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Yes, if your galaxy is embedded in an infinite series of infinitely thin spherical shells of the appropriate positive and negative masses spaced infinitely clos

Yes, if your galaxy is embedded in an infinite series of infinitely thin spherical shells of the appropriate positive and negative masses spaced infinitely close together, then you can make the rotation curve be anything you want it to be without adding any overall net mass. I checked, and the publish date is not in fact the first of April.

m_dupont 2y ago
Agreed. This paper is pretty much pointless. Its extremely implausible, putting it generously, that such fine-tuned structures could even form. They mention at the end "this paper does not attempt to tackle the problem of structure formation" but that feels like a colossal understatement. I've studied physics myself, and I understand that sometimes people toy around with implausible theories solely for the sake of it but .... it seems like these peoples brainpower could be much better spent elsewhere.
- m_dupont 2y ago
  Also, strictly speaking these aren't topological defects. They are massless spherical shells, i.e. a sphere embedded in 3d space and thus can be continuously deformed to a point by sending the radius to zero. Spheres are literally the second homotopy group and the homotopy group of flat space is zero. Q.E.D
  - barfbagginus 2y ago
    The sense in which they are defects is that on the shell, the density function is a distribution and takes no real value. Distributions are singularities/non-functions that we get when we take certain limits of ordinary functions, or solve certain differential equations with generalized functions. The distribution used works roughly as if we overlapped two equal positive and negative mass shells. But there are some extra details that ensure there is a net inward force for matter situated on the final shell. We'd have to actually work the math to really understand why that force appears, without hand waving.
    - m_dupont 2y ago
      I don't disagree with what you've said about distributions, however I don't think that the fact that these shells are created from dirac delta functions is sufficient to call them topological defects. Topological defects are solutions to the underlying physical equations that are of a different homotopy class to the vacuum, and these simply aren't in a different homotopy class, as I can smoothly deform them to zero, by sending the radius to zero or sending alpha to zero. Argued another way: a point charge can be modelled as a dirac delta charge distribution, but nobody would argue that a point charge is a topological defect
      - barfbagginus 2y ago
        Okay, fair point, I was not understanding that topological defects are points in solution space that are not path connected to the vacuum solution. I'm learning as I'm going! Now I also realize that the paper seems to say that the both the ordinary dirac shell solution and their modified shell are TDs, without proving it. I'd like to work up to proving whether the collapsing modified shell really does homotopy into into a point and then fade away into the vacuum. But first, I'm struggling with > nobody would argue that a point charge is a topological defect It's actually not clear to me that the point mass is not a TD! Let's try to write the homotopy sending a point mass Mδ₀ to the vacuum solution 0. Let t vary from 0 to 1. Then a possible homotopy is h(t) = (1-t)Mδ₀ This gives us h(0) = Mδ₀ h(1) = 0 The problem is that I don't know how to prove continuity of h. First, I don't know how to compute even the continuity of neighboring Delta functions for t < 1. But that feels intuitively like it should be continuous. On the other hand, I REALLY don't know how to prove continuity at t = 0, since the function seems to spontaneously collapse from a distribution with a kind of pseudoinfinite value at the origin, into a regular function with the value 0 at the origin. Using the notation of inner products and test functions, can we prove that it's continuous both for t < 1 and t = 1? I know that's a bit more technical than we usually get here on HN! I truly appreciate the help!
        m_dupont 2y ago
        Okay we are are getting a little lost in definitions here, but nonetheless. You can solve the above by remembering that the dirac delta is the limit of a series of functions. If you take your delta to be lim a -> 0 N(0, a) where N() is the normal distribution, then you can see that in your above equation, you then have two limits. lim t -> 0 and lim a -> 0. By swapping the order of the two (which is a dubious operation), you can send t to zero first then a to zero, and the result is zero. So in one way, it can be deformed to zero, in another way it can't, because it's 0 times infinity . However, the thing to focus on is that dirac deltas aren't actually valid points in the solution space of partial differential equations. They aren't functions, and they aren't actually physically real. Come to think of it, that would probably exclude them from being TDs a-priori. Because a TD must be a solution to an underlying physical equation, and that solution must be deformable to zero. But if it's not a solution to a PDE (because it doesn't live in any valid hilbert space), then it can't be a TD.
0xWTF 2y ago
I think this paper is an effect of there being not much to do in theoretical physics these days. I mean, there's plenty of work that can be done, but I think it was Feynman who said all physics that could ever affect humanity was described by the 1963. Since the Higgs was discovered in 2012, we've covered pretty much everything in the Standard Model. I'm sure there's some intellectually interesting problems left to solve, but it's hard to believe they'll affect us. Nevertheless, societies still need physicists, and those physicists, by the nature of how incredibly challenging the "industrial" problems are (e.g. designing next generation fusion weapons), need challenging problems. Indeed, they crave them. And so you're sort of stuck solving progressively more esoteric and even contrived problems.
- aebtebeten 2y ago
  In https://news.ycombinator.com/item?id=40572593 https://news.ycombinator.com/item?id=40572593 Arnold, his tongue in cheek (and channelling his inner GJC), suggests a tripartite division of maths according to the industrial form which supports the pure researches: - Celestial Mechanics (support: the descendants of von Braun): dynamical systems, linear algebra, topology, variational calculus, symplectic geometry - Cryptography (support: CIA, SVR, etc.): number theory, algebraic geometry over finite fields, algebra, combinatorics, and computers. - Hydrodynamics (support: boomer builders): complex analysis, partial derivative eqn's, Lie groups/algebras, cohomology theory, HPC. Might there be one small village of pure mathematicians still holding out against the industrial support? (resident druid: Dunforkix) Lagniappe: https://www.smbc-comics.com/comic/apply https://www.smbc-comics.com/comic/apply
- belter 2y ago
  > And so you're sort of stuck solving progressively more esoteric and even contrived problems. Good Science spreads into Technology where you least expect: https://youtu.be/-OkwGDKoY0o?t=2715 https://youtu.be/-OkwGDKoY0o?t=2715 Even the study of Black hole horizons could a have a direct and immediate usefulness for your Fluid Mechanics Engineers: https://youtu.be/-OkwGDKoY0o?t=2851 https://youtu.be/-OkwGDKoY0o?t=2851
- dwheeler 2y ago
  I disagree. We still cannot combine the key theories of the quantum and relativistic world. We don't understand whatvwe currently call dark matter and dark energy. However, it's not clear how to make progress in these areas.
barfbagginus 2y ago
You're off on two points; number of shells and the mass of the shells. First, the number of shells is finite. This actually gives us one of the points where we can test this theory: Namely, if there are discrete shells, their effect on gravitational lensing will introduce artifacts that we may be able to detect. We would observe the edge of a shell as a shearing zone in the gravitational lensing around a massive structure, where the amount of lensing jumps dramatically, significantly magnifying the cosmic background in the shear zone. These would look like rings around gravitational lensing sources. This discrete lensing artifact - these rings that might appears - could let us reject some dark matter theories, which predict smoother lensing with no rings. Second the shells themselves do not have positive or negative mass. They have density functions which are a sum of a delta distribution and the derivative of a delta distribution. On the shell singularity itself, the density function breaks down - from being a function into being a distribution. So we can't say it's positive or negative mass - it's a singularity without any mass. The reason we invoke positive and negative shells is because we can think of distributions as limits of ordinary functions. We can think of the topological defect shell -as if- it was the limit of a pair of positive and negative physical mass shells of positive thickness as they merged together into a shell of zero thickness. In the limit, there is no longer any externally observable mass at any enclosed radius. There's no longer any finite mass density function that properly describes the singular shell's density - the density has become a distribution.

