In this paper, we will use the mathematical discipline known as Geometric Algebra to show that the three spatial dimensions create a mathematical framework with 8 degrees of freedom. We will check that these degrees are sufficient to explain different disciplines of Physics (Dirac Equation, Gravitation, comparison with E8 theory…).

There is a discipline in mathematics that is called Geometric Algebra [1] [2] also known as Clifford Algebras. One curious thing of this Algebra is that if you consider a certain number of spatial dimensional (a certain number of independent vectors), automatically appear other dimensions (or if you want to call them, new degrees of freedom or other entities other than vectors).

In fact, the total number of degrees of freedom in an n-dimensional (understanding n as the number of special dimensions or independent vectors) in Geometric Algebra is:

$\text{Total number of degrees of freedom}=2^n$

If we consider that our world has three spatial dimensions (in Geometric Algebra it is called Cl_3,0), we will have:

$\text{Total number of degrees of freedom}=2^3=8$

And in fact, we can check that this is true:

In three dimensions, we have three independent vectors $\hat{x}$, $\hat{y}$ and $\hat{z}$ ( Figure 1 ):

In geometric algebra, these three vectors create 5 other entities.

The first other three entities are the bivectors. The bivectors are created multiplying perpendicular vectors. The result of this product is the bivector, an independent entity from the vectors that represent oriented planes. For example, the $\hat{x}\hat{y}$ bivector ( Figure 2 ):

There are three independent bivectors: $\hat{x}\hat{y},\hat{y}\hat{z}$ and $\hat{z}\hat{x}$.

Another appearing entity is the trivector. It is formed by the product of the three independent vectors (and represent an oriented element of volume) ( Figure 3 ).

We can check that $\hat{x}\hat{y}\hat{z}=-\hat{y}\hat{x}\hat{z}$.

One important thing of the trivector is that in three dimensions there is only “one trivector”. I mean, it can be bigger or smaller or with opposite direction (this means it can be escalated by a real scalar—positive or negative), but the trivector itself as basis or unit trivector is always the same. You can check Annex A1 to check wheat I am talking about.

Another special property of the bivectors and the trivectors is that the square of a bivector or a trivector is −1. This you can check in all the papers of GA [1] - [6] . And the square of a vector is 1. Always talking in Euclidean metric. If this is not the case, you can check [7] [8] .

That the square of the bivectors and the trivectors is −1, means that they are a clear candidate for the imaginary unit i in certain circumstances. And we will see that this property is key for the trivector in the next chapter.

The last entity exiting in Geometric Algebra are the scalars (the numbers). They exist in their own space (are not linear as vectors, surface as bivectors or volume as trivector).

So, in total you can check that we have 8 entities when we have three spatial dimensions: 3 vectors, three bivectors, one trivector and the scalars.

But why are they “degrees of freedom”?

Ok, I will define another concept, the multivector. A multivector is just a sum of all the commented previous entities. This is, for example:

$A={\alpha }_0+{\alpha }_1\hat{x}+{\alpha }_2\hat{y}+{\alpha }_3\hat{z}+{\alpha }_4\hat{x}\hat{y}+{\alpha }_5\hat{y}\hat{z}+{\alpha }_6\hat{z}\hat{x}+{\alpha }_7\hat{x}\hat{y}\hat{z}$

Being ${\alpha }_i$ scalars. This means the multivector (whatever it represents) it has eight degrees of freedom (from ${\alpha }_0$ to ${\alpha }_7$). Its meaning can vary a lot depending on the context or the discipline we are talking about.

For example, let us check the position multivector ( Figure 4 ):

This multivector has 8 coordinates (8 degrees of freedom corresponding to the scalar, the three space vectors, the three bivectors and the trivector):

$R=r_0+r_x\hat{x}+r_y\hat{y}+r_z\hat{z}+r_{xy}\hat{x}\hat{y}+r_{yz}\hat{y}\hat{z}+r_{zx}\hat{z}\hat{x}+r_{xyz}\hat{x}\hat{y}\hat{z}$ (1)

We can see that the vector a in the figure corresponds to the linear position of the particle or to the rigid body center of mass:

$a=r_x\hat{x}+r_y\hat{y}+r_z\hat{z}$ (2)

So, we can simplify the representation of the multivector as:

$R=r_0+a+r_{xy}\hat{x}\hat{y}+r_{yz}\hat{y}\hat{z}+r_{zx}\hat{z}\hat{x}+r_{xyz}\hat{x}\hat{y}\hat{z}$ (3)

Now let’s go to the bivectors. In Figure 4 you can see that there is a bivector b^c that represents the orientation of a preferred plane in the particle/rigid body. This is, if you select a preferred plane solidary to the particle/rigid body, it tells us the orientation of this plane at a certain time. To define this orientation, you need a coefficient per basis bivector (the same as to define a vector you need the sum of three basis vectors, for bivectors works the same). So:

$b^c=r_{xy}\hat{x}\hat{y}+r_{yz}\hat{y}\hat{z}+r_{zx}\hat{z}\hat{x}$ (4)

Introducing in R:

$R=r_0+a+b^c+r_{xyz}\hat{x}\hat{y}\hat{z}$ (5)

You can see that in a unique multivector R we are having the position and the orientation in the same expression. We have the sufficient degrees of freedom in the expression of the multivector R to give all this information just in one entity (the multivector R).

There are two other components $r_0$ (the scalar) and $\hat{x}\hat{y}\hat{z}$ (the trivector) that I will explain in the next chapter.

For information, this realm of Geometric Algebra that considers three spatial dimensions, and the eight degrees of freedom (or eight types of elements) created by them is called Geometric Algebra Cl_3,0.

And for orthonormal bases in Euclidean metric, the following rules apply in Cl_3,0 [3] - [5] [8] :

${\hat{x}}^2=\hat{x}\hat{x}=1$

${\hat{y}}^2=\hat{y}\hat{y}=1$

${\hat{z}}^2=\hat{z}\hat{z}=1$

$\hat{x}\hat{y}=-\hat{y}\hat{x}$

$\hat{y}\hat{z}=-\hat{z}\hat{y}$

$\hat{z}\hat{x}=-\hat{x}\hat{z}$

${\left(\hat{x}\hat{y}\right)}^2=\hat{x}\hat{y}\hat{x}\hat{y}=-\hat{y}\hat{x}\hat{x}\hat{y}=-1$

${\left(\hat{y}\hat{z}\right)}^2=\hat{y}\hat{z}\hat{y}\hat{z}=-\hat{z}\hat{y}\hat{y}\hat{z}=-1$

${\left(\hat{z}\hat{x}\right)}^2=\hat{z}\hat{x}\hat{z}\hat{x}=-\hat{x}\hat{z}\hat{z}\hat{x}=-1$

${\left(\hat{x}\hat{y}\hat{z}\right)}^2=\hat{x}\hat{y}\hat{z}\hat{x}\hat{y}\hat{z}=-1$

I am not going to explain a lot here and the reason is because what you are going to hear is very difficult to believe and digest. You can check papers [3] - [6] to check all the info that corroborates what I am going to tell now.

In Geometric Algebra, it is not necessary that the time is a fourth dimension of the space-time (the classical 3 space dimensions and one 4^th time dimension).

In Geometric Algebra, the time can be the 8^th degree of freedom of the 8 degrees of freedom (or dimensions created by the GA itself). The time is emerging as one of the dimensions that appear automatically when the three spatial dimensions exist.

