Quark Soup

@Didier - "But it really is: either it is hidden in the covariant derivative, or equivalently in the Christoffel symbols" YES! Exactly the point I was trying to make! If that is the case, then why is 0 (flat, Minkowski) favored over 1 (for example) for the RHS?

@Didier "acceleration isn't something that is defined on the whole manifold" This is useful. Can a manifold posses constant curvature? If so, how is the constant curvature not going to translate into constant (ubiquitous) acceleration?

@Didier Sorry, but doesn't the path depend on the tensors on the manifold? All I'm asking is "why can't I imagine a ubiquitous, non-zero acceleration field" and, if I did, what would the Geodesic Equation look like for such a manifold.

@A.ThomasYerger 'acceleration' is the second derivative of position with respect to time. Are these concepts really beyond the realm of mathematics? Let me rephrase, it appears to me that 'force' keeps on being dragged back into the conversation. I just want to talk about the mathematical properties of an arbitrary manifold with a particular vector field.

"If you equipped your spacetime with some field that exerts a force" This has nothing to do with force at all, so I'll repeat my question: Why can't I imagine a manifold with an acceleration field of $\{1, 1, 1, 1\}$ at every point. Why must we drag physics back into the conversation?

@A.ThomasYerger An acceleration tensor field is a vector field made up of accelerations. Please explain to me why that's difficult to imagine. Is there some rule of Differential Geometry that prevents me from imagining such a field on a manifold?

@A.ThomasYerger If my manifold has an acceleration tensor field of $\{1, 1, 1, 1\}$ at every point, then which equation above will give me the correct answer when I try to calculate the shortest distance between two points on this manifold?

I just realized, you don't really understand geometry, do you? You believe that there could be a curvature tensor field everywhere, but you don't believe there could be an accompanying acceleration tensor field everywhere. Happy to chat with you more when you can connect the dots.

"The first equation describes a geodesic" It describes a geodesic, it doesn't describe all geodesics. That's the point. The second equation makes no assumptions about the global acceleration of the manifold. The first equation does. That's why the OP's equations don't match:$$$$Geodesic Equation #1: assumption: $$0=\ddot{x}^{m} + \Gamma_{ij}^{m} \dot{x}^{i} \dot{x}^{j}.$$ Geodesic Equation #2: No assumption $$a^m=\ddot{x}^{m} + \Gamma_{ij}^{m} \dot{x}^{i} \dot{x}^{j}.$$

And that was the mistake that the OP made: assuming the acceleration tensor field was zero. He, like you, assumed that objects at rest remain at rest and that's why the two equations didn't reconcile. Remove that assumption and everything works.

You do set it to zero in the first equation.$$0 = \ddot{x}^{m} + \Gamma_{ij}^{m} \dot{x}^{i} \dot{x}^{j}.$$What justification do you have for this? Is there some rule of Differential Geometry that prevents the acceleration tensor field from taking on any value? Why do you assume that objects on an arbitrary manifold don't accelerate?

What justification do you have for setting $a^m=0$?

Please have a look at the formula you just pasted above (two comments above). I can't get it to compile.

Someone who has more familiarity with the Dodelson derivation answered a related question about the physics behind this integral. physics.stackexchange.com/questions/621902/… He evaluated the integral with spherical coordinates. I'm still analyzing the formula, but it seems to have the form that the author of the book uses. There's a single 'p' (momentum) term that eventually is converted into mass. That's why I'm trying to find an integral that results in a single 'p' term.

I think we agree. Thank you.

OK. I think I see the issue. Is there a way to analyze this spherically (3-sphere) such that, instead of p1, p2, p3, we have a single value for p?

Wonderful. We agree up to that point. Where does that leave us regarding a MMa version of the formula?

The source is Dodelson, Modern Cosmology, pg 61.

The formula $$Integrate[Integrate[Integrate[1, p], p], p]$$gives $\frac{p^3}{6}$ which is an analytical answer that seems to agree with the rest of the calculations on the page from which I'm quoting this formula. I'm wondering: is this answer is compatible with yours?

Why do I have to distinguish between $p$ and $p_1$, $p_2$, $p_3$? That is, why can't I assume that all three differentials - $dp_1$, $dp_2$, $dp_3$ - are simply $dp$?

"The definition of $U^0$ is $\frac{dx^0}{d\tau^0}$" That's it. That's what I was missing. Many thanks.

OK. We agree up to this point. Then what's the rational for making $U=\{1,0,0,0\}$? If we're working in SI units, then the point that I'm missing in a major way is, why isn't $U=\{c,0,0,0\}$? That's the velocity of something that's not moving in SI units.

Here's the line element with $x^0$=ct.$$d\tau^2=-c^2dt^2+\frac{a(t)^2}{1-k\space r^2}dr^2+a(t)^2r^2d\theta^2+a(t)^2r^2sin(\theta)^2d\phi^2$$The inverse metric is: $$g_{\mu\nu}= \begin{bmatrix} -\frac{1}{c^2}&0&0&0\\ 0&\frac{1-k\space r^2}{a(t)^2}&0&0\\ 0&0&\frac{1}{a(t)^2r^2}&0\\ 0&0&0&\frac{csc(\theta)^2}{a(t)^2r^2} \end{bmatrix}$$In your earlier comment, you say that $g^{00}=−1$ when $x^0=c t$. This is the part I don't follow. What have I got wrong?

Here's a reference slideshare.net/KemalAkin/cosmology-65752111 (Page 12-13) that agrees with the metric above. I'm having trouble following your math.

