[barfbagginus]

I'm doing way more than confusing things - all this stuff is mainly new to me!

I'll try to understand your comment over the next few days

BUT what do you think about that weird business with the s scaling factor that supposedly rescales the Delta function so that it takes a finite value at 0? I can't yet prove what kind of animal s is supposed to be. It's like some kind of funky infinitesimal.

[nyssos]

> BUT what do you think about that weird business with the s scaling factor that supposedly rescales the Delta function so that it takes a finite value at 0? I can't yet prove what kind of animal s is supposed to be. It's like some kind of funky infinitesimal.

Strictly speaking, it's nonsense. What they're probably actually doing is taking the limit of the result for shells of nonzero thickness as thickness goes to zero.

[barfbagginus]

That's not what the s factor controls. It's a scaling factor on the intensity of the dirac shell itself, not the radius or positioning.

s acts something like the 1/d, a reciprocal of the dirac delta function. But this requires some careful technical attention. For example it might only invert the Dirac Delta functio on its support at the origin, and leave it zero elsewhere. But even then that notion is problematic. We might require some generalization of the distributions to allow reciprocals of distributions like this.

Another idea is that it maybe be are working in the limit as s goes to zero, like you suggest, but what they are attenuating is the intensity of the shell (without ever literally inverting the delta function)

Read the paper and try to work it out if you know a bit about distributions. It's a fun mystery right now!

[nyssos]

Sorry, my wording was a bit unclear: what I meant was that they were probably taking the limit of

    \int_{0}^{r} s_{i} \* f_{i}(R - r)/r^2 + f_{i}'(R-r)/r

with respect to some suitable sequence of functions `f_{i}` (narrower and narrower gaussians, for instance), for which `s` is actually well-defined.