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Eureka! I have got it.

$$E-mc^2= \frac12mv^2$$ The interpretation goes like this. Total energy- rest energy= kinetic energy. But we all know Total Energy = Potential Energy + Kinetic Energy. Since $$E= mc^2 + \frac12 mv^2.$$ So the rest energy must be potential energy which is $mc^2$. Which means Einsteins formula gives us potential energy.

Please tell me if I am right.

Kyle Kanos
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avito009
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    [@tag removed after comment deletion-dmckee] Although the "interpretation" presented by the OP is not an appropriate one, it seems like the OP is only trying to learn Special Relativity. It doesn't seem too farfetched to assume that what OP meant by "interpretation" is his/her understanding of Special Relativity. If the OP were trying to just publish his/her own interpretation and not to understand what Special Relativity teaches us then the OP would not have asked if he/she is right or not. –  Jun 23 '17 at 13:00
  • Related post by OP: https://physics.stackexchange.com/q/340950/2451 – Qmechanic Jun 23 '17 at 13:17
  • It is more $E~=~mc^2~+~\frac{1}{2}\gamma mv^2$ for $\gamma~=~1/\sqrt{1~-~v^2/c^2}$. – Lawrence B. Crowell Jun 23 '17 at 14:01
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    @LawrenceB.Crowell Er ... no. Not that either. In terms of the Lorentz factor and mass the energy is $E = \gamma m c^2 = mc^2 + (\gamma - 1) m c^2$, so that the kinetic energy is $(\gamma - 1) m c^2 \ne \gamma \frac{1}{2} m v^2$. The taking only the leaning non-zero term in $v/c$ (use the binomial approximation for $[1 - (v/c)^2]^{-1/2}$) you get the classical approximation $E \approx mc^2 + \frac{1}{2} m v^2$. – dmckee --- ex-moderator kitten Jun 23 '17 at 19:06

1 Answers1

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The division of energy into kinetic and potential terms that you learned in intro mechanics is only part of the picture. The energy of mass is a third, distinct category.1

When we write $E = \gamma m c^2 = mc_2 + (\gamma - 1) mc^2$ we actually expressing the energy of relativistic mass absent any interactions (i.e. without potential energy).2

The full expression for the energy of a mass in a potential well $V(x)$ is \begin{align*} E &= \gamma m c^2 + V(x)\\ &= m c^2 + (\gamma - 1) m c^2 + V(x)\\ &= E_\text{mass} + E_\text{kinetic}(v) + E_\text{potential}(x) \;. \end{align*}

1 Well, sorta. The mass of compound objects can be composed in part of the kinetic and potential energies of it's parts, but fundamental particles can have intrinsic mass. One of the features of relativity is that the mass of a system is not the sum of the masses of its parts, a fact which is abundantly clear if you write the mass in terms of four-momenta $m^2 c^4 = \mathbf{p}^2 = E^2 - (\vec{p}c)^2$ and add the four momenta of two constituents moving in different directions.

2 Important and often overlooked point: potential energies arise from interactions and are therefore a feature of systems not of individual objects. And yeah, in intro mechanics we say "the potential energy of a book on a desk", but that energy relies on the presence of the Earth: take away the planet and the book isn't subject to gravitational force and therefore has no tendency to fall...

dmckee --- ex-moderator kitten
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