Are you saying that the actual formula changes based on the kind of signature used?

You're absolutely right about the sign on the formula. I transcribed that wrong bouncing between three different sources for the formula. Thank you. However, I'm not following your point about the signature. Isn't the signature of the metric coded in the inverse metric, $g^{\mu\nu}$? I shouldn't have to change this formula if the signature of my metric changes, should I?

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.

@J.Murray - I have done the calculation. The density of baryons is $1.03\times 10^{−26}\space kg\space m^{−3}$. 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.

$A$ 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.

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.

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.

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.

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.

@G.Smith - Thank you. This is progress. Now, can you point me to a place to research "Lorentzian to first order". I've got Gravitation but haven't found a discussion of 'locality to the first order'. It looks like a derivative, but a derivative of what? The metric, the Christoffel, the scale factor?

@G.Smith - I have no doubt that my understand is incorrect, but to correct it, I need to find out exactly which of my concepts is flawed. Are you saying 1.) That in my example there's a scale beyond which the particles are no longer accelerating away from each other, 2.) The $\lambda$ in Einstein's equation disappears at some point or 3.) In the language of differential geometry, 'flat' spacetime can still result in acceleration between two test particles.

@G.Smith - Yes, that's what I read in the books. How can that be if my two test particles accelerate away from each other no matter how small a scale I chose? Or are you saying that there's a limit beyond which the $\lambda$ is no longer a factor in the Einstein equation?

@G.Smith - "Lorentzian to first order in normal coordinates" I think I understand what 'normal coordinates' are, but I still don't know what you mean by 'first order'. This implies a derivative, but a derivative of which function, exactly? The metric, the scale factor, the Christoffel symbol? An example would be useful.

@G.Smith - I'm trying to develop an intuitive understanding of 'locality'. I'm trying to understand what Gravitation means by "The local geometry of spacetime is Lorentzian everywhere". From reading the comments here, it appears to mean "remove all the other influences (e.g. a galaxy, rocks, elementary particles, anything that could cause space to curve locally) and take a section of space that is small compared to the global curvature.

@ValterMoretti - Yes, I understand how that works on a sphere, but please apply that logic to the example here: No matter how small I make my experiment, the two test particles will move away from each other at the same acceleration according to Einstein's equation. How is that flat on any scale?

@JohnRennie - Thank you. That's the best definition I've seen so far. But I still have an issue with it. If the space in which my experiment takes place is curved (as in this example), but my instrument isn't sensitive enough to measure the curvature (that is, my experiment results agree with Newton's Second Law of Motion), what justification do I have to call that space Lorentzian? Unless this is some sort of Quantum Mechanics logic, my instruments don't change the reality of the geometry.

@JohnRennie - The statement from Gravitation is that a 'local' patch of spacetime is Lorentzian. I follow your argument about the Christoffel symbol vanishing at a 'point', but we're not talking about 'points', unless the definition of 'local' is something that has a zero length. I don't know how you'd even test for Lorentzian geometry if you didn't have space between your test particles.

From Gravitation, Page 21: "Statement of fact: The geometry of spacetime is locally Lorentzian everywhere" I've read this several times and I've yet to find room for ambiguity in that statement. Perhaps we have a different definition of everywhere?

@AdrianHoward - You said "The expansion of space does not affect gravitationally bound systems such as a galaxy or our solar system". The only way this statement can be valid is if you take the four-force equation:$$\Sigma F^{\mu}= m\left(\frac{d}{d\tau}U^{\mu} + \Gamma^{\mu}{}_{\nu\eta}U^{\nu}U^{\eta}\right)$$and assume that the Christoffel symbols vanish (i.e. Flat Space), leaving you with $$\Sigma F = ma$$ You're the one who implied that the local geometry of spacetime was flat.

@AdrianHoward - No evidence? How can you look at the Tully Fisher relationship and say there's no evidence? If spacetime were flat, then when you looked at galaxies, you'd see a relation between the luminosity (proxy for mass) and the product of the radius and the square of the tangential velocity. You don't. Explain to me how the failure of this prediction isn't overwhelming evidence that your model is wrong?

@AdrianHoward - As I mentioned in the question, any experiment you cite takes place in curved space. If your experiment tells you that space is flat, then your experiment is giving you the wrong answer. If your experiment shows a small acceleration, then you need to filter out the effects of the Earth, the moon, the sun, the planets and the galactic core. You seem very confident of your answer, so please show me, even a thought experiment, that removes these influences.

@AccidentalTaylorExpansion - Thank you for the illustration but I'm not trying to 'cook up' anything. I'm trying to figure out how the sum-of-forces equation works in curved space. If it's fair to say that objects at rest remain at rest in flat space, why isn't it fair to say that objects at rest accelerate in curved space?

@Dale - If I placed marble in front of me and saw it move away at a constant acceleration - and I ruled out every known force as the cause - could I reasonably conclude that I was in negatively curved space of constant curvature?

@AccidentalTaylorExpansion - "The acceleration $\frac{d^2\theta}{d\tau^2}$ on its own doesn't make much sense" Why not? If I had a planet that was positively charged, and a moon that was negatively charged, why wouldn't $\frac{d^2\theta}{d\tau^2}$ represent the acceleration due to electromagnetism while $A^{\mu}$ represent the acceleration due to the curvature of spacetime?

@AccidentalTaylorExpansion - holy crap! That connects a bunch of things. Thanks!