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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).

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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 $B^2$.

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    $\begingroup$ The chapter 7 of MTW discusses the issues with a different theories of gravity formulated in flat space-time. I highly recommend the OP looks at this examples to see how treating gravitation like a 4-vector potential is inherently flawed and gives many wrong predictions. $\endgroup$ – dj_mummy Sep 27 '13 at 6:45
  • $\begingroup$ @BenCrowell when you say electromagnetism is inherently relativistic, do you mean that if the electromagnetic field follows the same wave equation in Newtonian physics, the laws of electromagnetism are not conserved in all frames of reference? $\endgroup$ – Timothy Apr 4 '18 at 2:25
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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).

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  • $\begingroup$ The problem with this is that a free scalar particle coupled to a the EM field is the only scenario where this is applicable (as far as I know). A lot of the other field configurations like a general SU(N) gauge field and other interactions etc., cannot be reduced to the analysis of the Hamiltonian of a single particle. If you could show otherwise, that would be wonderful. $\endgroup$ – dj_mummy Sep 27 '13 at 6:34
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If you read Einstein's original paper here,

https://www.fourmilab.ch/etexts/einstein/E_mc2/www/

it is an easy read if you skip over some of the Maths but even that isn't too hard.

You will find that the E in his equation (or rather K for kinetic energy in this original version) is in fact potential energy. He does not refer to this directly but does refer to Inertia in the very last sentence. Inertia is the opposite of potential energy.

Is this something that everyone is aware of?

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