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In $E=mc^2$ what type of energy is $E$? For instance, you can find the kinetic energy of an electron by using $\frac{1}{2}mv^2$ but you can also find the energy from $E=mc^2$. How are those two different besides the values?

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It's usually called the "rest energy". In some sense this is a potential energy I suppose, but not the sort that can be converted to the kinetic energy of the same particle. An electron at rest has some rest energy, but it cannot lose some rest energy and gain some velocity; the rest energy is tied up in the mass (hence $E=mc^2$). The rest energy comes into play when particles are created and destroyed. For instance, in electron-positron annihilation:

$$e^{-} + e^{+} \rightarrow \gamma + \gamma$$

the two photons produced by the annihilation have a total energy equal to the total rest energy of the electron-positron pair.

Qualitatively, conservation of energy for such particle interactions is:

$$E_{\rm rest}({\rm input}) + E_{\rm kinetic}({\rm input}) = E_{\rm rest}({\rm output}) + E_{\rm kinetic}({\rm output})$$

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  • $\begingroup$ The typical representation of E=mc^2 can be misleading. As you say, the energy of a particle is the rest energy + kinetic energy + potential energy... If we always wrote something like $$\mbox{E}=mc^{2}+\frac{1}{2}\left( mv^{2} \right)$$ or the relativistic equivalent, this question would not come up so often. Though I think it is a little late to get $$\mbox{E}=mc^{2}$$ out of the public consciousness. $\endgroup$ – C. Towne Springer May 21 '14 at 2:31
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It is a kind of potential energy, since the object/particle is at rest in the formula $E = mc^2$. In the case where the object is moving, you would have both Kinetic and Potential energies, which is (ignoring some special relativistic effects) the sum of the two.

if the object has some momentum $p$ (i.e, velocity) then we have the real formula of $E = \sqrt{ (m_{0} c^2)^2 + (p c^2)}$.

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  • $\begingroup$ This somewhat misleadingly implies that potential energy is any different than any other type of energy. $\endgroup$ – Brandon Enright May 20 '14 at 23:17

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