This is, the basis vector of the time is not a separate vector $\hat{t}$ but it is the trivector $\hat{x}\hat{y}\hat{z}$ already commented. The main reasons to consider this are:

So, you will check that from this point on, we will consider always the trivector as the basis vector of time. This does not mean that the trivector could not mean other things depending on the context (sometimes, it could be related to spin [7] or to the electromagnetic trivector see chapter 7). The same than a vector can sometimes represent a position, others a force etc. The trivector is just a tool that has certain properties, and these properties match perfectly with the properties of what we perceive as time.

Anyhow, that the trivector represents at the same time the volume and the time could be a hint that somehow, they are related. And the time could be a kind of measurement of the continuous creation of volume in the universe (you can check different mechanisms of creation of volume by the masses in the universe in [10] [11] ).

After this shock, we continue with the other pending item of the previous chapter, this is, r₀. The meaning of this element is more obscure. As I have commented, the scalars in the multivectors are a kind of scalation factor that affects all the magnitudes that are multiplied by it.

So, it could be related to a kind of scalation in the metric appearing in non-Euclidean metrics (kind of local Ricci scalar or trace of the metric tensor). See [7] for example.

Another simpler interpretation for r₀, is that the scalars appear when we multiply or divide vectors (or bivectors or the trivector) by themselves. So, sometimes it is necessary a degree of freedom to accommodate these results when they appear. For example, in [4] [5] the current density through time, sometimes is accompanied by the trivector and other times is just a scalar depending on the operations that have been performed before.

In the papers [9] [12] it was already made a direct relation between a spinor in matrix formalism (a complex 4-vector) with a multivector in Geometric Algebra Cl_3,0. It was used the Dirac equation, leading for a one-to-one map between these two worlds.

But things could be even more simple. Let, us consider this spinor:

$\psi =\left(\begin{array}{c}{\psi }_1\\ {\psi }_2\\ {\psi }_3\\ {\psi }_4\end{array}\right)$

If we want to project it, in the normally considered space-time 4 dimensions, we can do it pre-multiplying by a raw vector composed by its basis vectors:

$\left(\begin{array}{cccc}\hat{x}& \hat{y}& \hat{z}& \hat{t}\end{array}\right)\left(\begin{array}{c}{\psi }_1\\ {\psi }_2\\ {\psi }_3\\ {\psi }_4\end{array}\right)=\hat{x}{\psi }_1+\hat{y}{\psi }_2+\hat{z}{\psi }_3+\hat{t}{\psi }_4={\psi }_1\hat{x}+{\psi }_2\hat{y}+{\psi }_3\hat{z}+{\psi }_4\hat{t}$

Where the last step can be done because scalars and vectors commute. Considering that the time is the trivector, as commented in chapter 3, and using the straight-forward convention:

$\hat{t}=\hat{x}\hat{y}\hat{z}$

We will get:

${\psi }_1\hat{x}+{\psi }_2\hat{y}+{\psi }_3\hat{z}+{\psi }_4\hat{x}\hat{y}\hat{z}$

(For information, in other papers [3] - [6] [8] [12] it is used the opposite convention $\hat{t}=\hat{z}\hat{y}\hat{x}$).

With this move, there is little gain, as we have just associated the four components of the column vector to a dimension. And, if consider that each component ${\psi }_i$ is a complex number, we are in the darkness again.

The solution is simple. Let’s make the same thing but now, knowing that each component is a complex number and taking action on that.

Again, we have:

$\psi =\left(\begin{array}{c}{\psi }_1\\ {\psi }_2\\ {\psi }_3\\ {\psi }_4\end{array}\right)$

Now, let’s decompose each component in the corresponding complex number:

$\psi =\left(\begin{array}{c}{\psi }_{1r}+i{\psi }_{1i}\\ {\psi }_{2r}+i{\psi }_{2i}\\ {\psi }_{3r}+i{\psi }_{3i}\\ {\psi }_{4r}+i{\psi }_{4i}\end{array}\right)$

Now, following the rules that the imaginary unit represents a bivector when it is any specific direction/orientation related and it corresponds to the trivector when this is not the case [3] - [6] . For a general complex number, we are always in the second case. So:

$\psi =\left(\begin{array}{c}{\psi }_{1r}+{\psi }_{1i}\hat{x}\hat{y}\hat{z}\\ {\psi }_{2r}+{\psi }_{2i}\hat{x}\hat{y}\hat{z}\\ {\psi }_{3r}+{\psi }_{3i}\hat{x}\hat{y}\hat{z}\\ {\psi }_{4r}+{\psi }_{4i}\hat{x}\hat{y}\hat{z}\end{array}\right)$

Now, projecting again to the usually considered space-time dimensions:

$\begin{array}{l}\left(\begin{array}{cccc}\hat{x}& \hat{y}& \hat{z}& \hat{t}\end{array}\right)\left(\begin{array}{c}{\psi }_{1r}+{\psi }_{1i}\hat{x}\hat{y}\hat{z}\\ {\psi }_{2r}+{\psi }_{2i}\hat{x}\hat{y}\hat{z}\\ {\psi }_{3r}+{\psi }_{3i}\hat{x}\hat{y}\hat{z}\\ {\psi }_{4r}+{\psi }_{4i}\hat{x}\hat{y}\hat{z}\end{array}\right)\\ =\hat{x}{\psi }_{1r}+\hat{x}{\psi }_{1i}\hat{x}\hat{y}\hat{z}+\hat{y}{\psi }_{2r}+\hat{y}{\psi }_{2i}\hat{x}\hat{y}\hat{z}+\hat{z}{\psi }_{3r}+\hat{z}{\psi }_{3i}\hat{x}\hat{y}\hat{z}+\hat{t}{\psi }_{4r}+\hat{t}{\psi }_{4i}\hat{x}\hat{y}\hat{z}\end{array}$

Substituting again and operating:

$\hat{t}=\hat{x}\hat{y}\hat{z}$

$\begin{array}{l}{\psi }_{1r}\hat{x}+{\psi }_{1i}\hat{y}\hat{z}+{\psi }_{2r}\hat{y}-{\psi }_{2i}\hat{x}\hat{z}+{\psi }_{3r}\hat{z}+{\psi }_{3i}\hat{x}\hat{y}+{\psi }_{4r}\hat{x}\hat{y}\hat{z}+{\psi }_{4i}\hat{x}\hat{y}\hat{z}\hat{x}\hat{y}\hat{z}\\ ={\psi }_{1r}\hat{x}+{\psi }_{1i}\hat{y}\hat{z}+{\psi }_{2r}\hat{y}+{\psi }_{2i}\hat{z}\hat{x}+{\psi }_{3r}\hat{z}+{\psi }_{3i}\hat{x}\hat{y}+{\psi }_{4r}\hat{x}\hat{y}\hat{z}-{\psi }_{4i}\end{array}$

You can see that we have arrived to the already commented multivector of 8 degrees of freedom (8 dimensions if you want) but starting from the three spatial dimensions. No special magic or hidden tiny dimension is necessary. Just geometric. The three special dimensions lead to three vectors, three bivectors and one trivector (plus the scalars), giving a total of 8 dimensions.

