@J.Delaney You did not address your response to me, so I was not notified. On radius, your are arguing a totally different point from the one in my original comment. Sure any coordinate can be defined in different ways, this does not affect physics. When you define the r coordinate as a reduced circumference (the textbook definition you prefer), it no longer represents the actual radial interval.

There is nothing wrong with doing this, as long as you understand the consequences, but many people don’t. For example, it is trivial to calculate and is well known anyway that the radial interval inside the horizon is timelike. This means that the radial “distance” from the horizon to the origin is not a distance in space, but a period of time. The spatial radial distance between the horizon and the origin is zero.

This is a fact, you can do this math yourself. So once you are at the horizon, you movement toward the origin is a movement in time or, in other words, just waiting. You can still move in space, but not in the radial direction (in the Schwarzschild coordinates). So, while your argument on defining the radial coordinate is neither wrong nor new, it is irrelevant to my comment.

The spatial radial distance from the horizon to the origin is zero mathematically, so space wise the horizon is at the origin. One could argue that this depends on the choice of coordinates, but so is the definition of what “radial” means, and in spherically symmetric and static Schwarzschild coordinates the spatial radius of the horizon is zero.

On the horizon being a boundary. I will try to explain, but common misconceptions about the horizon are bigger than the amount of time I have on dispelling them. So if you don’t get my explanation and want to go with the flow, it is up to you.

Since we are working in the Schwarzschild coordinates, notice that the horizon does not exist. It is a future null infinity, as observed from outside. Also note that the Schwarzschild original paper did not imply the existence of the inner spacetime “inside” the horizon.

Yes, the final formula works both inside and outside, but these are two different regions disconnected at the horizon. Furthermore, the inner spacetime is not static, it is a spatial spherinder (the 3D hypersurface of a 4D hypercylinder of the spherical type - as opposed to the cubinder) rapidly collapsing in time in the radial direction.

So the boundary conditions for the Schwarzschild solution are asymptotically at the horizon for both inner and outer regions. These regions are “stitched” together at the horizon by an arbitrary assumption that the horizon is crossable. The fact that you can cross the horizon does not follow from the solution, but is an assumption.

There is a lot of arguments claiming the validity of this assumption, but also serious quantum arguments against it (the information paradox, firewall, etc.). In any case its validity is irrelevant here, because, even if it is valid, the horizon still is a boundary condition for the so,Union. When you say “spherically symmetric”, it refers specifically to the horizon.

Finally, just a quick note on the naive arguments in elementary textbooks regarding coordinate transformations supposedly handling the horizon issue. These arguments are flowed simply because no mathematically valid (differentiable with a differentiable inverse) coordinate transformation exists that removes the horizon singularity. This is self evident and well known.

As a side note, your statement that the source of gravity is the stress-energy tensor is a common misconception. This tensor is in the equations, but it can be zero in gravity. It is zero anywhere in vacuum, so it is responsible only for how gravity changes inside a mass. Specifically, it adds the Ricci curvature breaking the conservation of the infinitesimal spacetime volume.

In physical terms this means that, while the time dilation always equals the length contraction in vacuum, this equality is broken inside a material medium by its stress-energy tensor. The Weyl curvature actually responsible for gravity is caused by the boundary conditions in vacuum.

*Typo: … for the solution.

* flawed

@Safesphere: GR is a well posed initial value problem in 3+1 dimensions. Given initial conditions specified on an appropriate (Cauchy) time slice the future spacetime is completely determined by the equations of motion, i.e. the Einstein equation. In particular, boundary conditions do not play a role.

The Schwarzschild solution is somewhat pathological in that its Cauchy slices go through (at least) regions I and III. This is a silly consequence for taking an eternal black hole. It is physically more sensible look at a solution that includes formation of the black hole, like the (infalling) Vaidya metric. This has nice well defined Cauchy slices which provide initial conditions, that will form an horizon in their future.

lastly, as also mentioned in TimRias's comment, you can define boundary conditions on any Cauchy surface you like. If you want, you can define a coordinate system where the surface of the earth is at r=0, and similarly claim that this is the "source" of earth's gravity. But this will be meaningless precisely because this is an arbitrary choice.

@safesphere

@TimRias GR is a problem of differential equations with boundary conditions just like anything else in physics. It can indeed be seen as an initial value problem under physically realistic conditions, as you mentioned, but not necessarily. For example, again as you mentioned, the Schwarzschild solution does not fit this definition, but it still is the boundary conditions problem. And specifically this boundary condition is the spherical symmetry at the horizon (asymptotically).

@J.Delaney So on the r coordinate your objection is the "conventional definition". You should be aware that in math $\equiv$ does not mean "equals", but means "defined as", so I have explicitly defined what I meant. Plus I stated time and again that I refer to the "spacelike" or "spatial" radius. I did not use the "conventional" definition of the radius and I clearly stated it. So the problem you see is not in my comment, but in your misunderstanding of it.

@J.Delaney Your second error is in the definition of "conventional". On this site, "conventional" refers to things like that gravity is caused by spacetime curvature rather than by magical unicorns. Here "conventional" does not mean the Spanish Inquisition of having to adhere to any specific "definitions" or widespread misconception. So respectfully, if you want to police this site, please feel free to run in moderator elections, but otherwise please refrain from opinion-based objections.

@J.Delaney Per the Birkhoff theorem, any aligned sphere in the spherically symmetric solution is the source of gravity outside this sphere. The smallest sphere in the Schwarzschild solution is the horizon, because its spatial radius is zero. So the horizon is the source of gravity outside.

@J.Delaney On the inside, the horizon (asymptotically) is the Cauchy surface and thus is the source of gravity there.

@safesphere As I said, you are free to use whatever definitions you want as long as you explicitly define them, but you didn't. Saying that $r^2\equiv x^2+y^2+z^2$ certainly doesn't mean anything because it also holds for the conventional coordinates. Also, "spacelike" can also refer to the "t" coordinate inside the horizon.

If you think there is a better definition for a "source" of gravity other than the conventional one, you are free suggest it. I doubt if you can come up with a consistent definition, but you are free to try (nothing you said so far can be regraded as a general definition)

@J.Delaney There is also a deep physical meaning for why the horizon is the source of gravity in Schwarzschild and why astrophysical black holes (non-vacuum) are physically indistinguishable form Schwarzschild black holes ("vacuum"), assuming no rotation (or vacuum Kerr with rotation). In other words, what happens to matter at the horizon, but this is beyond the scope of this conversation.

@J.Delaney Yes, I used the Schwarzschild definition with r=0 at the horizon and I clearly stated this. So again, you are free to limit your thinking by the shores of undergraduate textbooks, but please stop imposing your opinions on others. On the "source of gravity", perhaps a better term is the "cause of gravity" aligned with the Cauchy surface approach. As I stated in my original comment, "The singularity is the future result of gravity, not its cause."

I believe this matter is settled since it is about definitions, not physics. Physics is conventional GR, definitions of coordinates are a personal choice and its preference is opinion based.

PS. On t being spacelike inside, yes, but it is not radial. See my diagram here: math.stackexchange.com/questions/2929400

It is more accurate to say that in GR all coordinates systems are equally valid. In any case, yes it is about definitions, so I'll take it as an admission on you part that your original comment on the Schwarzschild coordinates was wrong, even if you don't want to explicitly say it. That's settles the issue.