In this paper, the electromagnetic field strength and the Lorentz force in Geometric Algebra Cl_3,0 will be calculated, and their equivalent in the tensor covariant formalism will be compared, explaining the different conclusions.

Afterward, an expanded version of the formulation will be commented on, includingith all its insights, including possible implications in the Dirac Equation.

A very complete introduction regarding Geometric Algebra and its use in physics and engineering can be found in [1] - [3] . Also, in [4] [5] you can find explicit examples of the power of Geometric Algebra to explain different physical disciplines (in this case Quantum Mechanics). The reference [6] will be the main one used to obtain this relation between covariant formalism and Geometric Algebra regarding electromagnetic field strength and the Lorentz force.

If you do not know anything regarding geometric algebra, I strongly recommend you [1] . You have completed your Geometric Algebra study in [2] .

Geometric Algebra Cl_3,0 will be used. This means it has three basis vectors with positive signature and zero basis vectors with negative signature. In Geometric Algebra Cl_3,0, includes three vectors:

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

In this case, the orthonormal basis will be considered (see Figure 1 ), and, in Appendix A1, the differences in the calculations if they were not orthonormal will be explained.

The square of these vectors in Geometric Algebra is its norm to the square. The norm of a vector is a scalar (not a vector anymore). As we have considered the basis as orthonormal, its square is the scalar 1.

${\hat{x}}^2=\hat{x}\hat{x}={\Vert \hat{x}\Vert }^2=1$ (1)

${\hat{y}}^2=\hat{y}\hat{y}={\Vert \hat{y}\Vert }^2=1$ (2)

${\hat{z}}^2=\hat{z}\hat{z}={\Vert \hat{z}\Vert }^2=1$ (3)

In the nomenclature Cl_3,0, the 3 stands for the number of vectors which square is positive and the 0 for the number of basis vectors which squares is negative. In this case, no basis vectors have a negative square (also known as negative signature), so all of them have a positive square (positive signature), that equals +1 in an orthonormal basis.

The basis vectors can be multiplied by each other (this operation is called Geometric Product). For orthonormal or orthogonal bases, this product follows the anticommutative property, this is:

$\hat{x}\hat{y}=-\hat{y}\hat{x}$ (4)

$\hat{y}\hat{z}=-\hat{z}\hat{y}$ (5)

$\hat{z}\hat{x}=-\hat{x}\hat{z}$ (6)

This combination of two vectors via this product is called a bivector. The bivector instead of representing a vector (an oriented segment), it represents an oriented plane. So $\hat{x}\hat{y}$ represents the plane xy with its normal in a certain direction. And $\hat{y}\hat{x}$ represents the same plane xy but with its normal in the opposite direction.

In Geometric Algebra we do not talk about normal vectors anymore. Instead, we talk about the orientation of a theoretical rotation in that plane. See Figure 2 for a visual explanation. Also, in [1] [2] and [3] , you can find more information about the meaning or interpretation of the bivectors.

If we multiply the three vectors, we obtain the trivector (also called pseudoscalar in the literature [1] [2] ):

$\hat{x}\hat{y}\hat{z}=-\hat{y}\hat{x}\hat{z}=\hat{y}\hat{z}\hat{x}=-\hat{z}\hat{y}\hat{x}=\hat{z}\hat{x}\hat{y}=-\hat{x}\hat{z}\hat{y}$ (6.1)

You can check that the same relations as in Equations (4)-(6) apply. So, every time you swap the position of two vectors you have to put a minus sign (or multiply by −1, as you prefer).

The meaning of the trivector is an oriented volume. The same is true for the bivector, which is a plane with two possible orientations. The trivector is a volume with two possible orientations. You can see visual representation in Figure 3 . Again in [1] [2] and [3] you can find more information regarding trivectors.

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

One of the most surprising characteristics of Geometric Algebra is that you can mix scalars with vectors, bivectors and trivectors. And they can be represented as a sum.

The same is done with polynomials or complex numbers, that is, to leave the sum among different components indicated, it is done in Geometric Algebra. This type of element in Geometric Algebra that has different components as scalars, vectors, bivectors etc., is called a multivector. So, the A element in the example above is a multivector.

In a multivector, the vectors and the trivector are called odd-grade elements. The reason is because they are composed of one vector or three vectors (odd grade number).

In a multivector, the scalars and the bivectors are called even grade elements. The reason is because the elements have 0 vectors (the scalars) or 2 vectors (the bivectors). We consider the 0 and 2 even for this purpose.

To have a clear insight into how the operations are performed in Euclidean Geometric Algebra, you can check references [1] - [3] .

For non-Euclidean metric, you can check Appendix A1.

If we multiply a bivector by itself (applying (1) to (6)):

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

$\hat{y}\hat{z}\hat{y}\hat{z}=-1$

$\hat{z}\hat{x}\hat{z}\hat{x}=-1$

We see that the result is −1. The same happens with the trivector:

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

In Geometric Algebra the imaginary or complex numbers are not used and are not necessary. The reason is that there are elements that are already in fact the square root of −1, as the bivectors or the trivector. Instead of using imaginary numbers, in Geometric Algebra Cl_3,0 these will be substituted by bivectors and trivectors with geometric meaning.

The imaginary unit i was defined as “something unknown” (whatever it is) that is the square root of −1. Now, that we have elements that are in fact, known, and are the square root of −1 (the bivectors and the trivector), we can be more specific and use these elements to play this role. We can use this conversion from the i imaginary unit into bivectors and trivector, mainly in Quantum Mechanics, not used in this paper, but yes in [4] [5] .

When the i does not have any preferred spatial direction will be related to the trivector $\hat{x}\hat{y}\hat{z}$.This happens for example when the i appears related to mass, energy or time.

If the i is related to something with a preferred direction like speed or momentum, normally the i is related to a bivector. It will not be used in this paper, but yes in [4] [5] .

Summing up, even if we are in Cl_3,0 with the three basis vectors with positive signature (positive square), the algebra itself has created two types of elements more (bivectors and trivector) which square is negative.

In a multivector we have these two types of elements depending on its square, the scalars and the vectors which square is +1 (positive signature) and the bivectors and the trivector which square is −1 (negative signature).

Another interesting property in Geometric Algebra is that you can take the inverse a vector. We can calculate its value for a basis vector the following way. We start with Equation (1):

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

We premultiply both equations by the inverse of $\hat{x}$:

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

By definition, the product of the inverse of an element by itself is equal to 1.