Even the time is not an ad-hoc dimension. It (the time trivector) emerges naturally from the three spatial dimensions. Somehow, the human being perceives the odd-grade dimensions (vectors of space and the trivector of time) as real dimensions. And the even-grade dimensions (scalars and bivectors) otherwise. Probably as orientations or forces or relations between entities. Just as we perceive the visible light with our eyes and the infrared light as heat, somehow, we perceive differently the dimensions (or degrees of freedom if you want) of the world.

If you want to check a proper formal relation between a spinor in matrix algebra and in GA algebra using the Dirac equation you can check [3] [12] to arrive to:

Dirac Equation in GA:

$\left(\hat{x}\hat{y}\hat{z}\frac{\partial }{\partial t}-\hat{y}\hat{z}\frac{\partial }{\partial x}-\hat{z}\hat{x}\frac{\partial }{\partial y}-\hat{x}\hat{y}\frac{\partial }{\partial z}\right)\psi -m{\psi }_{even}\hat{z}+m{\psi }_{odd}\hat{z}=0$

where:

${\psi }_{even}={\psi }_0+\hat{x}\hat{y}{\psi }_{xy}+\hat{y}\hat{z}{\psi }_{yz}+\hat{z}\hat{x}{\psi }_{zx}$

${\psi }_{odd}=\hat{x}{\psi }_x+\hat{y}{\psi }_y+\hat{z}{\psi }_z+\hat{x}\hat{y}\hat{z}{\psi }_{xyz}$

$\psi ={\psi }_{even}+{\psi }_{odd}={\psi }_0+\hat{x}{\psi }_x+\hat{y}{\psi }_y+\hat{z}{\psi }_z+\hat{x}\hat{y}{\psi }_{xy}+\hat{y}\hat{z}{\psi }_{yz}+\hat{z}\hat{x}{\psi }_{zx}+\hat{x}\hat{y}\hat{z}{\psi }_{xyz}$

If the wavefunction solution in Matrix Algebra is defined as:

$\psi =\left(\begin{array}{c}{\psi }_1\\ {\psi }_2\\ {\psi }_3\\ {\psi }_4\end{array}\right)=\left(\begin{array}{c}{\psi }_{1r}+i{\psi }_{1i}\\ {\psi }_{2r}+i{\psi }_{2i}\\ {\psi }_{3r}+i{\psi }_{3i}\\ {\psi }_{4r}+i{\psi }_{4i}\end{array}\right)$

There is a one-to-one mapping of both representations:

${\psi }_{1r}=-{\psi }_y$

${\psi }_{1i}=-{\psi }_x$

${\psi }_{2r}={\psi }_{xyz}$

${\psi }_{2i}={\psi }_z$

${\psi }_{3r}=-{\psi }_{yz}$

${\psi }_{3i}={\psi }_{zx}$

${\psi }_{4r}={\psi }_{xy}$

${\psi }_{4i}={\psi }_0$

So, we have seen that considering only three spatial dimensions, and the time as the trivector appearing naturally in the Geometric Algebra, is sufficient to accommodate the Dirac Equation. This means, not an added extra-time dimension is needed in this algebra. Time emerges naturally from the three spatial dimensions of the Geometric Algebra Cl_3,0 and is coherent with the results.

In [6] you can find a complete mapping of Electromagnetism and Lorentz force between classical tensor notation to Geometric Algebra Cl_3,0 realm, where specifically the Lorentz Force equation is represented as:

$\frac{\text{d}p}{\text{d}\tau }=qFU$ (6)

where:

$\frac{\text{d}p}{\text{d}\tau }=\frac{\text{d}{p}_0}{\text{d}\tau }+\frac{\text{d}{p}_{yz}}{\text{d}\tau }\hat{x}+\frac{\text{d}{p}_{zx}}{\text{d}\tau }\hat{y}+\frac{\text{d}{p}_{xy}}{\text{d}\tau }\hat{z}+\frac{\text{d}{p}_x}{\text{d}\tau }\hat{y}\hat{z}+\frac{\text{d}{p}_y}{\text{d}\tau }\hat{z}\hat{x}+\frac{\text{d}{p}_z}{\text{d}\tau }\hat{x}\hat{y}+\frac{\text{d}{p}_{xyz}}{\text{d}\tau }\hat{x}\hat{y}\hat{z}$ (7)

$F=E_x\hat{x}+E_y\hat{y}+E_z\hat{z}+B_x\hat{y}\hat{z}+B_y\hat{z}\hat{x}+B_z\hat{x}\hat{y}$ (8)

$U=U_{xyz}\hat{x}\hat{y}\hat{z}+U_x\hat{y}\hat{z}+U_y\hat{z}\hat{x}+U_z\hat{x}\hat{y}$ (9)

Getting the following equations:

$\frac{\text{d}{p}_x}{\text{d}\tau }=q\left(E_x{U}_{xyz}-B_y{U}_z+B_z{U}_y\right)$ (10)

$\frac{\text{d}{p}_y}{\text{d}\tau }=q\left(E_y{U}_{xyz}+B_x{U}_z-B_z{U}_x\right)$ (11)

$\frac{\text{d}{p}_z}{\text{d}\tau }=q\left(E_z{U}_{xyz}-B_x{U}_y+B_y{U}_x\right)$ (12)

$\frac{\text{d}{p}_{xyz}}{\text{d}\tau }=q\left(E_x{U}_x+E_y{U}_y+E_z{U}_z\right)$ (13)

That corresponds one to one with the Lorentz Force in the covariant formalism:

$\frac{\text{d}{p}_4}{\text{d}\tau }=q\left(E_x{u}^1+E_y{u}^2+E_z{u}^3\right)$ (14)

$\frac{\text{d}{p}_1}{\text{d}\tau }=q\left(-E_x{u}^4-B_z{u}^2+B_y{u}^3\right)$ (15)

$\frac{\text{d}{p}_2}{\text{d}\tau }=q\left(-E_y{u}^4+B_z{u}^1-B_x{u}^3\right)$ (16)

$\frac{\text{d}{p}_3}{\text{d}\tau }=q\left(-E_z{u}^4-B_y{u}^1+B_x{u}^2\right)$ (17)

With the following equivalences when comparing Geometric Algebra notation to covariant tensor notation [4] , so all the considerations taken into account in Geometric Algebra notation (for example, considering time as the trivector of the Cl_3,0 Geometric Algebra) are substantiated:

$u^4=U_{xyz}{u}^1=U_x{u}^2=U_y{u}^3=U_z$ (18)

$\frac{\text{d}{p}_{xyz}}{\text{d}\tau }=\frac{\text{d}{p}_4}{\text{d}\tau }\frac{\text{d}{p}_x}{\text{d}\tau }=-\frac{\text{d}{p}_1}{\text{d}\tau }\frac{\text{d}{p}_y}{\text{d}\tau }=-\frac{\text{d}{p}_2}{\text{d}\tau }\frac{\text{d}{p}_z}{\text{d}\tau }=-\frac{\text{d}{p}_3}{\text{d}\tau }$ (19)

Similarly, in [5] we make a complete comparison of Maxwell laws from classical tensor notation to Geometric Algebra Cl_3,0. Specifically, the Maxwell laws in Geometric Algebra are put together in one just equation:

$\nabla F=J$

where:

$\nabla =\frac{\partial }{\partial x}\hat{x}+\frac{\partial }{\partial y}\hat{y}+\frac{\partial }{\partial z}\hat{z}+\frac{\partial }{\partial t}$

