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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?

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marked as duplicate by Qmechanic Jan 25 at 19:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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.

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    $\begingroup$ I didn't see your answer until after I finished submitting mine! $\endgroup$ – Hal Hollis Jan 25 at 19:17
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    $\begingroup$ @HalHollis Well, they end up complementing each other anyway, so I think it works. $\endgroup$ – probably_someone Jan 25 at 19:19
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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 $E^2=p^2c^2+m^2c^4$

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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

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

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    $\begingroup$ 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. $\endgroup$ – Undead Jan 25 at 21:42
  • $\begingroup$ @Undead, oops! Not sure what happened there, fixing now. $\endgroup$ – Hal Hollis Jan 25 at 21:48

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