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.
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.)
Of course not. See above.
You are misstating this. See below.
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.
I have no idea what you are talking about here. Perhaps rethinking your understanding in the light of the above will help.
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.
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).
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.
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.
That isn’t really a problem. Many different mathematical solutions to many different differential equations can describe the same physics.
You are right. I forgot that in standard Schwarzschild coordinates ##r## becomes the time-like coordinate for ##r<r_{\text{S}}##.
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?
According to the manifold's definition I had posted, which I repeat here for convenience:
Since this distance has no physical meaning I would not call that space “Cartesian”.
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.
You could do that if you had a chart that covers the whole manifold otherwise you cannot.
That is not what a metric is!
This is also not true, even if you ignore the imprecisions you want ##ds##.
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.
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.
"Rectilinear" in the sense that you could consider each one a coordinate axis in an ##E_n## space.
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?
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.
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.
Exactly, for instance in situations where the curvature gradient is very steep, like in proximity of a neutron star.
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.
Strictly speaking, ##r## is a coordinate, not a parameter.
Correct, you have to compute ##r## from the observed distance. (Although in the case of the solar system the correction is very small.)
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.
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.)
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.
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##.
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
Yes.
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 }##.
With a square root over ##g_{rr}##, as @ergospherical says, yes.
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##.
If you're doing the integral that has been described, you don't, at least not in terms of measured radial distance.
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".
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.
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.
Sorry, I hope you've got the gist of my question anyway.
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?
How do you get that formula? Is it the ##ds## with ##dt = d\theta = d\phi = 0##?
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##?
”Euclidean space” includes the presumption of the existence of a metric. ##\mathbb R## does not.
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.
Then you will have to do the math to get r correctly.
Right. To my understanding, calling a space “Cartesian” or “Euclidean” introduces a notion of distance that is explicitly absent in a topological space.
How would you compute r from the measured radar distance?
With the integral without the square root.