$F=E_x\hat{x}+E_y\hat{y}+E_z\hat{z}+B_x\hat{y}\hat{z}+B_y\hat{z}\hat{x}+B_z\hat{x}\hat{y}$

$J=J_x\hat{x}+J_y\hat{y}+J_z\hat{z}+J_0$

This is:

$\begin{array}{l}\left(\frac{\partial }{\partial x}\hat{x}+\frac{\partial }{\partial y}\hat{y}+\frac{\partial }{\partial z}\hat{z}+\frac{\partial }{\partial t}\right)\left(E_x\hat{x}+E_y\hat{y}+E_z\hat{z}+B_x\hat{y}\hat{z}+B_y\hat{z}\hat{x}+B_z\hat{x}\hat{y}\right)\\ =J_x\hat{x}+J_y\hat{y}+J_z\hat{z}+J_0\end{array}$

Operating and obtaining an equation for each element of the Geometric Algebra (scalars, vectors, bivectors and trivector) we get eight equations:

$\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=J_0$

$\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}+\frac{\partial B_z}{\partial t}=0$

$-\frac{\partial E_z}{\partial x}+\frac{\partial E_x}{\partial z}+\frac{\partial B_y}{\partial t}=0$

$\frac{\partial B_x}{\partial x}+\frac{\partial B_y}{\partial y}+\frac{\partial B_z}{\partial z}=0$

$-\frac{\partial B_y}{\partial x}+\frac{\partial B_x}{\partial y}+\frac{\partial E_z}{\partial t}=J_z$

$\frac{\partial B_z}{\partial x}-\frac{\partial B_x}{\partial z}+\frac{\partial E_y}{\partial t}=J_y$

$\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}+\frac{\partial B_x}{\partial t}=0$

$-\frac{\partial B_z}{\partial y}+\frac{\partial B_y}{\partial z}+\frac{\partial E_x}{\partial t}=J_x$

which are exactly the Maxwell laws in classical tensor notation if we consider the following relations between both realms (Geometric Algebra and classical tensor notation):

$J_x=-J^x$

$J_y=-J^y$

$J_z=-J^z$

$J_0=J^t$

So, all the considerations taken into account in Geometric Algebra notation (for example, considering time as the trivector of the Cl_3,0 Geometric Algebra) are substantiated. You can check all the details in [5] .

More about the trivector. We have commented that the trivector is the basis vector of time, but it is involved in more tricky things.

In the papers [4] [5] when considering the electromagnetic field, we had the vectors and the bivectors as the electric and the magnetic field. But we saw that also the electromagnetic trivector could affect the particles. Probably just rotating them or creating a zitterbewegung movement. The issue is that the thing could go even further and be (the trivector) the creator of the Electromagnetic Field itself.

Check it this way. Imagine that there is an electromagnetic trivector field everywhere acting on currents (vectors) and on magnetic fields (bivectors).

If we have a current in direction $\hat{x}$, the “omnipresent” trivector acts (it is post-multiplied) on it, creating a bivector:

$\left(\hat{x}\right)\left(\hat{x}\hat{y}\hat{z}\right)=\hat{x}\hat{x}\hat{y}\hat{z}=\hat{y}\hat{z}$

(We are in the convention of the trivector $\hat{x}\hat{y}\hat{z}$, in the case of $\hat{z}\hat{y}\hat{x}$ we should premultiply).

This means, the omnipresent trivector converts a current (a “vector-directed” charge) in a bivector (magnetic field).

The opposite can also happen. A magnetic bivector field (under certain circumstances as varying during time) could create vector-directed currents:

$\left(\hat{x}\hat{y}\right)\left(\hat{x}\hat{y}\hat{z}\right)=\hat{x}\hat{y}\hat{x}\hat{y}\hat{z}=-\hat{x}\hat{x}\hat{y}\hat{y}\hat{z}=-\hat{z}$

This is nothing but the Maxwell equations or the Lorentz force. But with a difference. If we lived in a world with an omnipresent trivector but in with opposite direction, the Maxwell laws and the Lorentz force would be inverted. The right-hand rule would be transformed to the left-hand rule when calculating the direction of a magnetic field created by a current and vice-versa.

$\left(\hat{x}\right)\left(\hat{z}\hat{y}\hat{x}\right)=\hat{x}\hat{z}\hat{y}\hat{x}=\hat{z}\hat{y}=-\hat{y}\hat{z}$

The magnetic field would be the opposite as before.

The same with a current:

$\left(\hat{x}\hat{y}\right)\left(\hat{z}\hat{y}\hat{x}\right)=\hat{x}\hat{y}\hat{z}\hat{y}\hat{x}=-\hat{x}\hat{x}\hat{y}\hat{z}\hat{y}=-\hat{y}\hat{z}\hat{y}=\hat{z}$

So, the question is clear, can we create a “world” with a trivector acting in the opposite direction than the one existing and check if this inversion of the handed-rules could happen?

In principle, we should be able. We can create in a laboratory a current with the shape of a trivector ( Figure 5 ):

and then an opposite current with exactly the same shape ( Figure 6 ).

We can see that the currents and the created magnetic fields are opposite in Figure 5 and Figure 6 . (Current $-\hat{x}$ in opposition of $\hat{x}$, and magnetic field bivector $\hat{y}\hat{z}$ in opposition with magnetic bivector $\hat{z}y$ for example.)

But we can check that the trivector created in both cases is the same. It is the trivector $\hat{x}\hat{y}\hat{z}$. You can check on the right in Figure 6 . the arithmetic operation to arrive to that result or see it geometrically, as follows.

If we rotate Figure 6 by $\hat{y}$ axis clockwise direction (90˚), we get ( Figure 7 ):

Now, we rotate by $\hat{x}$ axis 180˚ ( Figure 8 ):

You can check that the trivector created in Figure 8 and in Figure 5 (same as Figure 9 ) is the same.

The name of the axes has changed but the physical trivector created in Figure 5 (same as Figure 9 ) and Figure 8 is the same. Meaning the physical effect (if it exists) of this trivector should be the same in both cases. If you do not understand that both trivectors are the same, you can check Annex A1 for a better explanation.

This means, if we superimpose Figure 5 (same as Figure 9 ) and Figure 6 :

We are superimposing two conductors with opposite direction of currents. We see that the vectors (currents) cancel. Also, the bivectors (magnetic fields) cancel.

But the trivector, is not cancelled. In fact, it is doubled. Both conductors create the same trivector and one is added to the other. Remind that the trivector of Figure 6 is the same as the trivector of Figure 8 (that is the same of Figure 5 and Figure 9 ).

In classical electromagnetism Figure 10 is just a loss of power with no effects. You are consuming electricity to create two opposite currents that do not create anything by themselves. All the effects (magnetic fields) cancel.

In Geometric algebra, it is not the case. It is different a situation with these currents than a situation without them.

If we have a conductor $\hat{v}$ with no other currents involved, it will create a magnetic field $M_{\hat{v}}$ ( Figure 11 ):

Considering that the magnetic field is created by the current with omnipresent trivector we have:

$M_{\hat{v}}=\hat{v}\hat{x}\hat{y}\hat{z}$

But if we have created an artificial trivector in the acting area of $\hat{v}$, the result could be different ( Figure 12 ):

If we consider that the environmental omnipresent trivector is the same value and in the same direction as the ones we have created, we will have:

${M^{\prime }}_{\hat{v}}=\hat{v}\left(\hat{x}\hat{y}\hat{z}+2\hat{x}\hat{y}\hat{z}\right)=3\hat{v}\hat{x}\hat{y}\hat{z}$

If we consider that the ones, we have created have the same value as the environmental one but in opposite direction, we will have:

${M^{\prime }}_{\hat{v}}=\hat{v}\left(\hat{x}\hat{y}\hat{z}-2\hat{x}\hat{y}\hat{z}\right)=-\hat{v}\hat{x}\hat{y}\hat{z}$

This means the Maxwell laws would be inversed, getting an opposite magnetic field than expected.

