Can a (conservative) four-force be derived from a scalar potential? In special relativity, the four-force is the derivative of the four-momentum with respect to proper time (a Lorentz scalar):
$$f^\mu = \frac{\text{d}P^\mu}{\text{d}\tau} = c \frac{\text{d}P^\mu}{\text{d}s}$$
Assuming that the force is associated with a static (with respect to a specific frame) force field, we should be able to find the force as the derivative of a potential (Lorentz scalar):
$$f^\mu = -g^{\mu\nu}\nabla_\nu \Phi$$
But, this lead to a very dubious "conservation law" different from energy conservation. To see this, multiply both sides of the equation of motion by $U^\mu$ the four-velocity:
$$ \frac{\text{d}P^\mu}{\text{d}\tau} U_\mu = -g^{\mu\nu}(\nabla_\nu \Phi)U_\mu$$
then, we arrive at:
$$\frac{\text{d}}{\text{d}\tau}\left( \frac{m}{2}U^\mu U_\mu+ \Phi \right) = \frac{\text{d}}{\text{d}\tau}\left( -\frac{m}{2}c^2+ \Phi \right) =0 $$
But this is nothing like the conservation of energy since the potential only depends on the coordinates. My question is, does the above mean that it is simply impossible to think that there are conservative forces in relativity that derive from a scalar potential?
 A: Your calculation is consistent, a bit misleading. You should view it as a constraint on the possibilities of $\Phi$ since the force $f$ must be spacelike. This is the case here since $\Phi$ has only spatial dependence in a certain frame.
Also, for the existence of $\Phi$, you need the additional assumption that $\nabla f$ is symmetric, the 4-vector analogue of having a vanishing curl for 3d vector fields.
There is another setting to define a “force deriving from a field”. More generally, in relativity, the Lagrangian formalism is more appropriate. The closest analogue of your heuristic approach is the following action:
$$
S=\int (-m-\Phi)d\tau
$$
From which the Euler-Lagrange equations give:
$$
\frac{d}{d\tau}((m+\Phi)U)=\nabla \Phi \\
P=(m+\Phi)U\\
f^\mu=\Phi_{,\nu}U^\nu U^\mu-\Phi^{,\mu}
$$
Note the velocity dependence of the force to ensure the automatic orthogonality to $U$. Now if $\Phi$ is invariant under some time-like translation, say along the 4-vector $V$, Noether’s theorem (or simply projecting the above equation with $V$) gives you the conservation law:
$$
E = P^\mu V_\mu
$$
which you can interpret as energy since it is the time component of the momentum 4-vector in a frame moving along $V$.
The trick to make $f$ depend on $U$, loosen the definition of $P$, and appeal to Noether’s theorem for the conservation law isn’t specific to this example and is very general. A more useful example would be electromagnetism, although it is slightly off topic here since the force derives from a 4-vector field $A$ and not a scalar field. In this case, you’ll have:
$$
f^\mu=qF^{\mu\nu}U_\nu
$$
with $F^{\mu\nu}=A^{\nu,\mu}-A^{\mu,\nu}$ the Faraday tensor which also gives a velocity dependent force. The new momentum this time is $P=mU-qA$ and similarly in the case of a time-like translation invariant field, you have the conservation of $P^\mu V_\mu$ which is the relativistic analogue of what you’ve seen in non-relativistic EM.
Hope this helps and tell me if you need more details.
