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)?
 A: I'll expose what I've understood.
In classical mechanics, $E=T+U$. Since for a free particle in SR, $E=\sqrt{p^2+m^2}$ (here $c=1$). We could try to introduce potential energy as: $E-U=\sqrt{p^2+m^2}$. But this would not be a covariant equation. 
So we have to use the 4-vector $Q^\mu=(U, \textbf{Q})$, which is the potential four-momentum while $\textbf{Q}$ is called just the potential momentum.
If we subtract $Q^\mu$ to $p^\mu=(E,\textbf{p})$, we get:
$E-U=\sqrt{(\textbf{p}-\textbf{Q})^2+m^2}$
The potential momentum is closely related to the Aharonov-Bohm effect.
This way to introduce potential energy the one used in gauge theories. There are two more possible ways: for gravity, using space-time curvature or supposing that the potential energy is a scalar field (Higgs field).
A: 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 $B^2$.
A: In special relativity, the equation of motion in Galilean coordinates are:
$$\frac{\text{d}\boldsymbol{p}}{\text{d}t} - \boldsymbol{F} = \frac{\text{d}}{\text{d}t}\left(\frac{m\boldsymbol{v}}{\sqrt{1-v^2/c^2}}\right) - \frac{\partial V_0}{\partial \boldsymbol{x}}  = 0$$
These are the Euler-Lagrange equations:
$$\frac{\text{d}}{\text{d}t}\left(\frac{\partial \mathcal{L}}{\partial \boldsymbol{v}}\right) - \frac{\partial \mathcal{L}}{\partial \boldsymbol{x}}  = 0$$
for a Lagrangian given by:
$$\mathcal{L} = -mc^2\sqrt{1-\frac{v^2}{c^2}}-V_0$$
In totally general coordinates, we would have:
$$\mathcal{L} = -mc^2\sqrt{-g_{\mu\nu}\frac{\text{d}x^\mu}{\text{d}x^0}\frac{\text{d}x^\mu}{\text{d}x^0}}- V_\alpha \frac{\text{d}x^\alpha}{\text{d}x^0}$$
where the "potential" is now a Lorentz vector potential. The action will be:
$$S_{AB} = -mc \int_A^B \text{d}s - \int_A^B \frac{V_\alpha}{c} \text{d}x^\alpha$$
which is Lorentz invariant and the equations of motion will be:
$$m\left(\frac{1}{c}\frac{\text{d}u^\alpha}{\text{d}\tau} + \Gamma^\alpha_{\mu\nu}u^\mu u^\nu \right) = \frac{1}{c}F^\alpha$$
where $\text{d}\tau = \text{d}s/c$ is proper time (a Lorentz scalar). The energy (timelike-component of the energy-impulse vector) will be:
$$\mathcal{H} = \frac{\partial \mathcal{L}}{\partial \dot{x}^a}\dot{x}^a -\mathcal{L}=
E = \sqrt{m^2c^4 + \|\boldsymbol{p}c-\boldsymbol{V}\|^2} + V_0$$