Of course, this is an extreme case. What we would expect in a real experiment is a little difference between $M_{\hat{v}}$ and ${M^{\prime }}_{\hat{v}}$. And we would notice that this difference changes the sign if we inverse the currents in our artificial trivector.

In fact, I have tried this experiment with very low power and with the cables that convey the opposite currents completely twisted. No difference was observed. Only when the cables were not perfectly twisted, some difference in the magnetic field could be noticed but completely explainable with the magnetic field created by parallel conductors [13] .

The issue is that a high-power experiment should be necessary to counterbalance whatever the trivector environmental value has. It is like trying to detect gravitational effects between two football balls in the surface of the earth. The gravitational force of the earth will shadow whatever gravitational effects between masses in its surface. Only a very precise, controlled experiment can check minimal effects of a force/field when a much bigger force/field of the same type is present during the experiment.

If the environmental trivector is something ad-hoc of the universe or depends somehow on other parameters (mainly the quantity of matter—in opposition of antimatter—present in the area spreading the “matter laws” as right-handed rule etc.) is something that could be studied if finally, this effect of the trivector is found. And its effects can be compared in different situations, (surface of earth, space etc.).

It is important to notice that the trivector here is just considered as another element of the electromagnetic field, separated from other forces. But as we will see later, a field that is everywhere affecting all the particles has all the chances to be also related to gravity or other interactions. Also, to be noted that the trivector also represent volume, time, spin [7] . Somehow, there is a relation between all these elements mathematically that could imply a real relation in the physical world.

In the paper [7] gravity is considered as a non-Euclidean metric that can be managed in GA via the scalar products among the basis vectors. This means, we have to define the products:

${\hat{x}}^2={\Vert \hat{x}\Vert }^2=g_{xx}$

${\hat{y}}^2={\Vert \hat{y}\Vert }^2=g_{yy}$

${\hat{z}}^2={\Vert \hat{z}\Vert }^2=g_{zz}$

$\hat{x}\hat{y}=2g_{xy}-\hat{y}\hat{x}$

$\hat{y}\hat{z}=2g_{yz}-\hat{z}\hat{y}$

$\hat{z}\hat{x}=2g_{zx}-\hat{x}\hat{z}$

So, we have only 6 variables to be defined, to define the metric. This is an issue, because in General Relativity (as time is considered a 4^th dimension), we need 10 parameters to define the metric tensor [9] [14] . The metric tensor has 16 parameters, but it is symmetric leading to 10 [9] .

But it has to be considered two things more:

It should be studied if this extra parameter is really necessary or again there is a degree of freedom in the definition of the metric leaving that 6 is sufficient. Or it could be that in GA also the following product should be defined (but it should have been already defined with the previous relations except something I cannot see):

$\hat{x}\hat{y}\hat{z}=2g_{xyz}-\hat{y}\hat{x}\hat{z}$

A detailed study of the gravity using Cl_3,0 is done in [15] [16] where also the Einstein’s equations are adapted to Geometric Algebra Cl_3,0 leading to the following equation when the metric is orthogonal. As you can check it is equal to the original equation just with the added element $-\frac{{\hslash }^2}{m^2{c}^2}R$.

$\frac{8\pi G}{c^4}{T}_{\mu \nu }\left(1-\frac{{\hslash }^2}{m^2{c}^2}R\right)=R_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }R+\text{Λ}{g}_{\mu \nu }$

This element acts reducing the gravitational field created (a kind of dark energy). In [15] [16] it is calculated its value for an empty space in the universe giving it value as:

$R=1.603\text{E}-52{\text{m}}^{-2}$

While the cosmological constant (the effects of the dark energy) has a very similar value of:

$\text{Λ}=1.1056\text{E}-52{\text{m}}^{-2}$

Making this R a candidate for the dar energy of the universe.

In [17] [18] they are calculated the effects of the strong force interaction using geometric algebra Cl_3,0 with the following conclusion. If we consider the original $\psi$:

$\psi ={\psi }_0+{\psi }_x\hat{x}+{\psi }_y\hat{y}+{\psi }_z\hat{z}+{\psi }_{yz}\hat{y}\hat{z}+{\psi }_{zx}\hat{z}\hat{x}+{\psi }_{xy}\hat{x}\hat{y}+{\psi }_{xyz}\hat{x}\hat{y}\hat{z}$

The new ${\psi }^{\prime }$ obtained when applying each of the Gell-Mann matrices ${\lambda }_i$ is:

${\psi }^{\prime }=\left({\lambda }_1\to \psi \right)={\psi }_0+{\psi }_y\hat{x}+{\psi }_x\hat{y}+{\psi }_{zx}\hat{y}\hat{z}+{\psi }_{yz}\hat{z}\hat{x}+{\psi }_{xyz}\hat{x}\hat{y}\hat{z}$

${\psi }^{\prime }=\left({\lambda }_2\to \psi \right)={\psi }_0+{\psi }_{zx}\hat{x}-{\psi }_{yz}\hat{y}-{\psi }_y\hat{y}\hat{z}+{\psi }_x\hat{z}\hat{x}+{\psi }_{xyz}\hat{x}\hat{y}\hat{z}$

${\psi }^{\prime }=\left({\lambda }_3\to \psi \right)={\psi }_0+{\psi }_x\hat{x}-{\psi }_y\hat{y}+{\psi }_{yz}\hat{y}\hat{z}-{\psi }_{zx}\hat{z}\hat{x}+{\psi }_{xyz}\hat{x}\hat{y}\hat{z}$

${\psi }^{\prime }=\left({\lambda }_4\to \psi \right)={\psi }_0+{\psi }_z\hat{x}+{\psi }_x\hat{z}+{\psi }_{xy}\hat{y}\hat{z}+{\psi }_{yz}\hat{x}\hat{y}+{\psi }_{xyz}\hat{x}\hat{y}\hat{z}$

${\psi }^{\prime }=\left({\lambda }_5\to \psi \right)={\psi }_0+{\psi }_{xy}\hat{x}-{\psi }_{yz}\hat{z}-{\psi }_z\hat{y}\hat{z}+{\psi }_x\hat{x}\hat{y}+{\psi }_{xyz}\hat{x}\hat{y}\hat{z}$

${\psi }^{\prime }=\left({\lambda }_6\to \psi \right)={\psi }_0+{\psi }_z\hat{y}+{\psi }_y\hat{z}+{\psi }_{xy}\hat{z}\hat{x}+{\psi }_{zx}\hat{x}\hat{y}+{\psi }_{xyz}\hat{x}\hat{y}\hat{z}$

${\psi }^{\prime }=\left({\lambda }_7\to \psi \right)={\psi }_0+{\psi }_{xy}\hat{y}-{\psi }_{zx}\hat{z}-{\psi }_z\hat{z}\hat{x}+{\psi }_y\hat{x}\hat{y}+{\psi }_{xyz}\hat{x}\hat{y}\hat{z}$

$\begin{array}{l}{\psi }^{\prime }=\left({\lambda }_8\to \psi \right)={\psi }_0+\frac{1}{\sqrt{3}}{\psi }_x\hat{x}+\frac{1}{\sqrt{3}}{\psi }_y\hat{y}-\frac{2}{\sqrt{3}}{\psi }_z\hat{z}+\frac{1}{\sqrt{3}}{\psi }_{yz}\hat{y}\hat{z}\\ +\frac{1}{\sqrt{3}}{\psi }_{zx}\hat{z}\hat{x}-\frac{2}{\sqrt{3}}{\psi }_{xy}\hat{x}\hat{y}+{\psi }_{xyz}\hat{x}\hat{y}\hat{z}\end{array}$

So, all the transformation of the Gell-Mann matrices (Strong Force interaction) can be translated to Geometric Algebra transformations.

