"I'm not talking about the escape velocity formula. I'm talking about the Newtonian gravitational potential, which is a property of a single body. This is basic physics - the potential is given by GM/R, where M is the mass of the gravitating body. It tells you what the force would be if you were to put a particle of mass m in the gravitational field (multiply by m). But the potential itself is defined as a property of a single body - and that makes sense, every mass should feel the same gravitational field, no?"
The Newtonian potential is defined as the work done per unit mass against the gravitational field. Newton's gravitational field involves the a priori interaction of two
masses. Your claim that you are not talking about escape velocity is specious. The Newtonian escape velocity expression appears in the so-called "Schwarzschild solution" whether you like it or not, as Crothers amplifies in his latest paper. Escape velocity involves two bodies: one body escapes from another body. But this is a concept that is impossible in a spacetime that is alleged to be a one-body problem (the "Schwarzschild solution"). Furthermore, "Schwarzschild's solution" is NOT Schwarzschild's solution at all. Schwarzschild's actual solution does not contain Newtonian relations and contains only one singularity and so it forbids black holes. Here is Schwarzschild's paper:
ramparts "Platitudinous drivel! My lord, we're using big words now, aren't we? You just ignore my claim that defining r so that we recover the formula for area until the very end, when you dismiss it offhand. I'd be curious about an explanation as to why there is a "real" r, and how you define it. Wouldn't the "real" definition of r have to be the areal radius as well? Otherwise the surface area of a sphere at radius r would not be 4 pi r^2, and that would just be weird."
It is platitudinous drivel. It is irrefutable that the quantity 'r' in the so-called "Schwarzschild solution" is the inverse square root of the Gaussian curvature of the spherically symmetric geodesic surface in the spatial section. The area relation follows from this, but does not yield the geometric identity of 'r' in the metric. Gaussian curvature is intrinsic to a surface. It is clear that you don;t understand the geometry. Here is a full explanation, from first principles:
And here is the relevant calculation using the Riemann tensor of the 1st kind:
Second, Minkowski spacetime has no counterpart to (-,+,-,-) and so (-,+,-,-) is unrelated to the problem.
ramparts "Sure there is - as you suggest in a bit, just switch the definitions of r and t. These things are just coordinates, you can define them however you like - the point is that there exist coordinates (in this case, switching r and t) such that the spacetime inside the event horizon is locally Minkowski."
Nonsense. The spatial section of Minkowski spacetime is a positive-definite quadratic form (ordinary Euclidean 3-space). The coordinates r, theta and phi are spacelike and must remain spacelike. Minkowski spacetime, written with signature (+,-,-,-) can't change signature to (-,+,-,-). Minkowski spacetime written with signature (+,+,+,-) can't change to (+,+,-,+). The derivation of the so-called "Schwarzschild's solution" reflects this fixed signature by construction, but the astrophysical magicians violate their own initial construction and allow the signature to change from (+,-,-,-) to (-,+,-,-), making r the timelike quantity and t the spacelike quantity, in which case all the components of the metric tensor are functions of the timelike quantity r, giving a non-static solution to a static problem. Marvelous!
Third, the change of signature to (-,+,-,-) cause the quantities t and r to exchange their roles, so that t becomes spacelike and r becomes timelike. Hence, the change to (-,+,-,-) makes all the components of the metric tensor functions of the now timelike quantity r, making the metric timelike, in direction violation of the required static condition at the outset - i.e. it produces a non-static solution to a static problem!
ramparts: "That's more or less what I just said, I think."
No it isn't.
ramparts: "That said, you raise an interesting point - the Schwarzschild solution is stationary with respect to t, but t becomes spacelike inside the horizon. I'll have to give that some thought. The one thing I will say is that the metric isn't supposed to lack dependence on the timelike variable, it's supposed to lack dependence on t. Hmm, interesting point though."
What? The solution is supposed to be static - in other words all the components of the metric tensor must be functions of a spacelike quantity, NOT of a timelike quantity: otherwise it is not a static solution. In the so-called "Schwarzschild solution r is spacelike and t timelike. But for 0 <= r < 2m, r becomes timelike and t becomes spacelike. The metric is no longer static. It does not matter what label you use of the spacelike and timelike quantities, what is important is what the components of the metric tensor are. For a static solution all the components of the metric tensor must be functions of a spacelike quantity, NOT a timelike quantity.