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

So,

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

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

The inverse of a basis vector in an orthonormal basis is the vector itself. So:

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

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

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

When we have to take the inverse a product of vectors (bivectors or trivectors) you can check in [2] that apart from inverting each element you have to reverse the order of them, this way:

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

or

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

Remember that every time you swap two vectors, you add a minus sign (4) to (6). To convert $\hat{z}\hat{y}\hat{x}$ into $-\hat{x}\hat{y}\hat{z}$ you have two make three swaps, that is the reason of the final negative sign.

We shall use the convention that the division by a vector is to postmultiply by the inverse of that vector. This means, for example, if we want to do the following operation, this will be the result:

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

Remind that we are always talking about orthonormal bases. To get a hint about not orthonormal bases, you should check Annex A1.

There is another operation we can make in Geometric Algebra that is the reversion of a multivector. I will represent this with a line above the multivector. This operation reverses all the internal order of bivectors and trivectors. As an example:

$A=3+5\hat{x}+3\hat{y}+4\hat{y}\hat{z}-2\hat{x}\hat{y}\hat{z}$

$\begin{array}{c}\bar{A}=\overline{\left(3+5\stackrel{^}{x}+3\stackrel{^}{y}+4\stackrel{^}{y}\stackrel{^}{z}-2\stackrel{^}{x}\stackrel{^}{y}\stackrel{^}{z}\right)}\\ =3+5\hat{x}+3\hat{y}+4\hat{z}\hat{y}-2\hat{z}\hat{y}\hat{x}\\ =3+5\hat{x}+3\hat{y}-4\hat{y}\hat{z}+2\hat{x}\hat{y}\hat{z}\end{array}$

You can see that it is similar to a conjugate in complex numbers. It changes the sign of the elements which square is −1 (in this case, are the bivectors and the trivector).

With this, we can define the reverse product. It consists of the product of a multivector by the reverse of itself.

The main characteristic of the reverse product is that if the multivector only has one type of elements with positive square and only one type of elements of negative square, the result of this product is a scalar.

This means, if the multivector only has scalars (positive square) and bivectors (negative square) the reverse product of the multivector will be a scalar. The same if it only has vectors (positive square) and bivectors (negative square). Or vectors (positive square) and trivector (negative square).

But when the multivector has scalars and vectors (both positive square) and a bivector, for example, the result could be not scalar. The same if it has scalars and both bivectors and the trivector (both negative square).

This is, the multivector has to have only scalars or vectors (not both) mixed with only bivectors or trivectors (not both).

Let’s see some examples. B only has vectors (positive square) and the trivector (negative square), the result must be scalar:

$B=5\hat{x}+3\hat{y}+4x\hat{y}\hat{z}$

$\bar{B}=\overline{\left(5\stackrel{^}{x}+3\stackrel{^}{y}+4x\stackrel{^}{y}\stackrel{^}{z}\right)}=5\hat{x}+3\hat{y}-4x\hat{y}\hat{z}$

$\begin{array}{c}B\bar{B}=\left(5\hat{x}+3\hat{y}+4\hat{x}\hat{y}\hat{z}\right)\left(5\hat{x}+3\hat{y}-4\hat{x}\hat{y}\hat{z}\right)\\ =25+15\hat{x}\hat{y}-20\hat{x}\hat{x}\hat{y}\hat{z}+15\hat{y}\hat{x}+9-12\hat{y}\hat{x}\hat{y}\hat{z}+20\hat{x}\hat{y}\hat{z}\hat{x}+12\hat{x}\hat{y}\hat{z}\hat{y}-16\hat{x}\hat{y}\hat{z}\hat{x}\hat{y}\hat{z}\\ =25+15\hat{x}\hat{y}-20\hat{y}\hat{z}-15\hat{x}\hat{y}+9+12\hat{x}\hat{y}\hat{y}\hat{z}-20\hat{x}\hat{y}\hat{x}\hat{z}-12\hat{x}\hat{y}\hat{y}\hat{z}+16\hat{x}\hat{y}\hat{x}\hat{z}\hat{y}\hat{z}\end{array}$

We sum the scalars, we see that the elements in xy sum zero, we square to +1 the vectors that are the same and consecutive and we continue swapping vectors to try to simplify:

$\begin{array}{l}=34-20\hat{y}\hat{z}+12\hat{x}\hat{z}+20\hat{x}\hat{x}\hat{y}\hat{z}-12\hat{x}\hat{z}-16\hat{x}\hat{x}\hat{y}\hat{z}\hat{y}\hat{z}\\ =34-20\hat{y}\hat{z}+12\hat{x}\hat{z}+20\hat{y}\hat{z}-12\hat{x}\hat{z}+16\hat{x}\hat{x}\hat{y}\hat{y}\hat{z}\hat{z}\\ =34+16=50\end{array}$

The result, 50, is a scalar as we expected.

Another example. C has only scalars and the trivector, the result should be scalar:

$C=5+4\hat{x}\hat{y}\hat{z}$

$\bar{C}=\overline{\left(5+4\stackrel{^}{x}\stackrel{^}{y}\stackrel{^}{z}\right)}=5-4\hat{x}\hat{y}\hat{z}$

$C\bar{C}=\left(5+4\hat{x}\hat{y}\hat{z}\right)\left(5-4\hat{x}\hat{y}\hat{z}\right)=25-20\hat{x}\hat{y}\hat{z}+20\hat{x}\hat{y}\hat{z}-16\hat{x}\hat{y}\hat{z}\hat{x}\hat{y}\hat{z}$

The elements in xyz sum zero. Swapping vectors in the last element we get:

$=25+16\hat{x}\hat{y}\hat{x}\hat{z}\hat{y}\hat{z}=25-16\hat{x}\hat{x}\hat{y}\hat{z}\hat{y}\hat{z}=25+16\hat{x}\hat{x}\hat{y}\hat{y}\hat{z}\hat{z}=25+16=31$

31 is a scalar as expected.

