I'm not mangling anything. The "force" of gravity relates to the first derivative, the tidal force relates to the second derivative. The former is in essence the slope, the latter the change in slope - the curvature.
When written out long-hand, the geodesic equations can be written in terms of the metric, the partial derivatives of the metric (with respect to coordinates), the first and second derivatives of the coordinates of the geodesic (in terms of the affine parameter). Since this is sufficient to precisely describe the bending of starlight, the shape and periods of orbits, the advance of the perihelion of Mercury, it would seem that first derivatives of the metric do contain information corresponding to the Newtonian view of gravity as a force. But you can't equate the two, because the metric may have non-zero partial derivatives and still not have anything that Newton would call a force. One simple example has already been given. Another example is:
$$ds^2 = -du^2 \; + \; 2 du \, dv \; + \; 2 a \sin^2 \theta \, dv \, d\phi \; + \; ( v^2 + a^2 \cos^2 \theta) d\theta^2 \; + \; ( v^2 + a^2 ) \sin^2 \theta \, d \phi $$
So equating "force" with coordinate derivatives of the metric can't be right with the general freedom to choose coordinates granted by Einstein in section one of his 1914 document cited previously. Yet there is a relation, it's just bound up in choice of coordinates so that we can always choose orthonormal Cartesian coordinates where the "force" of gravity vanishes at the space-time origin. These are called free-fall coordinates or a local Lorentz frame.
Tidal force, as we have seen, relates to the Riemann curvature tensor, which is composed from the metric, and its partial derivatives of first and second order. Since the spacetime described by the above metric has Riemann curvature tensor everywhere uniformly zero, it describes Minkowski geometry, which is a solution to the Einstein field equations. So even though the coordinates are peculiar, the spacetime is not curved. On the other hand, if spacetime itself is curved, then we can't make the Riemann curvature tensor go away by choice of coordinates, but there are spacetimes and choices of coordinates where the second derivatives were all zero but the first derivatives were not. Example, if $$g_{\mu\nu}$$ is diagonal and has no second derivatives, then
$$R_{0101} = g^{00} ( \partial_{1} g_{00} \partial_{1} g_{00} + \partial_{0} g_{00} \partial_{0} g_{11} ) + g^{11} ( \partial_{0} g_{11} \partial_{0} g_{11} + \partial_{1} g_{00} \partial_{1} g_{11} ) - g^{22} \partial_{2} g_{00} \partial_{2} g_{11} - g^{33} \partial_{3} g_{00} \partial_{3} g_{11} $$
which need not be zero. So second-order effects like tidal forces are not solely associated with second derivatives.
What then should we make of your claim that "slope" is responsible for first-order gravitational effects? Such a claim is either based on a particular and parochial choice of a coordinate system, in which case you have the burden to show you have not conflated coordinate choice effects with physics, or you have a coordinate-free way to describe how the space-time metric changes with position. Einstein introduces the Christoffel symbols from late 19th century mathematics to build a type of derivative on manifolds that doesn't include effects from choice of coordinates. But unfortunately for your position, $$\nabla_{\alpha} g_{\beta\gamma} = 0$$, so there is no "slope" of the metric in evidence. Similar statements hold in more modern mathematics of GR. These same Christoffel symbols are used to describe geodesics which remain unchanged under change of coordinates. And geodesics are the topic of this thread.
So obviously, if there is no coordinate-free way to get a non-zero "slope" from the metric, then there can be no non-zero second derivative. So what is the Riemman curvature tensor in coordinate-free language? It is the commutator of the second coordinate-free derivative of smooth vector fields.
$$R_{\beta\gamma\delta}^{\alpha} u^{\beta} v^{\gamma} w^{\delta} = v^{\gamma} w^{\delta} \left( \nabla_{\gamma} \nabla_{\delta} u^{\beta} - \nabla_{\delta} \nabla_{\gamma} u^{\beta} \right) $$.
In "slope" you have proposed an unworkable scheme and one that seems based on mere analogy with the rubber sheet model.
Nor am I misteaching anything, so please withdraw your warning.
I would disagree, but PMs are the required format for discussion individual moderator actions.
The illustration above gets the point home. A light beam does not curve because "it follows the spacetime curvature".
The illustration is not to scale, labeled, or derived from theoretical principle. Also, physics is not about "because" principles but about the behavior of observable phenomena; physics cannot distinguish two "because" reasons that lead to predictions of the same behavior. But since the geodesic principle applies both to massive particles and light, I choose it for parsimony. Also, 100 years of GR including Einstein's 1914 paper. Moreover, you have not distinguished your particular views from the shape of a null geodesic. The only way you can support your view is by calculating both, showing all your assumptions and approximations, finding a difference between the two and finding that physical observation confirms you view over mine. The only way you can support your view is physics is by showing it is
better physics.
Not when you don't understand why light curves and end up dismissing Einstein.
Calculation and unification go a long way towards demonstrating I have a good understanding of the behavior of light in a gravitational field and since my textbook doesn't contradict Einstein (except for his typos, which were many) I don't think the thousands of students of GR like myself have dismissed Einstein. This post and your warnings, are about your behavior, not Einstein's.
As I said in
this thread, it’s
no good using gravity to explain gravity, that’s circular. But the [rubber-sheet] picture isn’t totally wrong. Imagine you’ve placed a whole lot of parallel-mirror light-clocks in an equatorial slice through the Earth and the surrounding space. When you plot all the clock rates, your plot resembles the rubber-sheet picture because clocks go slower when they’re lower.
So you don't care if the rubber sheet model fails to reproduce the dynamics of even Newtonian gravity, you like it because it has axial symmetry and has increasing slope nearer the center which makes it vaguely resembles the Newtonian potential and a plot of clock rates at fixed spatial coordinates in the Schwarzschild coordinates and the related, but not identical, embedding diagram. But the only way you know this is by graphing the Newtonian potential or making GR calculations from first principles. So it's a cartoon like your picture of light bending. It explains nothing because it is wrong in principle and the details. Further, the rubber sheet requires the phenomenon of gravity to work in the first place.
Then the curvature you can see relates to Riemann curvature which relates to curved spacetime.
Do you mean the visible curvature of the 2-D surface of the rubber sheet has its own Riemann curvature which you can compare and contrast with the Riemann curvature of 4-dimensional space-time? You certainly can't equate them. You might even introduce the concept of Gaussian curvature. But that's only useful as a toy prerequisite in teaching preliminaries to GR. If that was enough, wouldn't Einstein have stopped there rather than to master tensor calculus on curved manifolds?
Even if you built a toy that gave exactly the right trajectories in the limit of low velocities, had exactly the right profile as your plot of clock rates or was an exactly physical model of the embedding diagram, such a toy would be based on the Schwarzschild geometry, and there is more to GR than the Schwarzschild geometry.
And yes, you measured those clock rates, so yes, it’s a curvature in your metric... How about if I posted a fresh version of that and you tried to explain where it was wrong? You and I both know you won't be able to, and that instead you'll spout platitudes like "cherry picking" or "out of context", and claim that I'm misteaching even though I'm referring to what Einstein said.
In addition to confusing "spacetime curvature," a property of spacetime, with "curved spacetime", spacetime with a non-zero value of that property, I believe you are misusing "platitudes". Just because being warned by moderation and given specific instances of where you take tiny quotes out of context doesn't provoke thought in you doesn't mean they were done without deliberation and causing others to think about them.