# Mathematical and Theoretical Physics

Mini Review Volume 1 Issue 5

# The universal vector

#### Paul TE Cusack

Correspondence:

Received: August 01, 2018 | Published: September 13, 2018

citation: Cusack PTE. The universal vector. Open Acc J Math Theor Phy. 2018;1(5):186-190. DOI: 10.15406/oajmtp.2018.01.00032

# Abstract

This paper takes Astrotheology mathematics and puts some of it in terms of linear algebra. All of physics can be described by two vectors. The dot product and cross product has important results. Here we describe the Gravity Equation. And we provide the coordinates of the universal and what they mean.

Keywords: linear algebra, astrotheology, gravity, mass, space

# Introduction

In this paper, we provide calculations on Astrotheology from Linear Algebra.1 Einstein was wrong about no absolute space and time and these calculations show how. Gravity, Mass, Density, and the zero vectors are used in calculations that show that the universe can be modelled as a tupple. We begin with gravity.

Product

$=4\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\sqrt{3}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\pi$

$=6.52$

Dot Product

Sum

$=4+3+\sqrt{4}+\pi$

$=118.72$

Mass in the periodic table of the elements.

Cross Product

${x}^{n}\int x=1$

${x}^{n}{x}^{2}/2=1$

${x}^{2n}=2$

Let n=4

${x}^{8}=2$

$x=1.0905=1/917$

Dot Product /Cross Product

$\theta =117.02$

${x}^{n}\int x=6.52=118$

$n=4$

${x}^{2n}=-/776$

$x=126.7$

$=\rho$

$0.917i=1.5$

$i={M}^{2+n}/G$

$i={i}^{6}/G$

$G=\left[1/i\right]{M}^{6}$

$=1.618{\left(4.486\right)}^{6}$

$=131866$

$=1+1/\pi$

$G=1/t\text{\hspace{0.17em}}+\text{\hspace{0.17em}}1/T$

$G=1/t+t$

$G=1+t\right]/t$

$G1\text{\hspace{0.17em}}+\text{\hspace{0.17em}}1/1=2=dM/dt$

$G=1/t\text{\hspace{0.17em}}+\text{\hspace{0.17em}}1/T$

$=1/1\text{\hspace{0.17em}}+\text{\hspace{0.17em}}1/251\text{\hspace{0.17em}}=\text{\hspace{0.17em}}1+1/\left(1/4\right)\text{\hspace{0.17em}}=\text{\hspace{0.17em}}5$

${x}^{2}-x-1=5$

Roots 3, -2 = Eigenvalues.

$2\left(2\right)\left(1\right)/\left({1}^{2}\right)$

$=4$

$|D|$

$=Max$

$=Ma$

$=F$

Golden mean

${x}^{2}-x-1=0$

${\left(x+iy\right)}^{2}-\left(x+iy\right)-1=0$

Reduces to the bilinear form

${x}^{2}-{y}^{2}=1$

Rotation matrix

Let

$\pi =\Omega$

$a11=x$

$a22=y$

${x}^{2}-{y}^{2}-1=0$

$a11=Ax$

$a22=Cy$

$A{x}^{2}+2Bxy+C{y}^{2}=D$

$\left(1\right){x}^{2}+2\left(-1\right)xy+1{y}^{2}=0$

${x}^{2}-2xy+{y}^{2}=0$

$0=0$

${x}^{2}-2xy+{y}^{2}=0$

$\left(x-y\right)\left(x-y\right)=0$

$x=y$

$a=v$

$P=F$

Figure 1 The universal vector space.

396,0
412,142
462,142
462,396
396,396
110,254
254,0
0,0
−−−−−
2492, 1330

$\left[T,s\right]$

Multiple be Operator Matrix

$=\left[-1/2\right]$

Multiple by $\left[T,s\right]$

[Emin, Antigravity]

Subspace & the zero vector.

is Perpendicular to

This point lies on the y axis on the E-t golden mean parabola.

$T=1/freq.=1/\left(1/\pi \right)=\pi$

$E=t=1$

$=\sqrt{3}/2$

$=0.866$

$\left(3,4,0\right)$Perpendicular to $\left(4,3,0\right)$

(-x,y,z) Perpendicular to (y,x,z)

$-t=E$

$z=z$

${L}_{1}=\sqrt{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}={L}_{2}=\sqrt{\left(-{x}^{2}+{y}^{2}+{z}^{2}\right)}$

(Likewise, for vectors in the III And IV Quadrant.