New example. D has scalars but has a mix of bivectors and trivectors (the result could be not a scalar):

$D=5+2\hat{x}\hat{y}+3\hat{x}\hat{y}\hat{z}$

$\bar{D}=\overline{\left(5+2\stackrel{^}{x}\stackrel{^}{y}+4\stackrel{^}{x}\stackrel{^}{y}\stackrel{^}{z}\right)}=5-2\hat{x}\hat{y}-3\hat{x}\hat{y}\hat{z}$

$\begin{array}{c}D\bar{D}=\left(5+2\hat{x}\hat{y}+3\hat{x}\hat{y}\hat{z}\right)\left(5-2\hat{x}\hat{y}-3\hat{x}\hat{y}\hat{z}\right)\\ =25-10\hat{x}\hat{y}-15\hat{x}\hat{y}\hat{z}+10\hat{x}\hat{y}-4\hat{x}\hat{y}\hat{x}\hat{y}-6\hat{x}\hat{y}\hat{x}\hat{y}\hat{z}\\ +15\hat{x}\hat{y}\hat{z}-6\hat{x}\hat{y}\hat{z}\hat{x}\hat{y}-9\hat{x}\hat{y}\hat{z}\hat{x}\hat{y}\hat{z}\end{array}$

We see that the terms in $\hat{x}\hat{y}$ and $\hat{x}\hat{y}\hat{z}$ vanish. Also, we swap some vectors:

$\begin{array}{l}=25-4\hat{x}\hat{y}\hat{x}\hat{y}-6\hat{x}\hat{y}\hat{x}\hat{y}\hat{z}-6\hat{x}\hat{y}\hat{z}\hat{x}\hat{y}-9\hat{x}\hat{y}\hat{z}\hat{x}\hat{y}\hat{z}\\ =25+4\hat{x}\hat{x}\hat{y}\hat{y}+6\hat{x}\hat{x}\hat{y}\hat{y}\hat{z}+6\hat{x}\hat{y}\hat{z}\hat{y}\hat{x}\\ =25+4+6\hat{z}-6\hat{x}\hat{y}\hat{y}\hat{z}\hat{x}\\ =29+6\hat{z}-6\hat{x}\hat{z}\hat{x}\\ =29+6\hat{z}+6\hat{x}\hat{x}\hat{z}\\ =29+6\hat{z}+6\hat{z}\\ =29+12\hat{z}\end{array}$

We see that the result is not a scalar as we had both bivectors and the trivector (both of negative signature) in the same multivector.

One important thing to comment about the reverse product is that it acts very similar to the scalar product of an element with himself (the square) in the bra-ket notation of Dirac Algebra [5] .

In the bracket notation of Dirac algebra, when you want to calculate the square of a complex function or vector, you multiply this function or vector by the conjugate of itself, so you always get a real scalar result. This reverse product makes the same, you multiply a multivector by a version of itself where the sign of different elements of this multivector have changed with the aim of obtaining a real scalar as a result.

We have seen that the Geometric Algebra has some elements called multivectors that are composed of scalars, vectors, bivectors and a trivector. In fact, although the Geometric Algebra Cl_3,0 has only three basis vectors, it has really 8 degrees of freedom. A general multivector in Geometric Algebra could have the form (being all the coefficients ${\alpha }_i$ real scalars):

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

This means that although we have only three special dimensions (x, y and z) we have really 8 degrees of freedom (or 8 expanded dimensions in a meta sense) coming from this original three special dimensions.

These eight degrees of freedom are represented by these 8 ${\alpha }_i$ scalars. These scalars are always real. As commented, we do not need imaginary numbers in Geometric Algebra as we have two types of elements (the bivectors and the trivector) which square is −1 and fulfills this necessity.

One comment for the people that have some experience in Geometric Algebra used in Physics. If you are new to Geometric Algebra, please do not read it, so you do not start running.

In most of the literature regarding the use of Geometric Algebra both in Quantum Mechanics and General Relativity the Cl_1,3 or Cl_3,1 is used. This means there are three basis vectors (the spatial dimensions) with one signature and another one (the time) with the opposite signature.

The issue is that these 4 dimensions expand to 16 degrees of freedom. However, in reality, only the sub-even algebra of these 16 degrees of freedom is used (only 8 degrees of the 16 possible are used). So why is this Cl_1,3 or Cl_3,1 used in the first place? We know that with Cl_3,0 we already have the 8 degrees of freedom we need.

The need of Cl_1,3 and Cl_3,1 is to accommodate the time dimension in Geometric Algebra. However, this will be explained in the next chapter as to why this is not necessary anymore.

We have $\hat{x}$, $\hat{y}$ and $\hat{z}$. But we do not have $\hat{t}$. As I have commented in some papers already [3] - [5] , we can use the trivector as the basis vector for the dimension of time. Does this mean that the dimension of time does not exist? No, the dimension of time has its own freedom (its own scalar coefficient t) but the basis vector $\hat{t}$ that accompanies this coefficient is a combination of the space vectors.

I know, it is very difficult to believe but if you continue reading the next chapters, you will see that this works perfectly.

In fact, we can work with the following definition:

${\hat{t}}^{-1}=\hat{x}\hat{y}\hat{z}$ (7)

The reason of why we define the inverse of the basis vector instead of the basis vector itself, will be shown later. Anyhow, following the rules in chapter 5 you can see that for an orthonormal basis (not in general for other bases):

$\hat{t}={\left(\hat{x}\hat{y}\hat{z}\right)}^{-1}={\hat{z}}^{-1}{\hat{y}}^{-1}{x}^{-1}=\hat{z}\hat{y}\hat{x}=-\hat{x}\hat{y}\hat{z}$ (7.1)

So:

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

So, a general multivector will be of the type:

$\begin{array}{c}A={\alpha }_0+{\alpha }_x\hat{x}+{\alpha }_y\hat{y}+{\alpha }_z\hat{z}+{\alpha }_{xy}\hat{x}\hat{y}+{\alpha }_{yz}\hat{y}\hat{z}+{\alpha }_{zx}\hat{z}\hat{x}+{\alpha }_{xyz}\hat{x}\hat{y}\hat{z}\\ ={\alpha }_0+{\alpha }_x\hat{x}+{\alpha }_y\hat{y}+{\alpha }_z\hat{z}+{\alpha }_{xy}\hat{x}\hat{y}+{\alpha }_{yz}\hat{y}\hat{z}+{\alpha }_{zx}\hat{z}\hat{x}-{\alpha }_{xyz}\hat{t}\end{array}$

Even, we reorder putting the time consecutive to the spatial dimensions we would have:

$A={\alpha }_0+{\alpha }_x\hat{x}+{\alpha }_y\hat{y}+{\alpha }_z\hat{z}-{\alpha }_{xyz}\hat{t}+{\alpha }_{xy}\hat{x}\hat{y}+{\alpha }_{yz}\hat{y}\hat{z}+{\alpha }_{zx}\hat{z}\hat{x}$

