# Connection between the spectra of a family of matrices and a modelization of particles' scattering?

In the excellent book "Numerical Computing with MATLAB" by Cleve B. Moler (SIAM 2004), [Moler is the "father" of Matlab], one finds, on pages 298-299, the following graphical representation (fig. 10.12 ; I have reconstructed it with minor changes ; Matlab program below) with the following explanations (personal adaptation of the text):

Let $$n$$ be a fixed integer $$>1$$. Let $$t\in (0,1)$$ ; let $$A_t$$ be the $$n \times n$$ matrix with entries

$$A_{t,i,j}=\dfrac{1}{i-j+t}$$

Consider Fig. 1, gathering the spectra of matrices $$A_t$$, for $$n=11$$ and $$0.1 \leq t \leq 0.9$$ with steps $$\Delta t=0.005$$.

Figure 1. Interpretation : the 11 rightmost points correspond to the spectrum of matrix $$A_t$$ for $$t=0.1$$.

There is a striking similarity with particles' scattering by a nucleus situated at the origin, with hyperbolic trajectories ; see for example this reference.

My question is :

How matrices $$A_t$$ can be connected to a model of particle scattering ?

My attempts (mostly unsuccessful) :

• Consider $$A_t$$ as a Cauchy matrix ($$x_i=i$$, $$y_j=j-t$$) (notations of https://en.wikipedia.org/wiki/Cauchy_matrix) with its different properties, in particular displacement equation. See as well the answers to this question.

• Connect $$A_t$$ with a quadratic form defined by its moment's matrix $$\int_C z^{i}\bar{z}^{j}z^{t-1}dz$$, $$z^{t-1}$$ playing the rôle of a weight function. But for which curve $$C$$ ?

• Prove that the eigenvalues are situated on hyperbolas.

• Make many simulations (see figures below).

Matlab program for Fig. 1 :

 n=11;
[I,J]=meshgrid(1:n,1:n);
E=[];
for t=0.1:0.005:0.9
A=1./(I-J+t);
E=[E,eig(A)];
end;
plot(E,'.k');axis equal


Addendum (November 22, 2018) : Here is a second figure that provides a "bigger" view, with $$n=20$$ and $$0.005 \leq t \leq 0.995$$. The eigenvalues corresponding

• to $$t=0.005$$ are grouped into the rightmost big red blob on the right,

• to $$t=0.995$$ are blue filled dots (quasi-circle with radius $$\approx 110$$).

Figure 2 : [enlarged version of Fig. 1 ; case $$n=15$$] Everything happens as if a planar wave enters at $$t=0$$ from the right, is slowed down by nucleus' repulsion, then scattered as a circular wave...

Fig. 3 : Two cases are gathered here, both for $$t=0.995$$ : $$n=51$$ (empty circles) and $$n=100$$ (stars). One can note that the eigenvalues are very close to the $$(n+1)$$-th roots of unity. For this value of $$t$$, the radii are given by experimental formula : $$R_n=200(1-12.4/n+212/n^2-3110/n^3)$$.

I am grateful to @AmbretteOrrisey who did very interesting remarks, in particular by giving the following polar representation for the trajectory of particles : [citation follows ; more details can be found in his/her answer] "The polar equation of the trajectory of a particle being deflected by a point charge is

$$r=\frac{2a^2}{\sqrt{b^2+4a^2}\sin\theta -b}\tag{1}$$

where $$a$$ is the impact parameter which is the closest approach to the nucleus were the path undeviated; $$b$$ is the closest approach of a head-on ( $$a=0$$) particle with repulsion operating."

Figure 4 displays a reconstruction result that takes into account the fact that (n+1)th roots of unity give asymptotic directions.

