From Einstein's own words, we can infer that identifying gravity with the coordinate-dependent connection is less than ideal. It seems to me that Einstein probably would have preferred to use the more modern, coordinate-independent tensorial notion had the modern tensor notation been available at the time. Einstein used explicit connections because that was the notation of his time. This may be a case of notion following notation.
I don't think he would. Though the concept of connection itself was not explicated until a few years after GTR (Levi-Civita, et. al.), its identification with the gravitational field not only jives with his early publications on GTR, but later throughout, e.g., in "The Meaning of Relativity". He definitely saw the gravitational field as explicitly coordinate-dependent, that even for a rotating frame in Minwkoski spacetime, the proper interpretation of the "inertial forces" (aka "fictitious forces") like the centrifugal or Coriolis is that they represent a genuine gravitational field. For example, he's a excerpt of a letter copied from "
Einstein from 'B' to 'Z'", in which he objects to von Laue's identification of gravity with curvature:
Einstein's to von Laue said:
It is true that case the Riklm vanishes, so that one might say: "There is no gravitational field present." However, what characterizes the existence of a gravitational field from the empirical standpoint is the non-vanishing of the Γiik, not the nonvanishing of the Riklm. If one does not think in such intuitive ways, one cannot comprehend why something like curvature should have anything to do with gravitational in the first place. In any case no reasonable man would have hit upon something in that way. The key to understanding of the equality of inertial and gravitational mass would have been missing.
It's true that many authors on books on relativity didn't like Einstein's view because the connection coefficients do not form a tensorial, but he was definitely aware of such criticisms (some of which came before even 1920s), and still maintained that it is only necessary for (dx
α + Γ
αμνdx
μdx
ν)/dλ² to transform covariantly.
Now, I'm certainly not an expert of relativity (probably Sol is the only in this thread who is, unless I'm forgetting someone), but to me Einstein's view makes the a lot of sense. The gravitational field is "fictitious" in the same sense as centrifugal and Coriolis forces are; GTR says it's not really a force. Freefalling frames are inertial, and 'inertial' means constant velocity. In GTR generalized as an affine geodesic: a curve such that parallel-transporting the four-velocity along itself does not change it. So a choice of connection coefficients seems the most natural. Sure, they don't form a single tensor, but that's also why they can be locally transformed away, which is an important property for the equivalence principle.
However, if there's a modern view, it's probably best expressed by MTW:
MTW §16.5 said:
"I know how to measure the electromagnetic field using test charges; what is the analogous procedure for measuring the gravitational field?" This question has, at the same time, many answers and none.
It has no answers because nowhere has a precise definition of the term "gravitational field" been given--nor will one be given. Many different mathematical entities are associated with gravitation: the metric, the Riemann curvature tensor, the Ricci curvature tensor, the curvature scalar, the covariant derivative, the connection coefficients, etc. Each of these plays an important role in gravitational theory, and non is so much more central than the others that it deserves the name "gravitational field." Thus it is that throughout this book the terms "gravitational field" and "gravity" refer in a vague, collective sort of way to all of these entities. Another, equivalent term used for them is the "geometry of spacetime."
And perhaps reducing the expression to essentially "whichever one makes the most sense in context" might be better. But beyond with the historical issues and interpretational nitpicks, I agree with you.
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There’s a distinction to make. If you fall into a supermassive black hole, the tidal forces are weak so you don’t get spaghettified. The gravitational field is so nearly uniform that locally it looks uniform. Think of the bowling ball analogy. The gradient at your feet is so similar to the gradient at your head, that there’s no curvature between the two gradients. Just like a man falling off a ladder, you don’t feel a tug at your feet, but you still fall in, and passing light still bends. In extremis you’ve still got curved spacetime even without Reimann curvature.
Making a black hole supermassive is completely equivalent to making yourself smaller. Of course you experience less spaghettification if you're small compared to the black hole. It's just another instance of "if you look close enough, it looks more and more flat."
I think I read somewhere that the gedanken situation for this is falling towards a massive plane.
Makes sense. Constant acceleration should be what's expected based on Newtonian gravity, and in GTR, you can make a patch near the horizon of a Schwarzschild black hole that's closely approximated by the Rindler chart of Minwkoski spacetime, i.e., constant acceleration. On smaller and smaller scale (or black hole larger compared to you), that approximation would become better and better.
I don't understand why you think this is any sort of problem that's being overlooked, or how you resolve it in light of:
There is no such thing as a uniform gravitational field.
...
It's very simple:
if you're adopting Einstein's own interpretation, as you've implicitly claimed to have, then the components of the gravitational field are the connection coefficients. He says so explicitly in his paper on GTR (referenced and translated above) and affirms it several times. The connection coefficients determine inertial motion. So it follows that gravity is (physically) inertia and (mathematically) the connection. Do particles in flat spacetime have inertia? Yes. Does flat spacetime have a connection? Yes.
If you are surrounded by homogeneous space you don't fall down. If the space remains homogeneous but is expanding you can say that spacetime is curved, but there's no actual gravity in the space. The gravitational potential is the same everywhere. It varies over time, but it's always the same above you and below you. You don’t fall toward the centre of the universe.
At this point I think
W.D.Clinger's criticisms are relevant. For one, you're apparently confusing spatial curvature with spacetime curvature.
If space is expanding you can say it's inhomogeneous over time, and that the spacetime is curved. But it's like the whole of space is like the space you go through when you zoom through the gap between two stars. There's gravitational potential in spades. But you don't fall down. Your path is straight.
You're just restating the fact that one can always find a locally inertial frame. That's true. And if you're far from gravitating bodies, it becomes better and better. So what?
It's not just compatible with GTR, it's downright required by it. (And actually, something essentially the same was proven by Newton himself.)
Divide up the energy of those two stars and smear it evenly thoughout a smaller universe. There's gravitational potential in spades. But you don't fall down, and nor do galaxies.
Even in a Newtonian universe, a uniform fluid of stars would gravitate. That's obvious from Newton's shell theorem: for any given (arbitrary!) center, a bit of matter at a distance r would feel a gravitational force determined by the mass enclosed by a sphere of that radius. So an initially static configuration would collapse, and since the center is arbitrary, completely uniformly throughout space: every point sees everything else collapsing towards it. Time-reverse it or give it an outward initial velocity, and you get uniform, homogeneous expansion governed by gravity instead.
This very closely mimics how it happens in some FLRW cosmologies in GTR.