We can even recall the ${\alpha }_{xyz}$ as ${\alpha }_t$:

${\alpha }_t={\alpha }_{xyz}$

This leads to,

$\begin{array}{c}A={\alpha }_0+{\alpha }_x\hat{x}+{\alpha }_y\hat{y}+{\alpha }_z\hat{z}+{\alpha }_{xy}\hat{x}\hat{y}+{\alpha }_{yz}\hat{y}\hat{z}+{\alpha }_{zx}\hat{z}\hat{x}-{\alpha }_t\hat{t}\\ ={\alpha }_0+{\alpha }_x\hat{x}+{\alpha }_y\hat{y}+{\alpha }_z\hat{z}-{\alpha }_t\hat{t}+{\alpha }_{xy}\hat{x}\hat{y}+{\alpha }_{yz}\hat{y}\hat{z}+{\alpha }_{zx}\hat{z}\hat{x}\end{array}$

You can see that the ${\alpha }_t$ assures that the time has its own freedom compared to the spatial dimensions. But we do not need an original dimension more to accommodate it, it appears naturally in Geometric Algebra. We shall use the more convenient definition we put in the beginning for a multivector:

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

And we shall explain how to work with these multivectors when time is involved.

If you have worked with Geometric Algebra before in Cl_1,3 or Cl_3,1, I give you in advance the following relations we shall use. If you do not what we are talking about, just skip the following equations and continue reading:

$i=I={\sigma }_1{\sigma }_2{\sigma }_3={\gamma }_0{\gamma }_1{\gamma }_2{\gamma }_3={\hat{t}}^{-1}=\hat{x}\hat{y}\hat{z}$ (7.2)

${\sigma }_1={\gamma }_1{\gamma }_0=\hat{x}$ (7.3)

${\sigma }_2={\gamma }_2{\gamma }_0=\hat{y}$ (7.4)

${\sigma }_3={\gamma }_3{\gamma }_0=\hat{z}$ (7.5)

${\gamma }_0\to {\hat{t}}^{-1}=\hat{x}\hat{y}\hat{z}$ (7.6)

As commented before, not always the i will be equal to the trivector, but sometimes to the bivectors also [5] . But in this paper, it will not be necessary to make the distinction. You can check more things regarding $\hat{t}$ as a composition of special vectors in Annex A2.

It seems that the odd-grade elements of the multivector, the vectors and the trivector, are the ones that for whatever reason we perceive as dimensions, three dimensions of space and one of time. And the bivectors and the scalars probably we perceive them in another form like forces, scalation of metrics (GR?) etc.

Regarding non-orthonormal bases or non-Euclidean metric, you can find more information in Annex A1.

Now, we shall start with the work. In this chapter, we shall not use Geometric Algebra. We shall just use the covariant formulation of the Classical Electromagnetism. We shall obtain the applicable equations in that formulation so we can compare them with the ones we shall obtain with Geometric Algebra Cl_3,0. In [6] we can see that Electromagnetic tensor is defined as:

$F_{\alpha \beta }=\left(\begin{array}{cccc}0& \frac{E_x}{c}& \frac{E_y}{c}& \frac{E_z}{c}\\ -\frac{E_x}{c}& 0& -B_z& B_y\\ -\frac{E_y}{c}& B_z& 0& -B_x\\ -\frac{E_z}{c}& -B_y& B_x& 0\end{array}\right)$

As it is normally done in these cases, we can normalize the units so c = 1 and ℏ = 1, leading to:

$F_{\alpha \beta }=\left(\begin{array}{cccc}0& E_x& E_y& E_z\\ -E_x& 0& -B_z& B_y\\ -E_y& B_z& 0& -B_x\\ -E_z& -B_y& B_x& 0\end{array}\right)$ (8)

And the Lorentz Force is defined for this case as:

$\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }=q{F}_{\alpha \beta }{u}^{\beta }$ (9)

where $u^{\beta }$ is the four velocity and $\tau$ is the proper time. And $q$ is the electric charge of the particle subject to the force. $p_{\alpha }$ is the momentum of the particle, so $\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }$ is the variation of the momentum with respect to proper time.

For the covariant formulation, we shall use the indexes 4, 1, 2, 3 for time, x, y, and z dimensions respectively. We shall use for time 4 instead of 0 to avoid a misunderstanding with another element that will appear later. So:

$\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }=\left(\begin{array}{c}\frac{\text{d}{p}_4}{\text{d}\tau }\\ \frac{\text{d}{p}_1}{\text{d}\tau }\\ \frac{\text{d}{p}_2}{\text{d}\tau }\\ \frac{\text{d}{p}_3}{\text{d}\tau }\end{array}\right)$ (10)

$u^{\beta }=\left(\begin{array}{c}{u}^4\\ u^1\\ u^2\\ u^3\end{array}\right)$ (11)

So, the Equation (9) gets the following form:

$\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }=q{F}_{\alpha \beta }{u}^{\beta }$ (9)

$\left(\begin{array}{c}\frac{\text{d}{p}_4}{\text{d}\tau }\\ \frac{\text{d}{p}_1}{\text{d}\tau }\\ \frac{\text{d}{p}_2}{\text{d}\tau }\\ \frac{\text{d}{p}_3}{\text{d}\tau }\end{array}\right)=q\left(\begin{array}{cccc}0& E_x& E_y& E_z\\ -E_x& 0& -B_z& B_y\\ -E_y& B_z& 0& -B_x\\ -E_z& -B_y& B_x& 0\end{array}\right)\left(\begin{array}{c}{u}^4\\ u^1\\ u^2\\ u^3\end{array}\right)$ (12)

Making the operations, we get the following result:

$\frac{\text{d}{p}_4}{\text{d}\tau }=q\left(E_x{u}^1+E_y{u}^2+E_z{u}^3\right)$ (13)

$\frac{\text{d}{p}_1}{\text{d}\tau }=q\left(-E_x{u}^4-B_z{u}^2+B_y{u}^3\right)$ (14)

$\frac{\text{d}{p}_2}{\text{d}\tau }=q\left(-E_y{u}^4+B_z{u}^1-B_x{u}^3\right)$ (15)

$\frac{\text{d}{p}_3}{\text{d}\tau }=q\left(-E_z{u}^4-B_y{u}^1+B_x{u}^2\right)$ (16)

These will be the equation we shall use for comparison with the Geometric Algebra Cl_3,0 result.

