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Agreed. This paper is pretty much pointless. Its extremely implausible, putting it generously, that such fine-tuned structures could even form. They mention at

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.

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.