In paper [19] , it is obtained the left and the right-handed representation (chirality) of the wavefunction using Geometric (real Clifford) Algebra Cl_3,0. Having the wavefunction $\psi$:

$\psi ={\psi }^0+{\psi }^1{e}_1+{\psi }^2{e}_2+{\psi }^3{e}_3+{\psi }^{23}{e}_{23}+{\psi }^{31}{e}_{31}+{\psi }^{12}{e}_{12}+{\psi }^{123}{e}_{123}$

In Chiral basis, the separation between left and right-handed elements is explicit:

${\psi }_L={\psi }^1{e}_1+{\psi }^2{e}_2+{\psi }^3{e}_3+{\psi }^{123}{e}_{123}$

${\psi }_R={\psi }^0+{\psi }^{23}{e}_{23}+{\psi }^{31}{e}_{31}+{\psi }^{12}{e}_{12}$

In Pauli/Dirac basis, this explicit separation is not possible, and the result is as follows:

$\begin{array}{c}{\psi }_L=\frac{1}{2}(\left({\psi }^3+{\psi }^0\right)+\left(+{\psi }^1+{\psi }^{31}\right)e_1+\left({\psi }^2-{\psi }^{23}\right)e_2+\left({\psi }^3+{\psi }^0\right)e_3\\ +\left(-{\psi }^2+{\psi }^{23}\right)e_{23}+\left({\psi }^1+{\psi }^{31}\right)e_{31}+\left({\psi }^{123}+{\psi }^{12}\right)e_{12}+\left({\psi }^{123}+{\psi }^{12}\right)e_{123})\end{array}$

$\begin{array}{c}{\psi }_R=\frac{1}{2}(\left(-{\psi }^3+{\psi }^0\right)+\left(+{\psi }^1-{\psi }^{31}\right)e_1+\left({\psi }^2+{\psi }^{23}\right)e_2+\left({\psi }^3-{\psi }^0\right)e_3\\ +\left({\psi }^2+{\psi }^{23}\right)e_{23}+\left(-{\psi }^1+{\psi }^{31}\right)e_{31}+\left(-{\psi }^{123}+{\psi }^{12}\right)e_{12}+\left({\psi }^{123}-{\psi }^{12}\right)e_{123})\end{array}$

Maning that also the chiral interactions (as the weak force interaction) can be translated to Geometric Algebra language.

A. Garret Lisi created the E8 theory [20] - [22] . The summary of this theory is that all the particles and forces existing in our world can be explained using a semi-regular figure of eight dimensions called the E8 polytope. I am very far to have the knowledge-comprehension to understand its depth. The idea is that transformations of particles into other ones and the different interactions among them could be explained as existing in different edges of the figure or via rotations of it.

For me, the incredible thing of this theory is that it has been created as an ad-hoc theory (not related with GA in a direct manner) but leading to the same conclusions, as we will see now.

In fact, this is not exactly correct, as yes there is a relation between both approaches, at least in an indirect manner. All the Lie groups SU(2), (3) (used in the E8 theory) have its direct correspondence with GA or Clifford Algebras. Anyhow, it is surprising how they lead to very similar conclusions anyhow.

Ones to be remarked:

$\left(\hat{x}\hat{y}\hat{z}\frac{\partial }{\partial t}-\hat{y}\hat{z}\frac{\partial }{\partial x}-\hat{z}\hat{x}\frac{\partial }{\partial y}-\hat{x}\hat{y}\frac{\partial }{\partial z}-m+\hat{x}\frac{\partial }{\partial r_{yz}}+\hat{y}\frac{\partial }{\partial r_{zx}}+\hat{z}\frac{\partial }{\partial r_{xy}}+\frac{\partial }{\partial r_0}\right)$

$\begin{array}{l}\left({\psi }_0+\hat{x}{\psi }_x+\hat{y}{\psi }_y+\hat{z}{\psi }_z+\hat{x}\hat{y}{\psi }_{xy}+\hat{y}\hat{z}{\psi }_{yz}+\hat{z}\hat{x}{\psi }_{zx}+\hat{x}\hat{y}\hat{z}{\psi }_{xyz}\right)\\ \times \left(\hat{x}\hat{y}\hat{z}\frac{\partial }{\partial t}-\hat{y}\hat{z}\frac{\partial }{\partial x}-\hat{z}\hat{x}\frac{\partial }{\partial y}-\hat{x}\hat{y}\frac{\partial }{\partial z}-m+\hat{x}\frac{\partial }{\partial r_{yz}}+\hat{y}\frac{\partial }{\partial r_{zx}}+\hat{z}\frac{\partial }{\partial r_{xy}}+\frac{\partial }{\partial r_0}\right)=0\end{array}$

This type of equation (with other multivectors) could lead to the 512 (or 248 free parameters). In this case, this equation as such is never used as the last element of the product is eliminated (as was de facto done in original Dirac equation). See chapter 4 and [3] [12] .

The only point I would try to do another way is the transformation that is done in [22] from the imaginary unit i to matrices. I would clearly exchange the imaginary units and the SU matrices to their equivalents in GA (mainly vectors, bivectors or trivectors) to simplify the calculations but more important to simplify the geometric understanding of the model.

In Geometric Algebra mathematical framework, the three spatial dimensions, expand naturally to entities with 8 degrees of freedom. It has been shown that one of these degrees of freedom (the trivector) can be considered to be the time (so no ad-hoc extra dimension is necessary). Even the issue of the negative signature of time is solved this way, as the square of the trivector is negative by definition.

It has been shown that all this can be checked experimentally looking for the electromagnetic trivector, an entity that should exist according to GA and could be checked experimentally.

Also, some comments regarding the similarities with E8 theory are given. Mainly that E8 theory considers 8 dimensions, exactly the same as emerging naturally in this paper.

But not only that, also some similarities regarding how gravity can be understood, and others are presented.

Bilbao, 3^rd June 2023 (viXra-v1).

Bilbao, 10^th June 2023 (viXra-v2).

Bilbao, 13^th June 2023 (viXra-v2.1).

Bilbao, 16^th June 2023 (viXra-v2.2).

Bilbao, 17^th June 2023 (viXra-v2.3).

Bilbao, 24^th June 2023 (viXra-v2.4).

Bilbao, 25^th June 2023 (viXra-v2.5).

Bilbao, 25^th June 2023 (viXra-v2.6).

Bilbao, 25^th June 2023 (viXra-v2.7).

Bilbao, 25^th July 2023 (viXra-v4.0).

