Jonathan Scott said:
In Schwarzschild's original model, where the physical origin's location in terms of the exterior radial coordinate is effectively r=2Gm/c2 (although to avoid the infinities this has to be taken as an extrapolated limit), this does not arise.
Again, keep in mind that what you are calling the spatial "origin" here is not a point in space. We have already covered that.
Secondly, merely doing a coordinate transformation does NOT change the physical results, as the predictions of GR are independent of the choice of coordinates. The only way you can change the situation is to artificially add in a boundary to spacetime at the event horizon.
And lastly,
even if we put in this boundary infalling matter will STILL hit this boundary in finite time. So even in this contrived case, this following comment is wrong:
Jonathan Scott said:
That's not relevant if you hit the mass before you reach where the event horizon would have been.
Trying to narrow down where the misunderstanding is occurring, I think this next comment is quite illuminating.
Jonathan Scott said:
All of this is about just one thing - a constant of integration in Schwarzschild's solution, which Schwarzschild explicitly assumed to be one thing, indicating that the mass is physically located at one point, and Hilbert explicitly changed to a different value
Yes, regarding the integration constant statement. But the conclusions you draw from this are incorrect. You seem to be missing some very important points here.
#1) There is no freedom to choose this integration constant. This integration constant comes about from solving the vacuum equation
outside the mass. Once you consider the source equations (ie. the non-vacuum region), this FIXES the integration constant. You are instead trying to argue we solve just the vacuum region and are
completely free to choose the boundary values. You are incorrect here.
Do you understand this?
This is why I keep bringing up examples of starting with a non-collapsed star (so we don't have to deal with arguments over singularities), and then allow it to evolve according to GR and watch it collapse. The result is NOT what you claim. Because this explicitly fixes the boundary term to something physical, it makes it easier to see where your error is. You keep ignoring these explicit examples.#2) If you want to consider his choice in integration constant, (or equivalently here, his choice of coordinates) you need to use \infty >r>-2GM/c^2. Otherwise, even though he started with a continuous coordinate chart (x,y,z,t), he somehow ended up with the "origin" not referring to a single point in space.
I am having trouble disentangling what mistake Schwarzschild made versus what is actually just a misunderstanding in your interpretation of his solution. So to help this along, can you please answer the following (hopefully this will help me and others understand your point of view better):
Post here explicitly what you are calling Schwarzschild's solution, using his coordinate chart, and specifying the valid ranges for each coordinate variable.
Jonathan Scott said:
People who are familiar with Hilbert's version obviously find it difficult to distinguish between what parts of GR derive from this additional assumption and what parts are based on Einstein's Field Equations. For example, with Schwarzschild's position, the fact that the "point" has finite area is not relevant, because a finite mass could not be compressed to a point and that area is merely a hypothetical limit.
I must stress that I'm not sure either way about this. I find Schwarzschild's assumption more plausible and comfortable than Hilbert's, and I don't find any obvious inconsistency in it now
There is no freedom for another "assumption" here. Again, what you feel is a freedom to set this integration constant, is actually completely fixed when considering the non-vacuum (ie. source term) parts. That is why I keep brining up the explicit examples of calculating a collapsing star.
For an analogy. Consider simple electrodynamics. In vacuum we have:
\nabla^2 V = 0
Outside a static spherical charge distribution, due to the spherical symmetry we get:
\frac{1}{r^2} \frac{\partial}{\partial r} (r^2 \frac{\partial}{\partial r} V(r)) = 0
We can solve this and find generally that
\frac{\partial}{\partial r} (r^2 \frac{\partial}{\partial r} V(r)) = 0
\frac{\partial}{\partial r} V(r) = C_1 / r^2
V(r) = -C_1 / r + C_2
Note we can make a coordinate transformation (scaling and/or translating r):
V(r') = a_1 / (r' + a_0) + C_2
If we demand the boundary condition V -> 0 as r-> infinity, we get
V(r') = a_1 / (r' + a_0)
This satisfies the equations outside the spherically symmetric charge distribution.
You seem to think we are free to choose a_0 and a_1 however we want, and this amount to a mere assumption. This is not the case. Considering the source equations, provides one constraint and in effect fixes one constant in terms of the other. The remaining constant arises merely due to coordinate choice ... in which case we NEED to consider r' < 0 if we choose a_0 > 0, to cover all space.
Once the total charge in the distribution is specified, there is NO freedom "to make another assumption". And just like in the GR situation, the internal solution is irrelevant here (merely the total "charge"/mass is needed to fix the "integration constant" / boundary term).
Does this simple E&M analogy help (hopefully)?
Birkhoff's theorem says there is a
unique physical solution in the GR case we are considering. You already agreed to this, but paradoxically keep claiming there is another physical solution. I'm trying my best, but this makes it impossible to really understand your claims.