Mathematical Foundations and the Singularity

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1.8.4 - Relativistic Mechanics and the Unified Field

The form defining the Unified Field yields powerful physics right out of the gate. For instance, physics struggles to explain the location of the antimatter from the universe’s birth, where antimatter has the property of being opposite in charge to its matter partner. If antimatter and matter were to collide, they would mutually annihilate, producing intense energy in the process. Far-fetched explanations have been proposed, e.g., it has been said that an asymmetry existed at the moment of the universe’s birth, favoring the numbers of matter versus antimatter. These kinds of explanations are a direct affront to the net-zero principle, and, for this reason, they are in no way valid. Examples of matter/antimatter pairs are the electron/positron (�, �), and the proton/antiproton (�, �), where we use the bar convention over the particle to denote antimatter. Mathematically, antimatter rotates in the opposite direction on the infinite sphere of tangent spaces comprising the Unified Field, which are the vacuum singularities at the Planck scale. Matter and antimatter share a mutual boundary on these infinite spheres of tangent spaces, but move around in opposite directions, accounting for the dynamics of the virtual particles of the vacuum. Exponentiation yields a radial extension away from the boundary of the singularity that can move inward or outward depending on the magnitude of the real-valued component. We present generalized mathematical terms for form as �, and � as its anti-form partner. In �-dimensions, the Lie algebra for the form is defined as �(�) = �(� + ��), and the anti-form

Exponentiation of the Lie algebra of the form yields

and for the anti-form

and for � ≠ 0, the radial component of ���5�(�)6 will be less than one, while the radial component of

greater than one. The pair represents the symmetric splitting of the identity element, and mutual annihilation occurs if the form/anti-form pair collides, given the result of their product

Mathematically, the answer to the question of antimatter’s whereabouts is that it is present in the Unified Field; only it exists at radial distances of the Plank length and below. Important to realize is that the distance from zero to the Plank length holds just as much information as the distance of the Plank length to infinity does. Depicted in the following figure, we show a plot of the function � % for � ∈ ℝ, which adheres to the net-zero principle given that it is a symmetric split of the zerodimensional identity, unitary in volume and located at the function’s vertical intercept.

The real exponential portion � % of the anti-form expression

= � $ = � % � &' is inversely proportional to its form counterpart � % . But it is critical to envision that when bound to the spacetime geometry of the hyperbolae, the magnitude of the radial extension from the origin is equal in the parametric representation (���ℎ(�) , ���ℎ(�)), � ∈ ℝ, depicted in Figure 1.8.6, for � = ±√2. Limiting this concept to only matter and antimatter obscures the true nature of mathematics at work. Extending the terminology to form and anti-form is the appropriate conceptual treatment, given that matter and antimatter are one manifestation of the multi-faceted Unified Field Antimatter must have all the accouterments that matter does, including the geometric space it sits in, forces, and relativistic implications of its mass, to name just a few. So, the mathematical framework which defines anti-form is as complete and flawless as its form counterpart.

Next, we consider a form in the guise of a flowing point in spacetime as the expression, ���5�(�)6 = � � = � , � &, (1 8 74) where � &- represents the form’s wave properties, and � , represents the radial extension beyond the Plank distance of the unit singularity’s boundary and encodes its relativistic properties. Lie algebras that are not axis-based, meaning solely real or imaginary, will cease to be located on the unit singularity’s boundary once exponentiated.

371
as �(�) = �( � ��), where �, � ∈ ℝ!"×" .
���5�(�)6 = � $ = � % � &' , (1 8 71)
���5�(�)6 = � $ = � % � &' , (1 8 72)
���5�(�)6 will be
���5�(�)6 × ���5�(�)6 = � * = � (1 8 73)
���5�(�)6

We define the following Lie algebra, Α, as

, representing the hyperbolic Lie algebra, and

as

, representing the imaginary spherical Lie algebra. Given that the algebras commute,

