Comment by barfbagginus | Hacker News Reader

Comment by barfbagginus | Hacker News Reader

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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!

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.

I think that dirac delta solutions let us model point particles. We can lift differential operators onto the space of D' of distributions. This lets us extend the pde solution set with distributions.

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!

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