Now, we shall try to replicate the same result but with Geometric Algebra Cl_3,0. We shall convert all the elements in Equation (9) to a Geometric Algebra Cl_3,0 form.

$\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }=q{F}_{\alpha \beta }{u}^{\beta }$ (9)

We start with the 4-velocity $u$. Normally the vector associated with velocity is a special one ( $\hat{x}$, $\hat{y}$ or $\hat{z}$) or a combination of them. But, as we saw in [5] the vectors of a momentum (that is the same as velocity multiplied by a scalar, the mass), were bivectors instead of vectors.

Let’s recheck it here why using velocity instead.

The units of velocity are space divided by time. In SI units:

$v_{units}=\frac{\text{space}}{\text{time}}=\frac{\text{m}}{\text{s}}$

If we consider a velocity in the direction of the x axis the vectors would be:

$v_{x\_vectors}=\frac{\hat{x}}{\hat{t}}=\hat{x}{\left(\hat{t}\right)}^{-1}=\hat{x}\hat{x}\hat{y}\hat{z}=\left(1\right)\hat{y}\hat{z}=\hat{y}\hat{z}$ (17)

where we have used the division by a vector convention commented in chapter 5, the equivalence of time with trivector commented in (7) and the square of a basis vector commented in (1).

You can see that the vectors of a velocity are a bivector instead of a vector. In fact, they are the complementary bivector of the vector that we would normally use. In this case, if the direction is x, the bivector is $\hat{y}\hat{z}$.

So, we have the bivectors for the three spatial directions. We need the vector for the time direction. In this case, for that we shall use the trivector as we have considered in Equation (7.6), where ${\gamma }_0$ is the time as considered in Space Time Algebra. You can see in Annex 2 more info about this.

So, the four-velocity vector in Geometric Algebra Cl_3,0 would be:

$U=U_{xyz}\hat{x}\hat{y}\hat{z}+U_x\hat{y}\hat{z}+U_y\hat{z}\hat{x}+U_z\hat{x}\hat{y}$ (18)

where the $U_{xyz}$ is the component through time as already commented.

So, the relation of $U$ with $u^{\beta }$ would be:

$u^4=U_{xyz}{u}^1=U_x{u}^2=U_y{u}^3=U_z$ (18.1)

Coming back to Equation (9):

$\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }=q{F}_{\alpha \beta }{u}^{\beta }$ (9)

Now, we shall convert the electromagnetic Field strength $F_{\alpha \beta }$ to Geometric Algebra.

This has already been studied (for example in [1] and [2] ) with the following result:

$F=E+IB=E+\hat{x}\hat{y}\hat{z}B=E_x\hat{x}+E_y\hat{y}+E_z\hat{z}+B_x\hat{y}\hat{z}+B_y\hat{z}\hat{x}+B_z\hat{x}\hat{y}$ (19)

As this has already been studied and validated, I will not enter it in detail here. Anyhow, I will comment that if you want to follow a process like we did in (17), it won’t exactly work. The units of the electric field are acceleration (factored by the scalars mass and charge).

If we check acceleration:

$a_{units}=\frac{\text{m}}{{\text{s}}^{\text{2}}}$

$a_{x\_vectors}=\frac{\hat{x}}{{\hat{t}}^2}=\hat{x}{\left({\hat{t}}^2\right)}^{-1}=\hat{x}{\left({\hat{t}}^{-1}\right)}^2=\hat{x}\left(-1\right)=-\hat{x}$

So, considering that the sign is a convention, we can say that yes, the electric field has the vector in its direction as vector.

But if we try to do the same with the magnetic field, it will not work. The magnetic field units are acceleration divided by velocity (factored by scalars as mass and charge).

$\frac{a_{units}}{v_{units}}=\frac{\frac{\text{m}}{{\text{s}}^{\text{2}}}}{\frac{\text{m}}{\text{s}}}=\frac{\text{1}}{\text{s}}={\text{s}}^{-1}$

We see that the unit would be the trivector or three units of space vectors if you want. But in any case, would be the trivector or a space vector (if the other two cancel acc (1) to (3)) but never a bivector.

It is clear that the vector units of the magnetic field are the bivector. In fact, the magnetic field is the most “bivector” field I can think about, but this cannot be obtained using the “trick” of the measurement units.

Anyhow, as commented Equation (19) has been validated already in the literature [1] [2] .

The only pending point in Equation (9) is the left side:

$\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }=q{F}_{\alpha \beta }{u}^{\beta }$ (9)

The units of $\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }$ would be momentum/time. Momentum is a velocity factored by the mass scalar, so its units would be the same as the four-velocity (the bivectors). But, if they are divided by time (multiplied by the trivector), the result should be vectors.

But as we have commented before with the magnetic field, this “trick” does not seem to work all the time, and this is one of the cases. We shall see that $\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }$ keeps the same units as the momentum (the bivectors) and the division by time is not considered. Probably because the proper time is considered scalar in opposition to the real time coordinate that would be the trivector? It is not clear. Anyhow, we shall demonstrate with calculations that this is like that.

Anyhow, we shall consider that we do not know anything of this, and we shall define the multivector $\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }$ with all its possible components. But as knowing, the final result, I will exchange the nomenclature of some elements. But the calculations would be the same, I will just exchange some names for convenience with the result we shall obtain:

$\frac{\text{d}p}{\text{d}\tau }=\frac{\text{d}{p}_0}{\text{d}\tau }+\frac{\text{d}{p}_{yz}}{\text{d}\tau }\hat{x}+\frac{\text{d}{p}_{zx}}{\text{d}\tau }\hat{y}+\frac{\text{d}{p}_{xy}}{\text{d}\tau }\hat{z}+\frac{\text{d}{p}_x}{\text{d}\tau }\hat{y}\hat{z}+\frac{\text{d}{p}_y}{\text{d}\tau }\hat{z}\hat{x}+\frac{\text{d}{p}_z}{\text{d}\tau }\hat{x}\hat{y}+\frac{\text{d}{p}_{xyz}}{\text{d}\tau }\hat{x}\hat{y}\hat{z}$ (20)

So now, we have all the elements to reproduce Equation (9) in Geometric Algebra Cl_3,0 using (18)(19) and (20)

$\frac{\text{d}{p}_{\alpha }}{\text{d}\tau }=q{F}_{\alpha \beta }{u}^{\beta }$ (9)

$\frac{\text{d}p}{\text{d}\tau }=qFU$ (21)