Bilbao, 26^th July 2023 (viXra-v4.1).

Bilbao, 27^th July 2023 (viXra-v4.2).

Bilbao, 27^th July 2023 (viXra-v4.3).

Bilbao, 5^th August 2023 (viXra-v4.4).

Bilbao, 15^th March 2025 (viXra-v4.5).

To my family and friends. To Paco Menéndez and Juan Delcán. To Cine3d [25] , To the current gods of Geometric Algebra: David Hestenes, Chris Doran, Anthony Lasenby, Garret Sobczyk, Joy Christian and many others. To A. Garret Lisi and J.O Weatherall.

If we lived in only one dimension (if we were unidimensional animals in a unidimensional world), we would know only one vector. And always the same. It could change of course, its magnitude or its direction. This means it can be escalated by a scalar real number (positive or negative) but the “original unit” vector will always be the same ( Figure 13 ).

If we lived in two dimensions, we would know only one bivector. The same, it can be escalated or even change its orientation. But the bivector will always represent the same plane. The same as before, that a vector in a unidimensional world represented always the same straight line.

In Figure 14 all the colors (the dotted grey line, the purple, lines, the orange lines, the original red and green lines) represent the same bivector. Those rotations do not mean anything for the bivector. It is the same bivector, as long as it is in the same plane. If you want more info regarding this, you can check [1] [2] [5] [7] [8] .

Now, we come to our world. We are three-dimensional beings in a three-dimensional world.

These two trivectors are the same and even have the same orientation ( Figure 15 , Figure 16 ):

Why? Because we can rotate them in our world to get to the same trivector. The same as we have seen in the two-dimensional world, where all the bivectors were the same because they were in the same plane. All the trivectors in our world are the same (again, magnitude and orientation aside) because they are just rotations of the same trivector in the same 3-D world. A 4-D entity, yes, it could rotate our trivector in a way that we cannot imagine and create a new trivector. But we, poor 3-D mortals will see always the same trivector (magnitude and orientation aside) our whole life.

The transformation of one into another can be done with the rotations commented in chapter 7.

Starting from ( Figure 17 ):

If we rotate above figure by $\hat{y}$ axis clockwise direction (90˚), we get ( Figure 18 ):

Now, we rotate above figure by $\hat{x}$ axis 180˚ ( Figure 19 ):

We have arrived to the same trivector (and in this case, even with the same orientation) as the first figure, that seemed completely opposite in the beginning ( Figure 20 ):

You can check that above both figures represent the same physical entity, even if the nomenclature of the vectors is different.

In [26] , I already commented that the quantum entanglement [23] could be explained as caused by a hidden field omnipresent everywhere. The “random” value (or alternative value) of this field at a given moment would affect both particles in the same way at a certain time, giving the impression that both particles are talking to each other. While in reality, it is the field itself which provokes this “coherent” behavior between both particles without the necessity of any faster than light information.

In [27] Joy Christian, already proposes a disproof of Bell’s theorem [28] using bivectors and the trivector (what he calls I), as the anti-commutative properties of the geometric algebra are sufficient to make his demonstration.

Making a much less formal way of explaining it (my style), we can go with an omnipresent field (that could be formed by vectors, bivectors or the trivector).

Let’s imagine a particle “Amber” which has an internal property (it could be orientation of spin, or whatever even not still known property) that is defined by bivectors:

$\alpha \hat{y}\hat{z}+\beta \hat{z}\hat{x}+\gamma \hat{x}\hat{y}$

Another particle “Blue” quantum entangled with “Amber” (for example its twin in a double particle creation-annihilation process), would have this property as:

$-\alpha \hat{y}\hat{z}-\beta \hat{z}\hat{x}-\gamma \hat{x}\hat{y}$

This is, the opposite of Amber, as quantum entanglement states.

Imagine that there exists an omnipresent (meaning omnipresent as it has the same value in a very large area of the space) trivector field that has an alternative value given by:

$\delta \left(t\right)\hat{x}\hat{y}\hat{z}$

where δ is an alternative scalar that could have a form like this or whatever alternative or random equation depending on time (and only at very large, large scales on space coordinates, so we can consider depending only on time in our experiment):

$\delta =\sin \left(t\right)\text{or}\delta =5\cos \left(t\right)-\sin \left(4t\right)$

or whatver other equation depending on time.

If the omnipresent trivector field acts in the commented property of the particles post-multiplying by it, we would get for “Amber”:

$\left(\alpha \hat{y}\hat{z}+\beta \hat{z}\hat{x}+\gamma \hat{x}\hat{y}\right)\left(\delta \hat{x}\hat{y}\hat{z}\right)=\alpha \delta \hat{y}\hat{z}\hat{x}\hat{y}\hat{z}+\beta \delta \hat{z}\hat{x}\hat{x}\hat{y}\hat{z}+\gamma \delta \hat{x}\hat{y}\hat{x}\hat{y}\hat{z}=-\alpha \delta \hat{x}-\beta \delta \hat{y}-\gamma \delta \hat{z}$

If you do not understand how this product is done (why have some vectors disappear?) and where these negative signs come from, you can check [1] to [8] .

If we do the same for the particle “Blue”:

$\left(-\alpha \hat{y}\hat{z}-\beta \hat{z}\hat{x}-\gamma \hat{x}\hat{y}\right)\left(\delta \hat{x}\hat{y}\hat{z}\right)=\alpha \delta \hat{x}+\beta \delta \hat{y}+\gamma \delta \hat{z}$

Now, if we want to measure whatever property observing in the x axis (let’s consider that this means, for example, we have to post-multiply by the x unit vector):

In Amber:

$\left(-\alpha \delta \hat{x}-\beta \delta \hat{y}-\gamma \delta \hat{z}\right)\left(\hat{x}\right)=-\alpha \delta +\beta \delta \hat{x}\hat{y}-\gamma \delta \hat{z}\hat{x}$

In Blue:

$\left(\alpha \delta \hat{x}+\beta \delta \hat{y}+\gamma \delta \hat{z}\right)\left(\hat{x}\right)=\alpha \delta -\beta \delta \hat{x}\hat{y}+\gamma \delta \hat{z}\hat{x}$

Whatever is the result we want to measure (the scalar αδ or the bivectors in xy or zx plane) we see that the result in Amber and Blue are opposite. This is true as far as δ—that depends on time—is “approximately” the same in both cases. This means, the measurements are really done in both particles at “approximately” the same time.

For the distant observers the result seems “magical”, as the result of the observable measured in one particle is “magically” related to the other one. But in fact, the only issue is that they have measured when a “hidden” field, that is acting on both particles, had the same value at a certain moment. As the field is “hidden”, the result seems “random” and the same for both particles. But it is not random, it is just “unknown” until measured.

Imagine that we measure at a moment when the specific value of $\delta$ is $-{\delta }_0$, we will have a result with different signs as before but, they will always be the opposite in both particles. The “magical” entanglement is always kept, as far the field acting on them keeps being the same. This is, they are measured at “approximately” the same time and in an area where the field does not depend in space coordinates (the field has to be the same or “smooth” in very large areas).