=

the following expressions are equal

which represents the product of two generators, one hyperbolic and the other complex spherical. The product

be expressed in terms of hyperbolic functions

or the circular trigonometric functions

The energy-momentum relation in physics is defined as

which holds for a system or a body, for the momentum �, the speed of light �

mass In spacetime, all massive bodies must have some motion, but with the ideological case of a massive body at rest, the energy-momentum relation reduces to the well-known energy-mass equivalence

372
Figure 1.8.6 – Plots of the exponential function � % and the parametric form (���ℎ(�) , ���ℎ(�)), � ∈ ℝ
Α = K� 0 0 0 0 � 0 � 0M , (1 8 75) for � ∈ ℝ
�Ω
�Ω = K�� 0 0 0 0 �� 0 �� 0 M , (1.8.76) for � ∈ ℝ
[Α, �Ω]
� . � &/ = � .!&/ = K� , 0 0 0 ���ℎ(�) ���ℎ(�) 0 ���ℎ(�) ���ℎ(�)M T� &- 0 0 0 ���(�) ����(�) 0 ����(�) ���(�) U , (1.8.77)
� .!&/
� .!&/ = T� ,!&- 0 0 0 ���ℎ(� + ��) ���ℎ(� + ��) 0 ���ℎ(� + ��) ���ℎ(� + ��)U , (1.8.78)
� .!&/ = T� ,!&- 0 0 0 ���(� ��) ����(� ��) 0 ����(� �� ) ���(� ��) U (1 8 79)
0,
can
�0 = (�� )0 + (�* � 0 )0 (1 8 80)
, also known as the rest
� = �* � 0 , (1 8 81)
, the invariant mass �*

and for massless particles, the relation simplifies to

The energy-momentum relation of a massive object is related to the Lorentz factor by the expression

and the scalar momentum, |�|, is defined as

Using the hyperbolic Lorentz factor equivalence, we find

The scalar momentum, |�|, is related to the hyperbolic

Considering how the two hyperbolic functions relate to relativistic mechanics leads to the following relativistic equivalence for the exponential function

and solving for energy, we find

We again consider the idealistic case for a massive object at the rest ideal (� = 0), where the energy relation becomes � = �* � 0 . After substituting this relation into the expression above, we find

and given that � = 0, we have that

Returning to the hyperbolic generator, � . , from our generator product,

we substitute our relativistic equivalences, yielding

373
� = �� (1 8 82)
� = �* � 0 �, (1.8.83)
|�| = �* �� (1 8 84)
� = ���ℎ(�) = � �1 � 0 , (1 8 85)
� = �* � 0 � = �* � 0 ���ℎ(�) . (1.8.86)
���ℎ(�)
|�| = �* � ���ℎ(�) , (1.8.87)
� = tanh(�), is the following ���ℎ(�) = ���ℎ(�) ���ℎ(�) = �� = b� � c � → �* � ���ℎ(�) = �* � b� � c � → �* � ���ℎ(�) = �* �� = |�| (1 8 88)
which yields the following expression for energy
function as the equivalence
and the derivation using the rapidity term
� , = � b1 + � � c �1 � 0 , (1.8.89)
� = � , �* � 0 b1 + � � c (1 8 90)
� , = �1 � 0 b1 + � � c �1 � 0 = 1 + � � , (1.8.91)
� , = 1 → � = 0. (1.8.92)
� .!&/ , � . = K� , 0 0 0 ���ℎ(�) ���ℎ(�) 0 ���ℎ(�) ���ℎ(�)M , (1.8.93)
� . = ⎝ ⎜ ⎜ ⎜ ⎛� b1 + � � c �1 � 0 0 0 0 � �1 � 0 b� � c � �1 � 0 0 b� � c � �1 � 0 � �1 � 0 ⎠ ⎟ ⎟ ⎟ ⎞ , (1 8 94)

� (�* � 0 ) ⁄ term out, gives us

The matrix product above is a form representing a flowing point in spacetime and will have an inverse anti-form counterpart, representing a flowing anti-point in anti-spacetime, found by the evaluation of the following inverse

