Paul Bird's Scientific Research

The Fourier transform the continuous variable "position" is the continuous variable "momentum". The Fourier transform of the interval around a finite dimension is an integer charge. Therefore particles of different charges are simply the different modes around a finite dimension. In position space we should also talk about position in compact space. Everything becomes geometrical. In momentum space we should talk about charge. Can we extend this to spin also? Is spin part of the position or momentum space picture? If X (a,b) = ((x+t,y+iz),(y-iz,x-t)) then X (a,b) X (-a,-b) = x^2+y^2+z^2-t^2. Perhaps this is the better momentum picture? What do a and b mean?  ++ -- +- -+ Ψ *(-a)( D (a,b)+i A (a,b)) Ψ (b) + ( D (a,b) A (c,d)- D (c,d) A (a,b)).( D (-a,-b) A (-c,-d)- D (-c,-d) A (-a,-b)) +m Ψ L(a) Ψ R(-a)                  = D(a,b)A(c,d)D(a,b)A(c,d) - D(a,b)A(c,d)D(-c,-d)A(-a,-b) X*(a,b) = X(b,a)

If we have a particle (X,Y,Z,T) = (a(T),a(T),a(T),T) in comoving coordinates, in Milne coordinates this is: t=T cosh( a(T)*sqrt(3) ) x=T/sqrt(3) * sinh( sqrt(3)*a(T) ) which gives a parametric equation in T. We can plot these curves. e.g. for linear expansion we have the lines a(T) = v.T for various constants v.

What is it that can actually be known in science? I suggest that all questions of science boil down to the following question: "What is the proportion of possible histories in which A happened that B also happened?" As an example: "What is the proportion of possible histories in which I ROLLED THIS DIE such that IT ENDS UP WITH 6 ON TOP?" Heuristically, we can say that the answer to this is (approximately) 1/6. 'Approximately', since there may be other factors that we are not aware of, the die could be waited for example.  Note that implicitly we are considering only a very small subset of possible histories beginning at the big bang. Namely ones that lead to a life-bearing planet on which humans evolved and eventually produced the intelligent(ish) bipedal life-form referred to in the question as "I". Further we are only considering histories in which I roll a die for some reason, perhaps I am playing Snakes and Ladders. 

A string world sheet can be decomposed into a sum of Feynman diagrams (graphs). Likewise a membrane should be decomposed into a sum of graphs also (or spin-networks). Hence Loop Quantum Gravity should be considered as a membrane theory. The differences are that Loop Quantum Gravity is a gravity only theory with no supersymmetry. It only exists in 4 dimensions whereas M-Theory exists in 11 dimensions.

See also here A general rotated ellipse is \(f(x,y) = a x^2+2 b x y+c y^2+d x+e y - 1\) = X.M.X Nearest point (x,y).(df/dx, df/dy) = 0. Let this point be (X1,Y1) and (X2,Y2). X1^2+Y1^2 = R1^2  Therefore a circle radius between R1 and R2 intersects. A circle of radius >R1 overlaps. We need to find the relationship between roots of two quadratic simultaneous equations and their coefficients. It intersects a circle of radius 1/C if (x1^2+y1^2-1/C)(x2^2+y2^1-1/C)<0 or: (9*e^4+18*d^2*e^2-24*c*e^2+24*a*e^2-96*b*d*e+9*d^4+24*c*d^2-24*a*d^2+16*c^2-32*a*c+64*b^2+16*a^2)* C^2 + (-6*a*e^4+12*b*d*e^3-6*c*d^2*e^2-6*a*d^2*e^2+8*a*c*e^2-16*b^2*e^2-8*a^2*e^2+12*b*d^3*e+16*b*c*d*e+16*a*b*d*e-6*c*d^4-8*c^2*d^2+8*a*c*d^2-16*b^2*d^2)* C +a^2*e^4-4*a*b*d*e^3+2*a*c*d^2*e^2+4*b^2*d^2*e^2-4*b*c*d^3*e+c^2*d^4    <   0 (which is a 6th degree polynomial) To convert this to the case where we have two ellipses we need to transform: a*x^2+2*b*x*y+c*y^2+2*d*x+2*e*y +f =

We can create Turing machines with no internal states if we have multiple heads. The question is how many colours do we need? Assuming the 3 state, 2 colour Turing machine is the simplest. For 1 head: Not Possible with 1 state For 2 heads: <=6 colours For 3 heads: <=6 colours (4?) .... For 2 colours: >=2 heads ... For 6 colours: 2 heads For example this is Universal: 0101001001001[010]00101010 0101010001010[110]11010100 0101110000000[111]10101010 But is this? ( 4 colours, 2 heads )? CTAGAGCATGC[CGT]CGACTCAG Or equivalently this? 100101110[10]10101010 001011100[11]10100101 Restricing the idea so that the ends must be repeated symbols. Then the (6,4) is the simplest which corresponds to 2 heads with 24 colours . So at least ABCHSADBJSADHSA[CX]ADBSADBSAHJDS Can be made Universal using only 24 letters of the alphabet. Using the more tighter idea of a Turing machine without repeating patterns at the ends. Using (2,4) Turing machine then

Lets assume we have a diffeomorphic invariant action with: Scalars  Φ , Vectors A, Fermions  Ψ  and Gravitinos X. And dilatons c. Then we have terms: (e^8) ∂ Φ ∂ Φ (e^3) Ψ ∂ Ψ (e^12)( ∂ A ∂ A+A^4) c^2(e^9) X ∂ X c^2(e^12) ∂ e ∂ e m Ψ* Ψ Each term must equal -4. Conformal weights [c] = 5/3 [e] = -2/3 [ X ] = -7/6 [A] = 1 [ Ψ ] = -3/2 [ Φ ] = -1/3 [m] = -1 Interaction weights = 2-|spin|

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 are given by the

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: ( A

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]- [QR]) 2   = 0

  Sphere Packing   The sphere packing problem or kissing number problem can be stated as a set of quadratic inequalities: {A i nm x n x m  + C i   ≥ 0 } Can we find a general algorithm to solve these equations? (Don't see why not!) (Actually need for pair of equations in N variables.) One Equation - One variable  ✓ (1) A x 2 +C ≥ 0 has  no  solution if: C < 0 & A <= 0 Two Equation - One variable  ✓ (1) A x 2 +c ≥ 0 (2) D x 2 +f ≥ 0 has no solution if: A<0 & c>0 & D>0 & f<0 &  (D c -A f )>0 or vice versa  N Equations - One variable (SOLVED!)  ✓ (1) A x 2 +C ≥ 0 (2) D x 2 +F ≥ 0 (3) G x 2 +I ≥ 0 has NO solution if (and only if) one pair does not have a solution or any single equation has no solution. ( If all pairs have solution then all MUST have a solution ). <-- This only works in one variable!  One Equation - M variables  ✓ (1)  x T A x +c ≥ 0 Solution if: A 1 1 <=0 & A 22 <=0 & det(A)(