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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?

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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.

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In this context, the question that I posted yesterday seems to be relevant. That is, Is there anything like relativistic potential energy? If not, why? We know relativistic force along the direction of velocity, involving γ3. We also have the standard expression relating force and potential energy in Newtonian mechanics, F=−dV/dx where V is a function of x. I don't see any reason why this can not be applied for relativistic case too. Hence I ask, "can we get potential energy in the relativistic case using F=−dV/dx? If not, why?" My intuition tells me that there is no reason to find fault with this expression and it should apply for relativistic case also. If it's application leads to problems, this only shows that all is not well with relativity and it contains a flaw that needs to be eliminated. Please note that we should not think that the relativity theory is fully correct just because it predicts many things successfully.

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