So we apparently assume that, locally, we can just forget about the inertial acceleration - that the Christoffel symbols vanish because there is no detectable curvature - and evaluate the force as $F=ma$.
No, that's not true. Whether the Christoffel symbols vanish at a point is purely a matter...
Here's a link to Sofue's data: ioa.s.u-tokyo.ac.jp/~sofue/h-rot.htm Sofue and his data are referenced frequently in studies of galaxy rotation curves. In this paper - arxiv.org/abs/1510.05752 - he describes his methodology for constructing the rotation curves. Perhaps you can direct me to the part of this analysis that considers the global curvature of spacetime. In fact, I'd be very grateful it if you can find any construction of galaxy rotation curves that employ the global curvature in the model. You need to explain your last comment.
@GluonSoup Galaxy rotation curves are measurements. To construct them, one simply uses spectroscopic data to calculate the velocities of the visible matter as a function of radial distance from the center. At no point does use any particular model for spacetime. Putting that aside, if one computes the average density of a galaxy (~$10^{-20}$ kg/m$^3$ for the Milky Way), one can estimate the average curvature radius produced.
@GluonSoup For the Milky Way, it is about 150 times the size of the galaxy itself. As you will learn if you continue your study of GR, this means that the Newtonian gravity approximation is extremely good. While it's true that there would be small corrections from GR, they would be far too small to have a qualitative impact on the expected rotation curves in the absence of a dark matter halo.
The issue is: what is the global curvature of the spacetime that the the galaxy occupies. In every study I've found, they assume that this region of space is flat before they added the stars, gas and other materials. I'm asking you to please find me a study where the global curvature of spacetime is considered when analyzing the dynamics of a galaxy.
While it's true that there would be small corrections from GR, they would be far too small to have a qualitative Again, I can prove to you that this is assumption is wrong and the primary reason why LCDM doesn't agree with itself anymore.
@GluonSoup If you compute the same calculation with the average density of e.g. the local group (~$10^{-26}$ kg/m$^3$), you find a curvature radius of 150,000 times the size of the Milky way galaxy. If you have a sphere of radius 150 km, then a 1 m radius disk drawn on the surface is flat to a ludicrously good approximation; this is essentially the same approximation you make when you assume that the spacetime in which the Milky way sits would otherwise be flat.
flat to a ludicrously good approximation That's the problem, it is not absolutely flat. It only takes a tiny acceleration to explain galaxy rotation curves without Dark Matter.
I'm unable to. When you don't have a Ph. D. after your name or an .edu after your email address, the papers are rejected summarily. So how about you do the right thing? Take that Sofue data, remove the Dark Matter assumptions, and use the formula $$F=m(a+A)$$Where the acceleration due to curvature is $4.4\times 10^{-11}\space m\space s^{-2}$. Let me know what you find or retract that last statement in your answer.
@GluonSoup Where does that $A$ come from? In what direction is it oriented? What coordinate system are you using? What is the metric tensor you're considering here?
$A$ is derived above in the original post on this thread. It's the acceleration due to curvature (fictitious, I believe it's called). If there's a problem with the derivation, then please correct it. The actual value is extracted from SNe Ia and the Tully-Fisher Relationship, but you don't need to know that. If you just assume that $A$ is non-zero and that Dark Matter doesn't exist, then you can extract a value of $4.4\times 10^{-11}\space m\space s^{-2}$ from galaxy rotation curve data as well.
@GluonSoup Assuming you have done all the calculations right, take the Christoffel symbols you've derived and use them to compute the Einstein tensor. Via Einstein's equations, this translates to an energy/momentum density which is present at every point in spacetime. If you compare that energy/momentum density to the density of visible matter in the galaxy, you find that the former vastly outweighs the latter. This is the entire point of introducing dark matter.
@J.Murray - I have done the calculation. The density of baryons is $1.03\times 10^{ā26}\space kg\space m^{ā3}$. Using Friedman's equations, that gives us a radius of curvature of $17.27 Gpc$. The ubiquitous curving of spacetime eliminates the need for Dark Matter. Everything balances if you drop your unsupportable belief that the universe is globally flat. Go do the calculation. A high school kid could do it.
@GluonSoup You are missing the point - spacetime curvature corresponds to the presence of matter and energy via Einstein's equations. The density and distribution of matter and energy which is required to fit galactic rotation curves is not accounted for by visible matter, which implies either that Einstein's equations are not correct or that there is some distribution of matter and energy which is not visible.
@GluonSoup I suppose Iām not smart enough, then. Best of luck.
which implies either that Einstein's equations are not correct or that there is some distribution of matter and energy which is not visible Why are you incapable of acknowledging the third option: that the global geometry of spacetime around a galaxy is curved? I can prove this to an undergrad with an open mind, but it's impossible to prove this to an accomplished scientist who's absolutely convinced of things that just aren't true.
Discussion on answer by J. Murray: Ho…
Discussion on answer by J. Murray: Ho…