I checked, and the publish date is not in fact the first of April.
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.
- 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
Good Science spreads into Technology where you least expect: 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
However, it's not clear how to make progress in these areas.
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.
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.
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
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.
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
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!
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.
Conceptually this is how we would model a field with interacting particles using PDEs, and is as far as I know the reason we solve with distributions and consider them physical in the first place.
The question of whether dirac is homotopic to vacuum would then mean exhibiting a homotopy in D', which is the dual space to the underlying function space, and represents both functions and distributions as functionals on a space of test functions. The topology of D' is the weak-* topology. Given a net N_t of distributions, N_t converges to N if N_t(g) converges to N(g) for all test functions g.
So to make your proof tight, I propose we should lift it into D', where our family u_t of shrinking deltas is already a net. We'd then prove that u_t -> 0 converges for each test function.
I think the result would be much more satisfying and watertight, since your current proof does not really make up its mind about what it's conclusion is.
Now, I am not sure that the distributions need to be a hilbert spaces to solve PDEs in this sense. This is because distributions have derivatives and can be acted on by our lifted differential operators. These solutions are weak in the sense that they are solutions with respect to the test functions. But I also believe that the solutions are faithful, in the sense that any solution corresponding to a function in the base space will also be a solution for the original operator.
Also, if you need our space of distributions to be a Hilbert space, we can pass to distributions over sobolev n space, the space of functions with square integrable nth derivatives. The distributions here end up being hilbert spaces! Not only that but the square integrability condition is pretty mild and still leaves us a lot of very useful functions for physics. I think that some quantum theories use these kinds of Hilbert spaces. Lol, take that with a grain of salt - this is lore to me, not terra firma.
Anyhow, if you want to try to persuade yourself that we can't just throw away delta functions and other distributions, and then try to lift your proof into space of distributions, I think that would be very fun!
I'll try to do that as well!
Ps. Forgive me if I'm Tom Sawyering you or nerd sniping! Please also forgive me for any mistakes I'm making!