# Is $E=mc^2$ only partially true? [duplicate]

This question already has an answer here:

I've obviously seen that $$E=mc^2$$. But I've also seen that the equation $$E^2=(mc^2)^2+(pc)^2$$ (where $$p$$ = momentum) is true.

If you square $$E=mc^2$$, and subtract the result from the other equation, you get $$0=(pc)^2$$. In other words, momentum doesn't exist!

I assume combining these equations is a mistake. Is $$E=mc^2$$ only something that holds in certain circumstances? Is $$E^2=(mc^2)^2+(pc)^2$$ the more complete, correct version of $$E=mc^2$$?

I know I'm wrong -- where, exactly, is my mistake?

## marked as duplicate by Qmechanic♦Jan 25 at 19:35

There are two ways to answer this, based on two different definitions of mass. In most modern courses, $$m$$ refers specifically to the rest energy of the object (meaning how much energy it has at rest). In that view, $$E=mc^2$$ only applies to objects at rest, while the full formula is $$E^2=p^2c^2+m^2c^4$$ (you can also say $$E=\gamma mc^2$$, where $$\gamma$$ is the Lorentz factor).

There is also another definition, used occasionally in older textbooks, in which $$m$$ refers to the relativistic mass of the object. In this view, the relativistic mass and the rest energy $$m_0$$ are related by $$m=\gamma m_0$$, where $$\gamma$$ is the Lorentz factor. So, under this set of definitions, $$E=mc^2$$ is the only valid equation. Let me stress that this view of mass is largely obsolete at this point (there are a few answers on this site that explain some reasons why), but it's useful to keep in mind that it was used in the past, especially since it was prominent when $$E=mc^2$$ first entered the public consciousness.

• I didn't see your answer until after I finished submitting mine! – Hal Hollis Jan 25 at 19:17
• @HalHollis Well, they end up complementing each other anyway, so I think it works. – probably_someone Jan 25 at 19:19

You have got it. $$E=mc^2$$ is only true for particles at rest - i.e. particles with no momentum $$(p=0)$$! The general equation is indeed $$E^2=p^2c^2+m^2c^4$$

The answer depends on what the symbol $$m$$ represents.

If $$m$$ is the (non-zero) invariant mass of a particle, then $$E=mc^2$$ holds in an inertial reference frame (IRF) in which the particle is at rest. If the particle has speed $$v$$ in an IRF, then the expression for the energy is

$$E = \gamma mc^2 = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$

and this equation is consistent with the energy-momentum equation written as

$$E^2 = (pc)^2 + (mc^2)^2$$

However, and confusingly, it is sometimes the case that $$m$$ represents the (more or less outdated) relativistic mass $$\gamma m_0$$ where $$m_0$$ is the rest mass thus

$$E = mc^2 = \gamma m_0c^2 = \frac{m_0c^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$

and then the energy-momentum equation is written as

$$E^2 = (pc)^2 + (m_0c^2)^2$$

• I think that it's the notion of relativistic mass that is outdated not that of rest mass. Rest mass and invariant mass are the same thing. – Undead Jan 25 at 21:42
• @Undead, oops! Not sure what happened there, fixing now. – Hal Hollis Jan 25 at 21:48