Multiple be Operator Matrix

Multiple by $\left[T,s\right]$

Period T=251

$\left[T,s\right]$

Subspace on te zero vector.

is Perpendicular to

This point lies on the y axis on the E-t golden mean parabola.

$T=1/freq.=1/\left(1/\pi \right)=\pi$

$E=t=1$

$=\sqrt{3}/2$

$=0.866$

(3,4,0) Perpendicular to (4,3,0)

(-x,y,z) Perpendicular to (y,x,z)

$-t=E$

$z=z$

${L}_{1}=\sqrt{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}={L}_{2}=\sqrt{\left(-{x}^{2}+{y}^{2}+{z}^{2}\right)}$

(Likewise, for vectors in the III and IV Quadrant.

Because you can’t have negative K.E., there is no such thing as negative time, so there is no orthogonal vector to the zero vectors (Figure 2).

Figure 2 Orthogonal vectors.

Let $s=z=0$

$\theta =\pi$

$E=-t$

$E=t$

$s=s=0$

$E=1/t$

$1=-{t}^{2}$

$\text{\hspace{0.17em}}t=\sqrt{\left(-1\right)}=i$

$E=t$

$E=1/t$

${E}^{2}=1$

$E=±1$

$ℝ\epsilon ℂ\epsilon F=0$

$x+iy$

$y=0$

$x+iy=0$

$E={t}^{2}$

$-E={t}^{2}$

$t=\sqrt{-1}=i$

$E+t×t=0$

$E+{t}^{2}=1$

$1+{1}^{2}=2=dM/dt$

$E+{t}^{2}=0$

$1+\left(-1\right)=0$

0=0 True!

Since time is K.E., and K.E is time, we measure K.E. relative to something - the zero vector. We also know that Mass is the dot product of E and t (Figure 3).

Figure 3 The zero vector.

And we know that Momentum, P, or

So

$=0.785$

$=1/127.3$

$=1/\rho$

$\rho =Density$

This disproves Einstein’s Relativity. There is a stationary point in the universe. It is the Zero Vector. Space and time are absolute.2

So this is how the universe crystallized into existence.3

But

Density

Equation of a plane

$\left(r\cdot a\right)\cdot n=0$

$\left(r\cdot a\right)\cdot \left(b-a\right)×\left(c\cdot a\right)=0$

$a=\left(0,0,0\right)=ℤ$ position vector

$\left(b-a\right)=\left(3,0,0\right)-\left(0,0,\pi \right)=\left(3,0,-\pi \right)$

$\left(C-a\right)=\left(3,0,0\right)-\left(-2,0,0\right)=\left(5,0,0\right)$

Mass Gap

$=1/G=3/2$

$\left(r-a\right)\cdot n=0$

$=0$

$r\epsilon R$

r is orthogonal to n

${G}_{0}={t}^{2}$

$={0}^{2}=0$

$E=t=s={G}_{0}=0$

$M=-1$

# Cusack’s gravity equation

$=6.522$

$|\left[\begin{array}{cccc}4& 0& 0& 0\\ 0& 3& 0& 0\\ 0& 0& \sqrt{3}& 0\\ 0& 0& o& -\pi \end{array}\right]|={G}_{0}$

$\left[4,3,\sqrt{3},-\pi \right]\left[\begin{array}{c}1\\ 1\\ 1\\ -1\end{array}\right]=M$

$\left[|D|C\sqrt{C}\Omega \right]\left[\begin{array}{c}E\\ t\\ s\\ M\end{array}\right]=M$

Figure 4 Mass an the Ln function.

$=781.8=\pi /4=45°$

$=1/\rho$

Universal vector

For the angle of this universal vector

So the Universal Vector produces t=E=1 rad;;cuz (or PTEC)(Figure 5).

Figure 5 The universal vector and time.

The Norm

For the universal vector

$=\sqrt{{G}_{0}}$

But we know $G={t}^{2}$

$=t$

And,

$||\lambda t||=|\lambda ||||t||$

Characteristic Equation

${x}^{2}-x-1=0$

Let

# Conclusion

We see that the universe can be well modelled by linear algebra.

None.

# Conflict of interest

The author declares that there is no conflict of interest.

# References

1. Axler S. Linear Algebra Done Right. 3rd ed. Springer, USA; 2015.
2. Shilov G. Linear Algebra. USA: Dover; 1971.
3. Cusack P. Astrotheology, Cusack’s universe. J of Phys Math. 2016;7(2):8.