Paul Bird's Scientific Research

Particles from First Principles Introduction The idea behind the following is to put the discrete indices on the same footing as the continuous variables. With nothing more complicated than momentum conservation and charge conservation we shall attempt to deduce the Standard Model. Definitions Let define a field  Φ ( k , N ) where k  is a 4-vector of continuous values and N is a vector of discrete values of unknown (or even infinite) length. k shall be called the momentum vector and N shall be called the charge vector. The vectors N  could, for example, be vertices on a polytope or lattice with the property that each vector is the sum of at least one pair of other vectors. Neutral Fields Now we must also define some more (vector) functions at the special points where k =0 or N =0. Where there is no momentum of no-charge. The neutral fields are given by the Z functions, which can be contracted with a charge: Z ( k , 0 ). N The sigma functions ar...

Hyper-determinant With the Levi-Civita symbol we can define a determinant: ε^{a1,a2,a3,...} = 0 if any 2 of the indices the same. (1 otherwise) (Also, antisymmetric) Determinant of a 2-tensor is: det_2(A) = ε^{a}ε^{b}A^{a1,b1}A^{a2,b2}A^{a3,b3} .... We can define f^{a1,a2,a3,...} = 0 if any 3 of the indices the same. Determinant of a 3-tensor is: det_3(A) = f^{a}f^{b}f^{c}A^{a1,b1,c1}A^{a2,b2,c2}A^{a3,b3,c3} .... In the case where it is a 2x2x2 symmetric 3-tensor this should give the discriminant of a cubic. Let g_n{a1,a2,a3...} =0 if any no n of the indices are the same. Determinant of a n-tensor is: det_n(A) = g_n^{a}g_n^{b}g_n^{c}...A^{a1,b1,c1,...}A^{a2,b2,c2,...}A^{a3,b3,c3,...} .... Properties Assuming all components of A are independent (not true for symmetric tensors!). det_1( A ) = 0 by definition. d/dAin d/dAim det_2( A ) = 0 d/dAin d/dAim d/Aio det_3( A ) = 0 Inverses We can generalise the definition of an inverse for a 2-tensor: ...

Converting Inequalities to Equalities All strict and unstrict inequalities can be written as existence theorems on equalities: x!=0  <==>  ∃y{xy+1 =0} x>0   <==>  ∃y{xy 2 -1=0}    x<0   <==>  ∃y{xy 2 +1=0} x>=0  <==>  ∃y{x-y 2  =0}   x<=0  <==>  ∃ y{x+y 2  =0} Note that the first is the same as  ∃y{y=1/x}  which is true as long as we don't let x be infinite! Two equalities can be combined to a single equality like this: x=0 ∧ y=0  <==>  x 2 +y 2 = 0 A polynomial in a high degree can be written as a system in a lower degree: ax^6+bx^5+cx^4+dx^3+ex 2 +f = 0 <==>  y-x 2 =0 ∧ z-y 2 =0 ∧ azy+bzx+cz+dxy+ey+f = 0 <==> (y-x^2) 2 +(z-y^2) 2 +(azy+bzx+cz+dxy+ey+f) 2 = 0 Hence it turns out that any system of polynomials equalities and inequalities and can be written as a quartic in multiple variables.

Tarski-Seidenberg theorem States that a set of equalities and inequalities f(x1,x2,x3,....xn)>=0 can be reduced down to a set of equalities and inequalities f(x1,x2,x3,....xn-1)>=0 with one less variable. Thus all equations can be reduced to a set of equalities and inequalities on their coefficients. See also: cylindrical algebraic decomposition See:  Elimination These algorithms aren't very simple.

When does a quartic have a real root? This is a method for deciding if a polynomial has any real roots (or equivalently when it has no real roots). Start with a quartic polynomial. Ax 4 +4Bx 3 +6Cx 2 +4Dx+E = 0 or Q nmop X n X m X o X p = 0 The (det_4( Q )) discriminant is: also D=A^3.E-4.A^2.B.D-9.A^2.C^2+24.A.B^2.C-12.B^4 D' = (D + P^2) = d^2/dE^2  det_4(Q) also P = AC-B^2  = discriminant(  d^2/d x ^2 p( x ))   It has NO real roots when det_4( Q ) >0  and D >0 and P >0 or det_4( Q ) =0 and D =0 and P >0 Question How to get these polynomials as determinants of matrices? Or just find a simple algorithm. Guess : Are these the resultants of f (x) with it's derivatives? f '''''(x). How are D and P generalized? Extensions How to extend this to multi-variable polynomials? (This would solve kissing number problem). Obviously, we could have the coefficients depend on values of y; a(y), b(y), c(y), etc. Then we would ha...

Sphere Packing in Distance Geometry In distance geometry the only condition we have is that any (n+1)-simplex made from (n+2) points is zero in n dimensions. Let R_nm be the squared distance between point n and point m. The kissing number problem is then stated as: R 0n = 1 R nm > 1 Then we ask is there a solution in terms of R such that the above holds and also the Cayley-Menger determinants are all zero? In 2 dimensions with 7 circles this gives us R 0n = 1 R nm > 1 VolumeOfTetrahedron ( R_ab,R_ac,R_ad,R_bc,R_bd,R_cd) = 0 Which gives cubic equalities for the R. We could set: R nm = 1+(S nm )^2 This gives us a set of 6th degree polynomials in S. Then we should be able to use the resultant to see if they all have a simultaneous solution! But the resultant does not distinguish between real and imaginary solutions. :( e.g. ( x ^2+1)( x -3)=0 ( x ^2+1)( x -4)=0 Share an imaginary solution! (Only possible if share common factor?) What we need...

General Systems Polynomial Equations We want to find a solution to the general problem of systems of polynomial equations. (There is a simple algorithm to solve these problems but maybe a neater way can be found?) Firstly all systems of polynomial equations can be written in the following form: ∪ i=1..n {X T A i X = 0} with X a vector, A a matrix with X 0 =1. The solutions should all be written in terms of determinants { A i +..} . Or using the antisymmetric tensor. det(Q) = {Q} One equation - n variables For one equation a solution exists if:  {Q}  > 0 This can be written in terms of traces. One variable: 2 {Q}  = [Q] 2 -[Q 2 ] Two variables: 6{ Q}  = [Q] 3 -3[Q][Q 2 ]+2[Q 3 ] etc. The general formula from traces comes from it's expression as a determinant. Q Q Q... ε ε Two Equations Two equations - One(n?) Variable(s) A solution exists (common root) if (each has a solution as above): ([Q] 2 -[Q 2 ]) ([R] 2 -[R 2 ])  - ([Q][R]...