2026-07-28
Relativistic force as a change in an external parameter
In special relativity, force is defined as the rate of change of relativistic momentum. But if the particle's motion is already contained in the exponent of the external operator \((-\jmath)^b\), then a natural question arises: can force be derived directly from changes in the parameter \(b(t)\)? If so, the external operator will describe not only the particle's kinematics its velocity and momentum but also its relativistic dynamics under the influence of an external force.
Relativistic Force
Earlier, we obtained a geometric representation of the particle's mass, energy, and momentum. The internal parameter \(a\) specifies the frequency of the state and thus determines its energy scale, while the external parameter \(b\) distributes the energy between the mass and momentum projections. At rest, \(b=0\), so all the energy is accounted for by the mass component:
\[ \tag{1} E_0=h\nu=\hbar\omega=m_0c^2. \] Thus, the rest mass \(m_0\), which will be needed later, is not introduced as a new independent quantity. It is already obtained from the internal frequency of the state:
\[ \tag{2} \boxed{ m_0=\frac{h\nu}{c^2} =\frac{\hbar\omega}{c^2} }. \] Now let's pose the following problem: obtain the force directly through the rate of change of the external parameter \(b(t)\), and then compare the result with the standard formula of special relativity for one-dimensional motion:
\[ \tag{3} F =\frac{d}{dt}(\gamma m_0v) =\gamma^3m_0\, \a. \] where: \(\a = dv/dt\) is the acceleration. If both entries match, this will mean that the geometric parameter of the external motion not only specifies the particle's velocity but also correctly reproduces its longitudinal relativistic dynamics.
In the split-geometry model, the parameter \(b\) is included in the full state operator.
\[ \tag{4} J(t)=\jmath^{a(t)}(-\jmath)^{b(t)}, \qquad b=\frac{\arcsin\beta}{\pi}, \qquad \beta=\frac{v}{c}. \] The parameter \(b\) describes the external motion of the particle. Let us show that its rate of change exactly reproduces expression (3).
The relativistic momentum of a particle with rest mass \(m_0\) is
\[ \tag{5} p=\gamma m_0v, \qquad \gamma=\frac{1}{\sqrt{1-\beta^2}}. \] From the definition of \(b\) it follows
\[ \tag{6} \beta=\sin(\pi b), \qquad \gamma=\frac{1}{\cos(\pi b)}. \] Therefore, the momentum is directly expressed through the exponent of the external operator:
\[ \tag{7} p =m_0c\, \frac{\sin(\pi b)}{\cos(\pi b)} =m_0c\,\tan(\pi b). \] Differentiating it with respect to time, we obtain the force
\[ \tag{8} F =\frac{dp}{dt} =\frac{\pi m_0c}{\cos^2(\pi b)} \frac{db}{dt}. \] On the other hand,
\[ \tag{9} \frac{db}{dt} = \frac{\dot\beta} {\pi\sqrt{1-\beta^2}}. \] After substituting formulas (6) and (9) into formula (8), we find
\[ \tag{10} F = \frac{m_0c\dot\beta} {(1-\beta^2)^{3/2}}. \] Since \(c\dot\beta=dv/dt\), we finally obtain
\[ \tag{11} \boxed{ F = \frac{m_0\,\a} {(1-\beta^2)^{3/2}} = \gamma^3m_0\,\a }. \] where: \(\a = dv/dt\) is the acceleration. The resulting expression is completely identical to formula (3). Therefore, the change in the exponent \(b\) of the external operator \((-\jmath)^b\) is consistent with the dynamics of one-dimensional motion in special relativity.
However, it is important to understand the result correctly: the relation \(b=\arcsin\beta/\pi\) is already adopted in the model, so the obtained agreement is not an independent derivation of relativistic dynamics, but a rigorous test of the consistency of the chosen geometric parameterization with it.
The origin of mass as an energy projection is discussed in detail in this work.