Extending all the elements in (18) (19) and (20) in (21) we have:

$\begin{array}{l}\frac{\text{d}{p}_0}{\text{d}\tau }+\frac{\text{d}{p}_{yz}}{\text{d}\tau }\hat{x}+\frac{\text{d}{p}_{zx}}{\text{d}\tau }\hat{y}+\frac{\text{d}{p}_{xy}}{\text{d}\tau }\hat{z}+\frac{\text{d}{p}_x}{\text{d}\tau }\hat{y}\hat{z}+\frac{\text{d}{p}_y}{\text{d}\tau }\hat{z}\hat{x}+\frac{\text{d}{p}_z}{\text{d}\tau }\hat{x}\hat{y}+\frac{\text{d}{p}_{xyz}}{\text{d}\tau }\hat{x}\hat{y}\hat{z}\\ =q\left(E_x\hat{x}+E_y\hat{y}+E_z\hat{z}+B_x\hat{y}\hat{z}+B_y\hat{z}\hat{x}+B_z\hat{x}\hat{y}\right)\left(U_{xyz}\hat{x}\hat{y}\hat{z}+U_x\hat{y}\hat{z}+U_y\hat{z}\hat{x}+U_z\hat{x}\hat{y}\right)\end{array}$ (22)

Making the calculations we have:

(23)

Now, we can get the components that multiply each of the vectors, bivectors or trivector and create separate equations. For example, we get all the elements that multiply by $\hat{y}\hat{z}$ both in the left side and in the right side of the equation, leading to:

$\frac{\text{d}{p}_x}{\text{d}\tau }=q\left(E_x{U}_{xyz}-B_y{U}_z+B_z{U}_y\right)$ (24)

If we get now the elements that multiply by $\hat{z}\hat{x}$ we obtain:

$\frac{\text{d}{p}_y}{\text{d}\tau }=q\left(E_y{U}_{xyz}+B_x{U}_z-B_z{U}_x\right)$ (25)

And so on. Putting all together we have:

$\frac{\text{d}{p}_x}{\text{d}\tau }=q\left(E_x{U}_{xyz}-B_y{U}_z+B_z{U}_y\right)$ (24)

$\frac{\text{d}{p}_y}{\text{d}\tau }=q\left(E_y{U}_{xyz}+B_x{U}_z-B_z{U}_x\right)$ (25)

$\frac{\text{d}{p}_z}{\text{d}\tau }=q\left(E_z{U}_{xyz}-B_x{U}_y+B_y{U}_x\right)$ (26)

$\frac{\text{d}{p}_{yz}}{\text{d}\tau }=q\left(-E_y{U}_z+E_z{U}_y-B_x{U}_{xyz}\right)$ (27)

$\frac{\text{d}{p}_{zx}}{\text{d}\tau }=q\left(E_x{U}_z-E_z{U}_x-B_y{U}_{xyz}\right)$ (28)

$\frac{\text{d}{p}_{xy}}{\text{d}\tau }=q\left(E_y{U}_x+E_x{U}_y-B_z{U}_{xyz}\right)$ (29)

$\frac{\text{d}{p}_{xyz}}{\text{d}\tau }=q\left(E_x{U}_x+E_y{U}_y+E_z{U}_z\right)$ (30)

$\frac{\text{d}{p}_0}{\text{d}\tau }=q\left(-B_x{U}_x-B_y{U}_y-B_z{U}_z\right)$ (31)

Now, if we compare for example Equation (30) obtained with Geometric Algebra Cl_3,0 with Equation (13) obtained with covariant formulation:

$\frac{\text{d}{p}_{xyz}}{\text{d}\tau }=q\left(E_x{U}_x+E_y{U}_y+E_z{U}_z\right)$ (30)

$\frac{\text{d}{p}_4}{\text{d}\tau }=q\left(E_x{u}^1+E_y{u}^2+E_z{u}^3\right)$ (13)

We see that they are the same equation using the following relations:

$u^4=U_{xyz}{u}^1=U_x{u}^2=U_y{u}^3=U_z$ (18.1)

And:

$\frac{\text{d}{p}_{xyz}}{\text{d}\tau }=\frac{\text{d}{p}_4}{\text{d}\tau }$

Now, let us compare Equation (24) with (14) reordering the position of two terms:

$\frac{\text{d}{p}_x}{\text{d}\tau }=q\left(E_x{U}_{xyz}-B_y{U}_z+B_z{U}_y\right)$ (24)

$\frac{\text{d}{p}_1}{\text{d}\tau }=q\left(-E_x{u}^4+B_y{u}^3-B_z{u}^2\right)$ (14)

If we apply 18.1:

$u^4=U_{xyz}{u}^1=U_x{u}^2=U_y{u}^3=U_z$ (18.1)

We see that they are the same equation but with signs changed.

This means, they are the same if we consider:

$\frac{\text{d}{p}_x}{\text{d}\tau }=-\frac{\text{d}{p}_1}{\text{d}\tau }$ (18.2)

We could say that we define the momentum in the other direction and that would work. This is not exactly true, as the four-velocity model we have considered is in the same direction in both formulations. So, it would not be coherent. Another reason could be the Minkowski metric should be multiplying somewhere and change the signs. In GA I try not to use this trick, as I consider that the metric is implicit in in the basis vectors (see Annex A1 of this paper and papers [3] - [5] ).

The third option would be that is the variation of the momentum through time (not the momentum itself) which has another sign as convention in both formulations. And it could be related to the case I have commented regarding the time or the inverse of time (which changes sign). And sometimes, it is difficult to know which to use.

Anyhow, if we consider the sign as a convention (or as an error regarding directions through time that should be corrected) the equation is correct if we define:

$\frac{\text{d}{p}_x}{\text{d}\tau }=-\frac{\text{d}{p}_1}{\text{d}\tau }$ (18.2)

We consider that the variation of momentum has a different sign convention in both formulations.

