and a non-degenerate but indefinite fundamental form ("pseudo-metric") with signature (1,3) (or equivalently (3,1)) and a metric compatible connection. In standard GR in addition you assume that the connection is torsion free, and thus being uniquely determined by the usual symmetric Christoffel symbols. This makes the GR spacetime model a pseudo-Riemannian manifold.

 

[Pyter]

what'd you do with my precious square root

Sorry, I hope you've got the gist of my question anyway.

No, it doesn't. The coordinates are members of $\mathbb{R}$, but that is not the same as $E_1$ because $E_1$ implies a Euclidean metric whereas $\mathbb{R}$ does not.
[...]
Not even that is true. See above. There is no "metric of the single coordinate".

In the cited definition in post #117, the $x^n$ charts all map to $E_1$. Doesn't that mean a 1-D Euclidean space?

No, it's connected to through the metric. A radial line of coordinate increment $dr$ covers an actual physical distance of $dr / \sqrt{ 1 - 2M / r }$.

How do you get that formula? Is it the $ds$ with $dt = d\theta = d\phi = 0$?

Since $r$ is an "areal radius", a circular orbit at radius $r$ has circumference $2 \pi r$, and that is an actual physical distance, so you could, for example, obtain the radial distance between two circular orbits by measuring both their circumferences $C_1$ and $C_2$ and inverting $C = 2 \pi r$ to obtain $r_1$ and $r_2$.

So if the Earth was orbiting around a neutron star and I wanted an r value to plug into the metric, I could measure the length of Earth's orbit (how so?) and divide by $2\pi$?
But what if I'm on a spaceship and I only have a radar blip of the estimated distance from the star?

 

[Orodruin]

Doesn't that mean a 1-D Euclidean space?

”Euclidean space” includes the presumption of the existence of a metric. $\mathbb R$ does not.

How do you get that formula? Is it the ds with dt=dθ=dϕ=0?

Yes, it is the radial distance along the hypersurface of constant t-coordinate. This is often taken as the ”distance” by definition. Note that it is not what you would measure as a ”distance” if you try to measure it by bouncing a light signal off the object and multiplying the round trip time by c/2 (see my previous post) due to Shapiro delay.

But what if I'm on a spaceship and I only have a radar blip of the estimated distance from the star?

Then you will have to do the math to get r correctly.

 

[Dale]

a differentiable manifold is defined by an equivalence class of complete atlasses of compatible local maps of a Hausdorff topological space to R4 (as a topological space with the usual topology).

Right. To my understanding, calling a space “Cartesian” or “Euclidean” introduces a notion of distance that is explicitly absent in a topological space.

 

[vanhees71]

The standard topology of $\mathbb{R}^d$ is, however, induced by all $\mathrm{L}_p$-norms.

 

[Pyter]

Then you will have to do the math to get r correctly.

How would you compute r from the measured radar distance?

 

[Orodruin]

How would you compute r from the measured radar distance?

With the integral without the square root.