Is potential energy frame dependent in special relativity? In newtonian mechanics, As far as I'm aware, only kinetic energy is dependent on frames of reference, since kinetic energy is a function of velocity(squared) and velocity is dependent on frames of reference, therefore kinetic energy is frame dependent.
In newtonian mechanics, assuming that all observers use the same reference for zero potential energy($U=0$), it's frame independent . Potential energy of point particles in force fields, like gravity and electric field are given by: $U=constant\dfrac{\xi_1\xi_2}{r}$, where $\xi$ is either the gravitational or the electric charge of a particle. Since both charges and the distance $r$ between them is invariant under all frames of reference, therefore potentential energy is frame independent.
In SR, one expects that the potential energy of an object in a force field should also be a function of the charges and the distance between them. However since the distance $r$ between the charges is relative to the choice of the frame of reference(it's $r$ in a rest frame and $\dfrac{r}{\gamma}$ in a frame that's moving relative to the two charges, owing to lorentz contraction), therefore it seems to me that potential energy becomes frame dependent in SR. 
Is this the case?
 A: Yes potentials are frame dependent. Let us take the electric and magnetic fields as an example. The electric field can be written as: 
$$\vec E=-\frac{1}{c} \frac{\partial \vec A}{\partial t}- \nabla  \phi $$
Where $\vec A$ is a vector potential and $\phi$ a scalar potential. Like wise, the magnetic field can be written as:
$$\vec B=\nabla \times \vec A$$
Where $\vec A$ is the same potential as that that appears in the magnetic field. Associated with these potentials we have a four vector, called  the electromagnetic four-potential and given by:
$$A^\mu=(\phi, A_x, A_y, A_z)^T$$
Like all four vectors this has to transform via the Lorenz transform matrix (assuming the relative motion between the two frames occurs in the $x$ direction):
$$L=\begin{pmatrix} \gamma & -\beta \gamma &0 &0\\ -\beta \gamma & \gamma & 0 &0\\ 0&0&1&0\\ 0&0&0&1\end{pmatrix}$$
Such that:
$$A^{\mu'}=LA^\mu $$
So potential does depend on frame.
A: 
As far as I'm aware, [...] velocity is dependent on frames of reference

No: any one value of velocity is properly determined and attributed to one particlar object (e.g. participant $A$) wrt. to one particular suitable reference system (e.g. system $\mathcal S$); therefore written explicitly as the value $\mathbf v_{\mathcal S}[~A~].$ Any one such value is unambiguous, regardless of considerations of possible other reference systems.

As far as I'm aware, only kinetic energy is dependent on frames of reference

No: any one value of kinetic energy is properly determined and attributed to one particlar object (e.g. participant $A$) wrt. to one particular suitable reference system (e.g. system $\mathcal S$). Any one such value is unambiguous, regardless of considerations of possible other reference systems.
The same argument holds for total energy. Therefore:

Is potential energy frame dependent in special relativity?

Well, it may depend of how to define "potential energy" (or at least: "change of potential energy, between trials"). If the definition is strictly in terms of (total) energy and kinetic energy then it is explicitly not "frame dependent", by the arguments above.

In SR, one expects that the potential energy of an object in a force field should also be a function of the charges and the distance between them. 

Right, e.g. $U \propto \frac{1}{r}$.

However since the distance r between the charges is relative to the choice of the frame of reference

No: any one value of distance is properly determined and attributed to a pair of "ends" which are at rest wih respect to each other; thus identifying one ("their") appropriate reference system. Any one such value is unambiguous, regardless of considerations of possible other reference systems.