m_dupont 2y ago

Agreed. This paper is pretty much pointless. Its extremely implausible, putting it generously, that such fine-tuned structures could even form. They mention at the end "this paper does not attempt to tackle the problem of structure formation" but that feels like a colossal understatement. I've studied physics myself, and I understand that sometimes people toy around with implausible theories solely for the sake of it but .... it seems like these peoples brainpower could be much better spent elsewhere.

m_dupont 2y ago
Also, strictly speaking these aren't topological defects. They are massless spherical shells, i.e. a sphere embedded in 3d space and thus can be continuously deformed to a point by sending the radius to zero. Spheres are literally the second homotopy group and the homotopy group of flat space is zero. Q.E.D
- barfbagginus 2y ago
  The sense in which they are defects is that on the shell, the density function is a distribution and takes no real value. Distributions are singularities/non-functions that we get when we take certain limits of ordinary functions, or solve certain differential equations with generalized functions. The distribution used works roughly as if we overlapped two equal positive and negative mass shells. But there are some extra details that ensure there is a net inward force for matter situated on the final shell. We'd have to actually work the math to really understand why that force appears, without hand waving.
  - m_dupont 2y ago
    I don't disagree with what you've said about distributions, however I don't think that the fact that these shells are created from dirac delta functions is sufficient to call them topological defects. Topological defects are solutions to the underlying physical equations that are of a different homotopy class to the vacuum, and these simply aren't in a different homotopy class, as I can smoothly deform them to zero, by sending the radius to zero or sending alpha to zero. Argued another way: a point charge can be modelled as a dirac delta charge distribution, but nobody would argue that a point charge is a topological defect
    - barfbagginus 2y ago
      Okay, fair point, I was not understanding that topological defects are points in solution space that are not path connected to the vacuum solution. I'm learning as I'm going! Now I also realize that the paper seems to say that the both the ordinary dirac shell solution and their modified shell are TDs, without proving it. I'd like to work up to proving whether the collapsing modified shell really does homotopy into into a point and then fade away into the vacuum. But first, I'm struggling with > nobody would argue that a point charge is a topological defect It's actually not clear to me that the point mass is not a TD! Let's try to write the homotopy sending a point mass Mδ₀ to the vacuum solution 0. Let t vary from 0 to 1. Then a possible homotopy is h(t) = (1-t)Mδ₀ This gives us h(0) = Mδ₀ h(1) = 0 The problem is that I don't know how to prove continuity of h. First, I don't know how to compute even the continuity of neighboring Delta functions for t < 1. But that feels intuitively like it should be continuous. On the other hand, I REALLY don't know how to prove continuity at t = 0, since the function seems to spontaneously collapse from a distribution with a kind of pseudoinfinite value at the origin, into a regular function with the value 0 at the origin. Using the notation of inner products and test functions, can we prove that it's continuous both for t < 1 and t = 1? I know that's a bit more technical than we usually get here on HN! I truly appreciate the help!
      - m_dupont 2y ago
        Okay we are are getting a little lost in definitions here, but nonetheless. You can solve the above by remembering that the dirac delta is the limit of a series of functions. If you take your delta to be lim a -> 0 N(0, a) where N() is the normal distribution, then you can see that in your above equation, you then have two limits. lim t -> 0 and lim a -> 0. By swapping the order of the two (which is a dubious operation), you can send t to zero first then a to zero, and the result is zero. So in one way, it can be deformed to zero, in another way it can't, because it's 0 times infinity . However, the thing to focus on is that dirac deltas aren't actually valid points in the solution space of partial differential equations. They aren't functions, and they aren't actually physically real. Come to think of it, that would probably exclude them from being TDs a-priori. Because a TD must be a solution to an underlying physical equation, and that solution must be deformable to zero. But if it's not a solution to a PDE (because it doesn't live in any valid hilbert space), then it can't be a TD.

