# Equation of motion for a particle under a potential in special relativity

The equation of motion of a particle in Newtonian mechanics in 3D under an arbitrary potential $$U$$, is written as $$m\frac{\mathrm{d}^2 \mathbf{r}}{\mathrm{d} t^2}=-\nabla U.$$ Now, my question is, how can this be generalised to Special relativity? I know that the naive answer, $$m\frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=-\partial^{\mu} \Psi$$, where $$\Psi$$ is some relativistic generalisation of potential energy, cannot work, since every four force $$K^{\nu}$$ has to satisfy $$K^{\nu} \dot{x}_{\nu}=0$$, where dot indicates derivative with respect to proper time, so for this shows that the above naive generalisation cannot work, unless $$\Psi$$ is a constant, which makes it physically useless.

How can one solve this caveat, in order to obtain a physically useful generalisation that works in special relativity?

As you point out, if $$K$$ is the force 1-form and $$v$$ the velocity 4-vector, $$K(v) = 0$$. This means that we cannot hope to find a scalar field $$\Psi$$ on space-time that gives $$K$$ by exterior derivative, that is, $$K=\text d\Psi=(\text d_0\Psi, \text d_{(3)}\Psi)$$. To see this, assume that the spatial part of $$K$$ is $$\text d_{(3)}U$$. Then the temporal part must be of the form $$K_0 = \frac{\mathbf v\cdot\nabla U}{\gamma c}$$ which is not the derivative w.r.t. $$t$$ of $$U$$ in general.
One can already see this in electrodynamics, where the force 1-form is proportional to the contraction between Faraday's 2-form with the velocity 4-vector, viz. $$K= \iota_vF$$. Indeed, given that $$F$$ is a 2-form, $$K(v) = (\iota_v F)(v) = 0$$ because of the skew-symmetry.