This way, we have that these four equations are the same in both formulations, so the conversion to Geometric Algebra Cl_3,0 has worked (except this issue with the sign to be studied, that is solved using (18.2)):

$\frac{\text{d}{p}_{xyz}}{\text{d}\tau }=q\left(E_x{U}_x+E_y{U}_y+E_z{U}_z\right)$ (30)

$\frac{\text{d}{p}_x}{\text{d}\tau }=q\left(E_x{U}_{xyz}-B_y{U}_z+B_z{U}_y\right)$ (24)

$\frac{\text{d}{p}_y}{\text{d}\tau }=q\left(E_y{U}_{xyz}+B_x{U}_z-B_z{U}_x\right)$ (25)

$\frac{\text{d}{p}_z}{\text{d}\tau }=q\left(E_z{U}_{xyz}-B_x{U}_y+B_y{U}_x\right)$ (26)

That are equivalent to these in covariant formulation:

$\frac{\text{d}{p}_4}{\text{d}\tau }=q\left(E_x{u}^1+E_y{u}^2+E_z{u}^3\right)$ (13)

$\frac{\text{d}{p}_1}{\text{d}\tau }=q\left(-E_x{u}^4-B_z{u}^2+B_y{u}^3\right)$ (14)

$\frac{\text{d}{p}_2}{\text{d}\tau }=q\left(-E_y{u}^4+B_z{u}^1-B_x{u}^3\right)$ (14)

$\frac{\text{d}{p}_3}{\text{d}\tau }=q\left(-E_z{u}^4-B_y{u}^1+B_x{u}^2\right)$ (16)

Considering:

$u^4=U_{xyz}{u}^1=U_x{u}^2=U_y{u}^3=U_z$ (18.1)

And:

$\frac{\text{d}{p}_{xyz}}{\text{d}\tau }=\frac{\text{d}{p}_4}{\text{d}\tau }\frac{\text{d}{p}_x}{\text{d}\tau }=-\frac{\text{d}{p}_1}{\text{d}\tau }\frac{\text{d}{p}_y}{\text{d}\tau }=-\frac{\text{d}{p}_2}{\text{d}\tau }\frac{\text{d}{p}_z}{\text{d}\tau }=-\frac{\text{d}{p}_3}{\text{d}\tau }$ (32)

So, the above ones are (somehow) ok, but what about the rest?

$\frac{\text{d}{p}_{yz}}{\text{d}\tau }=q\left(-E_y{U}_z+E_z{U}_y-B_x{U}_{xyz}\right)$ (27)

$\frac{\text{d}{p}_{zx}}{\text{d}\tau }=q\left(E_x{U}_z-E_z{U}_x-B_y{U}_{xyz}\right)$ (28)

$\frac{\text{d}{p}_{xy}}{\text{d}\tau }=q\left(E_y{U}_x+E_x{U}_y-B_z{U}_{xyz}\right)$ (29)

$\frac{\text{d}{p}_0}{\text{d}\tau }=q\left(-B_x{U}_x-B_y{U}_y-B_z{U}_z\right)$ (31)

These ones do not appear in the covariant formulation. If we recall the definition we have used for $\frac{\text{d}p}{\text{d}\tau }$:

$\frac{\text{d}p}{\text{d}\tau }=\frac{\text{d}{p}_0}{\text{d}\tau }+\frac{\text{d}{p}_{yz}}{\text{d}\tau }\hat{x}+\frac{\text{d}{p}_{zx}}{\text{d}\tau }\hat{y}+\frac{\text{d}{p}_{xy}}{\text{d}\tau }\hat{z}+\frac{\text{d}{p}_x}{\text{d}\tau }\hat{y}\hat{z}+\frac{\text{d}{p}_y}{\text{d}\tau }\hat{z}\hat{x}+\frac{\text{d}{p}_z}{\text{d}\tau }\hat{x}\hat{y}+\frac{\text{d}{p}_{xyz}}{\text{d}\tau }\hat{x}\hat{y}\hat{z}$ (20)

The linear momentum has units of bivector, as we can see in (33) for example using the direction in x. We have used all the equivalences commented in (17)

$p_{x\_vectors}=\text{kg}\left(\text{scalar}\right)\cdot v_{x\_vectors}=\frac{\hat{x}}{\hat{t}}=\hat{x}{\left(\hat{t}\right)}^{-1}=\hat{x}\hat{x}\hat{y}\hat{z}=\left(1\right)\hat{y}\hat{z}=\hat{y}\hat{z}$ (33)

But the angular momentum is the linear momentum multiplied by a direction of space so its dimensions are vectors not bivectors, as we can see in (34). We consider for example a linear momentum in x, multiplied by a distance in y.

$L_{y\_px}=\text{m}\left(\text{distance}\text{in}y\right)\cdot p_{x\_vectors}=\hat{y}\hat{y}\hat{z}=\left(1\right)\hat{z}=\hat{z}$ (34)

So, in Equation (20) all the coefficients ( $\frac{\text{d}{p}_{yz}}{\text{d}\tau }$, $\frac{\text{d}{p}_{zx}}{\text{d}\tau }$ and $\frac{\text{d}{p}_{xy}}{\text{d}\tau }$) that multiply the single vectors ( $\hat{x}$, $\hat{y}$ or $\hat{z}$ ) could be acting modifying the angular momentum.

The issue is that if it is really acting to the real angular momentum of the particle, this effect should have been already manifested in experiments. So, the only possible explanation would be that this effect is oscillatory by its nature, the forces are changing during time, so the average trajectory of the particle is not affected, only locally in a kind of zitterbewegung (rapid oscillatory movement of the particle) that does not change the “macroscopic” trajectory. The average trajectory is not modified.

Another explanation would be that the angular momentum affected is not the one related to trajectory but the internal one (its own rotation status). This means that the force is affecting the orientation of the axes or the velocity of rotation etc.

Another possibility is that those elements that multiply the single vectors are not the linear momentum, they could be position, over acceleration or others. Anyhow, we always have to take into account that in reality should be very near to zero or oscillatory not changing the average trajectory as they have not been considered until now. They seem something implicit in the oscillatory movements or in the probabilistic nature of certain measurements.

Here, what we see is that the Geometric Algebra shows us that there are other equations that affect parameters that have not been considered in the past. Hidden variables? Let’s go to the next chapter.

