Potential energy in Special Relativity

In Special Relativity, the energy of a free particle is $E^2=p^2c^2+m^2c^4$.

But what would be the energy when there is potential energy?

If it's something like $E=\sqrt{p^2c^2+m^2c^4}+U$, what does it mean if a particle has zero or less energy?

Addendum 2013/09/26

The potential momentum is used only in gauge theories (like EM). But could it be used in SR+Newton's gravity, without introducing the concept of curvature (GR).

Answer

Let's start with Newtonian mechanics. Of the fundamental forces of nature, the only one that can be handled at all by Newtonian mechanics is gravity. Newtonian mechanics can't handle electromagnetism. Electromagnetism is inherently relativistic (i.e., Maxwell's equations only make sense in the context of SR, not Galilean relativity).

Now let's pass from the Newtonian approximation to SR. We lose the ability to model gravity, since that would require GR. We gain the ability to model electromagnetism. In electromagnetism, we don't really have a useful concept of a scalar potential energy $q\Phi$, where $\Phi$ is the electric potential. The reason for this is that although the charge $q$ is a relativistic scalar, the electrical potential $\Phi$ is not a relativistic scalar, it's the timelike component of a four-vector. The conserved energy in Maxwell's equations is not really the energy of a point particle in some external field, it's the energy of the electromagnetic field itself, which depends on energy densities proportional to $E^2$ and $B^2$.