The result above can be expressed as the matrix product � . � &/ , and by substituting the relativistic terms into the product, we arrive at the inverse matrix � (.!&/)

374 and factoring the
� . = � �1 � 0 ⎝ ⎜ ⎜ ⎛b1 + � � c 0 0 0 1 � � 0 � � 1⎠ ⎟ ⎟ ⎞ (1 8 95) In terms of a relativistic relationship, our generator product � .!&/ becomes � .!&/ = � �1 � 0 ⎝ ⎜ ⎜ ⎛b1 + � � c 0 0 0 1 � � 0 � � 1⎠ ⎟ ⎟ ⎞ T� &- 0 0 0 ���(�) ����(�) 0 ����(�) ���(�) U , (1.8.96) which,
� .!&/ = � �1 � 0 ⎝ ⎜ ⎜ ⎛� &- b1 + � � c 0 0 0 ���(�) + � � � ���(�) � � ���(�) + ����(�) 0 � � ���(�) + ����(�) ���(�) + � � � ���(�)⎠ ⎟ ⎟ ⎞ (1 8 97) Given the identity � = ���ℎ(�) = � �1 � 0 , (1 8 98) our product relation in terms of the Lorentz factor is � .!&/ = � ⎝ ⎜ ⎜ ⎛� &- b1 + � � c 0 0 0 ���(�) + � � � ���(�) � � ���(�) + ����(�) 0 � � ���(�) + ����(�) ���(�) + � � � ���(�)⎠ ⎟ ⎟ ⎞ (1 8 99)
after multiplication, becomes
5� .!&/ 6 2 = � (.!&/) → T� ,!&- 0 0 0 ���ℎ(� + ��) ���ℎ(� + ��) 0 ���ℎ(� + ��) ���ℎ(� + ��)U 2 = T� (,!&-) 0 0 0 ���ℎ(� + ��) ���ℎ(� + ��) 0 ���ℎ(� + ��) ���ℎ(� + ��) U . (1.8.100)
� (.!&/) = � �1 � 0 ⎝ ⎜ ⎜ ⎛b1 � � c 0 0 0 1 � � 0 � � 1 ⎠ ⎟ ⎟ ⎞ T� &- 0 0 0 ���(�) ����(�) 0 ����(�) ���(�) U = � �1 � 0 ⎝ ⎜ ⎜ ⎛� &- b1 � � c 0 0 0 ���(�) + � � � ���(�) b� � ���(�) + ����(�)c 0 b� � ���(�) + ����(�)c ���(�) + � � � ���(�) ⎠ ⎟ ⎟ ⎞

where the anti-form counterpart contains the inverse and hyperbolically derived relativistic factor

With matrix representations for

375 → � (.!&/) = � ⎝ ⎜ ⎜ ⎛� &- b1 � � c 0 0 0 ���(�) + � � � ���(�) b� � ���(�) + ����(�)c 0 b� � ���(�) + ����(�)c ���(�) + � � � ���(�) ⎠ ⎟ ⎟ ⎞ , (1.8.101)
b1 � � c , � = � � (1 8 102)
� .!&/ and � (.!&/) established, we can solve for the energy of the � .!&/ generator product � .!&/ = � �1 � 0 T� &- (1 + �) 0 0 0 ���(�) + � ����(�) � ���(�) + ����(�) 0 � ���(�) + ����(�) ���(�) + � ����(�)U , (1.8.103) for which we will use the rapidity term notation, � = �⁄�. The matrix portion from the expression above results from the product Β� &/ = T(1 + �) 0 0 0 1 � 0 � 1 U T� &- 0 0 0 ���(�) ����(�) 0 ����(�) ���(�) U , (1 8 104) which we denote as the product of the Β matrix with � &/ , and to find the energy matrix �, we can solve the following expression � = � .!&/ 5Β� &/ 6 2 �1 � 0 . (1.8.105) The 5Β� &/6 2 term is equal to 5Β� &/ 6 2 = � &/ Β 2 , (1 8 106) given that the inverse of the Lie group � &/ is simply � &/ � &/ = T� &- 0 0 0 ���(�) ����(�) 0 ����(�) ���(�) U , (1.8.107) and with substitution, the energy expression becomes � = � .!&/ � &/ Β 2 �1 � 0 (1 8 108) Since � .!&/ � &/ = � . , (1 8 109) we find the relativistic energy for our generator product is not dependent on the � &/ generator portion and the expression reduces to � = � . Β 2 �1 � 0 . (1.8.110) The � . , Β 2 matrices, using notation for the Lorentz factor � and the rapidity term �, are equivalent to � . = T�(1 + �) 0 0 0 � �� 0 �� � U , (1 8 111) Β 2 = ⎝ ⎜ ⎜ ⎜ ⎛ 1 (1 + �) 0 0 0 1 (1 �0 ) � (1 �0 ) 0 � (1 �0 ) 1 (1 �0 )⎠ ⎟ ⎟ ⎟ ⎞ , (1 8 112) leading to the � . Β 2
product result

