# What Happened to Potential Energy in E=m$$c^2[tex] 1. Nov 30, 2005 ### bhthiang I have been wondering why is potential energy not included in the calculation of total energy E=m[tex]c^2[tex]? I know that m[tex]c^2[tex] includes the rest energy and the kinetic energy. Suppose the particle or object is charged and is moving in a strong electric field, its potential energy will be sdgnificant wouldn't it? If we have to include the PE, how should we do it? Thanks 2. Dec 18, 2005 ### Tom Mattson Staff Emeritus The energy-momentum relation that you see in a standard general physics textbook is: [tex]E^2=(pc)^2+(mc^2)^2$$

It is typically advertised as the energy-momentum relation for a *free* particle (so $V=0$). If memory serves, Goldstein's "Classical Mechanics" treats relativistic mechanics with a potential, via the Lagrangian formalism.

3. Dec 30, 2005

### krab

Properly,
$$(E-V)^2=(pc)^2+(mc^2)^2$$
where V is the potential energy and m, the rest mass, does NOT contain the kinetic energy.

4. Dec 31, 2005

### krab

Properly,
$$(E-V)^2=(pc)^2+m^2c^4$$
where m is the rest mass and V the potential energy