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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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up vote 3 down vote accepted

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

The mass in F=ma is relativistic mass if the force is perpendicular to the velocity. If the force is parallel to the velocity, the mass is neither the rest mass or the relativistic mass, but the so-called "longitudinal mass" (deprecated term used by Einstein).

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+1 you're welcome to copy my expression to make this clear to the op and I'll delete my answer. If not, I'll modify my answer to include yours ;) – John McVirgo Feb 29 '12 at 16:07
Modify your answer, I'll delete mine. – Ron Maimon Feb 29 '12 at 16:37
I think our answers are sufficiently different and upvoted so I'll leave it. – John McVirgo Feb 29 '12 at 23:17

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