With this result, the relativistic energy of our generator product simplifies to

which given �1 � 0 is a scalar value, it can be moved to the left-hand side of the �-matrix, ultimately, we find a result consistent with our initial relativistic energy equivalence � =

generator product, the energy of an idealized massive object system, for

which is consistent with the expected result from relativistic mechanics.

To consider the relativistic energy-momentum for the generator product of a system comprised of a massless object, we first take the � = ���ℎ(�) function identity

Next, we multiply the matrix by the scalar-valued

terms will cancel, leading to the energy matrix, �,

For a system comprised of a massless object, the invariant mass will be equal to zero, �

= 0, and solving each entry in the � matrix above for our massless object system

we find the previously defined result energy-momentum identity

the energy matrix result has the consistent form

376 � . Β 2 = K� 0 0 0 � 0 0 0 �M (1 8 113)
� = � . Β 2 �1 � 0 = �1 � 0 K� 0 0 0 � 0 0 0 �M , (1.8.114)
� 0 �
, becomes � = �1 � 0 ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛1 n1 b0 � c0 o 0 0 0 1 n1 b0 � c0 o 0 0 0 1 n1 b0 � c0 o ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞ = �1 � 0 K1 0 0 0 1 0 0 0 1M , (1 8 115)
�*
Given this result for the relativistic energy of the
� = 0
� = ���ℎ(�) = � �1 � 0 , (1 8 116) and substitute this back into our matrix energy expression � = � . Β 2 �1 � 0 = �1 � 0 K� 0 0 0 � 0 0 0 �M , (1 8 117) yielding � = �1 � 0 ⎝ ⎜ ⎜ ⎜ ⎛ � �1 � 0 0 0 0 � �1 � 0 0 0 0 � �1 � 0 ⎠ ⎟ ⎟ ⎟ ⎞ (1 8 118)
�1 � 0 term, after which the �* � 0
� = K� 0 0 0 � 0 0 0 �M (1 8 119)
� = p(��)0 + (�* � 0 )0 , (1 8 120)
� = p(��)0 = ��, (1 8 121)
� = K�� 0 0 0 �� 0 0 0 ��M (1 8 122)
*
implying that

The matrix solution above for � is the result of the energy-momentum relationship � = p(��)0 + (�* � 0)0 , and taking (solely) this approach obscures the fundamental relationship that exists between energy and the spherical and hyperbolic geometries. For this reason, we start by considering the �

matrix

As we have seen previously, an isolated generator-coordinate system will evolve in time due to the generator operating on itself through its successive powers. We define a spherical time variable � to represent a mathematical calculative progression. Then, we can express the evolution of the calculative/spherical time-dependent generator system as

which will have the following matrix form

� is the wavenumber and � is the speed of light This equivalence can be seen by starting with the

identity

where the product of the wavelength � and the frequency � equals � The angular frequency, �, is related to frequency as

the wavenumber, �, to the wavelength, as

and the speed of light is related to the angular frequency and the wavenumber as the following identity equivalence