Figure 4 : Case $$n=201$$. An approximate reconstruction of $$21$$ among the $$n$$ trajectories (Matlab program below ; polar equation - adapted from (1) can be seen on line 5). Please note the "ad hoc" coefficient $$3.0592$$...

 n=201;
for k=1:20:n
d=pi*k/(2*(n+1));c=cos(d);
t=-d+0.01:0.01:d-0.01;
r=3.0592*(1-c)*exp(i*(t-d))./(cos(t)-c);
plot(r);plot(conj(r));
end;


Slightly related : http://bdpi.usp.br/bitstream/handle/BDPI/35181/wos2012-3198.pdf

I mention here a book gathering publications of Steve Moler [Milestones in Matrix Computation] (https://global.oup.com/academic/product/milestones-in-matrix-computation-9780199206810?cc=fr&lang=en&) and an article mentionning the analycity of the obtain curves : (https://www.math.upenn.edu/~kazdan/504/eigenv.pdf)

I'm don't think the curves would be hyoerbolæ anyway if it were a model for classical scattering.

The polar equation of the particle being deflected by a point charge is

$$r=\frac{2a^2}{\sqrt{b^2+4a^2}\sin\theta -b}$$

where $$a$$ is the impact parameter which is the closest approach to the nucleus were the path undeviated; $$b$$ is the closest approach of a head-on ( $$a=0$$) particle with repulsion operating, and $$\theta$$ is the angle between the radius vector of the particle (nucleus at origin) and the line joining the nucleus to point of closest approach with charge in place (not the axis of approach - ie the line through the nucleus that particle is moving parallel to & distance $$a$$ from when it is yet infinite distance away).

Is that a polar coördinate representation of a hyperbola? I suppose it ought to be, as it is an inverse-square force!

Anyway - it's likely I'll look into it will-I-nill-I, now.

Oh ... the diæræsis in "coördinate" & the ligature in "diæræsis" ... I'm hoping you don't mind that archaïck stuff too much - I love it ... but many would probably go ballistic if I used that kind of idiom more generally on here. They even complain about my italics & emphasis! The diæræsis over the second vowel in a conjunction of vowels was a device used to denote that the conjunction does not form a diphthong. It never was actually a very widely-used convention!

• No! that's it! θ is the angle the radiusvector of the particle makes with a line at a right angle to the line joining the nucleus (origin) to the point of closest approach, ie the direction of motion of the particle at it's point of closest aporoach. Coz the distance of closest aporoach is (b+√(b^2+4a^2))/2. That's why it's better really changing it to formulation in terms of cosine; but the integral in Gradsteyn & Ryzhik is in terms of sine. It's coming back now. Comments don't bump posts ... do they!? – AmbretteOrrisey Nov 22 '18 at 12:55
• No they don't, fortunately. Yes of course it's a hyperbola - I rembember now from celestial mechanics - a planet's orbit is 1/(1+ε.cosθ) with ε the eccentricity of the orbit (& scaled appropriately of course). So where were we - you wanted to prove that the curves in your graph are hyperbolæ, but couldn't. Please don't rely on my being able to do so - you've seen just now what kind of gaps & lapses I can be afflicted with. But I'll have a look ... and I shall, too, as this has indeed caught my interest ... you never know - I might get an inspiration! – AmbretteOrrisey Nov 22 '18 at 13:25
• Couple of questions: do the curves go off to ∞ in their respective directions as t tends to the limits of its range; and as t tends to 0 is it -π to +π that the imaginary parts extend over? ... kind of looks like it. – AmbretteOrrisey Nov 22 '18 at 14:11
• I think it's quite likely they are an abalogue of scattering, TBPH. Even this wave interpretation that's used for calculating the scattering by a point with a Yukawa potential: what you've shown here is reminiscent of that. You have y – AmbretteOrrisey Nov 22 '18 at 15:52
• to use the wave method with a Yukawa potential: if you try to do it classically you get an integral that's intractable - I've 'farmed it out' on here and the consensus seems to be that it is intractable. So all it needs, then is for the 'trajectories of the solutions in rhe complex plane to be hyperbolæ with focus all at the same point. That's not really a great deal to ask ... & I have a hunch they are that. So maybe scattdring could be gorm – AmbretteOrrisey Nov 22 '18 at 15:58