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^...

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|