Solving this equivalence for � and �, we find the following

which establishes that � &/6 = � &785

As already stated, the relativistic energy for a massless object is equal to

where the momentum can be expressed in terms of the wavenumber as

and in terms of the angular frequency

leading to the energy equivalence

We recall that the time-dependent Schrödinger’s equation we presented for our generator was

377
&/
� &/ = T� &- 0 0 0 ���(�) ����(�) 0 ����(�) ���(�) U (1 8 123)
5� &/ 65 = � &/6 , (1.8.124)
� &/6 = T� &- 0 0 0 ���(�) ����(�) 0 ����(�) ���(�) U5 = T� &-5 0 0 0 ���(��) ����(��) 0 ����(��) ���(��) U (1 8 125)
� &785 = T� &9:5 0 0 0 ���(���) ����(���) 0 ����(���) ���(���) U , (1 8 126) where
speed-of-light
� = ��, (1.8.127)
This matrix can also be expressed in the following equivalent form
� = 2��, (1 8 128)
� = 2� � , (1 8 129)
� = v2� � w b � 2�c = � � (1 8 130)
� = ��, (1 8 131) � = � � , (1.8.132)
� = ��, (1.8.133)
� = ℏ�, (1.8.134)
� = ℏ� � , (1 8 135)
� = ℏ� = ℏ�� (1 8 136)

which expresses the energy equivalence between the time-derivative for

of ℏ0

�(�, �), representing a massive object in space. The energy of the system of the �

generator will take the same form of the left-hand side of the time-dependent Schrödinger’s equation, which can be appreciated after defining our generator to be the wavefunction

The negative sign for the above results from our choice of a positive-valued angular frequency � Calculating the time derivative for the wavefunction gives us the result

found using the relativistic energy-momentum relationship

When the velocity in space is positive and less than the speed of light, the Lorentz factor related terms

for which the coordinates in hyperbolic space are related to the object’s velocity in spacetime. Our massless object system represents axial light, with a purely spherical generator geometry, meaning it does not have a velocity in spacetime. Solving the Lorentz factor related terms � and � for � = 0, we find

and taking the Lorentz factor matrix representation for

we plug into it our null-velocity Lorentz factor results, finding the result

which is consistent with the system we are describing.

378 �ℏ�(�, �)�� | = ℏ0 2� �(�, �)�� 0 | , (1 8 137)
�ℏ�(�, �), as equal to the second space-derivative
⁄(2�)
&/6
� &/6 = �(Ω, �) �ℏ}5�(Ω, �)��|6~ = �ℏ ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ T� &-5 0 0 0 ���(��) ����(��) 0 ����(��) ���(��) U ⎝ ⎜ ⎜ ⎛ � �� 0 0 0 0 � �� 0 � �� 0 ⎠ ⎟ ⎟ ⎞ ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ (1 8 138)
�ℏ}5�(Ω, �)��|6~ = �ℏ T��� &-5 0 0 0 �����(��) ����(��) 0 ����(��) �����(��) U , (1.8.139) which
�ℏ}5�(Ω, �)�� | 6~ = ℏ� T� &-5 0 0 0 ���(��) ����(��) 0 ����(��) ���(��) U , (1 8 140)
�(Ω, �) wavefunction generator is equal to the energy of the massless system
� = p(��)0 + (�* (0)� 0 )0 = �� = ℏ� (1 8 141)
� and �� are equal to � = 1 p1 tanh0 (�) = 1 n1 v� 0 � 0 w , (1 8 142) �� = � � Ö 1 p1 tanh0 (�)Ü = � √� 0 � 0 , (1.8.143)
can be easily rearranged to
which shows that the time-derivative of our
� = 1 n1 v00 � 0 w = 1, (1.8.144) �� = 0 p� 0 (0)0 = 0, (1 8 145)
� . � . = T�(1 + �) 0 0 0 � �� 0 �� � U , (1 8 146)
� * = K1 0 0 0 1 0 0 0 1M , (1 8 147)

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