0xWTF 2y ago

I think this paper is an effect of there being not much to do in theoretical physics these days. I mean, there's plenty of work that can be done, but I think it was Feynman who said all physics that could ever affect humanity was described by the 1963. Since the Higgs was discovered in 2012, we've covered pretty much everything in the Standard Model. I'm sure there's some intellectually interesting problems left to solve, but it's hard to believe they'll affect us. Nevertheless, societies still need physicists, and those physicists, by the nature of how incredibly challenging the "industrial" problems are (e.g. designing next generation fusion weapons), need challenging problems. Indeed, they crave them. And so you're sort of stuck solving progressively more esoteric and even contrived problems.

aebtebeten 2y ago
In https://news.ycombinator.com/item?id=40572593 https://news.ycombinator.com/item?id=40572593 Arnold, his tongue in cheek (and channelling his inner GJC), suggests a tripartite division of maths according to the industrial form which supports the pure researches: - Celestial Mechanics (support: the descendants of von Braun): dynamical systems, linear algebra, topology, variational calculus, symplectic geometry - Cryptography (support: CIA, SVR, etc.): number theory, algebraic geometry over finite fields, algebra, combinatorics, and computers. - Hydrodynamics (support: boomer builders): complex analysis, partial derivative eqn's, Lie groups/algebras, cohomology theory, HPC. Might there be one small village of pure mathematicians still holding out against the industrial support? (resident druid: Dunforkix) Lagniappe: https://www.smbc-comics.com/comic/apply https://www.smbc-comics.com/comic/apply
belter 2y ago
> And so you're sort of stuck solving progressively more esoteric and even contrived problems. Good Science spreads into Technology where you least expect: https://youtu.be/-OkwGDKoY0o?t=2715 https://youtu.be/-OkwGDKoY0o?t=2715 Even the study of Black hole horizons could a have a direct and immediate usefulness for your Fluid Mechanics Engineers: https://youtu.be/-OkwGDKoY0o?t=2851 https://youtu.be/-OkwGDKoY0o?t=2851
dwheeler 2y ago
I disagree. We still cannot combine the key theories of the quantum and relativistic world. We don't understand whatvwe currently call dark matter and dark energy. However, it's not clear how to make progress in these areas.

barfbagginus 2y ago

You're off on two points; number of shells and the mass of the shells. First, the number of shells is finite. This actually gives us one of the points where we can test this theory: Namely, if there are discrete shells, their effect on gravitational lensing will introduce artifacts that we may be able to detect. We would observe the edge of a shell as a shearing zone in the gravitational lensing around a massive structure, where the amount of lensing jumps dramatically, significantly magnifying the cosmic background in the shear zone. These would look like rings around gravitational lensing sources. This discrete lensing artifact - these rings that might appears - could let us reject some dark matter theories, which predict smoother lensing with no rings. Second the shells themselves do not have positive or negative mass. They have density functions which are a sum of a delta distribution and the derivative of a delta distribution. On the shell singularity itself, the density function breaks down - from being a function into being a distribution. So we can't say it's positive or negative mass - it's a singularity without any mass. The reason we invoke positive and negative shells is because we can think of distributions as limits of ordinary functions. We can think of the topological defect shell -as if- it was the limit of a pair of positive and negative physical mass shells of positive thickness as they merged together into a shell of zero thickness. In the limit, there is no longer any externally observable mass at any enclosed radius. There's no longer any finite mass density function that properly describes the singular shell's density - the density has become a distribution.