In Amber:

$\left(-\alpha \left(-{\delta }_0\right)\hat{x}-\beta \left(-{\delta }_0\right)\hat{y}-\gamma \left(-{\delta }_0\right)\hat{z}\right)\left(\hat{x}\right)=\alpha {\delta }_0-\beta {\delta }_0\hat{x}\hat{y}+\gamma {\delta }_0\hat{z}\hat{x}$

In Blue:

$\left(\alpha \left(-{\delta }_0\right)\hat{x}+\beta \left(-{\delta }_0\right)\hat{y}+\gamma \left(-{\delta }_0\right)\hat{z}\right)\left(\hat{x}\right)=-\alpha {\delta }_0+\beta {\delta }_0\hat{x}\hat{y}-\gamma {\delta }_0\hat{z}\hat{x}$

As commented in chapters 7 to 9, it could be that the big masses (for example the Earth) force the trivector to be of a certain value near their area of influence (the value of the trivector will depend on time, but negligibly will depend on space coordinates while they are in the area of influence). This would mean that all the experiments near the Earth would result as “entanglement” working well, as the predominant trivector is always the same. Probably at really very large distances where local gravities or other effects are really different, the “entanglement” measurements start to differ. One possible experiment will be to check in the surface of Earth and outside the Earth to check if the entanglement starts failing in the experiments with a higher rate than when they are done for both particles in the surface of Earth.

We have seen that if we can create an artificial trivector which direction is opposite to the existing trivector in space, Maxwell laws could work in reverse direction in its area of influence. We could invert the “right-handed” rule. Or even it could be that whatever effect that we are accustomed to see, works in the opposite way (reversion of time, reversion of entropy inside its area of influence)? Too much sci-fi.

What is real is that we could try to create this “opposite trivector” in a laboratory with the sufficient power to overwhelm the “omnipresent” one. Even, we could try to reduce the needed power creating a “coil of trivector”.

This is, the same that we create a coil of current to sum up its effect to create magnetic fields, we could do something similar to this to help the creation of the trivector ( Figure 21 ):

Adding consecutive trivectors, we can try to sum up its effects. This Figure 21 is very theoretical and not practical. In fact, whatever coil of two twisted cables that is rolled in a cylinder from left to right (or the opposite) creates a trivector. The longer the cylinder, the longer its effect. But it has to be rolled from one side to the other in the same direction all the time, you cannot randomly move from left to right or viceversa. You have to keep the direction of rolling always the same, to sum the effects.

You can check Figure 22 . The green and blue cables are twisted (but represented parallel in the figure for simplicity). The currents in blue and green cables are opposite.

According to Maxwell laws and classical electromagnetism, this configuration is useless. It is only a loss of power, as green and blue cables effect cancel.

In Geometric Algebra, the vectors (currents) and the bivectors (magnetic fields) cancel also. But the trivector acts in the same direction for both cables. As the opposite currents move also in an opposite spatial direction—the axial direction of the cylinder—so both cables act creating the same trivector, and its effects (trivector-wise) are summed.

This is, the individual coil created by each cable creates an opposite bivector in both cables—because their currents are opposite. But the direction of the rolling of the cable in the other dimension—the vector representing the axis of the cylinder—is also opposite. So, the product between the bivector and the vector in both cables leads to the same trivector (the product of two positives is equal to the product of two negatives).

This means, inside the cylinder, the Maxwell laws are altered, and the ${M^{\prime }}_{\hat{v}}$ created by the purple vector $\hat{v}$ in Figure 22 would be different than the magnetic field, $M_{\hat{v}}$ that would be created in empty space ( Figure 23 ). In the most extreme case, they could even have different direction. See chapter 7 for more information.

As a general comment, the Figure 22 in a real experiment very probably would be more practical in vertical position (the axis of the cylinder and the $\hat{v}$ conductor in vertical position, having the coil planes horizontal).

Also, another thing that could be measured inside the cylinder, apart from the magnetic field ${M^{\prime }}_{\hat{v}}$, would be the time. This means, we could put an atomic clock inside the cylinder and another one outside the cylinder. Both synchronized in the beginning of the experiment. And to check their values afterwards.

It could be that to keep the general symmetry of all the laws (including gravity, and the arrow of time/entropy, not only electromagnetism) these laws could also change (at least slightly) if an artificial trivector is modifying the omnipresent one (that normally would define how these laws work). So yes, it could be that we see a difference in the speed of time (in the most extreme sci-fi case a reversion of it, as it could happen with the Maxwell law right-hand rule, but this is just sci-fi, as commented). What we could really expect is a slight change in the measurement of the clocks—if the sufficient power or number of turns is applied.

Also, as commented in chapter 7, it is expected that the Earth field increases the effect of the omnipresent trivector. So, it is expected that the changes in time or in the magnetic field ${M^{\prime }}_{\hat{v}}$, provoked by the new created artificial trivector, would be bigger in space, outside the Earth orbit, than on the Earth surface. So, making this experiment outside the Earth surface would increase the possibilities of success.

In all the papers [3] - [6] [8] [12] , I have considered the Geometric Algebra Cl_3,0 but there is another possibility, that is Cl_0,3. The difference is in the square signature of the vectors and the trivector.

Considering always orthonormal bases, in Geometric Algebra Cl_3,0:

$1^2=1$

${\hat{x}}^2=\hat{x}\hat{x}=1$

${\hat{y}}^2=\hat{y}\hat{y}=1$

${\hat{z}}^2=\hat{z}\hat{z}=1$

${\left(\hat{x}\hat{y}\right)}^2=\hat{x}\hat{y}\hat{x}\hat{y}=-1$

${\left(\hat{y}\hat{z}\right)}^2=\hat{y}\hat{z}\hat{y}\hat{z}=-1$

${\left(\hat{z}\hat{x}\right)}^2=\hat{z}\hat{x}\hat{z}\hat{x}=-1$

${\left(\hat{x}\hat{y}\hat{z}\right)}^2=\hat{x}\hat{y}\hat{z}\hat{x}\hat{y}\hat{z}=-1$

On the other hand, in Geometric Algebra Cl_0,3:

$1^2=1$

${\hat{x}}^2=\hat{x}\hat{x}=-1$

${\hat{y}}^2=\hat{y}\hat{y}=-1$

${\hat{z}}^2=\hat{z}\hat{z}=-1$

${\left(\hat{x}\hat{y}\right)}^2=\hat{x}\hat{y}\hat{x}\hat{y}=-1$

${\left(\hat{y}\hat{z}\right)}^2=\hat{y}\hat{z}\hat{y}\hat{z}=-1$

${\left(\hat{z}\hat{x}\right)}^2=\hat{z}\hat{x}\hat{z}\hat{x}=-1$

${\left(\hat{x}\hat{y}\hat{z}\right)}^2=\hat{x}\hat{y}\hat{z}\hat{x}\hat{y}\hat{z}=1$

In both Algebras (as commented, in at least orthogonal bases) the anticommutative relations are held:

$\hat{x}\hat{y}=-\hat{y}\hat{x}$

$\hat{y}\hat{z}=-\hat{z}\hat{y}$

$\hat{z}\hat{x}=-\hat{x}\hat{z}$

I have made comparisons between both Algebras regarding the the Maxwell Equations, Lorentz Force (and I will do also for the Dirac equation). The difference is mainly (as expected) in signs. In general, they are equivalent just changing some sign conventions or changing the signs when assigning correspondences between Geometric Algebra and its equivalents in Tensor or matrix Algebras.

Anyhow, none of them match so perfectly right that you can consider that one of them is the correct one against the other.

Although it is not the one I have used, I see some advantages in the Cl_0,3 version.