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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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1 Answers1

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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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  • Thanks for the response. Yes, within the purview of special theory of relativity, one would get into problems if we define potential energy as in the non relativistic theory. My aim in posing the question is to see if I could get any support for my thinking that all is not well with the theory of relativity and it needs some kind of modification. – Sridhar Aug 26 '21 at 11:41