Following the philosophy of the previous chapter, let’s extend the Equation (22):

$\begin{array}{l}\frac{\text{d}{p}_0}{\text{d}\tau }+\frac{\text{d}{p}_{yz}}{\text{d}\tau }\hat{x}+\frac{\text{d}{p}_{zx}}{\text{d}\tau }\hat{y}+\frac{\text{d}{p}_{xy}}{\text{d}\tau }\hat{z}+\frac{\text{d}{p}_x}{\text{d}\tau }\hat{y}\hat{z}+\frac{\text{d}{p}_y}{\text{d}\tau }\hat{z}\hat{x}+\frac{\text{d}{p}_z}{\text{d}\tau }\hat{x}\hat{y}+\frac{\text{d}{p}_{xyz}}{\text{d}\tau }\hat{x}\hat{y}\hat{z}\\ =q\left(E_x\hat{x}+E_y\hat{y}+E_z\hat{z}+B_x\hat{y}\hat{z}+B_y\hat{z}\hat{x}+B_z\hat{x}\hat{y}\right)\left(U_{xyz}\hat{x}\hat{y}\hat{z}+U_x\hat{y}\hat{z}+U_y\hat{z}\hat{x}+U_z\hat{x}\hat{y}\right)\end{array}$ (22)

But considering the complete multivector $U$ (I put in bold the addings compared with Equation (18):

$U=U_{xyz}\hat{x}\hat{y}\hat{z}+U_x\hat{y}\hat{z}+U_y\hat{z}\hat{x}+U_z\hat{x}\hat{y}+U_{yz}\hat{x}+U_{zx}\hat{y}+U_{xy}\hat{z}+U_0$ (35)

So here we have added the components $U_{ij}$ of the initial angular momentum (internal as a rotation? Or oscillatory during the trajectory?) in the initial values of $U$. Also, we have introduced the scalar $U_0$ which will be commented later. As commented the $U_{ij}$ could represent other things (position?) that should be checked about. I will consider the angular momentum.

And in the electromagnetic Field I will add the two remaining components in Equation (19):

$F=E_x\hat{x}+E_y\hat{y}+E_z\hat{z}+B_x\hat{y}\hat{z}+B_y\hat{z}\hat{x}+B_z\hat{x}\hat{y}+B_{xyz}\hat{x}\hat{y}\hat{z}+E_0$ (36)

I have introduced the Electromagnetic trivector $B_{xyz}$ and the Electromagnetic scalar $E_0$:

Now, the Equation (19) should read:

$\begin{array}{l}\frac{\text{d}{p}_0}{\text{d}\tau }+\frac{\text{d}{p}_{yz}}{\text{d}\tau }\hat{x}+\frac{\text{d}{p}_{zx}}{\text{d}\tau }\hat{y}+\frac{\text{d}{p}_{xy}}{\text{d}\tau }\hat{z}+\frac{\text{d}{p}_x}{\text{d}\tau }\hat{y}\hat{z}+\frac{\text{d}{p}_y}{\text{d}\tau }\hat{z}\hat{x}+\frac{\text{d}{p}_z}{\text{d}\tau }\hat{x}\hat{y}+\frac{\text{d}{p}_{xyz}}{\text{d}\tau }\hat{x}\hat{y}\hat{z}\\ =q\left(E_x\hat{x}+E_y\hat{y}+E_z\hat{z}+B_x\hat{y}\hat{z}+B_y\hat{z}\hat{x}+B_z\hat{x}\hat{y}+B_{xyz}\hat{x}\hat{y}\hat{z}+E_0\right)\\ \times \left(U_{xyz}\hat{x}\hat{y}\hat{z}+U_x\hat{y}\hat{z}+U_y\hat{z}\hat{x}+U_z\hat{x}\hat{y}+U_{yz}\hat{x}+U_{zx}\hat{y}+U_{xy}\hat{z}+U_0\right)\end{array}$ (37)

If we operate, we get:

(38)

We can see that a lot of new elements appear, leaving the equations as:

$\frac{\text{d}{p}_x}{\text{d}\tau }=q\left(E_x{U}_{xyz}-B_y{U}_z+B_z{U}_y+E_y{U}_{xy}-E_z{U}_{zx}+B_x{U}_0+B_{xyz}{U}_{yz}+E_0{U}_x\right)$ (39)

$\frac{\text{d}{p}_y}{\text{d}\tau }=q\left(E_y{U}_{xyz}+B_x{U}_z-B_z{U}_x-E_x{U}_{xy}+E_z{U}_{yz}+B_y{U}_0+B_{xyz}{U}_{zx}+E_0{U}_y\right)$ (40)

$\frac{\text{d}{p}_z}{\text{d}\tau }=q\left(E_z{U}_{xyz}-B_x{U}_y+B_y{U}_x+E_x{U}_{zx}-E_y{U}_{yz}+B_z{U}_0+B_{xyz}{U}_{xy}+E_0{U}_z\right)$ (41)

$\frac{\text{d}{p}_{yz}}{\text{d}\tau }=q\left(-E_y{U}_z+E_z{U}_y-B_x{U}_{xyz}+E_x{U}_0-B_y{U}_{xy}+B_z{U}_{zx}-B_{xyz}{U}_x+E_0{U}_{yz}\right)$ (42)

$\frac{\text{d}{p}_{zx}}{\text{d}\tau }=q\left(E_x{U}_z-E_z{U}_x-B_y{U}_{xyz}+E_y{U}_0+B_x{U}_{xy}-B_z{U}_{yz}-B_{xyz}+E_0{U}_{zx}\right)$ (43)

$\frac{\text{d}{p}_{xy}}{\text{d}\tau }=q\left(E_y{U}_x+E_x{U}_y-B_z{U}_{xyz}+E_z{U}_0-B_x{U}_{zx}+B_y{U}_{yz}-B_{xyz}{U}_z+E_0{U}_{xy}\right)$ (44)

$\frac{\text{d}{p}_{xyz}}{\text{d}\tau }=q\left(E_x{U}_x+E_y{U}_y+E_z{U}_z+B_x{U}_{yz}+B_y{U}_{zx}+B_z{U}_{xy}+B_{xyz}{U}_0+E_0{U}_{xyz}\right)$ (45)

$\frac{\text{d}{p}_0}{\text{d}\tau }=q\left(-B_x{U}_x-B_y{U}_y-B_z{U}_z+E_x{U}_{yz}+E_y{U}_{zx}+E_z{U}_{xy}-B_{xyz}{U}_{xyz}+E_0{U}_0\right)$ (46)

As commented the elements in bold are new compared to the covariant formalism. So to be correct, they should be very small (almost neglectable to be not detected) or oscillatory provoking local effects but not in the average trajectory measured in the particle.

Let’s start comparing the original equations that had a map relation with the covariant formalism (24) to (26) and (30) with the new ones obtained for them (39) to (41) and (45):

$\frac{\text{d}{p}_x}{\text{d}\tau }=q\left(E_x{U}_{xyz}-B_y{U}_z+B_z{U}_y\right)$ (24)