# Whether $m$ in $E=mc^{2}$ and $F=ma$ are both relativistic mass?

I know that $m$ in $E=mc^{2}$ is the relativistic mass, but can $m$ in $F=ma$ can also be relativistic? If the answer is yes, then can you tell me whether this equation is valid $E=\frac{F}{a}c^{2}$? If not, can you tell why this is not valid?

Advance thanks for your help and please forgive me my english as it is my second language.

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Relativistic force is defined as $$\vec F = \frac {d} {dt} (\gamma m_o \vec v) = \frac {m_o\gamma^3} {c^2}\vec a\cdot\vec v + \gamma m_o\vec a$$ Although generally different, this becomes the same as your expression when $\vec a$ is perpendicular to $\vec v$ giving $\vec a\cdot\vec v = 0$.

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Here, $m=$ relativistic mass; $m_0=$rest mass

Not really. $F=ma$ only applies for a system with constant (relativistic) mass. The true equation is $\vec{F}=\frac{d\vec{p}}{dt}$, where $\vec{p}=m\vec{v}$ is the momentum. Since acceleration $\implies$ increase in relativistic mass, $F=ma$ is pretty useless here.

Anyways, in relativity, we tend to talk in terms of momentum and force, not acceleration.

So the second equation isn't valid except in special cases.

A better equation to use in place of the second one is: $$E^2=p^2c^2+m_0^2c^4$$

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IMO it's good practice to always write $m=\gamma m_0$. It gets rid of such 'classical mechanics' prejudices. –  Manishearth Feb 29 '12 at 17:58