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Recently, I found a YouTube video saying that $$E=mc^2$$ is false. It says that the real one is

$$ E^2=m^2 c^4 + p^2 c^2 $$

where $E$: energy, $m$: mass, $c$: speed of light and $p$: momentum.

Is there anything true in it? It says that if it was $E=mc^2$ then as light had only momentum it would not have any energy.

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    $\begingroup$ Look up the difference between rest mass and relativistic mass : en.wikipedia.org/wiki/Relativistic_mass . It is not the same "mass" in the two formulae $\endgroup$
    – anna v
    Commented Oct 15, 2013 at 6:51
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    $\begingroup$ @WilliamNeill Sometimes people use $m$ to mean relativistic mass and $m_0$ for rest mass. And FWIW, $E=mc^2$ works both ways, you can either say that $m$ is rest mass and call it a special case, or say that $m$ is relativistic mass. Both are correct. $\endgroup$
    – Manishearth
    Commented Oct 15, 2013 at 7:14
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/143652/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Dec 24, 2021 at 10:19

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In the early days of special relativity it was noted that the mass of an object appeared to increase as the speed of the object approached the speed of light. It was common to see the notation $m_0$ used for the rest mass and $m$ for the relativistic mass. In this sense the equation $E = mc^2$ is always true.

However the concept of relativistic mass is deprecated these days, principally because it has caused confusion to generations of physics students. These days we use only the symbol $m$, and it always refers to rest mass. So $E = mc^2$ is only true in the rest frame of the object, and we now use the equation you mention. Having said this, in everyday life speeds tend to be low compared to the speed of light. For example if you're calculating the energy released by a nuclear reactor then using $E = mc^2$, where $m$ is the mass deficit, is an excellent approximation.

As you say, the equation:

$$ E^2 = p^2c^2 + m^2c^4 $$

works for photons as well, while the old equation did not.

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  • $\begingroup$ John I think it might be helpful to add, as William Neill writes in his comment, that even using modern terminology, the equation $E=mc^2$ does hold in an object's rest frame. $\endgroup$
    – joshphysics
    Commented Oct 15, 2013 at 7:14
  • $\begingroup$ @joshphysics: yes, I agree. Now clarified. $\endgroup$
    – John Rennie
    Commented Oct 15, 2013 at 7:35

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