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Is there anything like relativistic potential energy? If not, why? We know relativistic force along the direction of velocity, involving $\gamma^3$. We also have the standard expression relating force and potential energy in Newtonian mechanics, $F = - \mathrm dV/\mathrm dx$ where $V$ is a function of $x$. I don't see any reason why this can not be applied for relativistic case too.

Hence I ask, "can we get potential energy in the relativistic case using $F = - \mathrm dV/\mathrm dx$? If not, why?"

jng224
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Sridhar
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Yes, one could get potential energy in special relativity using the relation you wrote. Also one can use it, the equation of motion for a relativistic particle (written in one spacial dimension for simplicity, but it is easily generalized to two and three) is $$\frac{m\ddot{x}(t)}{\bigg(1-\frac{\dot{x}(t)}{c²} \bigg)^{\frac{3}{2}}}=-U'(x(t)).$$ This is relativistically correct, and physically useful, it does not allow to surpass the speed of light. And an energy $E=\gamma mc²+ U(x)$ may be defined. The problem is that this isn't Lorentz covariant and must be recalculated in other inertial frame of reference. This is discussed on chapter 7 of Goldstein's book Classical Mechanics.

That the potential energy defined as $\vec{F}=-\nabla U$ is not invariant under Lorentz transformations is left as an excercise for the reader.

Bibliography

Goldstein, H. (1980). Classical Mechanics. (2nd ed). Addison-Wesley: Boston, MA.

Don Al
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